EI331 Signals and Systems Lecture 12 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 12 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University April 4, 2019

Contents 1. Filtering 2. DT Fourier Series 3. Properties of DT Fourier Series 1/33

Ideal Frequency-selective Filters Ideal lowpass filter � 1 , | ω | ≤ ω c H ( j ω ) = 0 , otherwise ω c : cutoff frequency H ( j ω ) 1 ω c ω − ω c 0 stopband passband stopband 2/33

Ideal Frequency-selective Filters Ideal highpass filter � 1 , | ω | ≥ ω c H ( j ω ) = 0 , otherwise ω c : cutoff frequency H ( j ω ) 1 ω c ω − ω c 0 passband stopband passband 3/33

Ideal Frequency-selective Filters Ideal bandpass filter � 1 , ω c 1 ≤ | ω | ≤ ω c 2 H ( j ω ) = 0 , otherwise ω c 1 : lower cutoff frequency ω c 2 : upper cutoff frequency H ( j ω ) 1 ω c 1 ω c 2 ω − ω c 2 − ω c 1 0 4/33

Simple RC Lowpass Filter v R ( t ) ODE + − RCdv C ( t ) + v C ( t ) = v S ( t ) R dt + v S ( t ) v C ( t ) C For input − v S ( t ) = e j ω t output v C ( t ) = H ( j ω ) e j ω t 1 Frequency response H ( j ω ) = 1 + RCj ω 5/33

Simple RC Lowpass Filter | H ( j ω ) | Frequency response 1 1 H ( j ω ) = 1 + RCj ω ω − 1 0 1 RC RC 1 | H ( j ω ) | = � 1 + ( RC ω ) 2 arg H ( j ω ) arg H ( j ω ) = − arctan( RC ω ) π 4 Nonideal lowpass filter 1 RC ω passes lower frequencies − 1 RC − π attenuates higher frequencies 4 Larger RC = ⇒ passes smaller range of lower frequencies 6/33

Simple RC Lowpass Filter h ( t ) Impulse response 1 RC h ( t ) = 1 RCe − t / RC u ( t ) 1 RCe t 0 RC Step response s ( t ) s ( t ) = ( h ∗ u )( t ) = ( 1 − e − t / RC ) u ( t ) 1 Time constant τ = RC 1 − 1 e • larger τ , more sluggish response t RC Tradeoff • larger τ , passes fewer higher frequencies, more sluggish response • smaller τ , passes more higher frequencies, faster response 7/33

Simple RC Highpass Filter v R ( t ) ODE + − RCdv R ( t ) + v R ( t ) = RCdv S ( t ) R dt dt + v S ( t ) v C ( t ) C For input − v S ( t ) = e j ω t output v R ( t ) = H ( j ω ) e j ω t j ω RC H ( j ω ) = Frequency response 1 + j ω RC 8/33

Simple RC Highpass Filter | H ( j ω ) | Frequency response 1 j ω RC H ( j ω ) = 1 + j ω RC ω − 1 0 1 RC RC | ω | RC � | H ( j ω ) | = arg H ( j ω ) 1 + ( RC ω ) 2 π 1 2 arg H ( j ω ) = arctan RC ω π 4 − 1 RC Nonideal highpass filter ω 1 RC passes higher frequencies − π 4 attenuates lower frequencies − π 2 Larger RC = ⇒ passes larger range of lower frequencies 9/33

Simple RC Highpass Filter h ( t ) Step response 1 s ( t ) = e − t / RC u ( t ) t 0 RC Impulse response 1 RCe h ( t ) = s ′ ( t ) = δ ( t ) − 1 − 1 RCe − t / RC u ( t ) s ( t ) Time constant τ = RC 1 • larger τ , more sluggish response 1 e t RC Observations • larger τ , passes more lower frequencies, more sluggish response • smaller τ , passes fewer lower frequencies, faster response 10/33

Contents 1. Filtering 2. DT Fourier Series 3. Properties of DT Fourier Series 11/33

