EI331 Signals and Systems Lecture 28 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 28 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University June 4, 2019

Contents 1. Analysis of DT LTI Systems by Z -transform 2. Block Diagram Representations 3. Unilateral Z -transform 1/38

DT System Function Recall the response of a DT LTI system to the input x [ n ] is y [ n ] = ( x ∗ h )[ n ] where h is the impulse response of the system. If x and h have z -transforms, the convolution property implies Y ( z ) = X ( z ) H ( z ) in their common ROC. If the ROC has a nonempty interior point, the system function (aka transfer function) H ( z ) uniquely determines h and hence system properties through the Laurent series expansion ∞ � h [ n ] z − n H ( z ) = n = −∞ 2/38

Causality Recall h is right-sided iff its ROC is the exterior of a circle, i.e. R 1 < | z | < ∞ ∞ � h [ n ] z − n H ( z ) = n = N 1 The following conditions are equivalent 1. N 1 ≥ 0 ∞ � h [ n ] z n is a convergent power series on | z | < 2. H ( z − 1 ) = 1 R 1 n = N 1 3. 0 is a removable singularity of H ( z − 1 ) 4. ∞ is removable singularity of H ( z ) z → 0 H ( z − 1 ) exists and is finite 5. lim z →∞ H ( z ) exists and is finite 1 6. lim 1 This is the more precise statement of ∞ ∈ ROC. 3/38

Causality An LTI system with system function H ( z ) is causal iff 1. the ROC is the exterior of a circle 2. lim z →∞ H ( z ) exists and is finite causal ⇐ ⇒ ROC is the exterior of a circle including ∞ An LTI system with rational system function H ( z ) = N ( z ) D ( z ) is causal iff 1. the ROC is | z | > | p | , where p is the outermost pole 2. deg D ≥ deg N Example. H ( z ) = z 3 − 2 z 2 − z 8 cannot be the system function of a z 2 + 1 4 z + 1 causal system. 4/38

Stability Recall an LTI system is stable iff its impulse response h ∈ ℓ 1 , i.e. ∞ � | h [ n ] | < ∞ n = −∞ i.e. H ( z ) converges absolutely on the unit circle | z | = 1 , so its ROC R 1 < | z | < R 2 must satisfy R 1 < 1 < R 2 . stable ⇐ ⇒ ROC includes the unit circle | z | = 1 A causal LTI system with rational system function H ( z ) is stable iff all its poles are inside the unit circle. 1 Example. A causal system with H ( z ) = 1 − az − 1 is stable iff | a | < 1 1 Example. A system with H ( z ) = 1 − az − 1 where | a | > 1 and ROC | z | < | a | is also stable, but it is noncausal. 5/38

Example � � Im H ( z ) = 1 z z − z − 3 z + 1 2 2 2 There are two poles p 1 = − 1 2 and p 2 = 3 2 . X X Re 1. | z | > 3 2 , causal, unstable 1 3 − 1 2 2 � 3 � n � � n h 1 [ n ] = 1 u [ n ] − 1 − 1 u [ n ] 2 2 2 2 2 < | z | < 3 1 2. 2 , noncausal, stable � 3 � n � � n h 2 [ n ] = − 1 u [ − n − 1 ] − 1 − 1 u [ n ] 2 2 2 2 3. | z | < 1 2 , noncausal, unstable � 3 � n � � n h 3 [ n ] = − 1 u [ − n − 1 ]+ 1 − 1 u [ − n − 1 ] 2 2 2 2 6/38

Linear Constant-coefficient Difference Equations LTI system with input and output related by N M � � a k y [ n − k ] = b k x [ n − k ] k = 0 k = 0 Take z -transform of both sides N M � � a k z − k Y ( z ) = b k z − k X ( z ) k = 0 k = 0 so � M k = 0 b k z − k H ( z ) = Y ( z ) X ( z ) = � N k = 0 a k z − k System function is always rational Difference equation does not specify ROC! Need additional conditions (e.g. stability, causality) to determine h [ n ] . 7/38

