Properties of sequences A sequence { x n } is periodic ⇔ ∃ k > 0 : ∀ n ∈ Z : x n = x n + k ∞ � absolutely summable ⇔ | x n | < ∞ n = −∞ ∞ � | x n | 2 square summable ⇔ < ∞ ⇔ “energy signal” n = −∞ � �� � “energy” k 1 � | x n | 2 0 < lim < ∞ ⇔ “power signal” 1 + 2 k k →∞ n = − k � �� � “average power” This energy/power terminology reflects that if U is a voltage supplied to a load resistor R , then P = UI = U 2 /R is the power consumed, and � P ( t ) d t the energy. It is used even if we drop physical units (e.g., volts) for simplicity in calculations. 15 Types of discrete systems A causal system cannot look into the future: y n = f ( x n , x n − 1 , x n − 2 , . . . ) A memory-less system depends only on the current input value: y n = f ( x n ) A delay system shifts a sequence in time: y n = x n − d T is a time-invariant system if for any d { y n } = T { x n } ⇐ ⇒ { y n − d } = T { x n − d } . T is a linear system if for any pair of sequences { x n } and { x ′ n } T { a · x n + b · x ′ n } = a · T { x n } + b · T { x ′ n } . 16
Example: M -point moving average system M − 1 y n = 1 x n − k = x n − M +1 + · · · + x n − 1 + x n � M M k =0 It is causal, linear, time-invariant, with memory. With M = 4: x y 0 17 Example: exponential averaging system ∞ � (1 − α ) k · x n − k y n = α · x n + (1 − α ) · y n − 1 = α k =0 With α = 1 It is causal, linear, time-invariant, with memory. 2 : x y 0 18
Example: accumulator system n � y n = x k k = −∞ It is causal, linear, time-invariant, with memory. x y 0 19 Example: backward difference system y n = x n − x n − 1 It is causal, linear, time-invariant, with memory. x y 0 20
Other examples Time-invariant non-linear memory-less systems: y n = x 2 n , y n = log 2 x n , y n = max { min {⌊ 256 x n ⌋ , 255 } , 0 } Linear but not time-invariant systems: � x n , n ≥ 0 y n = n < 0 = x n · u n 0 , y n = x ⌊ n/ 4 ⌋ y n = x n · ℜ (e ˙ ω j n ) Linear time-invariant non-causal systems: y n = 1 2( x n − 1 + x n +1 ) 9 x n + k · sin( πkω ) � y n = · [0 . 5 + 0 . 5 · cos( π k/ 10)] πkω k = − 9 21 Constant-coefficient difference equations Of particular practical interest are causal linear time-invariant systems of the form x n b 0 y n N � z − 1 y n = b 0 · x n − a k · y n − k − a 1 k =1 y n − 1 z − 1 − a 2 Block diagram representation y n − 2 of sequence operations: x ′ z − 1 n − a 3 x n + x ′ x n n y n − 3 Addition: Multiplication x n a ax n The a k and b m are by constant: constant coefficients. x n x n − 1 z − 1 Delay: 22
or x n x n − 1 x n − 2 x n − 3 z − 1 z − 1 z − 1 M � b 0 b 1 b 2 b 3 y n = b m · x n − m m =0 y n or the combination of both: a − 1 x n b 0 y n 0 z − 1 z − 1 b 1 − a 1 x n − 1 y n − 1 N M � � a k · y n − k = b m · x n − m z − 1 z − 1 k =0 m =0 b 2 − a 2 x n − 2 y n − 2 z − 1 z − 1 b 3 − a 3 x n − 3 y n − 3 The MATLAB function filter is an efficient implementation of the last variant. 23 Convolution Another example of a LTI systems is ∞ � y n = a k · x n − k k = −∞ where { a k } is a suitably chosen sequence of coefficients. This operation over sequences is called convolution and is defined as ∞ � { p n } ∗ { q n } = { r n } ⇐ ⇒ ∀ n ∈ Z : r n = p k · q n − k . k = −∞ If { y n } = { a n } ∗ { x n } is a representation of an LTI system T , with { y n } = T { x n } , then we call the sequence { a n } the impulse response of T , because { a n } = T { δ n } . 24
Convolution examples A B C D A ∗ B A ∗ C E F C ∗ A A ∗ E D ∗ E A ∗ F 25 Properties of convolution For arbitrary sequences { p n } , { q n } , { r n } and scalars a , b : ◮ Convolution is associative ( { p n } ∗ { q n } ) ∗ { r n } = { p n } ∗ ( { q n } ∗ { r n } ) ◮ Convolution is commutative { p n } ∗ { q n } = { q n } ∗ { p n } ◮ Convolution is linear { p n } ∗ { a · q n + b · r n } = a · ( { p n } ∗ { q n } ) + b · ( { p n } ∗ { r n } ) ◮ The impulse sequence (slide 13) is neutral under convolution { p n } ∗ { δ n } = { δ n } ∗ { p n } = { p n } ◮ Sequence shifting is equivalent to convolving with a shifted impulse { p n − d } = { p n } ∗ { δ n − d } 26
Proof: all LTI systems just apply convolution Any sequence { x n } can be decomposed into a weighted sum of shifted impulse sequences: ∞ � { x n } = x k · { δ n − k } k = −∞ Let’s see what happens if we apply a linear ( ∗ ) time-invariant ( ∗∗ ) system T to such a decomposed sequence: � � ∞ ∞ ( ∗ ) � � T { x n } = T x k · { δ n − k } = x k · T { δ n − k } k = −∞ k = −∞ � � ∞ ∞ ( ∗∗ ) � � = x k · { δ n − k } ∗ T { δ n } = x k · { δ n − k } ∗ T { δ n } k = −∞ k = −∞ = { x n } ∗ T { δ n } q.e.d. ⇒ The impulse response T { δ n } fully characterizes an LTI system. 27 Direct form I and II implementations a − 1 a − 1 x n b 0 y n x n b 0 y n 0 0 z − 1 z − 1 z − 1 b 1 − a 1 − a 1 b 1 x n − 1 y n − 1 = z − 1 z − 1 z − 1 b 2 − a 2 − a 2 b 2 x n − 2 y n − 2 z − 1 z − 1 z − 1 b 3 − a 3 − a 3 b 3 x n − 3 y n − 3 The block diagram representation of the constant-coefficient difference equation on slide 23 is called the direct form I implementation . The number of delay elements can be halved by using the commutativity of convolution to swap the two feedback loops, leading to the direct form II implementation of the same LTI system. These two forms are only equivalent with ideal arithmetic (no rounding errors and range limits). 28
Convolution: optics example If a projective lens is out of focus, the blurred image is equal to the original image convolved with the aperture shape (e.g., a filled circle): ∗ = image plane focal plane Point-spread function h (disk, r = as 2 f ): x 2 + y 2 ≤ r 2 1 � r 2 π , h ( x, y ) = x 2 + y 2 > r 2 0 , a Original image I , blurred image B = I ∗ h , i.e. �� I ( x − x ′ , y − y ′ ) · h ( x ′ , y ′ ) · d x ′ d y ′ B ( x, y ) = s f 29 Convolution: electronics example R U in U in C U out U out t Any passive network (resistors, capacitors, inductors) convolves its input voltage U in with an impulse response function h , leading to U out = U in ∗ h , that is � ∞ U out ( t ) = U in ( t − τ ) · h ( τ ) · d τ −∞ In the above example: � − t U in − U out = C · d U out 1 RC , RC · e t ≥ 0 , h ( t ) = 0 , t < 0 R d t 30
Adding sine waves Adding together sine waves of equal frequency, but arbitrary amplitude and phase, results in another sine wave of the same frequency: A 1 · sin( ωt + ϕ 1 ) + A 2 · sin( ωt + ϕ 2 ) = A · sin( ωt + ϕ ) Why? Think of A · sin( ωt + ϕ ) as the height of ω an arrow of length A , rotating 2 π times per second, A 2 with start angle ϕ (radians) at t = 0. A Consider two more such arrows, ϕ 2 of length A 1 and A 2 , A 1 with start angles ϕ 1 and ϕ 2 . ωt ϕ ϕ 1 A 1 and A 2 stuck together are as high as A , all three rotating at the same frequency. But adding sine waves as vectors ( A 1 , ϕ 1 ) and ( A 2 , ϕ 2 ) in polar coordinates is cumbersome: tan ϕ = A 1 sin ϕ 1 + A 2 sin ϕ 2 � A 2 1 + A 2 A = 2 + 2 A 1 A 2 cos( ϕ 2 − ϕ 1 ) , A 1 cos ϕ 1 + A 2 cos ϕ 2 31 Cartesian coordinates for sine waves Sine waves of any amplitude A and phase (start angle) ϕ can be represented as linear combinations of sin( ωt ) and cos( ωt ): A · sin( ωt + ϕ ) = x · sin( ωt ) + y · cos( ωt ) cos( ωt ) = sin( ωt + 90 ◦ ) where A A · sin( ϕ ) x = A · cos( ϕ ) , y = A · sin( ϕ ) ωt ϕ and tan ϕ = y A · cos( ϕ ) � x 2 + y 2 , A = x. Base: two rotating arrows with start angles 0 ◦ [height = sin( ω )] and 90 ◦ [height = cos( ω )]. Adding two sine waves as vectors in Cartesian coordinates is simple: f 1 ( t ) = x 1 · sin( ω ) + y 1 · cos( ω ) f 2 ( t ) = x 2 · sin( ω ) + y 2 · cos( ω ) f 1 ( t ) + f 2 ( t ) = ( x 1 + x 2 ) · sin( ω ) + ( y 1 + y 2 ) · cos( ω ) 32
Why are sine waves useful? 1) Sine-wave sequences form a family of discrete sequences that is closed under convolution with arbitrary sequences. Convolution of a discrete sequence { x n } with another sequence { y n } is nothing but adding together scaled and delayed copies of { x n } . (Think of { y n } decomposed into a sum of impulses.) If { x n } is a sampled sine wave of frequency f , so is { x n } ∗ { y n } ! The same applies for continuous sine waves and convolution. 2) Sine waves are orthogonal to each other � ∞ sin( ω 1 t + ϕ 1 ) · sin( ω 2 t + ϕ 2 ) d t “=” 0 −∞ ⇐ ⇒ ω 1 � = ω 2 ∨ ϕ 1 − ϕ 2 = (2 k + 1) π / 2 ( k ∈ Z ) They can be used to form an orthogonal function basis for a transform. The term “orthogonal” is used here in the context of an (infinitely dimensional) vector space, where the “vectors” are functions of the form f : R → R (or f : R → C ) and the scalar product is � ∞ defined as f · g = −∞ f ( t ) · g ( t ) d t . 33 1 0 sin(1t) ⋅ sin(2t) sin(1t) sin(2t) −1 0 1.5708 3.1416 4.7124 6.2832 t 34
Why are exponential functions useful? Adding together two exponential functions with the same base z , but different scale factor and offset, results in another exponential function with the same base: A 1 · z t · z ϕ 1 + A 2 · z t · z ϕ 2 A 1 · z t + ϕ 1 + A 2 · z t + ϕ 2 = ( A 1 · z ϕ 1 + A 2 · z ϕ 2 ) · z t = A · z t = Likewise, if we convolve a sequence { x n } of values . . . , z − 3 , z − 2 , z − 1 , 1 , z, z 2 , z 3 , . . . x n = z n with an arbitrary sequence { h n } , we get { y n } = { z n } ∗ { h n } , ∞ ∞ ∞ � � � z n − k · h k = z n · z − k · h k = z n · H ( z ) y n = x n − k · h k = k = −∞ k = −∞ k = −∞ where H ( z ) is independent of n . Exponential sequences are closed under convolution with arbitrary sequences. The same applies in the continuous case. 35 Why are complex numbers so useful? 1) They give us all n solutions (“roots”) of equations involving polynomials up to degree n (the “ √− 1 = j ” story). 2) They give us the “great unifying theory” that combines sine and exponential functions: 1 e j θ + e − j θ � � cos( θ ) = 2 1 � e j θ − e − j θ � sin( θ ) = 2j or � e j( ωt + ϕ ) + e − j( ωt + ϕ ) � cos( ωt + ϕ ) = 1 2 or ω ) n · e j ϕ ] ℜ (e j( ˙ ωn + ϕ ) ) = ℜ [(e j ˙ cos( ˙ ωn + ϕ ) = ω ) n · e j ϕ ] ℑ (e j( ˙ ωn + ϕ ) ) = ℑ [(e j ˙ sin( ˙ ωn + ϕ ) = Notation: ℜ ( a + j b ) := a , ℑ ( a + j b ) := b and ( a + j b ) ∗ := a − j b , where j 2 = − 1 and a, b ∈ R . Then ℜ ( x ) = 1 2 ( x + x ∗ ) and ℑ ( x ) = 2 j ( x − x ∗ ) for all x ∈ C . 1 36
We can now represent sine waves as projections of a rotating complex vector. This allows us to represent sine-wave sequences as exponential sequences with basis e j ˙ ω . A phase shift in such a sequence corresponds to a rotation of a complex vector. 3) Complex multiplication allows us to modify the amplitude and phase of a complex rotating vector using a single operation and value. Rotation of a 2D vector in ( x, y )-form is notationally slightly messy, but fortunately j 2 = − 1 does exactly what is required here: � � � � � � x 3 x 2 − y 2 x 1 = · y 3 y 2 x 2 y 1 ( x 3 , y 3 ) � x 1 x 2 − y 1 y 2 � = ( − y 2 , x 2 ) x 1 y 2 + x 2 y 1 ( x 2 , y 2 ) z 1 = x 1 + j y 1 , z 2 = x 2 + j y 2 ( x 1 , y 1 ) z 1 · z 2 = x 1 x 2 − y 1 y 2 + j( x 1 y 2 + x 2 y 1 ) 37 Complex phasors Amplitude and phase are two distinct characteristics of a sine function that are inconvenient to keep separate notationally. Complex functions (and discrete sequences) of the form ( A · e j ϕ ) · e j ωt = A · e j( ωt + ϕ ) = A · [cos( ωt + ϕ ) + j · sin( ωt + ϕ )] (where j 2 = − 1) are able to represent both amplitude A ∈ R + and phase ϕ ∈ [0 , 2 π ) in one single algebraic object A · e j ϕ ∈ C . Thanks to complex multiplication, we can also incorporate in one single factor both a multiplicative change of amplitude and an additive change of phase of such a function. This makes discrete sequences of the form x n = e j ˙ ωn eigensequences with respect to an LTI system T , because for each ˙ ω , there is a complex number (eigenvalue) H ( ˙ ω ) such that T { x n } = H ( ˙ ω ) · { x n } In the notation of slide 35, where the argument of H is the base, we would write H (e j ˙ ω ). 38
Recall: Fourier transform We define the Fourier integral transform and its inverse as � ∞ g ( t ) · e − 2 π j ft d t F{ g ( t ) } ( f ) = G ( f ) = −∞ � ∞ G ( f ) · e 2 π j ft d f F − 1 { G ( f ) } ( t ) = g ( t ) = −∞ Many equivalent forms of the Fourier transform are used in the literature. There is no strong consensus on whether the forward transform uses e − 2 π j ft and the backwards transform e 2 π j ft , or vice versa. The above form uses the ordinary frequency f , whereas some authors prefer the angular frequency ω = 2 π f : � ∞ h ( t ) · e ∓ j ωt d t F{ h ( t ) } ( ω ) = H ( ω ) = α −∞ � ∞ H ( ω ) · e ± j ωt d ω F − 1 { H ( ω ) } ( t ) = h ( t ) = β −∞ This substitution introduces factors α and β such that αβ = 1 / (2 π ). Some authors set α = 1 and β = 1 / (2 π ), to keep the convolution theorem free of a constant prefactor; others prefer the √ unitary form α = β = 1 / 2 π , in the interest of symmetry. 39 Properties of the Fourier transform If x ( t ) • − ◦ X ( f ) and y ( t ) • − ◦ Y ( f ) are pairs of functions that are mapped onto each other by the Fourier transform, then so are the following pairs. Linearity: ax ( t ) + by ( t ) • − ◦ aX ( f ) + bY ( f ) Time scaling: � f � 1 x ( at ) • − ◦ | a | X a Frequency scaling: � t � 1 | a | x • − ◦ X ( af ) a 40
