Lesson_171 - EEL 3135 Dr. Fred J. Taylor, Professor Lesson...

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Unformatted text preview: EEL 3135 Dr. Fred J. Taylor, Professor Lesson Title: z-Transform Lesson Number: 17 (Sections 7-1 and 7-2) Background: Given the discrete-time signal: [ ] [ ] k n k x n x K N k- = = ] [ 1. the authors define the so-called z-transform of x[n] to be: ( 29 [ ] k N k z k x z X- = = 2. It can be seen that that z-operator is positional with z-k corresponds to a spatial delay. Specifically, z-1 represent a one sample delay, z-2 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 z-transform must have a forward and reverse mapping. The z-transform 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 : z-Transform Suppose a discrete-time signal has a sampled response: x[k]={1, 2, 3, 4, 3, 2, 1} The z-transform of the signal is: X(z) = 1z-0 + 2z-1 +3z-2 +4z-3 +3z-4 +2z-5 +1z-6 end of example ---------------------------------------------- Example: z-Transform Compute the z-transform of a causal unit step, decaying exponential and delayed functions shown below (this example is accelerating your introduction to z-transforms): [ ] ] [ . 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: z-transforms. 1) The z-transform 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 [ k-L ], 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 [ k-5]) 0 1 2 3 4 5 6 7 1 x 4 [ k ] = x 3 [ k-2 0 1 2 3 4 5 6 7 EEL 3135 Dr. Fred J. Taylor, Professor Presenting a continuous-time, or analog signal , to an analog to digital converter (ADC) produces...
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Lesson_171 - EEL 3135 Dr. Fred J. Taylor, Professor Lesson...

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