Example Consider LTI system with input and output related by y [ n ] − 1 2 y [ n − 1 ] = x [ n ] + 1 3 x [ n − 1 ] System function H ( z ) = 1 + 1 3 z − 1 1 − 1 2 z − 1 Two possibilities for ROC: | z | > 1 2 and | z | < 1 2 . 1. If | z | > 1 2 � 1 � n � 1 � n ∞ 1 � Z − 1 z − n 2 z − 1 = ← − − → u [ n ] 1 − 1 2 2 n = 0 By linearity and time-shifting property, � 1 � n � 1 � n − 1 u [ n ] + 1 h [ n ] = u [ n − 1 ] 2 3 2 8/38

Example (cont’d) Consider LTI system with input and output related by y [ n ] − 1 2 y [ n − 1 ] = x [ n ] + 1 3 x [ n − 1 ] System function H ( z ) = 1 + 1 3 z − 1 1 − 1 2 z − 1 Two possibilities for ROC: | z | > 1 2 and | z | < 1 2 . 2. If | z | < 1 2 � 1 � n ∞ − 2 z 1 � Z − 1 2 n z n 2 z − 1 = 1 − 2 z = − ← − − → − u [ − n − 1 ] 1 − 1 2 n = 1 By linearity and time-shifting property, � 1 � n � 1 � n − 1 u [ − n − 1 ] − 1 h [ n ] = − u [ − n ] 2 3 2 9/38

Example (cont’d) Consider LTI system with input and output related by y [ n ] − 1 2 y [ n − 1 ] = x [ n ] + 1 3 x [ n − 1 ] If we use Fourier transform, then frequency response is H ( e j ω ) = 1 + 1 3 e − j ω 1 − 1 2 e − j ω and � 1 � n � 1 � n − 1 u [ n ] + 1 h [ n ] = u [ n − 1 ] 2 3 2 Why only one possibility? • Fourier transform method assumes stability, requiring that ROC of H ( z ) contain the unit circle, i.e. | z | > 1 2 • In general, not applicable to unstable systems 10/38

Example Consider LTI system with input and output related by y [ n ] − 3 4 y [ n − 1 ] + 1 8 y [ n − 2 ] = 2 x [ n ] System function 2 H ( z ) = 4 z − 1 + 1 1 − 3 8 z − 2 By partial fraction expansion 2 4 2 H ( z ) = 4 z − 1 ) = 2 z − 1 − ( 1 − 1 2 z − 1 )( 1 − 1 1 − 1 1 − 1 4 z − 1 Take inverse z -transform to obtain impulse response  � 1 � n u [ n ] − 2 � 1 � n u [ n ] , ROC: | z | > 1 h 1 [ n ] = 4   2 4 2 � 1 � n u [ − n − 1 ] − 2 � 1 � n u [ n ] , ROC: 1 4 < | z | < 1 h 2 [ n ] = − 4 2 4 2 � 1 � n u [ − n − 1 ] + 2 � 1 � n u [ − n − 1 ] ,   ROC: | z | < 1 h 3 [ n ] = − 4 2 4 4 11/38

Example (cont’d) � 1 � n u [ n ] . Find response to x [ n ] = 4 1 | z | > 1 X ( z ) = 4 z − 1 , 1 − 1 4 z -transform of response 2 1 Y ( z ) = H ( z ) X ( z ) = 4 z − 1 ) · 4 z − 1 , ( 1 − 1 2 z − 1 )( 1 − 1 1 − 1 NB. Output well-defined only for | z | > 1 2 and 1 4 < | z | < 1 2 ! Exercise. Verify that h 3 ∗ x is not well-defined. By partial fraction expansion (see slides 20-22 of Lecture 19) 4 2 8 Y ( z ) = − 4 z − 1 − 4 z − 1 ) 2 + 1 − 1 ( 1 − 1 1 − 1 2 z − 1 Take inverse z -transform to obtain response for first two cases. 12/38

1 Inverse Z -transform of ( 1 − az − 1 ) n Recall ∞ 1 � a n ζ n , 1 − a ζ = | a ζ | < 1 n = 0 and − 1 1 � a n ζ n , 1 − a ζ = − | a ζ | > 1 n = −∞ Repeat the derivation on slide 23 of Lecture 19, and let ζ = z − 1 , � n + m − 1 � 1 Z a n u [ n ] ← − − → ( 1 − az − 1 ) m , | z | > | a | m − 1 and � − n − 1 � 1 Z ( − 1 ) m a n u [ − n − m ] ← − − → ( 1 − az − 1 ) m , | z | < | a | m − 1 13/38

