Consider the signal\[ x\left [ n\right ] =\left ( \frac{1}{5}\right ) ^{n}u\left [ n-3\right ] \] Use eq. (10.3) \begin{equation} X\left [ z\right ] =\sum _{n=-\infty }^{n=\infty }x\left [ n\right ] z^{-n} \tag{10.3} \end{equation} to evaluate the Z-transform of this signal, and specify the corresponding region of convergence.
solution\[ X\left [ z\right ] =\sum _{n=-\infty }^{n=\infty }\left ( \frac{1}{5}\right ) ^{n}u\left [ n-3\right ] z^{-n}\] But \(u\left [ n-3\right ] \) is zero for \(n<3\) and \(1\) otherwise. Hence the above becomes\[ X\left [ z\right ] =\sum _{n=3}^{n=\infty }\left ( \frac{1}{5}\right ) ^{n}z^{-n}\] Let \(m=n-3\). When \(n=3,m=0\) therefore the above can be written as\begin{align*} X\left [ z\right ] & =\sum _{m=0}^{m=\infty }\left ( \frac{1}{5}\right ) ^{m+3}z^{-\left ( m+3\right ) }\\ & =\left ( \frac{z^{-1}}{5}\right ) ^{3}\sum _{m=0}^{m=\infty }\left ( \frac{1}{5}\right ) ^{m}z^{-m}\\ & =\frac{z^{-3}}{125}\sum _{m=0}^{m=\infty }\left ( \frac{1}{5}\right ) ^{m}z^{-m} \end{align*}
Renaming back to \(n\)\begin{equation} X\left [ z\right ] =\frac{z^{-3}}{125}\sum _{n=0}^{\infty }\left ( \frac{1}{5}\right ) ^{n}z^{-n}\tag{1} \end{equation} Now, looking at \(\sum _{n=0}^{n=\infty }\left ( \frac{1}{5z}\right ) ^{n}\) then assuming \(\left \vert 5z\right \vert >1\) and using the formula \(\sum _{n=0}^{n=\infty }a^{n}=\frac{1}{1-a}\), where \(a=\frac{1}{5z}\) in this case gives\[ \sum _{n=0}^{n=\infty }\left ( \frac{1}{5z}\right ) ^{n}=\frac{1}{1-\frac{1}{5}z^{-1}}\] Hence (1) becomes\[ X\left [ z\right ] =\frac{z^{-3}}{125}\left ( \frac{1}{1-\frac{1}{5}z^{-1}}\right ) \] The above shows a pole at \(\frac{1}{5}z^{-1}=1\) or \(z=\frac{1}{5}\) and a pole at \(z=0\). Since this is right handed signal, then the ROC is outside the outer most pole. Therefore ROC is\[ \left \vert z\right \vert >\frac{1}{5}\] Which means the region is outside a circle of radius \(\frac{1}{5}\). Since this ROC includes the unit circle, meaning a DTFT exist, it shows that this is a stable signal.
Using partial-fraction expansion and the fact that\[ a^{n}u\left [ n\right ] \overset{Z}{\longleftrightarrow }\frac{1}{1-az^{-1}}\qquad \left \vert z\right \vert >\left \vert a\right \vert \] Find the inverse Z-transform of \[ X\left ( z\right ) =\frac{1-\frac{1}{3}z^{-1}}{\left ( 1-z^{-1}\right ) \left ( 1+2z^{-1}\right ) }\qquad \left \vert z\right \vert >2 \] solution
Let\[ \frac{1-\frac{1}{3}z^{-1}}{\left ( 1-z^{-1}\right ) \left ( 1+2z^{-1}\right ) }=\frac{A}{\left ( 1-z^{-1}\right ) }+\frac{B}{\left ( 1+2z^{-1}\right ) }\] Hence \(A=\left ( \frac{1-\frac{1}{3}z^{-1}}{1+2z^{-1}}\right ) _{z^{-1}=1}=\frac{1-\frac{1}{3}}{1+2}=\frac{2}{9}\) and \(B=\left ( \frac{1-\frac{1}{3}z^{-1}}{\left ( 1-z^{-1}\right ) }\right ) _{z^{-1}=-\frac{1}{2}}=\frac{1-\frac{1}{3}\left ( -\frac{1}{2}\right ) }{\left ( 1-\left ( -\frac{1}{2}\right ) \right ) }=\frac{7}{9}\) Therefore the above becomes\[ X\left ( z\right ) =\frac{2}{9}\frac{1}{\left ( 1-z^{-1}\right ) }+\frac{7}{9}\frac{1}{\left ( 1+2z^{-1}\right ) }\] The pole of first term at \(z^{-1}=1\) or \(z=1\) and the pole for second term is \(2z^{-1}=-1\) or \(z=-2\). Since the ROC is outside the out most pole, then this is right handed signal. Hence\begin{align*} x\left [ n\right ] & =\frac{2}{9}u\left [ n\right ] +\frac{7}{9}\left ( -2\right ) ^{n}u\left [ n\right ] \\ & =\left ( \frac{2}{9}+\frac{7}{9}\left ( -2\right ) ^{n}\right ) u\left [ n\right ] \end{align*}
Which is valid when \(X\left ( z\right ) \) defined for \(\left \vert z\right \vert >2\) since this is the common region for \(\left \vert z\right \vert >1\) and \(\left \vert z\right \vert >2\) at the same time. We notice the ROC does not include the unit circle and hence it is not stable signal. This is confirmed by looking at the term \(\left ( -2\right ) ^{n}\) which grows with \(n\) with no limit.