Time shifting: X ( f ) · e − 2 π j f ∆ t x ( t − ∆ t ) • − ◦ Frequency shifting: x ( t ) · e 2 π j∆ ft • − ◦ X ( f − ∆ f ) Parseval’s theorem (total energy): � ∞ � ∞ | x ( t ) | 2 d t | X ( f ) | 2 d f = −∞ −∞ 41 Fourier transform example: rect and sinc The Fourier transform of the “rectangular function” if | t | < 1 1 1 2 1 if | t | = 1 rect( t ) = 2 2 0 0 otherwise 2 0 − 1 1 2 is the “(normalized) sinc function” � 1 e − 2 π j ft d t = sin π f 2 F{ rect( t ) } ( f ) = = sinc( f ) π f − 1 2 and vice versa F{ sinc( t ) } ( f ) = rect( f ) . Some noteworthy properties of these functions: ◮ � ∞ � ∞ −∞ sinc( t ) d t = 1 = −∞ rect( t ) d t 1 ◮ sinc(0) = 1 = rect(0) 0 ◮ ∀ n ∈ Z \ { 0 } : sinc( n ) = 0 − 3 − 2 − 1 0 1 2 3 42
Convolution theorem Convolution in the time domain is equivalent to (complex) scalar multiplication in the frequency domain: F{ ( f ∗ g )( t ) } = F{ f ( t ) } · F{ g ( t ) } r z ( r )e − j ωr d r = s x ( s ) y ( r − s )e − j ωr d s d r = � � � � Proof: z ( r ) = s x ( s ) y ( r − s )d s ⇐ ⇒ r t := r − s r y ( r − s )e − j ωr d r d s = s x ( s )e − j ωs � r y ( r − s )e − j ω ( r − s ) d r d s � � � s x ( s ) = s x ( s )e − j ωs � t y ( t )e − j ωt d t d s = s x ( s )e − j ωs d s · t y ( t )e − j ωt d t . � � � Convolution in the frequency domain corresponds to scalar multiplication in the time domain: F{ f ( t ) · g ( t ) } = F{ f ( t ) } ∗ F{ g ( t ) } This second form is also called “modulation theorem”, as it describes what happens in the frequency domain with amplitude modulation of a signal (see slide 50). The proof is very similar to the one above. Both equally work for the inverse Fourier transform: F − 1 { ( F ∗ G )( f ) } = F − 1 { F ( f ) } · F − 1 { G ( f ) } F − 1 { F ( f ) · G ( f ) } = F − 1 { F ( f ) } ∗ F − 1 { G ( f ) } 43 Dirac delta function The continuous equivalent of the impulse sequence { δ n } is known as Dirac delta function δ ( x ). It is a generalized function, defined such that � 1 0 , x � = 0 δ ( x ) = ∞ , x = 0 � ∞ δ ( x ) d x = 1 −∞ 0 x and can be thought of as the limit of function sequences such as � 0 , | x | ≥ 1 /n δ ( x ) = lim n/ 2 , | x | < 1 /n n →∞ or n √ π e − n 2 x 2 δ ( x ) = lim n →∞ The delta function is mathematically speaking not a function, but a distribution , that is an expression that is only defined when integrated. 44
Some properties of the Dirac delta function: � ∞ f ( x ) δ ( x − a ) d x = f ( a ) −∞ � ∞ e ± 2 π j xa d x = δ ( a ) −∞ ∞ ∞ 1 � � e ± 2 π j nxa = δ ( x − n/a ) | a | n = −∞ n = −∞ 1 δ ( ax ) = | a | δ ( x ) Fourier transform: � ∞ δ ( t ) · e − 2 π j ft d t e 0 F{ δ ( t ) } ( f ) = = = 1 −∞ � ∞ · e 2 π j ft d f F − 1 { 1 } ( t ) = 1 = δ ( t ) −∞ 45 Linking the Dirac delta with the Fourier transform The Fourier transform of 1 follows from the Dirac delta’s ability to sample inside an integral: g ( t ) = F − 1 ( F ( g ))( t ) � ∞ �� ∞ � g ( s ) · e − 2 π j fs · d s · e 2 π j ft · d f = −∞ −∞ � ∞ �� ∞ � e − 2 π j fs · e 2 π j ft · d f = · g ( s ) · d s −∞ −∞ � ∞ �� ∞ � e − 2 π j f ( t − s ) · d f = · g ( s ) · d s −∞ −∞ � �� � δ ( t − s ) So if δ has the property � ∞ g ( t ) = δ ( t − s ) · g ( s ) · d s −∞ then � ∞ e − 2 π j f ( t − s ) d f = δ ( t − s ) −∞ 46
� ∞ � 10 e 2 π j tf d f = δ ( t ) i =1 cos(2 π f i t ) ≈ δ ( t ) −∞ f 1 , . . . , f 10 ∈ [0 , 3] chosen uniformly at random 1 0.5 0 −0.5 −1 −4 −3 −2 −1 0 1 2 3 4 10 5 0 −5 −10 −4 −3 −2 −1 0 1 2 3 4 47 Sine and cosine in the frequency domain cos(2 π f 0 t ) = 1 2 e 2 π j f 0 t + 1 sin(2 π f 0 t ) = 1 2j e 2 π j f 0 t − 1 2 e − 2 π j f 0 t 2j e − 2 π j f 0 t 1 2 δ ( f − f 0 ) + 1 F{ cos(2 π f 0 t ) } ( f ) = 2 δ ( f + f 0 ) F{ sin(2 π f 0 t ) } ( f ) = − j 2 δ ( f − f 0 ) + j 2 δ ( f + f 0 ) ℜ ℜ 1 1 2 2 ℑ ℑ 1 1 2 j 2 j − f 0 f 0 f − f 0 f 0 f As any x ( t ) ∈ R can be decomposed into sine and cosine functions, the spectrum of any real-valued signal will show the symmetry X ( − f ) = [ X ( f )] ∗ , where ∗ denotes the complex conjugate (i.e., negated imaginary part). 48
Fourier transform symmetries We call a function x ( t ) odd if x ( − t ) = − x ( t ) even if x ( − t ) = x ( t ) and · ∗ is the complex conjugate, such that ( a + j b ) ∗ = ( a − j b ). Then X ( − f ) = [ X ( f )] ∗ x ( t ) is real ⇔ X ( − f ) = − [ X ( f )] ∗ x ( t ) is imaginary ⇔ x ( t ) is even ⇔ X ( f ) is even x ( t ) is odd ⇔ X ( f ) is odd x ( t ) is real and even ⇔ X ( f ) is real and even x ( t ) is real and odd ⇔ X ( f ) is imaginary and odd x ( t ) is imaginary and even ⇔ X ( f ) is imaginary and even x ( t ) is imaginary and odd ⇔ X ( f ) is real and odd 49 Example: amplitude modulation Communication channels usually permit only the use of a given frequency interval, such as 300–3400 Hz for the analog phone network or 590–598 MHz for TV channel 36. Modulation with a carrier frequency f c shifts the spectrum of a signal x ( t ) into the desired band. Amplitude modulation (AM): y ( t ) = A · cos(2 π tf c ) · x ( t ) X ( f ) Y ( f ) ∗ = f f f − f l 0 f l − f c f c − f c 0 f c The spectrum of the baseband signal in the interval − f l < f < f l is shifted by the modulation to the intervals ± f c − f l < f < ± f c + f l . How can such a signal be demodulated? 50
Sampling using a Dirac comb The loss of information in the sampling process that converts a continuous function x ( t ) into a discrete sequence { x n } defined by x n = x ( t s · n ) = x ( n/f s ) can be modelled through multiplying x ( t ) by a comb of Dirac impulses ∞ � s ( t ) = t s · δ ( t − t s · n ) n = −∞ to obtain the sampled function x ( t ) = x ( t ) · s ( t ) ˆ The function ˆ x ( t ) now contains exactly the same information as the discrete sequence { x n } , but is still in a form that can be analysed using the Fourier transform on continuous functions. 51 The Fourier transform of a Dirac comb ∞ ∞ � � e 2 π j nt/t s s ( t ) = t s · δ ( t − t s · n ) = n = −∞ n = −∞ is another Dirac comb � � ∞ � S ( f ) = F t s · δ ( t − t s n ) ( f ) = n = −∞ ∞ ∞ ∞ � � � f − n � � δ ( t − t s n ) e − 2 π j ft d t = t s · δ . t s n = −∞ n = −∞ −∞ s ( t ) S ( f ) − 2 t s − t s 0 t s 2 t s t − 2 f s − f s 0 f s 2 f s f 52
Sampling and aliasing sample cos(2 π tf) cos(2 π t(k ⋅ f s ± f)) 0 Sampled at frequency f s , the function cos(2 πtf ) cannot be distinguished from cos[2 πt ( kf s ± f )] for any k ∈ Z . 53 Frequency-domain view of sampling x ( t ) s ( t ) x ( t ) ˆ · = . . . . . . . . . . . . t t t 0 − 1 /f s 0 1 /f s − 1 /f s 0 1 /f s ˆ X ( f ) S ( f ) X ( f ) ∗ = . . . . . . . . . . . . f f f 0 − f s f s − f s 0 f s Sampling a signal in the time domain corresponds in the frequency domain to convolving its spectrum with a Dirac comb. The resulting copies of the original signal spectrum in the spectrum of the sampled signal are called “images”. 54
Discrete-time Fourier transform (DTFT) The Fourier transform of a sampled signal ∞ � x ( t ) = t s · ˆ x n · δ ( t − t s · n ) n = −∞ is � ∞ ∞ � x n · e − 2 π j f x ( t ) } ( f ) = ˆ f s n x ( t ) · e − 2 π j ft d t = t s · F{ ˆ X ( f ) = ˆ −∞ n = −∞ The inverse transform is � ∞ � f s / 2 X ( f ) · e 2 π j f ˆ ˆ f s m d f. X ( f ) · e 2 π j ft d f x ( t ) = ˆ or x m = −∞ − f s / 2 The DTFT is also commonly expressed using the normalized frequency ω = 2 π f ˙ f s (radians per sample), and the notation � X (e j ˙ ω ) = x n · e − j ˙ ωn n is customary, to highlight both the periodicity of the DTFT and its relationship with the z -transform of { x n } (see slide 123). 55 1 8 DTFT real DTFT imag 0.8 6 0.6 4 0.4 2 0.2 0 0 -2 -5 0 5 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 1 8 DTFT real DTFT imag 0.8 6 0.6 4 0.4 2 0.2 0 0 -2 -5 0 5 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 56
1 8 DTFT real DTFT imag 0.8 6 0.6 4 0.4 2 0.2 0 0 -2 -5 0 5 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 1 8 DTFT real DTFT imag 0.8 6 0.6 4 0.4 2 0.2 0 0 -2 -5 0 5 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 57 1 8 DTFT real DTFT imag 0.8 6 0.6 4 0.4 2 0.2 0 0 -2 -5 0 5 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 1 8 DTFT real DTFT imag 6 0.5 4 0 2 -0.5 0 -1 -2 -5 0 5 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 58
1 8 DTFT real DTFT imag 6 0.5 4 0 2 -0.5 0 -1 -2 -5 0 5 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 1 8 DTFT real DTFT imag 0.8 6 0.6 4 0.4 2 0.2 0 0 -2 -5 0 5 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 59 Properties of the DTFT The DTFT is periodic: X ( f ) = ˆ ˆ X (e j ˙ ω ) = X (e j( ˙ ω +2 π k ) ) X ( f + kf s ) or ∀ k ∈ Z Beyond that, the DTFT is just the Fourier transform applied to a discrete sequence, and inherits the properties of the continuous Fourier transform, e.g. ◮ Linearity ◮ Symmetries ◮ Convolution and modulation theorem: X (e j ˙ ω ) · Y (e j ˙ ω ) = Z (e j ˙ ω ) { x n } ∗ { y n } = { z n } ⇐ ⇒ and � π ω ′ = Z (e j ˙ ω ′ ) · Y (e j( ˙ ω ′ ) ) d ˙ X (e j ˙ ω − ˙ ω ) x n · y n = z n ⇐ ⇒ − π 60
Nyquist limit and anti-aliasing filters If the (double-sided) bandwidth of a signal to be sampled is larger than the sampling frequency f s , the images of the signal that emerge during sampling may overlap with the original spectrum. Such an overlap will hinder reconstruction of the original continuous signal by removing the aliasing frequencies with a reconstruction filter . Therefore, it is advisable to limit the bandwidth of the input signal to the sampling frequency f s before sampling, using an anti-aliasing filter . In the common case of a real-valued base-band signal (with frequency content down to 0 Hz), all frequencies f that occur in the signal with non-zero power should be limited to the interval − f s / 2 < f < f s / 2. The upper limit f s / 2 for the single-sided bandwidth of a baseband signal is known as the “Nyquist limit”. 61 Nyquist limit and anti-aliasing filters Without anti-aliasing filter With anti-aliasing filter single-sided Nyquist X ( f ) X ( f ) bandwidth limit = f s / 2 anti-aliasing filter 0 f − f s 0 f s f double-sided bandwidth ˆ ˆ reconstruction filter X ( f ) X ( f ) − 2 f s − f s 0 f s 2 f s f − 2 f s − f s 0 f s 2 f s f Anti-aliasing and reconstruction filters both suppress frequencies outside | f | < f s / 2. 62
Reconstruction of a continuous band-limited waveform The ideal anti-aliasing filter for eliminating any frequency content above f s / 2 before sampling with a frequency of f s has the Fourier transform � if | f | < f s 1 2 H ( f ) = = rect( t s f ) . if | f | > f s 0 2 This leads, after an inverse Fourier transform, to the impulse response � t � h ( t ) = f s · sin π tf s = 1 · sinc . π tf s t s t s The original band-limited signal can be reconstructed by convolving this with the sampled signal ˆ x ( t ), which eliminates the periodicity of the frequency domain introduced by the sampling process: x ( t ) = h ( t ) ∗ ˆ x ( t ) Note that sampling h ( t ) gives the impulse function: h ( t ) · s ( t ) = δ ( t ) . 63 Impulse response of ideal low-pass filter with cut-off frequency f s / 2: 0 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3 t ⋅ f s 64
Reconstruction filter example sampled signal interpolation result scaled/shifted sin(x)/x pulses 1 2 3 4 5 65 If before being sampled with x n = x ( t/f s ) the signal x ( t ) satisfied the Nyquist limit � ∞ x ( t ) · e − 2 π j ft d t = 0 for all | f | ≥ f s F{ x ( t ) } ( f ) = 2 −∞ � � then it can be reconstructed by interpolation with h ( t ) = 1 t t s sinc : t s � ∞ x ( t ) = h ( s ) · ˆ x ( t − s ) · d s −∞ � s � ∞ ∞ � 1 � = sinc · t s x n · δ ( t − s − t s · n ) · d s t s t s −∞ n = −∞ � s � ∞ ∞ � � = x n · sinc · δ ( t − s − t s · n ) · d s t s −∞ n = −∞ ∞ � t − t s · n � ∞ � � x n · sinc = = x n · sinc( t/t s − n ) t s n = −∞ n = −∞ ∞ x n · sin π ( t/t s − n ) � = π ( t/t s − n ) n = −∞ 66