Example Im Causal LTI system described by − 2 y [ n ] − 1 2 y [ n − 1 ] = x [ n ] + 2 3 3 x [ n − 1 ] X Re 1 1 2 Find response to x [ n ] = ( − 2 3 ) n u [ n ] Im H ( z ) = 1 + 2 3 z − 1 | z | > 1 2 z − 1 , 1 − 1 2 X Re 1 | z | > 2 X ( z ) = 3 z − 1 , − 2 1 3 1 + 2 3 � 1 � n 1 2 z − 1 , | z | > 1 Im Y ( z ) = 2 = ⇒ y [ n ] = u [ n ] 1 − 1 2 ROC of Y is larger than intersection of ROCs X Re of X and H due to pole-zero cancellation at 1 1 2 z = − 2 3 . 14/38

Example Suppose an LTI system satisfies the following, 2 ) n + 10 ( 1 1. output for input x 1 [ n ] = ( 1 6 ) n u [ n ] is y 1 [ n ] = [ a ( 1 3 ) n ] u [ n ] 2. output for input x 2 [ n ] = ( − 1 ) n is y 2 [ n ] = 7 4 ( − 1 ) n Want to find H ( z ) . 1 | z | > 1 X 1 ( z ) = 6 z − 1 , 1 − 1 6 a 10 | z | > 1 Y 1 ( z ) = 2 z − 1 + 3 z − 1 , 1 − 1 1 − 1 2 System function X 1 ( z ) = [( a + 10 ) − ( 5 + a 3 ) z − 1 ][ 1 − 1 6 z − 1 ] H ( z ) = Y 1 ( z ) , ( 1 − 1 2 z − 1 )( 1 − 1 3 z − 1 ) ⇒ H ( z ) = ( 1 − 2 z − 1 )( 1 − 1 6 z − 1 ) By 2, H ( − 1 ) = 7 4 = ⇒ a = − 9 = ( 1 − 1 2 z − 1 )( 1 − 1 3 z − 1 ) 15/38

Example (cont’d) Three possibilities for ROC of H ( z ) = ( 1 − 2 z − 1 )( 1 − 1 6 z − 1 ) ( 1 − 1 2 z − 1 )( 1 − 1 3 z − 1 ) • | z | < 1 3 • 1 3 < | z | < 1 2 • | z | > 1 2 By the convolution property, ROC of Y 1 contains the intersection ⇒ ROC of H is | z | > 1 of the ROCs of X 1 and H = 2 = ⇒ stable. Also H ( ∞ ) = 1 is finite = ⇒ causal 6 z − 1 + 1 H ( z ) = 1 − 13 3 z − 2 6 z − 1 + 1 1 − 5 6 z − 2 Difference equation y [ n ] − 5 6 y [ n − 1 ] + 1 6 y [ n − 2 ] = x [ n ] − 13 6 x [ n − 1 ] + 1 3 x [ n − 2 ] 16/38

Contents 1. Analysis of DT LTI Systems by Z -transform 2. Block Diagram Representations 3. Unilateral Z -transform 17/38

System Interconnections LTI systems in series connection Y = ( XH 1 ) H 2 = X ( H 1 H 2 ) = ( XH 2 ) H 1 H = H 1 H 2 H 1 ( z ) H 2 ( z ) y x H ( z ) commutative y H = H 2 H 1 x H 2 ( z ) H 1 ( z ) H ( z ) 18/38

System Interconnections LTI systems in systems in parallel connection h 1 y h = h 1 + h 2 x + h 2 h H 1 ( z ) y H = H 1 + H 2 x + H 2 ( z ) H ( z ) 19/38

System Interconnections LTI systems in systems in feedback connection + + y x h 1 - y = h 1 ∗ ( x − h 2 ∗ y ) h 2 h = ? h + + y Y = H 1 X − H 1 H 2 Y H 1 ( z ) x - H 2 ( z ) H = Y H 1 X = 1 + H 1 H 2 H ( z ) 20/38