EI331 Signals and Systems Lecture 19 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 19 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University April 30, 2019

Contents 1. Convolution Property of DTFT 2. Multiplication Property of DTFT 3. Systems Described by Linear Constant-coefficient Difference Equations 4. Sampling of DT Signals 5. DT Processing of CT Signals 1/43

Convolution Property of DTFT F → X ( e j ω ) H ( e j ω ) ( x ∗ h )[ n ] ← − − NB. Dual of multiplication property of CTFS FS x ( t ) y ( t ) ← − − → (ˆ x ∗ ˆ y )[ k ] NB. Similar to CTFT, applicable when formula is well-defined Proof. ( x ∗ h )[ n ] e − j ω n = F { x ∗ h } ( e j ω ) = � � � x [ m ] h [ n − m ] e − j ω n n ∈ Z n ∈ Z m ∈ Z � � h [ n − m ] e − j ω n = x [ m ] m ∈ Z n ∈ Z � x [ m ] e − j ω m H ( e j ω ) = m ∈ Z = X ( e j ω ) H ( e j ω ) 2/43

Frequency Response of LTI System LTI system T • fully characterized by impulse response h y = T ( x ) = h ∗ x • also fully characterized by frequency response H = F { h } , if H is well-defined ◮ BIBO stable system, h ∈ ℓ 1 ◮ other systems: accumulator h = u , ... Typically, convolution property implies Y = F { y } = HX = F { h } F { x } Instead of computing x ∗ h , can do y = F − 1 ( F { h } F { x } ) 3/43

Example Response of LTI system with impulse response h [ n ] = a n u [ n ] to input x [ n ] = b n u [ n ] , where | a | < 1 , | b | < 1 Method 1. Direct convolution y = x ∗ h Method 2. Solve difference equation with initial rest condition y [ n ] − ay [ n − 1 ] = b n u [ n ] Method 3. Fourier transform. 1 1 1 H = 1 − ae − j ω , X = 1 − be − j ω = ⇒ Y = ( 1 − ae − j ω )( 1 − be − j ω ) a − b ( a n + 1 − b n + 1 ) u [ n ] If a � = b , Y ( e j ω ) = a / ( a − b ) 1 − ae − j ω − b / ( a − b ) 1 1 − be − j ω , y [ n ] = If a = b , Y ( e j ω ) = d � a � da a n + 1 u [ n ] = ( n + 1 ) a n u [ n ] d , y [ n ] = 1 − ae − j ω da Or use Y ( e j ω ) = j a e j ω d � a � and differentiation property d ω 1 − ae − j ω 4/43

Example Response of LTI system with impulse response h [ n ] = a n u [ n ] to input x [ n ] = cos( ω 0 n ) , | a | < 1 . Frequency response 1 H ( e j ω ) = 1 − ae − j ω Method 1. Use eigenfunction property x [ n ] = 1 2 e j ω 0 n + 1 2 e − j ω 0 n e j ω 0 n y [ n ] = 1 2 H ( e j ω 0 ) e j ω 0 n + 1 2 H ( e − j ω 0 ) e − j ω 0 n = Re 1 − ae − j ω 0 � a sin ω 0 � 1 √ 1 − 2 a cos ω 0 + a 2 cos = ω 0 n − arctan 1 − a cos ω 0 5/43

Example Method 2. Use Fourier transform of X ∞ � X ( e j ω ) = [ πδ ( ω − ω 0 + 2 k π ) + πδ ( ω + ω 0 + 2 k π )] k = −∞ ∞ � Y ( e j ω ) = π H ( e j ( ω 0 − 2 k π ) ) δ ( ω − ω 0 + 2 k π ) k = −∞ ∞ � π H ( e − j ( ω 0 + 2 k π ) ) δ ( ω + ω 0 + 2 k π ) + k = −∞ ∞ = H ( e j ω 0 ) � πδ ( ω − ω 0 + 2 k π ) k = −∞ ∞ � + H ( e − j ω 0 ) πδ ( ω + ω 0 + 2 k π ) k = −∞ y [ n ] = 1 2 H ( e j ω 0 ) e j ω 0 n + 1 2 H ( e − j ω 0 ) e − j ω 0 n 6/43

Example: Ideal Bandstop Filter H hp ( e j ω ) ( − 1 ) n ( − 1 ) n H lp ( e j ω ) × × y x + H lp ( e j ω ) H lp ( e j ω ) 0 < W < π/ 2 1 ω π − π − 2 π − W 2 π 0 W H ( e j ω ) 1 π ω − π − W − 2 π 0 W 2 π π − W 7/43

