EI331 Signals and Systems Lecture 15 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 15 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University April 16, 2019

Contents 1. Recap 2. Properties of CT Fourier Transform 1/32 w

Recap: Fourier Transform for L 1 ( R ) Functions Fourier transform and its inverse as principal values � T x ( t ) e − j ω t dt F { x } ( j ω ) = lim T →∞ − T � W 1 F − 1 ( X )( t ) = lim X ( j ω ) e j ω t d ω 2 π W →∞ − W Theorem. (Pointwise inversion) 1. If x ∈ L 1 ( R ) satisfies Dirichlet conditions on all finite intervals, ( F − 1 ◦ F { x } )( t ) = x ( t + ) + x ( t − ) , ∀ t 2 2. If X ∈ L 1 ( R ) satisfies Dirichlet conditions on all finite intervals, ( F ◦ F − 1 { X } )( j ω ) = X ( j ω + ) + X ( j ω − ) , ∀ ω 2 3. If x ∈ C ( R ) ∩ L 1 ( R ) and F { x } ∈ L 1 ( R ) , then F − 1 ◦ F { x } = x 2/32 w

Recap: Fourier Transform for L 1 ( R ) Functions Fourier transform pairs case x ( t ) X ( j ω ) 1 e − at u ( t ) , 1 Re a > 0 a + j ω 2 a e − a | t | , 3 Re a > 0 a 2 + ω 2 a e − ω 2 e − at 2 , � π 3 a > 0 4 a 2 sin( ω T ) 1 u ( t + T ) − u ( t − T ) ω sin( ω c t ) 2 u ( ω + ω c ) − u ( ω − ω c ) π t 3/32 w

Recap: Fourier Transform for Generalized Functions Schwarz space S = S ( R ) of rapidly decreasing functions S = { φ ∈ C ∞ ( R ) : � φ � ℓ, k < ∞ , ∀ ℓ, k ∈ N } where � φ � ℓ, k = sup t ∈ R | t ℓ φ ( k ) ( t ) | . Note F { S } = S . Fourier transform of tempered distributions � ( x , φ ) � x ( ξ ) φ ( ξ ) d ξ R • If x n → x in distribution, then F { x } � lim n F { x n } in distribution ⇒ ( F { x } , φ ) � lim ( x , φ ) = lim n ( x n , φ ) = n ( F { x n } , φ ) , ∀ φ ∈ S ( F { x } , φ ) � ( x , F { φ } ) , • Alternatively, ∀ φ ∈ S 4/32 w

Recap: Fourier Transform for Generalized Functions Fourier transform pairs x ( t ) X ( j ω ) δ ( t ) 1 e − j ω t 0 δ ( t − t 0 ) 2 πδ ( ω ) 1 e j ω 0 t 2 πδ ( ω − ω 0 ) � � δ ( t − t 0 ) e − j ω t dt = e − j ω t 0 , e j ( ω 1 − ω 2 ) t dt = 2 πδ ( ω 1 − ω 2 ) R R � � � � X ( j ω ) e j ω t d ω = x ( τ ) e j ω ( t − τ ) d τ d ω = x ( τ ) 2 πδ ( t − τ ) d τ = 2 π x ( t ) R R R R 5/32 w

Recap: Fourier Transform for Periodic Functions Fourier series ∞ � x [ k ] e jk ω 0 t x ( t ) = ˆ k = −∞ Fourier transform ∞ � X ( j ω ) = 2 π ˆ x [ k ] δ ( ω − k ω 0 ) k = −∞ Equivalent representations • ˆ x [ k ] is amplitude of component at ω k = k ω 0 • ˆ x [ k ] δ ( ω − k ω 0 ) is density of component at ω k = k ω 0 Cf. discrete random variable x in probability � � � E f ( X ) = f ( x n ) p n = f ( x ) p ( x ) dx with p ( x ) = p n δ ( x − x n ) R n n 6/32 w

Recap: Relation between FS and FT • Aperiodic signal x with finite support of length < T • Periodic extension x T of x ∞ � x T ( t ) = x ( t − kT ) k = −∞ NB. In other words, x is one period of periodic signal x T x T [ k ] = 1 F x ← − − → X ˆ T X ( jk ω 0 ) FS ← − − → ˆ x T x T ∞ F � ← − − → X T X T ( j ω ) = 2 π ˆ x T [ k ] δ ( ω − k ω 0 ) x T k = −∞ ω 0 = 2 π ∞ � = ω 0 X ( jk ω 0 ) δ ( ω − k ω 0 ) T k = −∞ 7/32 w

