part8 - Part 8: Z-Transform Z-Transform l l Do we have...

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Part 8: Z-Transform Z-Transform l Do we have similar tools for discrete-time systems, as having Laplace Transform in continuous-time system? l Recall or if x(t)=0 with t<0 l Taking Laplace Transform, we have ( 29 ( 29 z X nT x → Z ( 29 ( 29 ( 29 -∞ = - = n s nT t x t x d ( 29 ( 29 ( 29 = - = 0 n s t x t x d ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 = - = - = - = - = - = 0 0 0 0 0 n n sT n t s n st s e x dt e t x dt e t x s X d d ( 29 ( 29 = - = 0 n n z nT x z X z -n
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Remarks l Coefficient x(nT) denotes the sample value at nT l indicates the sample occurs nT delay after the reference zero time l Multiplying X(z) by z means to advance the sequence by one sample interval T. l Multiplying X(z) by z -1 means to delay the sequence by one sample interval T n z - Properties of Z-Transform l Linearity l Delay l Shifting l Multiplication by n l Multiplication by r n l Convolution l Initial Value Theorem l Final Value Theorem ( 29 [ ] ( 29 z X z k n x k - = - Z ( 29 ( 29 [ ] ( 29 ( 29 z X a z X a n x a n x a 2 2 1 1 2 2 1 1 + = + Z ( 29 [ ] ( 29 m m m m dz z X d z n x n ) ( - = Z [ ] ( 29 ) 1 ( ) 1 ( ) 0 ( ) ( 1 - - - - = + - k zx x z x z z X z k n x k k k L Z ( 29 [ ] = r z X n x r n Z ( 29 ( 29 ( 29 ( 29 z X z X k n x k x k 2 1 0 2 1 = - = Z ( 29 ( 29 z X x z = lim 0 ( 29 ( 29 z X z n x z n ) 1 ( 1 1 - - =
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EE3118 Linear System & Signal Analysis (KST) Example: Consider the following discrete-time signal, which has a finite number of elements () ( ) () ( ) ( ) 2 2 , 1 , 2 0 , 3 , 2 2 , 4 3 = = = = = = T x T x x T x T x T x The z-transform of this sequence will be obtained as 2 1 0 2 3 2 2 3 2 4 + + + = z z z z z z z X Example: 5 2 1 12 5 3 + + z z z implies that samples have the value of 3, 5 and 12 at 1, 2 and 5 sample periods after the t=0 reference, respectively. Example: Z-transform of unit pulse n δ : [] 1 0 0 = = = = z z n n n n Z Example: Z-transform of unit step sample sequence ( ) 1 = n u with 0 n . 1 0 0 1 1 = = = = = z z z n u n u n n n n Z Example: Find the Z-transform of n Ar n x = r z r z Az rz A rz A z Ar n x n n n n n > = = = = = = 1 0 1 0 1 Z 3T 4T 0 -T -2T T 2T . . . x 1 Im(z) Re(z) x 1 Im(z) Re(z)
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EE3118 Linear System & Signal Analysis (KST) Linearity () () [] z X a z X a z n x a z n x a z n x a z n x a z n x a n x a n x a n x a n n n n n n n n n n 2 2 1 1 0 2 2 0 1 1 0 2 2 0 1 1 0 2 2 1 1 2 2 1 1 + = + = + = + = + = = = = = Z Delay () z X z z mT x z m mT x z mT x k n m z mT x z kT nT x kT nT x k m m k m k m k m k m n n = = = = = = < = = = = = 0 0 0 0 if 0 letting Q Z Shifting [] () () ( ) = = = = = = = = = + = = + = + 1 0 1 0 0 0 letting k m m k k k m m m m k k m m k k m m k n n z m x z X z z m x z m x z z m x z k n m z m x z k n x k n x Z
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EE3118 Linear System & Signal Analysis (KST) Convolution () ( ) () ( ) () ( ) () () z X z X z m x z k x z m x k x z k n x k x z k n x k x k n x k x n m k k kn k m n n nk k 2 1 0 2 0 1 00 2 1 2 1 2 1 0 2 1 = = = = = ∑∑ =
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part8 - Part 8: Z-Transform Z-Transform l l Do we have...

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