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

This preview shows pages 1–6. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 - - =
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)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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 = = = = = ∑∑ =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 18

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online