Reconstruction filters The mathematically ideal form of a reconstruction filter for suppressing aliasing frequencies interpolates the sampled signal x n = x ( t s · n ) back into the continuous waveform ∞ x n · sin π ( t/t s − n ) � x ( t ) = π ( t/t s − n ) . n = −∞ Choice of sampling frequency Due to causality and economic constraints, practical analog filters can only approximate such an ideal low-pass filter. Instead of a sharp transition between the “pass band” ( < f s / 2) and the “stop band” ( > f s / 2), they feature a “transition band” in which their signal attenuation gradually increases. The sampling frequency is therefore usually chosen somewhat higher than twice the highest frequency of interest in the continuous signal (e.g., 4 × ). On the other hand, the higher the sampling frequency, the higher are CPU, power and memory requirements. Therefore, the choice of sampling frequency is a tradeoff between signal quality, analog filter cost and digital subsystem expenses. 67 Band-pass signal sampling Sampled signals can also be reconstructed if their spectral components remain entirely within the interval n · f s / 2 < | f | < ( n + 1) · f s / 2 for some n ∈ N . (The baseband case discussed so far is just n = 0.) In this case, the aliasing copies of the positive and the negative frequencies will interleave instead of overlap, and can therefore be removed again with a reconstruction filter with the impulse response sin π tf s / 2 � 2 n + 1 � sin π t ( n + 1) f s sin π tnf s h ( t ) = f s · cos 2 π tf s = ( n + 1) f s − nf s . π tf s / 2 4 π t ( n + 1) f s π tnf s ˆ X ( f ) X ( f ) anti-aliasing filter reconstruction filter − 5 5 4 f s 0 4 f s f − f s 0 f s f − f s f s 2 2 n = 2 68
IQ sampling / downconversion / complex baseband signal Consider signal x ( t ) ∈ R in which only frequencies f l < | f | < f h are of interest. This band has a centre frequency of f c = ( f l + f h ) / 2 and a bandwidth B = f h − f l . It can be sampled efficiently (at the lowest possible sampling frequency) by downconversion : ◮ Shift its spectrum by − f c : y ( t ) = x ( t ) · e − 2 π j f c t ◮ Low-pass filter it with a cut-off frequency of B/ 2: � ∞ z ( t ) = B y ( τ ) · sinc(( t − τ ) B ) · d τ • − ◦ Z ( f ) = Y ( f ) · rect( f/B ) −∞ ◮ Sample the result at sampling frequency f s ≥ B : z n = z ( n/f s ) 69 X ( f ) δ ( f + f c ) ∗ − f c 0 f c f − f c 0 f c f ˆ Y ( f ) Z ( f ) Z ( f ) anti-aliasing filter sample − → − 2 f c − f c 0 f c f − 2 f c − f c 0 B f c f − B B 2 2 Shifting the center frequency f c of the interval of interest to 0 Hz (DC) makes the spectrum asymmetric. This leads to a complex-valued time-domain representation ( ∃ f : Z ( f ) � = [ Z ( − f )] ∗ = ⇒ ∃ t : z ( t ) ∈ C \ R ). 70
IQ upconversion / interpolation Given a discrete sequence of downconverted samples z n ∈ C recorded with sampling frequency f s at centre frequency f c (as on slide 69), how can we reconstruct a continuous waveform ˜ x ( t ) ∈ R that matches the original signal x ( t ) within the frequency interval f l to f h ? Reconstruction steps: ◮ Interpolation of complex baseband signal (remove aliases): ∞ � z ( t ) = ˜ z n · sinc( t · f s − n ) n = −∞ ◮ Upconvert by modulating a complex phasor at carrier frequency f c . Then discard the imaginary part (to reconstruct the negative frequency components of the original real-valued signal): � z ( t ) · e 2 π j f c t � x ( t ) = 2 ℜ ˜ ˜ �� �� �� � � � � = 2 ℜ ℜ z ( t ) ˜ + j ℑ z ( t ) ˜ · cos 2 π f c t + j sin 2 π f c t � � � � = 2 ℜ z ( t ) ˜ · cos 2 π f c t − 2 ℑ z ( t ) ˜ · sin 2 π f c t Recall that 2 ℜ ( c ) = c + c ∗ for all c ∈ C. 71 Example: IQ downconversion of a sine wave What happens if we downconvert the input signal x ( t ) = A · cos(2 π ft + φ ) = A 2 · e 2 π j ft +j φ + A 2 · e − 2 π j ft − j φ using centre frequency f c and bandwidth B < 2 f c with | f − f c | < B/ 2? After frequency shift: y ( t ) = x ( t ) · e − 2 π j f c t = A 2 · e 2 π j( f − f c ) t +j φ + A 2 · e − 2 π j( f + f c ) t − j φ After low-pass filter with cut-off frequency B/ 2 < f c < f + f c : z ( t ) = A 2 · e 2 π j( f − f c ) t +j φ After sampling: z n = A 2 · e 2 π j( f − f c ) n/f s +j φ 72
Software-defined radio (SDR) front end IQ downconversion in SDR receiver: ⊗ sample Q − 90 ◦ x ( t ) y ( t ) z ( t ) z n ⊗ sample I The real part ℜ ( z ( t )) is also known as “in-phase” signal (I) and the imaginary part ℑ ( z ( t )) as “quadrature” signal (Q). cos(2 π f c t ) IQ upconversion in SDR transmitter: ⊗ δ Q +90 ◦ x ( t ) ˜ z ( t ) ˜ z ( t ) ˆ z n ⊗ δ I In SDR, x ( t ) is the antenna voltage and z n appears on the digital interface with the microprocessor. cos(2 π f c t ) 73 Visualization of IQ representation of sine waves ⊗ Q Q − 90 ◦ x ( t ) y ( t ) z ( t ) ⊗ I cos(2 π f c t ) I Recall these products of sine and cosine functions: ◮ cos( x ) · cos( y ) = 1 2 cos( x − y ) + 1 2 cos( x + y ) ◮ sin( x ) · sin( y ) = 1 2 cos( x − y ) − 1 2 cos( x + y ) ◮ sin( x ) · cos( y ) = 1 2 sin( x − y ) + 1 2 sin( x + y ) Consider: (with x = 2 π f c t ) ◮ sin( x ) = cos( x − 1 2 π ) ◮ cos( x ) · cos( x ) = 1 2 + 1 2 cos 2 x ◮ sin( x ) · sin( x ) = 1 2 − 1 2 cos 2 x ◮ sin( x ) · cos( x ) = 0 + 1 2 sin 2 x ◮ cos( x ) · cos( x − ϕ ) = 1 2 cos( ϕ ) + 1 2 cos(2 x − ϕ ) ◮ sin( x ) · cos( x − ϕ ) = 1 2 sin( ϕ ) + 1 2 sin(2 x − ϕ ) 74
IQ representation of amplitude-modulated signals Assume voice signal s ( t ) contains only frequencies below B/ 2. Antenna signal amplitude-modulated with carrier frequency f c : x ( t ) = s ( t ) · A · cos(2 π f c t + ϕ ) After IQ downconversion with centre frequency f ′ c ≈ f c : z ( t ) = A 2 · s ( t ) · e 2 π j( f c − f ′ c ) t +j ϕ ℑ [ z ( t )] With perfect receiver tuning f ′ c = f c : z ( t ) = A 2 · s ( t ) · e j ϕ ℜ [ z ( t )] Reception techniques: ◮ Non-coherent demodulation (requires s ( t ) ≥ 0): s ( t ) = 2 A | z ( t ) | ◮ Coherent demodulation (requires knowing ϕ and f ′ c = f c ): s ( t ) = 2 A ℜ [ z ( t ) · e − j ϕ ] 75 IQ representation of frequency-modulated signals In frequency modulation, the voice signal s ( t ) changes the carrier frequency f c : f c ( t ) = f c + k · s ( t ) Compared to a constant-frequency carrier signal cos(2 π f c t + ϕ ), to allow variable frequency, we � need to replace the phase-accumulating term 2 π f c t with an integral 2 π f c ( t )d t . Frequency-modulated antenna signal: � t � � x ( t ) = A · cos 2 π · [ f c + k · s ( τ )]d τ + ϕ 0 � t � � = A · cos 2 π f c t + 2 π k · s ( τ )d τ + ϕ 0 After IQ downconversion from centre frequency f c : z ( t ) = A � t 2 · e 2 π j k 0 s ( τ )d τ +j ϕ Therefore, s ( t ) is proportional to the rotational rate of z ( t ). 76
Frequency demodulating IQ samples � t Determine s ( t ) from downconverted signal z ( t ) = A 2 · e 2 π j k 0 s ( τ )d τ +j ϕ . First idea: measure the angle ∠ z ( t ), where the angle operator ∠ is defined such that ∠ a e j φ = φ ( a, φ ∈ R , a > 0). Then take its derivative: 1 d s ( t ) = d t ∠ z ( t ) 2 π k Problem: angle ambiguity, ∠ works only for − π ≤ φ < π . Ugly hack: MATLAB function unwrap removes 2 π jumps from sample sequences Better idea: first take the complex derivative d z ( t ) = A � t 2 · 2 π j k · s ( t ) · e 2 π j k 0 s ( τ )d τ +j ϕ d t ℑ [ z ( t )] d z ( t ) then divide by z ( t ): d t /z ( t ) = 2 π j k · s ( t ) Other practical approaches: � � d z ( t ) ◮ s ( t ) ∝ ℑ · z ∗ ( t ) / | z ( t ) | 2 ℜ [ z ( t )] d t z ( t ) ◮ s ( t ) ∝ ∠ z ( t − ∆ t ) / ∆ t 77 Digital modulation schemes Pick z n ∈ C from a constellation of 2 n symbols to encode n bits: ASK BPSK QPSK 10 00 0 1 0 1 11 01 8PSK 16QAM FSK 101 1 111 100 00 0 01 110 000 11 10 010 001 00 01 11 10 011 78
Spectrum of a periodic signal A signal x ( t ) that is periodic with frequency f p can be factored into a single period ˙ x ( t ) convolved with an impulse comb p ( t ). This corresponds in the frequency domain to the multiplication of the spectrum of the single period with a comb of impulses spaced f p apart. x ( t ) x ( t ) ˙ p ( t ) = ∗ . . . . . . . . . . . . t t t − 1 /f p 0 1 /f p − 1 /f p 0 1 /f p ˙ X ( f ) X ( f ) P ( f ) = · . . . . . . f f f − f p 0 f p − f p 0 f p 79 Spectrum of a sampled signal A signal x ( t ) that is sampled with frequency f s has a spectrum that is periodic with a period of f s . x ( t ) s ( t ) x ( t ) ˆ · = . . . . . . . . . . . . t t t 0 − 1 /f s 0 1 /f s − 1 /f s 0 1 /f s ˆ X ( f ) S ( f ) X ( f ) ∗ = . . . . . . . . . . . . f f f 0 − f s f s − f s 0 f s 80
Continuous vs discrete Fourier transform ◮ Sampling a continuous signal makes its spectrum periodic ◮ A periodic signal has a sampled spectrum We sample a signal x ( t ) with f s , getting ˆ x ( t ). We take n consecutive samples of ˆ x ( t ) and repeat these periodically, getting a new signal ¨ x ( t ) with period n/f s . Its spectrum ¨ X ( f ) is sampled (i.e., has non-zero value) at frequency intervals f s /n and repeats itself with a period f s . x ( t ) and its spectrum ¨ Now both ¨ X ( f ) are finite vectors of length n . ¨ ¨ x ( t ) X ( f ) . . . . . . . . . . . . f − 1 f − 1 t f − n/f s 0 n/f s − f s − f s /n 0 f s /n f s s s 81 Discrete Fourier Transform (DFT) n − 1 n − 1 x k = 1 � � x i · e − 2 π j ik X i · e 2 π j ik X k = n · n n i =0 i =0 The n -point DFT multiplies a vector with an n × n matrix 1 1 1 1 · · · 1 e − 2 π j n − 1 e − 2 π j 1 e − 2 π j 2 e − 2 π j 3 1 · · · n n n n e − 2 π j 2( n − 1) e − 2 π j 2 e − 2 π j 4 e − 2 π j 6 1 · · · n n n n e − 2 π j 3( n − 1) F n = e − 2 π j 3 e − 2 π j 6 e − 2 π j 9 1 · · · n n n n . . . . . ... . . . . . . . . . . e − 2 π j n − 1 e − 2 π j 2( n − 1) e − 2 π j 3( n − 1) e − 2 π j ( n − 1)( n − 1) 1 · · · n n n n x 0 X 0 X 0 x 0 x 1 X 1 X 1 x 1 1 x 2 X 2 X 2 x 2 n · F ∗ F n · = , n · = . . . . . . . . . . . . x n − 1 X n − 1 X n − 1 x n − 1 82
Discrete Fourier Transform visualized x 0 X 0 x 1 X 1 x 2 X 2 x 3 X 3 · = x 4 X 4 x 5 X 5 x 6 X 6 x 7 X 7 The n -point DFT of a signal { x i } sampled at frequency f s contains in the elements X 0 to X n/ 2 of the resulting frequency-domain vector the frequency components 0, f s /n , 2 f s /n , 3 f s /n , . . . , f s / 2, and contains in X n − 1 downto X n/ 2 the corresponding negative frequencies. Note that for a real-valued input vector, both X 0 and X n/ 2 will be real, too. Why is there no phase information recovered at f s / 2? 83 Inverse DFT visualized X 0 x 0 X 1 x 1 X 2 x 2 1 X 3 x 3 8 · · = X 4 x 4 X 5 x 5 X 6 x 6 X 7 x 7 84
Fast Fourier Transform (FFT) n − 1 � � x i · e − 2 π j ik F n { x i } n − 1 � k = n i =0 i =0 n n 2 − 1 2 − 1 − 2 π j ik n/ 2 + e − 2 π j k − 2 π j ik � � = x 2 i · e x 2 i +1 · e n/ 2 n i =0 i =0 n n � 2 − 1 � � 2 − 1 � + e − 2 π j k n · k < n F n 2 { x 2 i } F n 2 { x 2 i +1 } k , i =0 i =0 2 k = n n � 2 − 1 � � 2 − 1 � + e − 2 π j k n · k ≥ n F n 2 { x 2 i } F n 2 { x 2 i +1 } , i =0 i =0 2 k − n k − n 2 2 The DFT over n -element vectors can be reduced to two DFTs over n/ 2-element vectors plus n multiplications and n additions, leading to log 2 n rounds and n log 2 n additions and multiplications overall, compared to n 2 for the equivalent matrix multiplication. A high-performance FFT implementation in C with many processor-specific optimizations and support for non-power-of-2 sizes is available at http://www.fftw.org/ . 85 Efficient real-valued FFT The symmetry properties of the Fourier transform applied to the discrete Fourier transform { X i } n − 1 i =0 = F n { x i } n − 1 i =0 have the form X ∗ ∀ i : x i = ℜ ( x i ) ⇐ ⇒ ∀ i : X n − i = i ∀ i : X n − i = − X ∗ ∀ i : x i = j · ℑ ( x i ) ⇐ ⇒ i These two symmetries, combined with the linearity of the DFT, allows us to calculate two real-valued n -point DFTs i } n − 1 i } n − 1 i } n − 1 i } n − 1 { X ′ i =0 = F n { x ′ { X ′′ i =0 = F n { x ′′ i =0 i =0 simultaneously in a single complex-valued n -point DFT, by composing its input as x i = x ′ i + j · x ′′ i and decomposing its output as i = 1 i = 1 X ′ 2( X i + X ∗ X ′′ 2j( X i − X ∗ n − i ) n − i ) where X n = X 0 . To optimize the calculation of a single real-valued FFT, use this trick to calculate the two half-size real-value FFTs that occur in the first round. 86
Fast complex multiplication Calculating the product of two complex numbers as ( a + j b ) · ( c + j d ) = ( ac − bd ) + j( ad + bc ) involves four (real-valued) multiplications and two additions. The alternative calculation α = a ( c + d ) ( a + j b ) · ( c + j d ) = ( α − β ) + j( α + γ ) with β = d ( a + b ) γ = c ( b − a ) provides the same result with three multiplications and five additions. The latter may perform faster on CPUs where multiplications take three or more times longer than additions. This “Karatsuba multiplication” is most helpful on simpler microcontrollers. Specialized signal-processing CPUs (DSPs) feature 1-clock-cycle multipliers. High-end desktop processors use pipelined multipliers that stall where operations depend on each other. 87 FFT-based convolution Calculating the convolution of two finite sequences { x i } m − 1 and { y i } n − 1 i =0 i =0 of lengths m and n via min { m − 1 ,i } � z i = x j · y i − j , 0 ≤ i < m + n − 1 j =max { 0 ,i − ( n − 1) } takes mn multiplications. Can we apply the FFT and the convolution theorem to calculate the convolution faster, in just O ( m log m + n log n ) multiplications? { z i } = F − 1 ( F{ x i } · F{ y i } ) There is obviously no problem if this condition is fulfilled: { x i } and { y i } are periodic, with equal period lengths In this case, the fact that the DFT interprets its input as a single period of a periodic signal will do exactly what is needed, and the FFT and inverse FFT can be applied directly as above. 88
