This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: EEL 3135 Dr. Fred J. Taylor, Professor Lesson Title: zTransform Lesson Number: 17 (Sections 71 and 72) Background: Given the discretetime signal: [ ] [ ] k n k x n x K N k = = ] [ 1. the authors define the socalled ztransform of x[n] to be: ( 29 [ ] k N k z k x z X = = 2. It can be seen that that zoperator is positional with zk corresponds to a spatial delay. Specifically, z1 represent a one sample delay, z2 is a two sample delay, and so forth. There is no other motivation for this action offered at this time. One must, however, be mindful of the fact that in order to qualify as a mathematical a transform, a ztransform must have a forward and reverse mapping. The ztransform pairs are defined to be: [ ] [ ] [ ] ( 29 z X z k x k n k x n x k N k Z K N k =  = = = ] [ 3. Example : zTransform Suppose a discretetime signal has a sampled response: x[k]={1, 2, 3, 4, 3, 2, 1} The ztransform of the signal is: X(z) = 1z0 + 2z1 +3z2 +4z3 +3z4 +2z5 +1z6 end of example  Example: zTransform Compute the ztransform of a causal unit step, decaying exponential and delayed functions shown below (this example is accelerating your introduction to ztransforms): [ ] ] [ . 4 ]). [ ] [ ( ] [ . 3 ], [ ] [ . 2 , all for 1 ] [ ] [ . 1 3 4 3 2 1 L k x k x N k u k u a k x k u a k x k k u k x k k = = = = = 1 EEL 3135 Dr. Fred J. Taylor, Professor Figure 1: ztransforms. 1) The ztransform of the step function x 1 [ k ]= u [ k ] can be computed as follows: = = + + + = = = 2 1 1 1 ..., 1 1 ] [ ) ( k k k k z z z z k x z X The infinite sum X 1 ( z ) can also be rewritten in closed form as 1 2 1 1 1 1 ... 1 ) ( = + + + = z z z z X 2) For the decaying exponential, x 2 [ k ]= a k u [ k ], and a<1, can be expressed as: = = = = + + + = = = = 1 2 2 1 1 2 2 1 1 ... 1 ) ( ] [ ) ( k k k k k k k az z a az az z a z k x z X 3) For the finite length decaying exponential, x 3 [ k ]= a k ( u [ k ] u [ k N ]), 1 1 1 1 3 3 1 1 1 1 1 ] [ ) ( = = = = = = = = = = = = az z a az z a az z a z a z a z a z a z a z k x z X N N N N N k N k k k k N N k k k k k k k k k k k k 4) For the finite length decaying exponential, x 4 [ k ]= x 3 [ kL ], 1 1 3 4 1 1 1 ) ( ) ( = = = az z a z az z a z z X z z X L N N L N N L L end of example  Traditional Approach {This discussion assumes that you have an introductory exposure to Laplace transforms as applied to ordinary differential equations (ODE) course} 2 1 0 1 2 3 4 5 6 7 x 1 [ k ] =u [ k ] 1 0 1 2 3 4 5 6 7 x 2 [ k ] =a k u [ k ] 1 x 3 [ k ] =a k ( u [ k ] u [ k5]) 0 1 2 3 4 5 6 7 1 x 4 [ k ] = x 3 [ k2 0 1 2 3 4 5 6 7 EEL 3135 Dr. Fred J. Taylor, Professor Presenting a continuoustime, or analog signal , to an analog to digital converter (ADC) produces...
View Full
Document
 Spring '08
 ?

Click to edit the document details