ztransform_06

# ztransform_06 - Z-Transform Extending the Discrete-Time...

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Z -Transform Extending the Discrete-Time Fourier Transform (DTFT) Just as the Fourier transform was extended to include a broader class of aperiodic, continuous-time signals, the DTFT can also be extended to include a broader class of aperiodic discrete-time signals. Extending the Fourier transform resulted in the Laplace transform. Extending the DTFT will result in a new transform called the z-transform. For aperiodic, discrete-time signals () () () jn j n xn xne Xe −Ω =−∞ ℑ = =  and 1 2 1 2 jj j n Xe e d π  =ℑ = It was shown that this transform exists if the signal is absolutely summable, i.e., n =−∞ <∞ Unfortunately, there are many signals that do not satisfy this condition. The goal is to extend the DTFT to include many signals that do not satisfy this condition. Consider the following signal ( ) ( ) n xn aun = Now For 1, this signal has a DTFT For 1, this signal does not have a DTFT a a < > First consider signals that do have a DTFT. Define a new quantity

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2 () () n n X zx n z =−∞ ± This is called the z-transform . Observe that if x(n) has a DTFT, then () jj n j n X ze xne Xe Ω− =−∞ == = Next define a new signal ( ) ( ) for 0 n r gn xnr r =≥ Here r is any positive real value. This signal will have a DTFT if n r nn gn xnr ∞∞ =−∞ =−∞ =< ∑∑ Its DTFT will then be () () () n n n j n j rr n G e g n ex n r n r e =−∞ =−∞ =−∞ = Now write the quantity z in polar form, i.e., , then substituting the polar j zr e = representation into the z-transform, it can be observed that ( ) ( ) r Xz r e Ge ΩΩ Is it possible for both to both have a DTFT? The answer is yes, as ( ) ( ) and r xn g n illustrated in the next example. Example : ( ) ( ) for 1 n xn aun a
3 The DTFT is () 00 1 for 1 1 n jj n n j n j nn n j j Xe xne ae a e ae a ae ∞∞ Ω− =−∞ = = −Ω == = < ∑∑ Next ( ) 1 1 1 1 for 1 1 n jn n j n j r j j Ge are a re a ar e r a ar e r −−Ω < > This result only requires that r is chosen to be larger than the magnitude of a . It does not depend on the magnitude of a being less than 1. Now both transforms in this example exist if and r is chosen to be greater than the a < . a Inverting the z-transform If exists, then it can be inverted in the usual way,’ ( ) j r 1 2 11 22 j n j j n rr r g n G e G eed G π ππ ΩΩ  =ℑ = Ω=  ∫∫ But () () 1 2 nj j n r g nx n r G Then ( ) 1 n n n j j r xn Ge e d Ge r e d −− =Ω =

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4 Now observe that substituting the representation of z in polar ( ) into X ( z ), gives j zr e = the result () ( ) j jj r e X zX r e G e ΩΩ = == Now consider r to be a constant chosen to insure the existence of , then ( ) j r Ge ( ) 1 1 j j re z dz jre d d dz dz =Ω = = Make these substitutions in the preceding expression for x ( n ) . Then ( ) 1 1 11 22 1 2 j j j j re n n r re re n re x n G e re d X z z z dz j Xzz d z j π ππ = = ∫∫
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ztransform_06 - Z-Transform Extending the Discrete-Time...

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