In the general case, measures have to be taken to prevent a wrap-over: F −1 [F(A) ⋅ F(B)] A B F −1 [F(A’) ⋅ F(B’)] A’ B’ Both sequences are padded with zero values to a length of at least m + n − 1. This ensures that the start and end of the resulting sequence do not overlap. 89 Zero padding is usually applied to extend both sequence lengths to the next higher power of two (2 ⌈ log 2 ( m + n − 1) ⌉ ), which facilitates the FFT. With a causal sequence, simply append the padding zeros at the end. With a non-causal sequence, values with a negative index number are wrapped around the DFT block boundaries and appear at the right end. In this case, zero-padding is applied in the center of the block, between the last and first element of the sequence. Thanks to the periodic nature of the DFT, zero padding at both ends has the same effect as padding only at one end. If both sequences can be loaded entirely into RAM, the FFT can be applied to them in one step. However, one of the sequences might be too large for that. It could also be a realtime waveform (e.g., a telephone signal) that cannot be delayed until the end of the transmission. In such cases, the sequence has to be split into shorter blocks that are separately convolved and then added together with a suitable overlap. 90
Each block is zero-padded at both ends and then convolved as before: ∗ ∗ ∗ = = = The regions originally added as zero padding are, after convolution, aligned to overlap with the unpadded ends of their respective neighbour blocks. The overlapping parts of the blocks are then added together. 91 Deconvolution A signal u ( t ) was distorted by convolution with a known impulse response h ( t ) (e.g., through a transmission channel or a sensor problem). The “smeared” result s ( t ) was recorded. Can we undo the damage and restore (or at least estimate) u ( t )? ∗ = ∗ = 92
The convolution theorem turns the problem into one of multiplication: � s ( t ) = u ( t − τ ) · h ( τ ) · d τ s = u ∗ h F{ s } = F{ u } · F{ h } F{ u } = F{ s } / F{ h } F − 1 {F{ s } / F{ h }} u = In practice, we also record some noise n ( t ) (quantization, etc.): � c ( t ) = s ( t ) + n ( t ) = u ( t − τ ) · h ( τ ) · d τ + n ( t ) Problem – At frequencies f where F{ h } ( f ) approaches zero, the noise will be amplified (potentially enormously) during deconvolution: u = F − 1 {F{ c } / F{ h }} = u + F − 1 {F{ n } / F{ h }} ˜ 93 Typical workarounds: ◮ Modify the Fourier transform of the impulse response, such that |F{ h } ( f ) | > ǫ for some experimentally chosen threshold ǫ . ◮ If estimates of the signal spectrum |F{ s } ( f ) | and the noise spectrum |F{ n } ( f ) | can be obtained, then we can apply the “Wiener filter” (“optimal filter”) |F{ s } ( f ) | 2 W ( f ) = |F{ s } ( f ) | 2 + |F{ n } ( f ) | 2 before deconvolution: u = F − 1 { W · F{ c } / F{ h }} ˜ 94
Vowel "A" sung at varying pitch 8000 7000 6000 Frequency (Hz) 5000 4000 3000 2000 1000 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Time [w,fs, bits] = auread('sing.au'); specgram(w,2048,fs); ylim([0 8e3]); xlim([0 4.5]); saveas(gcf, 'sing.eps', 'eps2c'); 95 Different vovels at constant pitch 8000 7000 6000 Frequency (Hz) 5000 4000 3000 2000 1000 0 0.5 1 1.5 2 2.5 3 3.5 4 Time 96
Spectral estimation cos(2 *[0:15]/16*4) 1 12 DTFT mag 10 DFT mag 0.5 8 0 6 4 -0.5 2 -1 0 0 5 10 15 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) cos(2 *[0:15]/16*4.2) 1 12 DTFT mag 10 DFT mag 0.5 8 0 6 4 -0.5 2 -1 0 0 5 10 15 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 97 We introduced the DFT as a special case of the continuous Fourier transform, where the input is sampled and periodic . If the input is sampled, but not periodic, the DFT can still be used to calculate an approximation of the Fourier transform of the original continuous signal. However, there are two effects to consider. They are particularly visible when analysing pure sine waves. Sine waves whose frequency is a multiple of the base frequency ( f s /n ) of the DFT are identical to their periodic extension beyond the size of the DFT. They are, therefore, represented exactly by a single sharp peak in the DFT. All their energy falls into one single frequency “bin” in the DFT result. Sine waves with other frequencies, which do not match exactly one of the output frequency bins of the DFT, are still represented by a peak at the output bin that represents the nearest integer multiple of the DFT’s base frequency. However, such a peak is distorted in two ways: ◮ Its amplitude is lower (down to 63.7%). ◮ Much signal energy has “leaked” to other frequencies. 98
35 30 25 20 15 10 5 0 16 0 5 10 15.5 15 20 25 30 15 input freq. DFT index The leakage of energy to other frequency bins not only blurs the estimated spectrum. The peak amplitude also changes significantly as the frequency of a tone changes from that associated with one output bin to the next, a phenomenon known as scalloping . In the above graphic, an input sine wave gradually changes from the frequency of bin 15 to that of bin 16 (only positive frequencies shown). 99 Windowing Sine wave Discrete Fourier Transform 300 1 200 0 100 −1 0 0 200 400 0 200 400 Sine wave multiplied with window function Discrete Fourier Transform 100 1 0 50 −1 0 0 200 400 0 200 400 100
The reason for the leakage and scalloping losses is easy to visualize with the help of the convolution theorem: The operation of cutting a sequence of the size of the DFT input vector out of a longer original signal (the one whose continuous Fourier spectrum we try to estimate) is equivalent to multiplying this signal with a rectangular function. This destroys all information and continuity outside the “window” that is fed into the DFT. Multiplication with a rectangular window of length T in the time domain is equivalent to convolution with sin( πfT ) / ( πfT ) in the frequency domain. The subsequent interpretation of this window as a periodic sequence by the DFT leads to sampling of this convolution result (sampling meaning multiplication with a Dirac comb whose impulses are spaced f s /n apart). Where the window length was an exact multiple of the original signal period, sampling of the sin( πfT ) / ( πfT ) curve leads to a single Dirac pulse, and the windowing causes no distortion. In all other cases, the effects of the convolution become visible in the frequency domain as leakage and scalloping losses. 101 Some better window functions 1 0.8 0.6 0.4 0.2 Rectangular window Triangular window 0 Hann window Hamming window 0 0.2 0.4 0.6 0.8 1 All these functions are 0 outside the interval [0,1]. 102
cos(2 *[0:15]/16*4.2) 1 12 DTFT mag 10 DFT mag 0.5 8 0 6 4 -0.5 2 -1 0 0 5 10 15 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) cos(2 *[0:15]/16*4.2).*hann(16) 1 12 DTFT mag 10 DFT mag 0.5 8 0 6 4 -0.5 2 -1 0 0 5 10 15 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 103 Rectangular window (64−point) Triangular window 20 20 Magnitude (dB) Magnitude (dB) 0 0 −20 −20 −40 −40 −60 −60 0 0.5 1 0 0.5 1 Normalized Frequency ( ×π rad/sample) Normalized Frequency ( ×π rad/sample) Hann window Hamming window 20 20 Magnitude (dB) Magnitude (dB) 0 0 −20 −20 −40 −40 −60 −60 0 0.5 1 0 0.5 1 Normalized Frequency ( ×π rad/sample) Normalized Frequency ( ×π rad/sample) 104
Numerous alternatives to the rectangular window have been proposed that reduce leakage and scalloping in spectral estimation. These are vectors multiplied element-wise with the input vector before applying the DFT to it. They all force the signal amplitude smoothly down to zero at the edge of the window, thereby avoiding the introduction of sharp jumps in the signal when it is extended periodically by the DFT. Three examples of such window vectors { w i } n − 1 i =0 are: Triangular window (Bartlett window): � � i � � w i = 1 − � 1 − � � n/ 2 � Hann window (raised-cosine window, Hanning window): � � i w i = 0 . 5 − 0 . 5 × cos 2 π n − 1 Hamming window : � � i w i = 0 . 54 − 0 . 46 × cos 2 π n − 1 105 Does zero padding increase DFT resolution? The two figures below show two spectra of the 16-element sequence s i = cos(2 π · 3 i/ 16) + cos(2 π · 4 i/ 16) , i ∈ { 0 , . . . , 15 } . The left plot shows the DFT of the windowed sequence x i = s i · w i , i ∈ { 0 , . . . , 15 } and the right plot shows the DFT of the zero-padded windowed sequence � s i · w i , i ∈ { 0 , . . . , 15 } x ′ i = 0 , i ∈ { 16 , . . . , 63 } where w i = 0 . 54 − 0 . 46 × cos (2 πi/ 15) is the Hamming window. DFT without zero padding DFT with 48 zeros appended to window 4 4 2 2 0 0 0 5 10 15 0 20 40 60 106
cos(2 *[0:15]/16*3.3) + cos(2 *[0:15]/16*4) 2 8 DTFT mag DFT mag 1 6 0 4 -1 2 -2 0 0 5 10 15 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) zero-padded to 64 samples 2 8 DTFT mag DFT mag 1 6 0 4 -1 2 -2 0 0 20 40 60 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 107 cos(2 *[0:15]/16*3.3) + cos(2 *[0:15]/16*4) 2 8 DTFT mag DFT mag 1 6 0 4 -1 2 -2 0 0 5 10 15 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) with 64 actual samples 2 35 DTFT mag 30 DFT mag 1 25 20 0 15 10 -1 5 -2 0 0 20 40 60 - -¾ -½ -¼ 0 ¼ ½ ¾ time-domain samples DTFT frequency (1 period) 108
Applying the discrete Fourier transform (DFT) to an n -element long real-valued sequence samples the DTFT of that sequence at n/ 2 + 1 discrete frequencies. The DTFT spectrum has already been distorted by multiplying the (hypothetically longer) signal with a windowing function that limits its length to n non-zero values and forces the waveform down to zero outside the window. Therefore, appending further zeros outside the window will not affect the DTFT. The frequency resolution of the DFT is the sampling frequency divided by the block size of the DFT. Zero padding can therefore be used to increase the frequency resolution of the DFT, to sample the DTFT at more places. But that does not change the limit imposed on the frequency resolution (i.e., blurriness) of the DTFT by the length of the window. Note that zero padding does not add any additional information to the signal. The DTFT has already been “low-pass filtered” by being convolved with the spectrum of the windowing function. Zero padding in the time domain merely causes the DFT to sample the same underlying DTFT spectrum at a higher resolution, thereby making it easier to visually distinguish spectral lines and to locate their peak more precisely. 109 Digital filters Filter: supresses (removes, attenuates) unwanted signal components. ◮ low-pass filter – suppress all frequencies above a cut-off frequency ◮ high-pass filter – suppress all frequencies below a cut-off frequency, including DC (direct current = 0 Hz) ◮ band-pass filter – suppress signals outside a frequency interval (= passband) ◮ band-stop filter (aka: band-reject filter) – suppress signals inside a single frequency interval (= stopband) ◮ notch filter – narrow band-stop filter, ideally suppressing only a single frequency For digital filters, we also distinguish ◮ finite impulse response (FIR) filters ◮ infinite impulse response (IIR) filters depending on how far their memory reaches back in time. 110
Window-based design of FIR filters Recall that the ideal continuous low-pass filter with cut-off frequency f c has the frequency characteristic � 1 � f � if | f | < f c H ( f ) = = rect 0 if | f | > f c 2 f c and the impulse response sin 2 π tf c h ( t ) = 2 f c = 2 f c · sinc(2 f c · t ) . 2 π tf c Sampling this impulse response with the sampling frequency f s of the signal to be processed will lead to a periodic frequency characteristic, that matches the periodic spectrum of the sampled signal. There are two problems though: ◮ the impulse response is infinitely long ◮ this filter is not causal, that is h ( t ) � = 0 for t < 0 111 Solutions: ◮ Make the impulse response finite by multiplying the sampled h ( t ) with a windowing function ◮ Make the impulse response causal by adding a delay of half the window size The impulse response of an n -th order low-pass filter is then chosen as h i = 2 f c /f s · sin[2 π ( i − n/ 2) f c /f s ] · w i 2 π ( i − n/ 2) f c /f s where { w i } is a windowing sequence, such as the Hamming window w i = 0 . 54 − 0 . 46 × cos (2 πi/n ) with w i = 0 for i < 0 and i > n . Note that for f c = f s / 4, we have h i = 0 for all even values of i . Therefore, this special case requires only half the number of multiplications during the convolution. Such “half-band” FIR filters are used, for example, as anti-aliasing filters wherever a sampling rate needs to be halved. 112