Contents 1. Convolution Property of DTFT 2. Multiplication Property of DTFT 3. Systems Described by Linear Constant-coefficient Difference Equations 4. Sampling of DT Signals 5. DT Processing of CT Signals 8/43

Multiplication Property of DTFT → 1 2 π X ⊛ Y = 1 � F X ( e j θ ) Y ( e j ( ω − θ ) ) d θ x [ n ] y [ n ] ← − − 2 π 2 π NB. Dual of periodic convolution property of CTFS FS x ⊛ y ← − − → T ˆ x ˆ y Proof. � 1 � � x [ n ] y [ n ] e − j ω n = � � F { xy } ( e j ω ) = X ( e j θ ) e j θ n d θ y [ n ] e − j ω n 2 π 2 π n ∈ Z n ∈ Z �� � = 1 � X ( e j θ ) y [ n ] e − j ( ω − θ ) n d θ 2 π 2 π n ∈ Z = 1 � X ( e j θ ) Y ( e j ( ω − θ ) ) d θ 2 π 2 π 9/43

Example x [ n ] = x 1 [ n ] x 2 [ n ] , where x 1 [ n ] = sin( 3 π n / 4 ) , x 2 [ n ] = sin( π n / 2 ) π n π n 2 π ˜ 2 π X 1 ⊛ X 2 = 1 1 X = X 1 ∗ X 2 periodic aperiodic X 1 ( e j ω ) 1 − π π ω − 2 π 2 π − π 0 π 2 2 X 2 ( e j ω ) 1 − π π ω − 2 π 2 π − 3 π 0 3 π 4 4 ˜ X 1 ( e j ω ) 1 ω − π 0 π 2 2 10/43

Example x [ n ] = x 1 [ n ] x 2 [ n ] , where x 1 [ n ] = sin( 3 π n / 4 ) , x 2 [ n ] = sin( π n / 2 ) π n π n ∞ 2 π ˜ τ 2 k π (˜ X 1 ∗ ˜ 2 π X 1 ⊛ X 2 = 1 1 1 � X = X 1 ∗ X 2 = X 2 ) 2 π k = −∞ periodic aperiodic aperiodic ˜ X 1 ( e j ω ) 1 ω − π 0 π 2 2 ˜ X 2 ( e j ω ) 1 ω − 3 π 0 3 π 4 4 2 π (˜ X 1 ∗ ˜ 1 Y = X 2 ) 1 / 2 π ω − π − 2 π − π 0 π 2 π 2 2 11/43

Example x [ n ] = x 1 [ n ] x 2 [ n ] , where x 1 [ n ] = sin( 3 π n / 4 ) , x 2 [ n ] = sin( π n / 2 ) π n π n ∞ 2 π ˜ τ 2 k π (˜ X 1 ∗ ˜ 2 π X 1 ⊛ X 2 = 1 1 1 � X = X 1 ∗ X 2 = X 2 ) 2 π k = −∞ periodic aperiodic aperiodic ˜ X 1 ( e j ω ) 1 ω − π 0 π 2 2 ˜ X 2 ( e j ω ) 1 ω − 3 π 0 3 π 4 4 2 π ( X 1 ⊛ X 2 ) 1 X = 1 / 2 π ω − π − 2 π − π 0 π 2 π 2 2 11/43

Example x [ n ] = x 1 [ n ] x 2 [ n ] , where x 1 [ n ] = sin( 3 π n / 4 ) , x 2 [ n ] = sin( π n / 2 ) π n π n ∞ 2 π ˜ τ 2 k π (˜ X 1 ∗ ˜ 2 π X 1 ⊛ X 2 = 1 1 1 � X = X 1 ∗ X 2 = X 2 ) 2 π k = −∞ periodic aperiodic aperiodic ˜ X 1 ( e j ω ) 1 ω − π 0 π 2 2 ˜ X 2 ( e j ω ) 1 ω − 3 π 0 3 π 4 4 2 π ( X 1 ⊛ X 2 ) 1 X = 1 / 2 π ω − π − 2 π − π 0 π 2 π 2 2 11/43