DT Periodic Signals Recall DT signal is periodic with period N if x [ n ] = x [ n + N ] , ∀ n ∈ Z x = τ N x or • fundamental period N : smallest positive period � 2 π N , if N > 1 • fundamental frequency 0 , if N = 1 N n = e jk ω 0 n is periodic with N [ n ] = e jk 2 π Complex exponential φ k • period N and fundamental period N gcd( N , k ) • fundamental frequency � 0 , if N | k ω k = ω 0 · gcd( k , N ) , otherwise always integer multiple of ω 0 = 2 π N 12/33

Finiteness of DT Fourier Basis Fourier series represent N -periodic signals in terms of harmonically related complex exponentials φ k N � � � c k e jk 2 π c k φ k c k φ k N n x = N , or x [ n ] = N [ n ] = k k k Key difference with CT case φ k + rN = φ k N , so only N distinct φ k N , Fourier basis is finite , i.e. N { φ k N : k ∈ Z } = { φ k N : k ∈ [ N ] } where [ N ] = { 0 , 1 , . . . , N − 1 } (can think [ N ] = { ¯ 0 , ¯ 1 , . . . , N − 1 } ) Proof. For r ∈ Z , N n = e jk 2 π N n e jr 2 π n = e jk 2 π N n = φ k [ n ] = e j ( k + rN ) 2 π φ k + rN N [ n ] N 13/33

Finiteness of DT Fourier Basis φ k N for N = 4 and k = 0 , 1 , . . . , 8 x x x n n n φ 0 N [ n ] = cos( 0 · n ) = 1 φ 1 φ 2 N [ n ] = cos( π n / 4 ) N [ n ] = cos( π n / 2 ) x x x n n n φ 3 φ 4 φ 5 N [ n ] = cos( 3 π n / 4 ) N [ n ] = cos( π n ) N [ n ] = cos( 5 π n / 4 ) x x x n n n φ 6 φ 7 φ 8 N [ n ] = cos( 3 π n / 2 ) N [ n ] = cos( 7 π n / 4 ) N [ n ] = cos( 2 π n ) = 1 14/33

Orthonormality of Harmonics DT Fourier series � � x [ k ] e jk 2 π x [ k ] φ k N n x = ˆ N , or x [ n ] = ˆ k ∈ [ N ] k ∈ [ N ] Summation can also be taken over any N successive integers. Find coefficients ˆ x [ k ] using orthonormality of harmonics. Define inner product between two signals with period N by � � x , y � = 1 x [ n ] y [ n ] N n ∈ [ N ] Same as inner product in C N up to factor N − 1 { φ k N : k ∈ [ N ] } is orthonormal system of functions � φ k N , φ m N � = δ km = δ [ k − m ] 15/33

Proof of Orthonormality of Harmonics { φ k N : k ∈ [ N ] } is orthonormal system of functions, i.e. � φ k N , φ m N � = δ km = δ [ k − m ] Proof. � � N − 1 N � = 1 N n = 1 e jk 2 π N n e − jm 2 π e j ( k − m ) 2 π � φ k N , φ m N n N N n = 0 n ∈ [ N ] If k = m , N − 1 � N � = 1 � φ k N , φ m 1 = 1 N k = 0 � n 2 a n = a n 1 − a n 2 + 1 If k � = m , since | k − m | ≤ N − 1 , e j ( k − m ) 2 π N � = 1 . By , 1 − a n = n 1 � N − 1 1 − e j ( k − m ) 2 π N N N � = 1 N n = 1 e j ( k − m ) 2 π � φ k N , φ m = 0 1 − e j ( k − m ) 2 π N N N n = 0 16/33

Fourier Coefficients Suppose x has period N and Fourier series representation � x [ k ] φ k x = ˆ N k ∈ [ N ] For m ∈ [ N ] , � � � x , φ m x [ k ] φ k N , φ m x [ k ] � φ k N , φ m N � = � ˆ N � = ˆ N � k ∈ [ N ] k ∈ [ N ] � = ˆ x [ k ] δ [ m − k ] = ˆ x [ m ] k ∈ [ N ] Can thick of ˆ x as N -periodic signal, since x [ m + rN ] � � x , φ m + rN � = � x , φ m ˆ N � = ˆ x [ m ] N but use only N successive values in Fourier series! 17/33