FIR low-pass filter design example z Plane Impulse Response 1 1 Imaginary Part Amplitude 0.5 0.5 30 0 −0.5 0 −1 −1 0 1 0 10 20 30 Real Part n (samples) 0 0 Phase (degrees) Magnitude (dB) −20 −500 −40 −1000 −60 −1500 0 0.5 1 0 0.5 1 Normalized Frequency ( ×π rad/sample) Normalized Frequency ( ×π rad/sample) order: n = 30, cutoff frequency ( − 6 dB): f c = 0 . 25 × f s / 2, window: Hamming 113 Filter performance An ideal filter has a gain of 1 in the pass-band and a gain of 0 in the stop band, and nothing in between. A practical filter will have ◮ frequency-dependent gain near 1 in the passband ◮ frequency-dependent gain below a threshold in the stopband ◮ a transition band between the pass and stop bands We truncate the ideal, infinitely-long impulse response by multiplication with a window sequence. In the frequency domain, this will convolve the rectangular frequency response of the ideal low-pass filter with the frequency characteristic of the window. The width of the main lobe determines the width of the transition band, and the side lobes cause ripples in the passband and stopband. 114
Low-pass to band-pass filter conversion (modulation) To obtain a band-pass filter that attenuates all frequencies f outside the range f l < f < f h , we first design a low-pass filter with a cut-off frequency ( f h − f l ) / 2. We then multiply its impulse response with a sine wave of frequency ( f h + f l ) / 2, effectively amplitude modulating it, to shift its centre frequency. Finally, we apply a window function: h i = ( f h − f l ) /f s · sin[ π ( i − n/ 2)( f h − f l ) /f s ] · cos[ π i ( f h + f l ) /f s ] · w i π ( i − n/ 2)( f h − f l ) /f s H ( f ) = ∗ − f h − f l f h − f l − f h + f l f h + f l f f f − f h − f l 0 f l f h 0 2 2 2 2 115 Low-pass to high-pass filter conversion (freq. inversion) In order to turn the spectrum X ( f ) of a real-valued signal x i sampled at f s into an inverted spectrum X ′ ( f ) = X ( f s / 2 − f ), we merely have to shift the periodic spectrum by f s / 2: X ′ ( f ) X ( f ) = ∗ . . . . . . . . . . . . − f s f s f f f − f s 0 f s − f s 0 f s 0 2 2 This can be accomplished by multiplying the sampled sequence x i with y i = cos π f s t = cos π i , which is nothing but multiplication with the sequence . . . , 1 , − 1 , 1 , − 1 , 1 , − 1 , 1 , − 1 , . . . So in order to design a discrete high-pass filter that attenuates all frequencies f outside the range f c < | f | < f s / 2, we merely have to design a low-pass filter that attenuates all frequencies outside the range − f c < f < f c , and then multiply every second value of its impulse response with − 1. 116
Finite impulse response (FIR) filter M � y n = b m · x n − m m =0 M = 3: x n x n − 1 x n − 2 x n − 3 z − 1 z − 1 z − 1 b 0 b 1 b 2 b 3 y n (see slide 23) Transposed implementation: x n b 3 b 2 b 1 b 0 z − 1 z − 1 z − 1 y n 117 Infinite impulse response (IIR) filter N M � � a k · y n − k = b m · x n − m Usually normalize: a 0 = 1 m =0 k =0 � M � N � � y n = b m · x n − m − a k · y n − k /a 0 m =0 k =1 Direct form I implementation: a − 1 x n b 0 y n 0 z − 1 z − 1 b 1 − a 1 x n − 1 y n − 1 z − 1 z − 1 b 2 − a 2 x n − 2 y n − 2 z − 1 z − 1 b 3 − a 3 x n − 3 y n − 3 118
Infinite impulse response (IIR) filter – direct form II � M � N � � y n = b m · x n − m − a k · y n − k /a 0 m =0 k =1 Direct form II: Transposed direct form II: a − 1 a − 1 x n b 0 y n x n b 0 y n 0 0 z − 1 z − 1 − a 1 b 1 b 1 − a 1 z − 1 z − 1 − a 2 b 2 b 2 − a 2 z − 1 z − 1 − a 3 b 3 b 3 − a 3 119 Polynomial representation of sequences We can represent sequences { x n } as polynomials: ∞ � x n v n X ( v ) = n = −∞ Example of polynomial multiplication: 3 v 2 ) (1 + 2 v + · (2 + 1 v ) 6 v 2 2 + 4 v + 2 v 2 3 v 3 + 1 v + + 8 v 2 3 v 3 = 2 + 5 v + + Compare this with the convolution of two sequences (in MATLAB): conv([1 2 3], [2 1]) equals [2 5 8 3] 120
Convolution of sequences is equivalent to polynomial multiplication: ∞ � { h n } ∗ { x n } = { y n } ⇒ y n = h k · x n − k k = −∞ ↓ ↓ � � � � ∞ ∞ � � h n v n x n v n H ( v ) · X ( v ) = · n = −∞ n = −∞ ∞ ∞ � � h k · x n − k · v n = n = −∞ k = −∞ Note how the Fourier transform of a sequence can be accessed easily from its polynomial form: ∞ � X (e − j ˙ ω ) = x n e − j ˙ ωn n = −∞ 121 Example of polynomial division: ∞ 1 � 1 − av = 1 + av + a 2 v 2 + a 3 v 3 + · · · = a n v n n =0 a 2 v 2 1 + av + + · · · 1 − av 1 1 − av av x n y n a 2 v 2 av − a 2 v 2 v a 2 v 2 a 3 v 3 a − · · · y n − 1 Rational functions (quotients of two polynomials) can provide a convenient closed-form representations for infinitely-long exponential sequences, in particular the impulse responses of IIR filters. 122
The z -transform The z -transform of a sequence { x n } is defined as: ∞ � x n z − n X ( z ) = n = −∞ Note that this differs only in the sign of the exponent from the polynomial representation discussed on the preceeding slides. Recall that the above X ( z ) is exactly the factor with which an exponential sequence { z n } is multiplied, if it is convolved with { x n } : { z n } ∗ { x n } = { y n } ∞ ∞ � � z n − k x k = z n · z − k x k = z n · X ( z ) ⇒ y n = k = −∞ k = −∞ 123 The z -transform defines for each sequence a continuous complex-valued surface over the complex plane C . For finite sequences, its value is defined across the entire complex plane (except possibly at z = 0 or | z | = ∞ ). For infinite sequences, it can be shown that the z -transform converges only for the region � � � � x n +1 x n +1 � � � � lim � < | z | < lim � � � � x n x n � � � n →∞ n →−∞ The z -transform identifies a sequence unambiguously only in conjunction with a given region of convergence . In other words, there exist different sequences, that have the same expression as their z -transform, but that converge for different amplitudes of z . The z -transform is a generalization of the discrete-time Fourier transform, which it contains on the complex unit circle ( | z | = 1): ∞ � t − 1 x ( t ) } ( f ) = X (e j ˙ ω ) = x n e − j ˙ ωn · F{ ˆ s n = −∞ ω = 2 π f where ˙ f s . 124
Properties of the z -transform If X ( z ) is the z -transform of { x n } , we write here { x n } • − ◦ X ( z ). If { x n } • − ◦ X ( z ) and { y n } • − ◦ Y ( z ), then: Linearity: { ax n + by n } • − ◦ aX ( z ) + bY ( z ) Convolution: { x n } ∗ { y n } • − ◦ X ( z ) · Y ( z ) Time shift: ◦ z k X ( z ) { x n + k } • − Remember in particular: delaying by one sample is multiplication with z − 1 . 125 Time reversal: ◦ X ( z − 1 ) { x − n } • − Multiplication with exponential: { a − n x n } • − ◦ X ( az ) Complex conjugate: { x ∗ ◦ X ∗ ( z ∗ ) n } • − Real/imaginary value: ◦ 1 2( X ( z ) + X ∗ ( z ∗ )) {ℜ{ x n }} • − ◦ 1 2j( X ( z ) − X ∗ ( z ∗ )) {ℑ{ x n }} • − Initial value: x 0 = lim z →∞ X ( z ) if x n = 0 for all n < 0 126
Some example sequences and their z -transforms: x n X ( z ) δ n 1 z 1 u n z − 1 = 1 − z − 1 z 1 a n u n z − a = 1 − az − 1 z nu n ( z − 1) 2 z ( z + 1) n 2 u n ( z − 1) 3 z e an u n z − e a � n − 1 � 1 e a ( n − k ) u n − k ( z − e a ) k k − 1 z 2 sin( ϕ ) + z sin( ˙ ω − ϕ ) sin( ˙ ωn + ϕ ) u n z 2 − 2 z cos( ˙ ω ) + 1 127 Example: What is the z -transform of the impulse response { h n } of the discrete system y n = x n + ay n − 1 ? y n = x n + ay n − 1 Y ( z ) = X ( z ) + az − 1 Y ( z ) Y ( z ) − az − 1 Y ( z ) = X ( z ) Y ( z )(1 − az − 1 ) = X ( z ) Y ( z ) 1 z X ( z ) = 1 − az − 1 = z − a Since { y n } = { h n } ∗ { x n } , we have Y ( z ) = H ( z ) · X ( z ) and therefore H ( z ) = Y ( z ) z X ( z ) = z − a We have applied here the linearity of the z -transform, and its time-shift and convolution properties. 128
z -transform of recursive filter structures a − 1 x n b 0 y n Consider the discrete system defined by 0 z − 1 z − 1 k m b 1 − a 1 � � a l · y n − l = b l · x n − l x n − 1 y n − 1 l =0 l =0 z − 1 z − 1 · · · · · · or equivalently · · · · · · ? z − 1 z − 1 k m b m − a k � � a 0 y n + a l · y n − l = b l · x n − l x n − m y n − k l =1 l =0 � m � k � � y n = a − 1 · b l · x n − l − a l · y n − l 0 l =0 l =1 What is the z -transform H ( z ) of its impulse response { h n } , where { y n } = { h n } ∗ { x n } ? 129 Using the linearity and time-shift property of the z -transform: k m � � a l · y n − l = b l · x n − l l =0 l =0 k m � � a l z − l · Y ( z ) = b l z − l · X ( z ) l =0 l =0 k m � � a l z − l = X ( z ) b l z − l Y ( z ) l =0 l =0 � m l =0 b l z − l H ( z ) = Y ( z ) X ( z ) = � k l =0 a l z − l H ( z ) = b 0 + b 1 z − 1 + b 2 z − 2 + · · · + b m z − m a 0 + a 1 z − 1 + a 2 z − 2 + · · · + a k z − k 130
a − 1 The z -transform of the impulse re- x n b 0 y n 0 sponse { h n } of the causal LTI system z − 1 z − 1 defined by b 1 − a 1 x n − 1 y n − 1 k m � � a l · y n − l = b l · x n − l z − 1 z − 1 · · · · · · l =0 l =0 · · · · · · with { y n } = { h n } ∗ { x n } is the z − 1 z − 1 b m − a k rational function x n − m y n − k H ( z ) = b 0 + b 1 z − 1 + b 2 z − 2 + · · · + b m z − m a 0 + a 1 z − 1 + a 2 z − 2 + · · · + a k z − k ( b m � = 0, a k � = 0) which can also be written as H ( z ) = z k � m z m · b 0 z m + b 1 z m − 1 + b 2 z m − 2 + · · · + b m l =0 b l z m − l l =0 a l z k − l = z k . a 0 z k + a 1 z k − 1 + a 2 z k − 2 + · · · + a k z m � k H ( z ) has m zeros and k poles at non-zero locations in the z plane, plus k − m zeros (if k > m ) or m − k poles (if m > k ) at z = 0. 131 This function can be converted into the form m m � � (1 − c l · z − 1 ) ( z − c l ) H ( z ) = b 0 = b 0 · z k − m · l =1 l =1 · a 0 k a 0 k � � (1 − d l · z − 1 ) ( z − d l ) l =1 l =1 where the c l are the non-zero positions of zeros ( H ( c l ) = 0) and the d l are the non-zero positions of the poles (i.e., z → d l ⇒ | H ( z ) | → ∞ ) of H ( z ). Except for a constant factor, H ( z ) is entirely characterized by the position of these zeros and poles. On the unit circle z = e j ˙ ω , H (e j ˙ ω ) is the discrete-time Fourier transform ω = π f/ f s of { h n } ( ˙ 2 ). The DTFT amplitude can also be expressed in ω to the zeros and poles: terms of the relative position of e j ˙ � � � m ω − c l | l =1 | e j ˙ b 0 � � | H (e j ˙ ω ) | = � · � � � k ω − d l | a 0 � l =1 | e j ˙ 132
Example: a single-pole filter Consider this IIR filter: Its z -transform 0 . 8 0 . 8 z x n 0 . 8 y n H ( z ) = 1 − 0 . 2 · z − 1 = z − 0 . 2 has one pole at z = d 1 = 0 . 2 and one z − 1 0 . 2 zero at z = 0. y n − 1 Amplitude | H ( z ) | : a 0 = 1, a 1 = − 0 . 2, b 0 = 0 . 8 2 1.75 x n = δ n ⇒ y n = 1.5 1.25 Impulse Response 0.8 1 0.75 0.6 Amplitude 0.5 0.4 0.25 0 0.2 1 0.5 1 0 0.5 0 0 2 4 0 −0.5 −0.5 n (samples) −1 −1 imaginary real 133 0 . 8 0 . 8 z H ( z ) = 1 − 0 . 2 · z − 1 = z − 0 . 2 (cont’d) Magnitude Response 1 0.95 2 1.75 0.9 1.5 Magnitude 1.25 0.85 1 0.75 0.8 0.5 0.25 0.75 0 1 0.5 1 0.7 0.5 0 0 −0.5 −0.5 −1 −1 imaginary real 0 0.2 0.4 0.6 0.8 Normalized Frequency ( ×π rad/sample) Run this LTI filter at sampling frequency f s and test it with sinusoidial input (frequency f , amplitude 1): x n = cos(2 π fn/f s ) Output: y n = A ( f ) · cos(2 π fn/f s + θ ( f )) What are the gain A ( f ) and phase delay θ ( f ) at frequency f ? Answer: A ( f ) = | H (e j2 π f/f s ) | θ ( f ) = ∠ H (e j2 π f/f s ) = tan − 1 ℑ{ H (e j π f/f s ) } ℜ{ H (e j π f/f s ) } Example: f s = 8 kHz, f = 2 kHz (normalized frequency f/ f s 2 = 0 . 5) ⇒ Gain A (2 kHz) = � 0 . 82+0 . 162 � 0 . 8 j � = 0 . 8 j( − j − 0 . 2) � = � 0 . 8 − 0 . 16 j � = | H (e j π / 2 ) | = | H (j) | = � � � � � � = 0 . 784 . . . � j − 0 . 2 ( j − 0 . 2)( − j − 0 . 2) 1+0 . 04 1 . 042 134
Visual verification in MATLAB: x n = 0:15; 1.5 y (time domain) fs = 8000; y (z−transform) f = 1500; x = cos(2*pi*f*n/fs); b = [0.8]; a = [1 -0.2]; 1 y1 = filter(b,a,x); z = exp(j*2*pi*f/fs); H = 0.8*z/(z-0.2); 0.5 A = abs(H); theta = atan(imag(H)/real(H)); y2 = A*cos(2*pi*f*n/fs+theta); plot(n, x, 'bx-', ... 0 n, y1, 'go-', ... n, y2, 'r+-') legend('x', ... −0.5 'y (time domain)', ... 'y (z-transform)') ylim([-1.1 1.8]) −1 0 5 10 15 135 z 1 H ( z ) = z − 0 . 7 = How do poles affect time domain? 1 − 0 . 7 · z − 1 z Plane Impulse Response 1 1 Imaginary Part Amplitude 0 0.5 −1 0 −1 0 1 0 10 20 30 Real Part n (samples) z 1 H ( z ) = z − 0 . 9 = 1 − 0 . 9 · z − 1 z Plane Impulse Response 1 1 Imaginary Part Amplitude 0 0.5 −1 0 −1 0 1 0 10 20 30 Real Part n (samples) 136
