[PPT].Lesson 07-Inverse+z_Transform-1

# [PPT].Lesson 07-Inverse+z_Transform-1 - News Audio Signal...

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Inverse z-Transform Challenge 06 esson 07: Inverse z- ansform Lesson 07: Inverse z transform Challenge 07 What’s it all about hat is an inverse z ansform? • What is an inverse z-transform? • How can it be computed? • What type of MATLAB support is available? Practice Exam to be on-line by 12:00. Consists of 5 question (actual Exam will have 2 questions) Lesson [07] News – Audio Signal Processing Lesson [07] hallenge 06 Challenge 06 The discrete-time (a.k.a., digital control) system shown below is to be studied and interpreted as a discrete-time transfer function G(z)=Y(z)/X(z). x(t) T=1 Zero-order hold s)=(1 - T /s Plant G p (s) s)=1/(s(s+1)) y(t) G 0 (s)=(1-e )/s G p (s)=1/(s(s+1)) Ideal sampler : What is G )? Q: What is G 0 (z)? … and for the overachiever, what is G(z)? Lesson [07] hallenge 06 Challenge 06 x(t) T=1 Zero-order hold G 0 (s)=(1-e -sT )/s Plant G p (s) G p (s)=1/(s(s+1)) y(t) eal Ideal sampler 0 T s In the continuous-time domain the impulse captures the signal and olds its value for a sample period n the discrete me holds its value for a sample period (i.e., ZOH) In the discrete-time domain, the ZOH becomes a Kronecker pulse k] 0 Lesson [07] impulse [k]. hallenge 06 Challenge 06 x(t) T=1 Zero-order hold G 0 (s)=(1-e -sT )/s Plant G p (s) G p (s)=1/(s(s+1)) y(t) eal Ideal sampler What is G 0 (z)? ince T =1 Since T s 1, G 0 (z)=Z(G 0 (s))=Z(( 1-e -sT ) 1/s ) =( - -1 /(z- A Kronecker impulse ( 1 z ) ( z/(z 1 )) =(1-z -1 ) (1/(1-z -1 )=1 Lesson [07] p hallenge 06 Challenge 06 x(t) T=1 Zero-order hold G 0 (s)=(1-e -sT )/s Plant G p (s) G p (s)=1/(s(s+1)) y(t) eal Ideal sampler What is G(z)? G(s) = G 0 (s) G p (s) = ((1-e -sT )/s) (1/(s(s+1)))= (1-e -sT ) 1/(s 2 (s+1)) What now? – compute Z(1/(s 2 (s+1))) Lesson [07] hallenge 06 Challenge 06 x(t) T=1 Zero-order hold G 0 (s)=(1-e -sT )/s Plant G p (s) G p (s)=1/(s(s+1)) y(t) eal Ideal sampler G(z) = Z [(1-e -sT ) 1/(s 2 (s+1)) ] Trick: Perform a partial fraction expansion (assumed legacy knowledge) = (1-z -1 ) Z ( 1/s 2 -1/s +1/(s+1) ) Lesson [07] hallenge 06 Challenge 06 x(t) T=1 Zero-order hold G 0 (s)=(1-e -sT )/s Plant G p (s) G p (s)=1/(s(s+1)) y(t) eal Ideal sampler Manipulating G(z) = Z 1-e -sT /(s 2 s+1))] = 1-z -1 Z 1/s 2 -1/s +1/(s+1) () [( )( ( ))] ( ) ( ) = (z-1)/z (z/(z-1) 2 - z/(z-1) + z/(z-e -1 )) (Simplify as needed) Lesson [07] Inverse z- Table 1: Primitive Signals and their z -Transform Time-domain z -transform Transform [ k ] 1 [ k m ] z m u [ k ] z /( z –1) Assume that the z- transform of elementary signals are known and ku [ k ] z /( z –1) 2 k 2 u [ k ] z ( z +1)/( z –1) 3 a k u [ k ] z /( z a ) cataloged. ka k u [ k ] az /( z a ) 2 k 2 a k u [ k ] az ( z + a )/( z a ) 3 n[ k in( What is common to their numerators? sin[ bk ] u [ k ] cos[ bk ] u [ k ] 1 ) cos( 2 ) sin( 2 b z z b z os( )) cos( ( 2 b z z exp[ akT s ]sin[ bkT s ] u [ kT s ] 1 ) cos( 2 b z z S S S aT S aT S aT e bT ze z bT ze 2 2 ) cos( 2 ) sin( They have a numerator “z” (except for [k]). his will become exp[ akT s ]cos[ bkT s ] u [ kT s ] a k sin( bkT ) u [ kT ] S S S aT S aT S aT e bT ze z bT e z z 2 2 ) cos( 2 )) cos( ( ) sin( bT az S This will become important later in the lesson.

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## This note was uploaded on 09/18/2011 for the course EEL 5718 taught by Professor Janisemcnair during the Fall '11 term at University of Florida.

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[PPT].Lesson 07-Inverse+z_Transform-1 - News Audio Signal...

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