Question 9.11: Total system response using the z transform and the DTFT A s......

The z transform of the response is

\mathbf{Y}(z)=\mathbf{H}(z)\mathbf{X}(z)={\frac{z}{(z-0.3)(z+0.8)}}\times{\frac{z}{z-1}},\ \left|z\right|\gt 1.

Expanding in partial fractions,

Y(z)={\frac{z^{2}}{(z-0.3)(z+0.8)(z-1)}}=-{\frac{0.1169}{z-0.3}}+{\frac{0.3232}{z+0.8}}+{\frac{0.7937}{z-1}},~|z|\gt 1

Therefore the total response is

y[n]=[-0.1169(0.3)^{n-1}+0.3232(-0.8)^{n-1}+0.7937]\mathrm{u[}n-1].

This problem can also be analyzed using the DTFT but the notation is significantly clumsier, mainly because the DTFT of a unit sequence is

{\frac{1}{1-e^{-j\Omega}}}+\pi\delta_{2\pi}(\Omega).

The system frequency response is

\mathrm{H}(e^{j\Omega})={\frac{e^{j\Omega}}{(e^{j\Omega}-0.3)(e^{j\Omega}+0.8)}}

The DTFT of the system response is

Y(e^{j\Omega})=\mathrm{H}(e^{j\Omega})\mathrm{X}(e^{j\Omega})=\frac{e^{j\Omega}}{(e^{j\Omega}-0.3)(e^{j\Omega}+0.8)}\times\left\lgroup\frac{1}{1-e^{-j\Omega}}+\pi\delta_{2\pi}(\Omega)\right\rgroup

or

\mathrm{Y}(e^{j \Omega})=\frac{e^{j 2\Omega}}{(e^{j\Omega}-0.3)(e^{j\Omega}+0.8)(e^{j\Omega}-1)}+\pi\frac{e^{j\Omega}}{(e^{j\Omega}-0.3)(e^{j\Omega}+0.8)}\delta_{2\pi}(\Omega)

Expanding in partial fractions

Y(e^{j\Omega})=\frac{-0.1169}{e^{j\Omega}-0.3}+\frac{0.3232}{e^{j\Omega}+0.8}+{\frac{0.7937}{e^{j\Omega}-1}}+\frac{\pi}{(1-0.3)(1+0.8)}\delta_{2\pi}(\Omega)

Using the equivalence property of the impulse and the periodicity of both \delta_{2\pi}(\Omega){\mathrm{~and~}}e^{j\Omega}

Y(e^{j\Omega})=\frac{-0.1163e^{-j\Omega}}{1-0.3e^{-j\Omega}}+\frac{0.3232e^{-j\Omega}}{1+0.8e^{-j\Omega}}+\frac{0.7937e^{-j\Omega}}{1-e^{-j\Omega}}+2.4933\delta_{2\pi}(\Omega)

Then, manipulating this expression into a form for which the inverse DTFT is direct

Y(e^{j\Omega})=\frac{-0.1169e^{-j\Omega}}{1-0.3e^{-j\Omega}}+\frac{0.3232e^{-j\Omega}}{1+0.8e^{-j\Omega}}+0.7937\left\lgroup\frac{e^{-j\Omega}}{1-e^{-j\Omega}}+\pi\delta_{2\pi}(\Omega)\right\rgroup

\underbrace{-0.7937\pi\delta_{2\pi}(\Omega)+2.4933\delta_{2\pi}(\Omega)}_{=0}

\mathrm{Y}(e^{j\Omega})=\frac{-0.1169e^{-j\Omega}}{1-0.3e^{-j\Omega}}+\frac{0.3232e^{-j\Omega}}{1+0.8e^{-j\Omega}}+0.7937\left\lgroup\frac{e^{-j\Omega}}{1-e^{-j\Omega}}+\pi\delta_{2\pi}(\Omega)\right\rgroup

And, finally, taking the inverse DTFT

y[n]=[-0.1169(0.3)^{n-1}+0.3232(-0.8)^{n-1}+0.7937]\mathbf{u}[n-1]

The result is the same but the effort and the probability of error are considerably greater.