z 1 H ( z ) = z − 1 = 1 − z − 1 z Plane Impulse Response 1 1 Imaginary Part Amplitude 0 0.5 −1 0 −1 0 1 0 10 20 30 Real Part n (samples) z 1 H ( z ) = z − 1 . 1 = 1 − 1 . 1 · z − 1 z Plane Impulse Response 20 1 Imaginary Part Amplitude 0 10 −1 0 −1 0 1 0 10 20 30 Real Part n (samples) 137 z 2 1 H ( z ) = ( z − 0 . 9 · e j π/ 6 ) · ( z − 0 . 9 · e − j π/ 6 ) = 1 − 1 . 8 cos( π/ 6) z − 1 +0 . 9 2 · z − 2 z Plane Impulse Response 2 1 Imaginary Part Amplitude 1 2 0 0 −1 −1 −1 0 1 0 10 20 30 Real Part n (samples) z 2 1 H ( z ) = ( z − e j π/ 6 ) · ( z − e − j π/ 6 ) = 1 − 2 cos( π/ 6) z − 1 + z − 2 z Plane Impulse Response 5 1 Imaginary Part Amplitude 2 0 0 −1 −5 −1 0 1 0 10 20 30 Real Part n (samples) 138
z 2 1 1 H ( z ) = ( z − 0 . 9 · e j π/ 2 ) · ( z − 0 . 9 · e − j π/ 2 ) = 1 − 1 . 8 cos( π/ 2) z − 1 +0 . 9 2 · z − 2 = 1+0 . 9 2 · z − 2 z Plane Impulse Response 1 1 Imaginary Part Amplitude 2 0 0 −1 −1 −1 0 1 0 10 20 30 Real Part n (samples) z 1 H ( z ) = z +1 = 1+ z − 1 z Plane Impulse Response 1 1 Imaginary Part Amplitude 0 0 −1 −1 −1 0 1 0 10 20 30 Real Part n (samples) 139 IIR filter design goals The design of a filter starts with specifying the desired parameters: ◮ The passband is the frequency range where we want to approximate a gain of one. ◮ The stopband is the frequency range where we want to approximate a gain of zero. ◮ The order of a filter is the number of poles it uses in the z -domain, and equivalently the number of delay elements necessary to implement it. ◮ Both passband and stopband will in practice not have gains of exactly one and zero, respectively, but may show several deviations from these ideal values, and these ripples may have a specified maximum quotient between the highest and lowest gain. ◮ There will in practice not be an abrupt change of gain between passband and stopband, but a transition band where the frequency response will gradually change from its passband to its stopband value. 140
IIR filter design techniques The designer can then trade off conflicting goals such as: small transition band, low order, low ripple amplitude or absence of ripples. Design techniques for making these tradeoffs for analog filters (involving capacitors, resistors, coils) can also be used to design digital IIR filters: Butterworth filters: Have no ripples, gain falls monotonically across the pass and transition band. Within the passband, the gain drops slowly down to � 1 − 1 / 2 ( − 3 dB). Outside the passband, it drops asymptotically by a factor 2 N per octave ( N · 20 dB/decade). Chebyshev type I filters: Distribute the gain error uniformly throughout the passband (equiripples) and drop off monotonically outside. Chebyshev type II filters: Distribute the gain error uniformly throughout the stopband (equiripples) and drop off monotonically in the passband. Elliptic filters (Cauer filters): Distribute the gain error as equiripples both in the passband and stopband. This type of filter is optimal in terms of the combination of the passband-gain tolerance, stopband-gain tolerance, and transition-band width that can be achieved at a given filter order. 141 IIR filter design in MATLAB The aforementioned filter-design techniques are implemented in the MATLAB Signal Processing Toolbox in the functions butter , cheby1 , cheby2 , and ellip . They output the coefficients a n and b n of the difference equation that describes the filter. These can be applied with filter to a sequence, or can be visual- ized with zplane as poles/zeros in the z -domain, with impz as an im- pulse response, and with freqz as an amplitude and phase spectrum. The commands and sptool fdatool provide interactive GUIs to design digital filters. MATLAB fdatool 142
Cascade of filter sections Higher-order IIR filters can be numerically unstable (quantization noise). A commonly used trick is to split a higher-order IIR filter design into a cascade of l second-order (biquad) filter sections of the form: x n b 0 y n z − 1 H ( z ) = b 0 + b 1 z − 1 + b 2 z − 2 − a 1 b 1 1 + a 1 z − 1 + a 2 z − 2 z − 1 − a 2 b 2 Filter sections H 1 , H 2 , . . . , H l are then applied sequentially to the input sequence, resulting in a filter b k, 0 + b k, 1 z − 1 + b k, 2 z − 2 l l � � H ( z ) = H k ( z ) = 1 + a k, 1 z − 1 + a k, 2 z − 2 k =1 k =1 Each section implements one pair of poles and one pair of zeros. Jackson’s algorithm for pairing poles and zeros into sections: pick the pole pair closest to the unit circle, and place it into a section along with the nearest pair of zeros; repeat until no poles are left. 143 Butterworth filter design example z Plane Impulse Response 1 1 Imaginary Part 0.5 Amplitude 0.5 0 −0.5 0 −1 −1 0 1 0 10 20 Real Part n (samples) 0 0 Phase (degrees) Magnitude (dB) −20 −50 −40 −60 −100 0 0.5 1 0 0.5 1 Normalized Frequency ( ×π rad/sample) Normalized Frequency ( ×π rad/sample) order: 1, cutoff frequency ( − 3 dB): 0 . 25 × f s / 2 144
Butterworth filter design example z Plane Impulse Response 1 1 Imaginary Part 0.5 Amplitude 0.5 0 −0.5 0 −1 −1 0 1 0 10 20 Real Part n (samples) 0 0 Phase (degrees) Magnitude (dB) −20 −200 −40 −400 −60 −600 0 0.5 1 0 0.5 1 Normalized Frequency ( ×π rad/sample) Normalized Frequency ( ×π rad/sample) order: 5, cutoff frequency ( − 3 dB): 0 . 25 × f s / 2 145 Chebyshev type I filter design example z Plane Impulse Response 1 1 Imaginary Part 0.5 Amplitude 0.5 0 −0.5 0 −1 −1 0 1 0 10 20 Real Part n (samples) 0 0 Phase (degrees) Magnitude (dB) −20 −200 −40 −400 −60 −600 0 0.5 1 0 0.5 1 Normalized Frequency ( ×π rad/sample) Normalized Frequency ( ×π rad/sample) order: 5, cutoff frequency: 0 . 5 × f s / 2, pass-band ripple: − 3 dB 146
Chebyshev type II filter design example z Plane Impulse Response 1 1 Imaginary Part 0.5 Amplitude 0.5 0 −0.5 0 −1 −1 0 1 0 10 20 Real Part n (samples) 0 100 Phase (degrees) Magnitude (dB) 0 −20 −100 −40 −200 −60 −300 0 0.5 1 0 0.5 1 Normalized Frequency ( ×π rad/sample) Normalized Frequency ( ×π rad/sample) order: 5, cutoff frequency: 0 . 5 × f s / 2, stop-band ripple: − 20 dB 147 Elliptic filter design example z Plane Impulse Response 1 1 Imaginary Part 0.5 Amplitude 0.5 0 −0.5 0 −1 −1 0 1 0 10 20 Real Part n (samples) 0 0 Phase (degrees) Magnitude (dB) −100 −20 −200 −40 −300 −60 −400 0 0.5 1 0 0.5 1 Normalized Frequency ( ×π rad/sample) Normalized Frequency ( ×π rad/sample) order: 5, cutoff frequency: 0 . 5 × f s / 2, pass-band ripple: − 3 dB, stop-band ripple: − 20 dB 148
Notch filter design example z Plane Impulse Response 1 1 Imaginary Part 0.5 Amplitude 0.5 0 −0.5 0 −1 −1 0 1 0 10 20 Real Part n (samples) 0 0 Phase (degrees) Magnitude (dB) −100 −20 −200 −40 −300 −60 −400 0 0.5 1 0 0.5 1 Normalized Frequency ( ×π rad/sample) Normalized Frequency ( ×π rad/sample) order: 2, cutoff frequency: 0 . 25 × f s / 2, − 3 dB bandwidth: 0 . 05 × f s / 2 149 Peak filter design example z Plane Impulse Response 1 1 Imaginary Part 0.5 Amplitude 0.5 0 −0.5 0 −1 −1 0 1 0 10 20 Real Part n (samples) 0 100 Phase (degrees) Magnitude (dB) 50 −20 0 −40 −50 −60 −100 0 0.5 1 0 0.5 1 Normalized Frequency ( ×π rad/sample) Normalized Frequency ( ×π rad/sample) order: 2, cutoff frequency: 0 . 25 × f s / 2, − 3 dB bandwidth: 0 . 05 × f s / 2 150
Random sequences and noise A discrete random sequence { x n } is a sequence of numbers . . . , x − 2 , x − 1 , x 0 , x 1 , x 2 , . . . where each value x n is the outcome of a random variable x n in a corresponding sequence of random variables . . . , x − 2 , x − 1 , x 0 , x 1 , x 2 , . . . Such a collection of random variables is called a random process . Each individual random variable x n is characterized by its probability distribution function P x n ( a ) = Prob( x n ≤ a ) and the entire random process is characterized completely by all joint probability distribution functions P x n 1 ,..., x nk ( a 1 , . . . , a k ) = Prob( x n 1 ≤ a 1 ∧ . . . ∧ x n k ≤ a k ) for all possible sets { x n 1 , . . . , x n k } . 151 Two random variables x n and x m are called independent if P x n , x m ( a, b ) = P x n ( a ) · P x m ( b ) and a random process is called stationary if P x n 1+ l ,..., x nk + l ( a 1 , . . . , a k ) = P x n 1 ,..., x nk ( a 1 , . . . , a k ) for all l , that is, if the probability distributions are time invariant. The derivative p x n ( a ) = P ′ x n ( a ) is called the probability density function , and helps us to define quantities such as the � ◮ expected value E( x n ) = ap x n ( a ) d a � ◮ mean-square value (average power) E( | x n | 2 ) = | a | 2 p x n ( a ) d a ◮ variance Var( x n ) = E[ | x n − E( x n ) | 2 ] = E( | x n | 2 ) − | E( x n ) | 2 ◮ correlation Cor( x n , x m ) = E( x n · x ∗ m ) ◮ covariance Cov( x n , x m ) = E[( x n − E( x n )) · ( x m − E( x m )) ∗ ] = E( x n x ∗ m ) − E( x n )E( x m ) ∗ Remember that E( · ) is linear, that is E( a x ) = a E( x ) and E( x + y ) = E( x ) + E( y ). Also, Var( a x ) = a 2 Var( x ) and, if x and y are independent, Var( x + y ) = Var( x ) + Var( y ). 152
A stationary random process { x n } can be characterized by its mean value m x = E( x n ) , its variance σ 2 x = E( | x n − m x | 2 ) = γ xx (0) ( σ x is also called standard deviation ), its autocorrelation sequence φ xx ( k ) = E( x n + k · x ∗ n ) and its autocovariance sequence γ xx ( k ) = E[( x n + k − m x ) · ( x n − m x ) ∗ ] = φ xx ( k ) − | m x | 2 A pair of stationary random processes { x n } and { y n } can, in addition, be characterized by its crosscorrelation sequence φ xy ( k ) = E( x n + k · y ∗ n ) and its crosscovariance sequence γ xy ( k ) = E[( x n + k − m x ) · ( y n − m y ) ∗ ] = φ xy ( k ) − m x m ∗ y 153 Deterministic crosscorrelation sequence For deterministic sequences { x n } and { y n } , the crosscorrelation sequence is ∞ � c xy ( k ) = x i + k y i . i = −∞ After dividing through the overlapping length of the finite sequences involved, c xy ( k ) can be used to estimate, from a finite sample of a stationary random sequence, the underlying φ xy ( k ). MATLAB’s xcorr function does that with option unbiased . If { x n } is similar to { y n } , but lags l elements behind ( x n ≈ y n − l ), then c xy ( l ) will be a peak in the crosscorrelation sequence. It is therefore widely calculated to locate shifted versions of a known sequence in another one. The deterministic crosscorrelation sequence is a close cousin of the convolution, with just the second input sequence mirrored: { c xy ( n ) } = { x n } ∗ { y − n } It can therefore be calculated equally easily via the Fourier transform: C xy ( f ) = X ( f ) · Y ∗ ( f ) Swapping the input sequences mirrors the output sequence: c xy ( k ) = c yx ( − k ). 154
Demonstration of covert spread-spectrum communication: n = randn(1,10000); pattern=2*round(rand(1,1000))-1; p1 = [zeros(1,2000), pattern, zeros(1,7000)]; p2 = [zeros(1,4000), pattern, zeros(1,5000)]; r = n + p1/3 - p2/3; figure(1) plot([n;p1/3-3;p2/3-4;r-6]'); figure(2) plot(conv(r,fliplr(pattern))); % or: plot(xcorr(r,pattern)); 155 Deterministic autocorrelation sequence Equivalently, we define the deterministic autocorrelation sequence in the time domain as ∞ � c xx ( k ) = x i + k x i . i = −∞ which corresponds in the frequency domain to C xx ( f ) = X ( f ) · X ∗ ( f ) = | X ( f ) | 2 . In other words, the Fourier transform C xx ( f ) of the autocorrelation sequence { c xx ( n ) } of a sequence { x n } is identical to the squared amplitudes of the Fourier transform, or power spectrum , of { x n } . This suggests, that the Fourier transform of the autocorrelation sequence of a random process might be a suitable way for defining the power spectrum of that random process. What can we say about the phase in the Fourier spectrum of a time-invariant random process? 156
Filtered random sequences Let { x n } be a random sequence from a stationary random process. The output ∞ ∞ � � y n = h k · x n − k = h n − k · x k k = −∞ k = −∞ of an LTI applied to it will then be another random sequence, characterized by ∞ � m y = m x h k k = −∞ and E( x n + k · x ∗ φ xx ( k ) = n ) ∞ � φ yy ( k ) = φ xx ( k − i ) c hh ( i ) , where � ∞ c hh ( k ) = i = −∞ h i + k h i . i = −∞ In other words: { φ yy ( n ) } = { c hh ( n ) } ∗ { φ xx ( n ) } { y n } = { h n } ∗ { x n } ⇒ | H ( f ) | 2 · Φ xx ( f ) Φ yy ( f ) = Similarly: { φ yx ( n ) } = { h n } ∗ { φ xx ( n ) } { y n } = { h n } ∗ { x n } ⇒ Φ yx ( f ) = H ( f ) · Φ xx ( f ) 157 White noise A random sequence { x n } is a white noise signal, if m x = 0 and φ xx ( k ) = σ 2 x δ k . The power spectrum of a white noise signal is flat: Φ xx ( f ) = σ 2 x . Application example: Where an LTI { y n } = { h n } ∗ { x n } can be observed to operate on white noise { x n } with φ xx ( k ) = σ 2 x δ k , the crosscorrelation between input and output will reveal the impulse response of the system: φ yx ( k ) = σ 2 x · h k where φ yx ( k ) = φ xy ( − k ) = E( y n + k · x ∗ n ). 158