Example: DT Modulation y [ n ] = x [ n ] c [ n ] , where c [ n ] = cos( ω c n ) X ( e j ω ) A ω M − ω M − 2 π 2 π C ( e j ω ) π π ω c − ω c − 2 π − ω c − 2 π − 2 π + ω c 0 2 π − ω c 2 π 2 π + ω c Y ( e j ω ) 2 π − ω c − ω M A 2 ω c − ω c − 2 π − 2 π 0 2 π ω c + ω M � ω c > ω M ⇒ ω M < π No overlap between replicas: = 2 ω c + ω M < π 12/43

Example: DT Demodulation z [ n ] = y [ n ] c [ n ] , where c [ n ] = cos( ω c n ) Y ( e j ω ) 2 π − ω c − ω M A 2 ω c − ω c − 2 π − 2 π 2 π 0 ω c + ω M C ( e j ω ) π π ω c − ω c − 2 π − ω c − 2 π + ω c 2 π − ω c 2 π + ω c − 2 π 0 2 π Z ( e j ω ) 2 2 ω c − ω M A 2 ω M − ω M − 2 π 2 π − W W Recover X by lowpass filtering Y with W ∈ ( ω M , 2 ω c − ω M ) 13/43

Contents 1. Convolution Property of DTFT 2. Multiplication Property of DTFT 3. Systems Described by Linear Constant-coefficient Difference Equations 4. Sampling of DT Signals 5. DT Processing of CT Signals 14/43

Linear Constant-coefficient Difference Equations Frequency response of LTI system described by N M � � a k y [ n − k ] = b k x [ n − k ] k = 0 k = 0 Method 1. Use eigenfunction property x [ n ] = e j ω n = ⇒ y [ n ] = H ( e j ω ) e j ω n Substitution into difference equation yields N M a k H ( e j ω ) e j ω ( n − k ) = � � b k e j ω ( n − k ) k = 0 k = 0 � M k = 0 b k e − j ω k H ( e j ω ) = � N k = 0 a k e − j ω k 15/43

Linear Constant-coefficient Difference Equations Method 2. Take Fourier transform of both sides � N � M � � � � F a k y [ n − k ] = F b k x [ n − k ] k = 0 k = 0 By linearity and time-shifting property N M � a k e − j ω k Y ( e j ω ) = � b k e − j ω k X ( e j ω ) k = 0 k = 0 By convolution property � M H ( e j ω ) = Y ( e j ω ) k = 0 b k e − j ω k X ( e j ω ) = � N k = 0 a k e − j ω k Frequency response is rational function of e − j ω 16/43

Example Consider LTI system described by y [ n ] − 3 4 y [ n − 1 ] + 1 8 y [ n − 2 ] = 2 x [ n ] Frequency response 2 H ( e j ω ) = 4 e − j ω + 1 1 − 3 8 e − j 2 ω By partial fraction expansion 2 4 2 H ( e j ω ) = 4 e − j ω ) = 2 e − j ω − ( 1 − 1 2 e − j ω )( 1 − 1 1 − 1 1 − 1 4 e − j ω Take inverse Fourier transform to obtain impulse response � n � n � 1 � 1 h [ n ] = 4 u [ n ] − 2 u [ n ] 2 4 17/43

Example � 1 � n u [ n ] . Find response to x [ n ] = 4 Fourier transform of response 2 1 Y ( e j ω ) = H ( e j ω ) X ( e j ω ) = 4 e − j ω ) · ( 1 − 1 2 e − j ω )( 1 − 1 1 − 1 4 e − j ω By partial fraction expansion 4 2 8 Y ( e j ω ) = − 4 e − j ω − 4 e − j ω ) 2 + 1 − 1 ( 1 − 1 1 − 1 2 e − j ω Take inverse Fourier transform to obtain response � n � n � n � � � 1 � 1 � 1 y [ n ] = − 4 − 2 ( n + 1 ) + 8 u [ n ] 4 4 2 18/43

Partial Fraction Expansion 2 H ( e j ω ) = 4 e − j ω + 1 1 − 3 8 e − j 2 ω Replace e − j ω by z , 2 2 A B H ( z − 1 ) = 8 z 2 = 4 z ) = 2 z + 1 − 3 4 z + 1 ( 1 − 1 2 z )( 1 − 1 1 − 1 1 − 1 4 z � � � � ( 1 − 1 2 � � 2 z ) H ( z − 1 ) A = = = 4 � � 1 − 1 � 4 z � � � z = 2 z = 2 � � � � ( 1 − 1 2 � � 4 z ) H ( z − 1 ) B = = = − 2 � � 1 − 1 � � 2 z � � z = 4 z = 4 A A NB. Terms take form ( s − r ) m for CT, but ( 1 − rz ) m for DT 19/43