DT Fourier Series Synthesis equation � � x [ k ] e jk 2 π x [ k ] φ k N n x [ n ] = ˆ N [ n ] = ˆ k ∈ [ N ] k ∈ [ N ] Analysis equation � N � = 1 x [ n ] e − jk 2 π x [ k ] = � x , φ k N n ˆ N n ∈ [ N ] No convergence issue since all sums are finite! 18/33

Example 5 ) = e j π 5 n + e − j π 2 e j 3 2 π 2 e − j 3 2 π x [ n ] = cos( 6 π 5 n + π 5 n , period N = 5 5 5 x [ 3 ] = 1 x [ − 3 ] = 1 2 e j π 5 ; 2 e − j π 5 ; ˆ ˆ x [ 2 ] = ˆ ˆ x [ 0 ] = ˆ x [ 1 ] = ˆ x [ 4 ] = 0 ˆ x [ k ] repeats with period N = 5 x [ n ] n − 5 0 5 1 1 2 2 | ˆ x [ k ] | − 3 − 2 0 2 3 7 8 k π 5 − 3 2 7 arg ˆ x [ k ] − 2 0 3 8 k − π 5 19/33

Example: Periodic Square Wave Periodic square wave with period N , in one period � 1 , − N 1 ≤ n ≤ N 1 x [ n ] = 0 , N 1 < n < N − N 1 n − N − N 1 0 N 1 N Fourier coefficients � N 1 x [ k ] = 1 e − jk 2 π N n ˆ N n = − N 1 x [ k ] = 2 N 1 + 1 If k is integer multiple of N , ˆ N 20/33

Example: Periodic Square Wave If k is not integer multiple of N , then e − jk 2 π N � = 1 . Using a n = a m − a M + 1 � M 1 − a n = m we obtain � N 1 N N 1 − e − jk 2 π e jk 2 π N ( N 1 + 1 ) x [ k ] = 1 N n = 1 e − jk 2 π ˆ 1 − e − jk 2 π N N N n = − N 1 2 ) − e − jk 2 π e jk 2 π N ( N 1 + 1 N ( N 1 + 1 2 ) = 1 e jk π N − e − jk π N N sin( k 2 π N ( N 1 + 1 2 )) = 1 sin( k π N ) N 21/33

Example: Periodic Square Wave sin( k 2 π N ( N 1 + 1 2 )) x [ k ] = 1 ˆ sin( k π N N ) N 1 = 2 N = 10 k N 1 = 2 N = 20 k N 1 = 2 N = 40 k 22/33

DT Fourier Series: Matrix Form Synthesis equation � � N − 1 N n = x [ k ] e jk 2 π where W N = e jk 2 π W kn x [ n ] = ˆ N ˆ x [ k ] , N k = 0 k ∈ [ N ]       1 1 1 . . . 1 x [ 0 ] ˆ x [ 0 ]   W N − 1   W 2 . . .   1 W N   x [ 1 ] ˆ x [ 1 ]   N N       W 2 ( N − 1 )   W 2 W 4 . . . x [ 2 ]  1  ˆ x [ 2 ]     N N N     =   . . . . . ... .   . . . . . .     . . . . . .           W 2 ( N − 2 ) W ( N − 1 )( N − 2 )   W N − 2 x [ N − 2 ] ˆ x [ N − 2 ] . . .  1  N N N W ( N − 1 ) 2 x [ N − 1 ] W 2 ( N − 1 ) x [ N − 1 ] ˆ W N − 1 . . . 1 N N N N , . . . , φ N − 1 F = ( φ 0 N , φ 1 ) N 23/33

DT Fourier Series: Matrix Form Synthesis equation � � N − 1 N n = x [ k ] e jk 2 π where W N = e jk 2 π W kn x [ n ] = ˆ N ˆ x [ k ] , N k = 0 k ∈ [ N ] Matrix form where F = ( φ 0 N , φ 1 N , . . . , φ N − 1 x = F ˆ x , ) N Analysis equation � � N − 1 x [ k ] = 1 N n = 1 x [ n ] e − jk 2 π ¯ W kn ˆ N x [ n ] N N n = 0 n ∈ [ N ] Matrix form x = F − 1 x = 1 N F H x ˆ where F H = (¯ F ) T is Hermitian transpose of F 24/33

Contents 1. Filtering 2. DT Fourier Series 3. Properties of DT Fourier Series 25/33