Contents 1. Recap 2. Properties of CT Fourier Transform 8/32 w

Properties of CT Fourier Transform Linearity F { ax + by } = a F { x } + b F { y } Time shifting F → e − j ω t 0 X ( j ω ) F { τ t 0 x } = E − t 0 F { x } x ( t − t 0 ) ← − − or where ( E a f )( s ) = e jas f ( s ) Proof. � � � x ( t − t 0 ) e − j ω t dt = x ( s ) e − j ω ( s + t 0 ) ds = e − j ω t 0 x ( s ) e − j ω s ds R R R Frequency shifting F e j ω 0 t x ( t ) F { E ω 0 x } = τ ω 0 ˆ or ← − − → X ( j ( ω − ω 0 )) x 9/32 w

Example: Multipath Effect Transmitter: x ( t ) x ( t ) y ( t ) Receiver: y ( t ) = a 0 x ( t ) + a 1 x ( t − τ ) Y ( j ω ) = a 0 X ( j ω ) + a 1 e − j ωτ X ( j ω ) = ( a 0 + a 1 e − j ωτ ) X ( j ω ) | X ( j ω ) | Let x ( t ) = u ( t + T ) − u ( t − T ) . 2 T X ( j ω ) = 2 sin( ω T ) π ω 0 T ω | Y ( j ω ) | Y ( j ω ) = ( a 0 + a 1 e − j ωτ ) 2 sin( ω T ) 4 T ω For a 0 = a 1 = 1 , 2 )sin( ω T ) Y ( j ω ) = 4 e − j ωτ/ 2 cos( ωτ π π ω τ T ω 10/32 w

Example: Modulation x ( t ) 1 Baseband signal: x ( t ) − T t T Modulated signal X ( j ω ) y ( t ) = x ( t ) cos( ω c t ) = 1 2 x ( t )( e j ω c t + e − j ω c t ) 2 T Y ( j ω ) = 1 2 X ( j ( ω − ω c )) + 1 2 X ( j ( ω + ω c )) π ω 0 T For x ( t ) = u ( t + T ) − u ( t − T ) , y ( t ) 1 Y ( j ω ) = sin(( ω − ω c ) T ) + sin(( ω + ω c ) T ) ω − ω c ω + ω c − T t T Y ( j ω ) T ω c ω − ω c 0 11/32 w

Properties of CT Fourier Transform Time reversal F F { Rx } = R F { x } or x ( − t ) ← − − → X ( − j ω ) Conjugation F F { x ∗ } = R F { x } x ∗ ( t ) → X ∗ ( − j ω ) or ← − − � ∗ = X ∗ ( − j ω ) Proof. F { x ∗ } ( j ω ) = R x ∗ ( t ) e − j ω t dt = R x ( t ) e j ω t dt � �� Symmetry • x even ⇐ ⇒ X even, x odd ⇐ ⇒ X odd • x real ⇐ ⇒ X ( − j ω ) = X ∗ ( j ω ) • x real and even ⇐ ⇒ X real and even • x real and odd ⇐ ⇒ X purely imaginary and odd 12/32 w

Example For a > 0 , x ( t ) � e − at , t > 0 1 x ( t ) = − e at , t < 0 O t − 1 Let y ( t ) = e − at u ( t ) , with | X ( j ω ) | 1 / a 1 Y ( j ω ) = a + j ω Since x ( t ) = y ( t ) − y ( − t ) . ω − a a O X ( j ω ) = Y ( j ω ) − Y ( − j ω ) arg X ( j ω ) − 2 j ω 1 1 π = a + j ω − a − j ω = 2 a 2 + ω 2 O ω − π 2 x real & odd, X purely imaginary & odd 13/32 w

Time and Frequency Scaling For a � = 0 , � j ω � F { S a x } = 1 → 1 F a F { x } or x ( at ) ← − − | a | S 1 | a | X a Proof. Change variables by τ = at . For a > 0 , � ∞ � ∞ � j ω � a τ d τ a = 1 x ( at ) e − j ω t dt = x ( τ ) e − j ω aX a −∞ −∞ For a < 0 , � ∞ � −∞ � j ω � a τ d τ a = − 1 x ( at ) e − j ω t dt = x ( τ ) e − j ω aX a −∞ ∞ 14/32 w