Demonstration of covert spread-spectrum radar: x = randn(1,10000) h = [0 0 0.4 0 0 0.3 0 0 0.2 0 0]; y = conv(x, h); figure(1) plot(1:length(x), x, 1:length(y), y-5); figure(2) c = conv(fliplr(x),y); stem(c(length(c)/2-20:length(c)/2+20)); 4000 2000 0 −2000 0 10 20 30 40 50 159 DFT averaging The above diagrams show different types of spectral estimates of a sequence x i = sin(2 π j × 8 / 64) + sin(2 π j × 14 . 32 / 64) + n i with φ nn ( i ) = 4 δ i . Left is a single 64-element DFT of { x i } (with rectangular window). The flat spectrum of white noise is only an expected value. In a single discrete Fourier transform of such a sequence, the significant variance of the noise spectrum becomes visible. It almost drowns the two peaks from sine waves. After cutting { x i } into 1000 windows of 64 elements each, calculating their DFT, and plotting the average of their absolute values, the centre figure shows an approximation of the expected value of the amplitude spectrum, with a flat noise floor. Taking the absolute value before spectral averaging is called incoherent averaging , as the phase information is thrown away. 160
The rightmost figure was generated from the same set of 1000 windows, but this time the complex values of the DFTs were averaged before the absolute value was taken. This is called coherent averaging and, because of the linearity of the DFT, identical to first averaging the 1000 windows and then applying a single DFT and taking its absolute value. The windows start 64 samples apart. Only periodic waveforms with a period that divides 64 are not averaged away. This periodic averaging step suppresses both the noise and the second sine wave. Periodic averaging If a zero-mean signal { x i } has a periodic component with period p , the periodic component can be isolated by periodic averaging : k 1 � ¯ x i = lim x i + pn 2 k + 1 k →∞ n = − k Periodic averaging corresponds in the time domain to convolution with a Dirac comb � n δ i − pn . In the frequency domain, this means multiplication with a Dirac comb that eliminates all frequencies but multiples of 1 /p . 161 Audiovisual data compression Structure of modern audiovisual communication systems: sensor + perceptual entropy channel signal ✲ ✲ ✲ ✲ sampling coding coding coding ❄ noise ✲ channel ❄ entropy perceptual channel human display ✛ ✛ ✛ ✛ senses decoding decoding decoding 162
Audio-visual lossy coding today typically consists of these steps: ◮ A transducer converts the original stimulus into a voltage. ◮ This analog signal is then sampled and quantized . The digitization parameters (sampling frequency, quantization levels) are preferably chosen generously beyond the ability of human senses or output devices. ◮ The digitized sensor-domain signal is then transformed into a perceptual domain . This step often mimics some of the first neural processing steps in humans. ◮ This signal is quantized again, based on a perceptual model of what level of quantization-noise humans can still sense. ◮ The resulting quantized levels may still be highly statistically dependent. A prediction or decorrelation transform exploits this and produces a less dependent symbol sequence of lower entropy. ◮ An entropy coder turns that into an apparently-random bit string, whose length approximates the remaining entropy. The first neural processing steps in humans are in effect often a kind of decorrelation transform; our eyes and ears were optimized like any other AV communications system. This allows us to use the same transform for decorrelating and transforming into a perceptually relevant domain. 163 Outline of the remaining lectures ◮ Quick review of entropy coding ◮ Transform coding: techniques for converting sequences of highly-dependent symbols into less-dependent lower-entropy sequences. • run-length coding • decorrelation, Karhunen-Lo` eve transform (PCA) • Discrete cosine transform ◮ Introduction to some characteristics and limits of human senses • perceptual scales and sensitivity limits • colour vision ◮ Quantization techniques to remove information that is irrelevant to human senses 164
Entropy coding review – Huffman 1 � Entropy: H = p ( α ) · log 2 1.00 p ( α ) 0 1 α ∈ A = 2 . 3016 bit 0.40 0.60 0 1 0 1 v w 0.25 0.20 0.20 u 0 1 0.35 Mean codeword length: 2.35 bit x 0.10 0.15 0 1 Huffman’s algorithm constructs an optimal code-word tree for a set of symbols with known probability distribution. It iteratively picks the two elements of the set with the smallest probability and combines them into y z a tree by adding a common root. The resulting tree goes back into the 0.05 0.05 set, labeled with the sum of the probabilities of the elements it combines. The algorithm terminates when less than two elements are left. 165 Entropy coding review – arithmetic coding Partition [0,1] according 0.0 0.35 0.55 0.75 0.9 0.95 1.0 to symbol probabilities: u v w x y z Encode text wuvw . . . as numeric value (0.58. . . ) in nested intervals: 1.0 0.75 0.62 0.5885 0.5850 z z z z z y y y y y x x x x x w w w w w v v v v v u u u u u 0.55 0.0 0.55 0.5745 0.5822 166
Arithmetic coding Several advantages: ◮ Length of output bitstring can approximate the theoretical information content of the input to within 1 bit. ◮ Performs well with probabilities > 0.5, where the information per symbol is less than one bit. ◮ Interval arithmetic makes it easy to change symbol probabilities (no need to modify code-word tree) ⇒ convenient for adaptive coding Can be implemented efficiently with fixed-length arithmetic by rounding probabilities and shifting out leading digits as soon as leading zeros appear in interval size. Usually combined with adaptive probability estimation. Huffman coding remains popular because of its simplicity and lack of patent-licence issues. 167 Coding of sources with memory and correlated symbols Run-length coding: ↓ 5 7 12 3 3 Predictive coding: encoder decoder f(t) g(t) g(t) f(t) − + predictor predictor P(f(t−1), f(t−2), ...) P(f(t−1), f(t−2), ...) Delta coding (DPCM): P ( x ) = x n � Linear predictive coding: P ( x 1 , . . . , x n ) = a i x i i =1 168
Old (Group 3 MH) fax code pixels white code black code ◮ Run-length encoding plus modified Huffman 0 00110101 0000110111 code 1 000111 010 2 0111 11 ◮ Fixed code table (from eight sample pages) 3 1000 10 ◮ separate codes for runs of white and black 4 1011 011 pixels 5 1100 0011 ◮ termination code in the range 0–63 switches 6 1110 0010 7 1111 00011 between black and white code 8 10011 000101 ◮ makeup code can extend length of a run by a 9 10100 000100 multiple of 64 10 00111 0000100 ◮ termination run length 0 needed where run 11 01000 0000101 length is a multiple of 64 12 001000 0000111 ◮ single white column added on left side before 13 000011 00000100 14 110100 00000111 transmission 15 110101 000011000 ◮ makeup codes above 1728 equal for black and 16 101010 0000010111 white . . . . . . . . . ◮ 12-bit end-of-line marker: 000000000001 (can 63 00110100 000001100111 be prefixed by up to seven zero-bits to reach 64 11011 0000001111 next byte boundary) 128 10010 000011001000 192 010111 000011001001 Example: line with 2 w, 4 b, 200 w, 3 b, EOL → . . . . . . . . . 1000 | 011 | 010111 | 10011 | 10 | 000000000001 1728 010011011 0000001100101 169 Modern (JBIG) fax code Performs context-sensitive arithmetic coding of binary pixels. Both encoder and decoder maintain statistics on how the black/white probability of each pixel depends on these 10 previously transmitted neighbours: ? Based on the counted numbers n black and n white of how often each pixel value has been encountered so far in each of the 1024 contexts, the probability for the next pixel being black is estimated as n black + 1 p black = n white + n black + 2 The encoder updates its estimate only after the newly counted pixel has been encoded, such that the decoder knows the exact same statistics. Joint Bi-level Expert Group: International Standard ISO 11544, 1993. Example implementation: http://www.cl.cam.ac.uk/~mgk25/jbigkit/ 170
Statistical dependence Random variables X, Y are dependent iff ∃ x, y : P ( X = x ∧ Y = y ) � = P ( X = x ) · P ( Y = y ) . If X, Y are dependent, then ⇒ ∃ x, y : P ( X = x | Y = y ) � = P ( X = x ) ∨ P ( Y = y | X = x ) � = P ( Y = y ) ⇒ H ( X | Y ) < H ( X ) ∨ H ( Y | X ) < H ( Y ) Application Where x is the value of the next symbol to be transmitted and y is the vector of all symbols transmitted so far, accurate knowledge of the conditional probability P ( X = x | Y = y ) will allow a transmitter to remove all redundancy. An application example of this approach is JBIG, but there y is limited to 10 past single-bit pixels and P ( X = x | Y = y ) is only an estimate. 171 Practical limits of measuring conditional probabilities The practical estimation of conditional probabilities, in their most general form, based on statistical measurements of example signals, quickly reaches practical limits. JBIG needs an array of only 2 11 = 2048 counting registers to maintain estimator statistics for its 10-bit context. If we wanted to encode each 24-bit pixel of a colour image based on its statistical dependence of the full colour information from just ten previous neighbour pixels, the required number of (2 24 ) 11 ≈ 3 × 10 80 registers for storing each probability will exceed the estimated number of particles in this universe. (Neither will we encounter enough pixels to record statistically significant occurrences in all (2 24 ) 10 contexts.) This example is far from excessive. It is easy to show that in colour images, pixel values show statistical significant dependence across colour channels, and across locations more than eight pixels apart. A simpler approximation of dependence is needed: correlation. 172
Correlation Two random variables X ∈ R and Y ∈ R are correlated iff E { [ X − E( X )] · [ Y − E( Y )] } � = 0 where E( · · · ) denotes the expected value of a random-variable term. Dependent but not correlated: 1 Correlation implies dependence, but dependence does not always lead to correlation (see example to the right). However, most dependency in audio- 0 visual data is a consequence of corre- lation, which is algorithmically much easier to exploit. −1 −1 0 1 Positive correlation: higher X ⇔ higher Y , lower X ⇔ lower Y Negative correlation: lower X ⇔ higher Y , higher X ⇔ lower Y 173 Correlation of neighbour pixels Values of neighbour pixels at distance 1 Values of neighbour pixels at distance 2 256 256 192 192 128 128 64 64 0 0 0 64 128 192 256 0 64 128 192 256 Values of neighbour pixels at distance 4 Values of neighbour pixels at distance 8 256 256 192 192 128 128 64 64 0 0 0 64 128 192 256 0 64 128 192 256 174
Covariance and correlation We define the covariance of two random variables X and Y as Cov( X, Y ) = E { [ X − E( X )] · [ Y − E( Y )] } = E( X · Y ) − E( X ) · E( Y ) and the variance as Var( X ) = Cov( X, X ) = E { [ X − E( X )] 2 } . The Pearson correlation coefficient Cov( X, Y ) ρ X,Y = � Var( X ) · Var( Y ) is a normalized form of the covariance. It is limited to the range [ − 1 , 1]. If the correlation coefficient has one of the values ρ X,Y = ± 1, this implies that X and Y are exactly linearly dependent, i.e. Y = aX + b , with a = Cov( X, Y ) / Var( X ) and b = E( Y ) − E( X ) . 175 Covariance Matrix For a random vector X = ( X 1 , X 2 , . . . , X n ) ∈ R n we define the covariance matrix � ( X − E( X )) · ( X − E( X )) T � Cov( X ) = E = (Cov( X i , X j )) i,j = Cov( X 1 , X 1 ) Cov( X 1 , X 2 ) Cov( X 1 , X 3 ) · · · Cov( X 1 , X n ) Cov( X 2 , X 1 ) Cov( X 2 , X 2 ) Cov( X 2 , X 3 ) · · · Cov( X 2 , X n ) Cov( X 3 , X 1 ) Cov( X 3 , X 2 ) Cov( X 3 , X 3 ) · · · Cov( X 3 , X n ) . . . . ... . . . . . . . . Cov( X n , X 1 ) Cov( X n , X 2 ) Cov( X n , X 3 ) · · · Cov( X n , X n ) The elements of a random vector X are uncorrelated if and only if Cov( X ) is a diagonal matrix. Cov( X, Y ) = Cov( Y, X ), so all covariance matrices are symmetric : Cov( X ) = Cov T ( X ). 176
Decorrelation by coordinate transform Neighbour−pixel value pairs Decorrelated neighbour−pixel value pairs 256 320 256 192 192 128 128 64 64 0 0 −64 0 64 128 192 256 −64 0 64 128 192 256 320 Probability distribution and entropy Idea: Take the values of a group of cor- correlated value pair (H = 13.90 bit) related symbols (e.g., neighbour pixels) as decorrelated value 1 (H = 7.12 bit) decorrelated value 2 (H = 4.75 bit) a random vector. Find a coordinate trans- form (multiplication with an orthonormal matrix) that leads to a new random vector whose covariance matrix is diagonal. The vector components in this transformed co- ordinate system will no longer be corre- lated. This will hopefully reduce the en- tropy of some of these components. −64 0 64 128 192 256 320 177 Theorem: Let X ∈ R n and Y ∈ R n be random vectors that are linearly dependent with Y = A X + b , where A ∈ R n × n and b ∈ R n are constants. Then E( Y ) = A · E( X ) + b A · Cov( X ) · A T Cov( Y ) = Proof: The first equation follows from the linearity of the expected-value operator E( · ), as does E( A · X · B ) = A · E( X ) · B for matrices A, B . With that, we can transform � ( Y − E( Y )) · ( Y − E( Y )) T � Cov( Y ) = E � ( A X − A E( X )) · ( A X − A E( X )) T � = E � A ( X − E( X )) · ( X − E( X )) T A T � = E � ( X − E( X )) · ( X − E( X )) T � · A T = A · E A · Cov( X ) · A T = 178
