Lesson_23 - EEL 3135: Dr. Fred J. Taylor, Professor Lesson...

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Unformatted text preview: EEL 3135: Dr. Fred J. Taylor, Professor Lesson Title: Inverse z-transform Lesson Number: 23 (Section 8-7 to 8-8) Background: The z-transform of many primitive signals, such as a k u [ k ], have been reduced to Tables (e.g., Table 1). It has been agued that given knowledge of the z-transform of a LTI systems input x [ k ] X ( z ), and an at-rest LTIs impulse response h [ k ] H ( z ), the output can be modeled using the convolution theorem, namely y [ k ] Y ( z )= H ( z ) X ( z ). The problem is then reduced to producing the time response y [ k ], which is given in terms of an inverse z-transform of Y(z). The inverse z-transform of some z- transform X ( z ) is theoretically defined to be: -- = = C k dz z z X j z X Z k x 1 1 ) ( 2 1 )) ( ( ] [ .1 where C is a restricted closed path which resides within the ROC of X ( z ) . Solving the integral equation can be a very tedious process. Fortunately there exists non-integral alternatives which are motivated below. A noteworthy point is that the numerators of the z-transforms reported in Table 1 have a z multiplier. This observation will be used to format the transform of a signal or system for inversion using a general purpose digital computer. Long Division The polynomial X ( z ) =N ( z )/ D ( z ) is said to be proper if N M and be strictly proper if N > M . Using long-division, X ( z ) =N ( z )/ D ( z ) can be expressed as shown below: ... ... ... 1 2 1 1 2 2 1 1 2 2 1 1 +- + + + + + + +----- z a b a b a a b z b z b b z a z a a . 2 The first few terms of the quotient are seen to correspond to the sample values of the inverse z- transform of X ( z ) . In the context of Equation 2, the sample values of x[k] are x [0] =b / a , x [1] = ( a b 1 a 1 b )/ a 2 , . Table 1: Primitive Signals and their z-Transform Time-domain z-transform [ 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 ] 1 ) cos( 2 ) sin( 2 +- b z z b z cos[ bk ] u [ k ] 1 ) cos( 2 )) cos( ( 2 +-- b z z b z z exp[ akT s ]sin[ bkT s ] u [ kT s ] S aT S S aT S S aT e bT ze z bT ze 2 2 ) cos( 2 ) sin( +- exp[ akT s ]cos[ bkT s ] u [ kT s ] S aT S S aT S S aT e bT ze z bT e z z 2 2 ) cos( 2 )) cos( ( +-- a k sin( bkT s ) u [ kT s ] 2 2 ) cos( 2 ) sin( a bT az z bT az S S +- a k cos( bkT s ) u [ kT s ] 2 2 ) cos( 2 )) cos( ( a bT az z bT a z z S S +-- 1 EEL 3135: Dr. Fred J. Taylor, Professor Example What is the inverse z-transform of X(z)=1/(1+z-1 )? 1-z-1 z-2 ... 1+z-1 1 1 z-1- z-1- z-1- z-2 z-2 or X(z)=1/(1+z-1 ) x[k]={ (-1) k }....
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Lesson_23 - EEL 3135: Dr. Fred J. Taylor, Professor Lesson...

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