Lesson+07-Inverse+z_Transform-1

Lesson+07-Inverse+z_Transform-1 - Inverse z-Transform...

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Lesson [07] Inverse z-Transform Challenge 06 Lesson 07: Inverse z-transform Challenge 07 What’s it all about • 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)
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Lesson [07] News – Audio Signal Processing
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Lesson [07] 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). Q: What is G 0 (z)? … and for the overachiever, what is G(z)? 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) Ideal  sampler
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Lesson [07] 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) Ideal  sampler In the discrete-time  domain, the ZOH  becomes a Kronecker  impulse  δ [k]. 0      T s In the continuous-time domain the impulse captures the signal and holds its value for a sample period (i.e., ZOH) 0
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Lesson [07] 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) Ideal  sampler A Kronecker impulse What is G 0 (z)? Since T s =1, G 0 (z)=Z(G 0 (s))=Z(( 1-e -sT ) × 1/s ) =( 1-z -1 ) × ( z/(z-1 )) =(1-z -1 ) × (1/(1-z -1 ) = 1
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Lesson [07] Challenge 06 What is G(z)? G(s) = G 0 (s) G p (s) = ((1-e -sT )/s) (1/(s(s+1)))= (1-e ) × 1/(s 2 (s+1)) What now? – compute Z(1/(s 2 (s+1))) 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) Ideal  sampler
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Lesson [07] Challenge 06 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) ) 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) Ideal  sampler
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Lesson [07] Challenge 06 Manipulating G(z) = Z [(1-e -sT )/(s 2 (s+1))] = (1-z ) × 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) 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) Ideal  sampler
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Lesson [07] Inverse z- Transform Assume that the z- transform of elementary signals are known and cataloged. What is common to their numerators? Table 1: Primitive Signals and their z -Transform Time-domain z -transform δ [ k ] 1 [ k m ] z m u [ k ] z /( z –1) ku [ k ] z /( z –1) 2 k 2 u [ k ] z ( z +1)/( z –1) 3 a k u [ k ] z /( z a ) ka k u [ k ] az /( z a ) 2 k 2 a k u [ k ] az ( z + a )/( z a ) 3 sin[ bk ] u [ k ] cos[ bk ] u [ k ] exp[ akT s ]sin[ bkT s ] u [ kT s ] exp[ akT s ]cos[ bkT s ] u [ kT s ] a k sin( bkT s ) u [ kT s ] a k cos( bkT s ) u [ kT s ] 1 ) cos( 2 ) sin( 2 + - b z z b z 1 ) cos( 2 )) cos( ( 2 + - - b z z b z z S S S aT S aT S aT e bT ze z bT ze 2 2 ) cos( 2 ) sin( + - S S S aT S aT S aT e bT ze z bT e z z 2 2 ) cos( 2 )) cos( ( + - - 2 2 ) cos( 2 ) sin( a bT az z bT az S S + - 2 2 ) cos( 2 )) cos( ( a bT az z bT a z z S S + - - They have a numerator “z” (except for δ [k]). This will become important later in the lesson.
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Lesson [07] Inverse z-Transform What are the accepted inverse z-transform methods?
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Lesson+07-Inverse+z_Transform-1 - Inverse z-Transform...

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