Quick review: eigenvectors and eigenvalues We are given a square matrix A ∈ R n × n . The vector x ∈ R n is an eigenvector of A if there exists a scalar value λ ∈ R such that Ax = λx. The corresponding λ is the eigenvalue of A associated with x . The length of an eigenvector is irrelevant, as any multiple of it is also an eigenvector. Eigenvectors are in practice normalized to length 1. Spectral decomposition Any real, symmetric matrix A = A T ∈ R n × n can be diagonalized into the form A = U Λ U T , where Λ = diag( λ 1 , λ 2 , . . . , λ n ) is the diagonal matrix of the ordered eigenvalues of A (with λ 1 ≥ λ 2 ≥ · · · ≥ λ n ), and the columns of U are the n corresponding orthonormal eigenvectors of A . 179 Karhunen-Lo` eve transform (KLT) We are given a random vector variable X ∈ R n . The correlation of the elements of X is described by the covariance matrix Cov( X ). How can we find a transform matrix A that decorrelates X , i.e. that turns Cov( A X ) = A · Cov( X ) · A T into a diagonal matrix? A would provide us the transformed representation Y = A X of our random vector, in which all elements are mutually uncorrelated. Note that Cov( X ) is symmetric. It therefore has n real eigenvalues λ 1 ≥ λ 2 ≥ · · · ≥ λ n and a set of associated mutually orthogonal eigenvectors b 1 , b 2 , . . . , b n of length 1 with Cov( X ) b i = λ i b i . We convert this set of equations into matrix notation using the matrix B = ( b 1 , b 2 , . . . , b n ) that has these eigenvectors as columns and the diagonal matrix D = diag( λ 1 , λ 2 , . . . , λ n ) that consists of the corresponding eigenvalues: Cov( X ) B = BD 180
B is orthonormal, that is BB T = I . Multiplying the above from the right with B T leads to the spectral decomposition Cov( X ) = BDB T of the covariance matrix. Similarly multiplying instead from the left with B T leads to B T Cov( X ) B = D and therefore shows with Cov( B T X ) = D that the eigenvector matrix B T is the wanted transform. The Karhunen-Lo` eve transform (also known as Hotelling transform or Principal Component Analysis ) is the multiplication of a correlated random vector X with the orthonormal eigenvector matrix B T from the spectral decomposition Cov( X ) = BDB T of its covariance matrix. This leads to a decorrelated random vector B T X whose covariance matrix is diagonal. 181 Karhunen-Lo` eve transform example I colour image green channel red channel blue channel The colour image (left) has m = r 2 pixels, each We can now define the mean colour vector of which is an n = 3-dimensional RGB vector 0 . 4839 m S c = 1 ¯ � ¯ S c,i , S = 0 . 4456 I x,y = ( r x,y , g x,y , b x,y ) T m 0 . 3411 i =1 The three rightmost images show each of these and the covariance matrix colour planes separately as a black/white m image. 1 � ( S c,i − ¯ S c )( S d,i − ¯ C c,d = S d ) We want to apply the KLT on a set of such R n m − 1 i =1 colour vectors. Therefore, we reformat the 0 . 0328 0 . 0256 0 . 0160 image I into an n × m matrix of the form 0 . 0256 0 . 0216 0 . 0140 C = 0 . 0160 0 . 0140 0 . 0109 r 1 , 1 r 1 , 2 r 1 , 3 · · · r r,r S = g 1 , 1 g 1 , 2 g 1 , 3 · · · g r,r [“ m − 1” because ¯ S c only estimates the mean] b 1 , 1 b 1 , 2 b 1 , 3 · · · b r,r 182
Karhunen-Lo` eve transform example I The resulting covariance matrix C has three eigenvalues 0.0622, 0.0025, and 0.0006: 0 . 0328 0 . 0256 0 . 0160 0 . 7167 0 . 7167 = 0 . 0622 0 . 0256 0 . 0216 0 . 0140 0 . 5833 0 . 5833 0 . 0160 0 . 0140 0 . 0109 0 . 3822 0 . 3822 0 . 0328 0 . 0256 0 . 0160 − 0 . 5509 − 0 . 5509 = 0 . 0025 0 . 0256 0 . 0216 0 . 0140 0 . 1373 0 . 1373 0 . 0160 0 . 0140 0 . 0109 0 . 8232 0 . 8232 0 . 0328 0 . 0256 0 . 0160 − 0 . 4277 − 0 . 4277 = 0 . 0006 0 . 0256 0 . 0216 0 . 0140 0 . 8005 0 . 8005 0 . 0160 0 . 0140 0 . 0109 − 0 . 4198 − 0 . 4198 It can thus be diagonalized as 0 . 0328 0 . 0256 0 . 0160 = C = U · D · U T = 0 . 0256 0 . 0216 0 . 0140 0 . 0160 0 . 0140 0 . 0109 0 . 7167 − 0 . 5509 − 0 . 4277 0 . 0622 0 0 0 . 7167 0 . 5833 0 . 3822 0 . 5833 0 . 1373 0 . 8005 0 0 . 0025 0 − 0 . 5509 0 . 1373 0 . 8232 0 . 3822 0 . 8232 − 0 . 4198 0 0 0 . 0006 − 0 . 4277 0 . 8005 − 0 . 4198 (e.g. using MATLAB’s singular-value decomposition function svd ). 183 Karhunen-Lo` eve transform example I Before KLT: We finally apply the orthogonal 3 × 3 transform matrix U , which we just used to diagonalize the covariance matrix, to the entire image: S 1 ¯ ¯ ¯ S 1 · · · S 1 T = U T · S 2 ¯ ¯ ¯ S − S 2 · · · S 2 S 3 ¯ ¯ ¯ S 3 · · · S 3 red green blue S 1 ¯ ¯ ¯ S 1 · · · S 1 After KLT: S 2 ¯ ¯ ¯ + S 2 · · · S 2 S 3 ¯ ¯ ¯ S 3 · · · S 3 The resulting transformed image u 1 , 1 u 1 , 2 u 1 , 3 · · · u r,r u v w T = v 1 , 1 v 1 , 2 v 1 , 3 · · · v r,r Projections on eigenvector subspaces: w 1 , 1 w 1 , 2 w 1 , 3 · · · w r,r consists of three new “colour” planes whose pixel values have no longer any correlation to the pixels at the same coordinates in another plane. [The bear disappeared from the last of these ( w ), which represents mostly some of the v = w = 0 w = 0 original green grass in the background.] 184
Spatial correlation The previous example used the Karhunen-Lo` eve transform in order to eliminate correlation between colour planes. While this is of some relevance for image compression, far more correlation can be found between neighbour pixels within each colour plane. In order to exploit such correlation using the KLT, the sample set has to be extended from individual pixels to entire images. The underlying calculation is the same as in the preceeding example, but this time the columns of S are entire (monochrome) images. The rows are the different images found in the set of test images that we use to examine typical correlations between neighbour pixels. In other words, we use the same formulas as in the previous example, but this time n is the number of pixels per image and m is the number of sample images. The Karhunen-Lo` eve transform is here no longer a rotation in a 3-dimensional colour space, but it operates now in a much larger vector space that has as many dimensions as an image has pixels. To keep things simple, we look in the next experiment only at m = 9000 1-dimensional “images” with n = 32 pixels each. As a further simplification, we use not real images, but random noise that was filtered such that its amplitude spectrum is proportional to 1 /f , where f is the frequency. The result would be similar in a sufficiently large collection of real test images. 185 Karhunen-Lo` eve transform example II Matrix columns of S filled with samples of 1 /f filtered noise . . . Covariance matrix C Matrix U with eigenvector columns 186
Matrix U ′ with normalised KLT Matrix with Discrete Cosine Transform base vector columns eigenvector columns Breakthrough: Ahmed/Natarajan/Rao discovered the DCT as an excellent approximation of the KLT for typical photographic images, but far more efficient to calculate. Ahmed, Natarajan, Rao: Discrete Cosine Transform. IEEE Transactions on Computers, Vol. 23, January 1974, pp. 90–93. 187 Discrete cosine transform (DCT) The forward and inverse discrete cosine transform N − 1 C ( u ) s ( x ) cos (2 x + 1) uπ � S ( u ) = � 2 N N/ 2 x =0 N − 1 C ( u ) S ( u ) cos (2 x + 1) uπ � s ( x ) = � 2 N N/ 2 u =0 with � 1 u = 0 √ C ( u ) = 2 1 u > 0 is an orthonormal transform: � 1 N − 1 · C ( u ′ ) cos (2 x + 1) u ′ π u = u ′ C ( u ) cos (2 x + 1) uπ � = � � u � = u ′ 0 2 N 2 N N/ 2 N/ 2 x =0 188
DCT base vectors for N = 8: 7 6 5 4 u 3 2 1 0 0 1 2 3 4 5 6 7 x 189 Discrete cosine transform – 2D The 2-dimensional variant of the DCT applies the 1-D transform on both rows and columns of an image: S ( u, v ) = C ( u ) C ( v ) · � � N/ 2 N/ 2 N − 1 N − 1 s ( x, y ) cos (2 x + 1) uπ cos (2 y + 1) vπ � � 2 N 2 N x =0 y =0 s ( x, y ) = N − 1 N − 1 C ( u ) C ( v ) · S ( u, v ) cos (2 x + 1) uπ cos (2 y + 1) vπ � � � � 2 N 2 N N/ 2 N/ 2 u =0 v =0 A range of fast algorithms have been found for calculating 1-D and 2-D DCTs (e.g., Ligtenberg/Vetterli). 190
Whole-image DCT 2D Discrete Cosine Transform (log10) Original image 4 50 50 3 100 100 2 150 150 1 200 200 250 250 0 300 300 −1 350 350 −2 400 400 −3 450 450 −4 500 500 100 200 300 400 500 100 200 300 400 500 191 Whole-image DCT, 80% coefficient cutoff 80% truncated 2D DCT (log10) 80% truncated DCT: reconstructed image 4 50 50 3 100 100 2 150 150 1 200 200 250 250 0 300 300 −1 350 350 −2 400 400 −3 450 450 −4 500 500 100 200 300 400 500 100 200 300 400 500 192
Whole-image DCT, 90% coefficient cutoff 90% truncated 2D DCT (log10) 90% truncated DCT: reconstructed image 4 50 50 3 100 100 2 150 150 1 200 200 250 250 0 300 300 −1 350 350 −2 400 400 −3 450 450 −4 500 500 100 200 300 400 500 100 200 300 400 500 193 Whole-image DCT, 95% coefficient cutoff 95% truncated 2D DCT (log10) 95% truncated DCT: reconstructed image 4 50 50 3 100 100 2 150 150 1 200 200 250 250 0 300 300 −1 350 350 −2 400 400 −3 450 450 −4 500 500 100 200 300 400 500 100 200 300 400 500 194
Whole-image DCT, 99% coefficient cutoff 99% truncated 2D DCT (log10) 99% truncated DCT: reconstructed image 4 50 50 3 100 100 2 150 150 1 200 200 250 250 0 300 300 −1 350 350 −2 400 400 −3 450 450 −4 500 500 100 200 300 400 500 100 200 300 400 500 195 Base vectors of 8 × 8 DCT v 0 1 2 3 4 5 6 7 0 1 2 3 u 4 5 6 7 196
Psychophysics of perception Sensation limit (SL) = lowest intensity stimulus that can still be perceived Difference limit (DL) = smallest perceivable stimulus difference at given intensity level Weber’s law Difference limit ∆ φ is proportional to the intensity φ of the stimulus (except for a small correction constant a , to describe deviation of experimental results near SL): ∆ φ = c · ( φ + a ) Fechner’s scale Define a perception intensity scale ψ using the sensation limit φ 0 as the origin and the respective difference limit ∆ φ = c · φ as a unit step. The result is a logarithmic relationship between stimulus intensity and scale value: φ ψ = log c φ 0 197 Fechner’s scale matches older subjective intensity scales that follow differentiability of stimuli, e.g. the astronomical magnitude numbers for star brightness introduced by Hipparchos ( ≈ 150 BC). Stevens’ power law A sound that is 20 DL over SL is perceived as more than twice as loud as one that is 10 DL over SL, i.e. Fechner’s scale does not describe well perceived intensity. A rational scale attempts to reflect subjective relations perceived between different values of stimulus intensity φ . Stanley Smith Stevens observed that such rational scales ψ follow a power law: ψ = k · ( φ − φ 0 ) a Example coefficients a : brightness 0.33, loudness 0.6, heaviness 1.45, temperature (warmth) 1.6. 198
RGB video colour coordinates Hardware interface (VGA): red, green, blue signals with 0–0.7 V Electron-beam current and photon count of cathode-ray displays are roughly proportional to ( v − v 0 ) γ , where v is the video-interface or control-grid voltage and γ is a device parameter that is typically in the range 1.5–3.0. In broadcast TV, this CRT non-linearity is compensated electronically in TV cameras. A welcome side effect is that it approximates Stevens’ scale and therefore helps to reduce perceived noise. Software interfaces map RGB voltage linearly to { 0 , 1 , . . . , 255 } or 0–1. How numeric RGB values map to colour and luminosity depends at present still highly on the hardware and sometimes even on the operating system or device driver. The new specification “sRGB” aims to standardize the meaning of an RGB value with the parameter γ = 2 . 2 and with standard colour coordinates of the three primary colours. http://www.w3.org/Graphics/Color/sRGB, IEC 61966 199 YUV video colour coordinates The human eye processes colour and luminosity at different resolutions. To exploit this phenomenon, many image transmission systems use a colour space with a luminance coordinate Y = 0 . 3 R + 0 . 6 G + 0 . 1 B and colour (“chrominance”) components V = R − Y = 0 . 7 R − 0 . 6 G − 0 . 1 B U = B − Y = − 0 . 3 R − 0 . 6 G + 0 . 9 B 200
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