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Determine the z-transform of the signals (a) x(n)=(cosω0n)u(n) (b) x(n)=(sinωk,n)u(n)
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Solution:
(a) By using Euler's identity,
the signal x(n) can be expressed as,
x(n)=(cosω0n)μ(n)=12ejmannu(n)+12e−jωpnu(n)
that,
X(z)=12Z{ejmovnu(n)}+12Z{e−jmovnu(n)}
If we set α=e±jω0(|α|=|e±ja0|=1) in (3.2.2).
we obtain,
ejman)nu(n)∴⟷11−e/ω0z−1 ROC: |z|>1
and
Thus
X(z)=1211−ejω0z−1+1211−e−jω1z−1ROC:|z|>1
(b) From Euler's identity.
Thus,
x(z)=12j(11−esinz−1−1−11−e−tanz−1) ROC: |z|>1
Time shifting. If,
x(n)z⟷X(z)
then,
x(n−k)∼⟷z−kX(z)
The ROC of z−kX(z) is the same as that of X(z) except for z=0 if k>0 and z=∞ if k<0. The proof of this property follows immediately from the definition of the z-transform.
The properties of linearity and time shifting are the key features that make the z-transform extremely useful for the analysis of discrete-time LTI systems.
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