Time and Frequency Scaling � j ω � F { S a x } = 1 → 1 F a F { x } or x ( at ) ← − − | a | S 1 | a | X a compression in time ⇐ ⇒ stretching in frequency stretching in time ⇐ ⇒ compression in frequency e.g. audio: faster playback = ⇒ higher pitch √ π X ( j ω ) F | a | e − 1 x ( t ) = e − ( at ) 2 a ) 2 4 ( ω ← − − → X ( j ω ) = x ( t ) a = 1 / 2 a = 1 a = 2 ω t 15/32 w

Differentiation F F x ′ ( t ) x ( k ) ( t ) → ( j ω ) k X ( j ω ) ← − − → j ω X ( j ω ) , ← − − Proof. Differentiate under integral sign � � x ( t ) = 1 ⇒ x ′ ( t ) = 1 X ( j ω ) e j ω t d ω = j ω X ( j ω ) e j ω t d ω 2 π 2 π R R y = x ′ is output of differentiator with frequency response � δ ′ ( t ) e − j ω t dt = − d dte − j ω t � H ( j ω ) = F { δ ′ } ( j ω ) = 0 = j ω � � R • Y ( j ω ) = H ( j ω ) X ( j ω ) | H ( j ω ) | • amplifies high frequencies • suppresses low frequencies • DC completely eliminated ω 0 16/32 w

Integration � t → Y ( j ω ) = 1 F y ( t ) = x ( τ ) d τ ← − − j ω X ( j ω ) + π X ( 0 ) δ ( ω ) −∞ Since x ( t ) = y ′ ( t ) , by differentiation property ⇒ Y ( j ω ) = 1 X ( j ω ) = j ω Y ( j ω ) = j ω X ( j ω ) for ω � = 0 Intuitively, y ( t ) has DC component � T � T 1 1 � ¯ y = lim y ( t ) dt = lim x ( τ ) u ( t − τ ) d τ dt 2 T 2 T T →∞ T →∞ − T − T R � T � � � 1 d τ = 1 = x ( τ ) lim u ( t − τ ) dt 2 X ( 0 ) 2 T T →∞ − T R So Y also has component 2 π ¯ y δ ( ω ) = π X ( 0 ) δ ( ω ) 17/32 w

Integration Example. Unit step function u ( t ) � t 1 u ( t ) = δ ( τ ) d τ 1 2 −∞ O t U ( j ω ) = 1 j ω F { δ } ( j ω ) + π F { δ } ( 0 ) δ ( ω ) | U ( j ω ) | = 1 j ω + πδ ( ω ) π Example. Sign function ω O � 1 , t > 0 sgn( t ) = sgn( t ) − 1 t < 0 1 Since sgn = 2 u − 1 , O F { sgn } ( j ω ) = 2 U ( j ω ) − 2 πδ ( ω ) = 2 t − 1 j ω 18/32 w

Integration Example. Unit ramp function u − 2 ( t ) = tu ( t ) � t u − 2 ( t ) = u ( τ ) d τ −∞ Integration property suggests ω 2 + π F { u − 2 } ( j ω ) = 1 j ω U ( j ω ) + π U ( 0 ) δ ( ω ) = − 1 j ωδ ( ω ) + π U ( 0 ) δ ( ω ) But what’s U ( 0 ) = 1 j 0 + πδ ( 0 ) ? What’s π j ω δ ( ω ) ? Not well-defined! Integration property not applicable here! Will see F { u − 2 } = − 1 ω 2 + j πδ ′ ( ω ) Rule of thumb. Applicable when formula is well-defined 19/32 w

Example x 1 � 1 − 2 | t | � � � u ( t + T 2 ) − u ( t − T x ( t ) = 2 ) T O t − T T 2 2 � ω T � X 2 ( j ω ) = 2 2 ] = − 8 T [ e j ω T 2 − 2 + e − j ω T T sin 2 x 1 = x ′ 4 2 + π X 2 ( 0 ) δ ( ω ) = − 8 sin 2 � ω T T T � X 1 ( j ω ) = X 2 ( j ω ) 2 4 O t j ω j ω T − T 2 − 2 T + π X 1 ( 0 ) δ ( ω ) = 8 sin 2 � ω T � X ( j ω ) = X 1 ( j ω ) 4 x 2 = x ′′ j ω ω 2 T 2 2 T T X ( j ω ) T 2 O t − T T 2 2 − 4 ω T 0 20/32 w