Consider a left-sided sequence \(x[n]\) with z-transform\[ X\left ( z\right ) =\frac{1}{\left ( 1-\frac{1}{2}z^{-1}\right ) \left ( 1-z^{-1}\right ) }\]
solution
\begin{align*} X\left ( z\right ) & =\frac{z}{z\left ( 1-\frac{1}{2}z^{-1}\right ) \left ( 1-z^{-1}\right ) }\\ & =\frac{z}{\left ( z-\frac{1}{2}\right ) \left ( 1-z^{-1}\right ) }\\ & =\frac{z^{2}}{z\left ( z-\frac{1}{2}\right ) \left ( 1-z^{-1}\right ) }\\ & =\frac{z^{2}}{\left ( z-\frac{1}{2}\right ) \left ( z-1\right ) }\\ & =\frac{z^{2}}{z^{2}-\frac{3}{2}z+\frac{1}{2}} \end{align*}
One pols at \(z=\frac{1}{2}\) and one pole at \(z=1\).
\[ X\left ( z\right ) =\frac{z^{2}}{\left ( z-\frac{1}{2}\right ) \left ( z-1\right ) }\] To do partial fractions, the degree in numerator must be smaller than in the denominator, which is not the case here. Hence we start by factoring out a \(z\) which gives\begin{align*} X\left ( z\right ) & =z^{2}\left ( \frac{1}{\left ( z-\frac{1}{2}\right ) \left ( z-1\right ) }\right ) \\ & =z^{2}\left ( \frac{A}{z-\frac{1}{2}}+\frac{B}{z-1}\right ) \end{align*}
Hence\[ \frac{1}{\left ( z-\frac{1}{2}\right ) \left ( z-1\right ) }=\frac{A}{z-\frac{1}{2}}+\frac{B}{z-1}\] Therefore \(A=\left ( \frac{1}{\left ( z-1\right ) }\right ) _{z=\frac{1}{2}}=\frac{1}{\left ( \frac{1}{2}-1\right ) }=\allowbreak -2\) and \(B=\left ( \frac{1}{z-\frac{1}{2}}\right ) _{z=1}=\frac{1}{1-\frac{1}{2}}=2\). Hence the above becomes\begin{align*} X\left ( z\right ) & =z^{2}\left ( -\frac{2}{z-\frac{1}{2}}+\frac{2}{z-1}\right ) \\ & =2z^{2}\left ( -\frac{1}{z-\frac{1}{2}}+\frac{1}{z-1}\right ) \end{align*}
Pole at \(z=\frac{1}{2}\) and one at \(z=1\).
Writing the above as\[ X\left ( z\right ) =2z^{2}X_{1}\left ( z\right ) \] Where \(x_{1}\left [ n\right ] \iff X_{1}\left ( z\right ) \) where ROC for \(X_{1}\left ( z\right ) \) is inside the inner most pole (since left sided). Hence ROC for \(X_{1}\left ( z\right ) \) is \(\left \vert z\right \vert <\frac{1}{2}\). What is left is to find \(x_{1}\left [ n\right ] \) which is the inverse Z transform of \(\frac{-1}{z-\frac{1}{2}}+\frac{1}{z-1}\). We want to use \(a^{n}u\left [ n\right ] \overset{Z}{\longleftrightarrow }\frac{1}{1-az^{-1}}\) so rewriting this as\begin{align*} X_{1}\left ( z\right ) & =\frac{-1}{z-\frac{1}{2}}+\frac{1}{z-1}\\ & =\frac{-z^{-1}}{1-\frac{1}{2}z^{-1}}+\frac{z^{-1}}{1-z^{-1}} \end{align*}
Hence \begin{align} X\left ( z\right ) & =2z^{2}X_{1}\left ( z\right ) \nonumber \\ & =2z^{2}\left ( \frac{-z^{-1}}{1-\frac{1}{2}z^{-1}}+\frac{z^{-1}}{1-z^{-1}}\right ) \nonumber \\ & =2z\left ( \frac{-1}{1-\frac{1}{2}z^{-1}}+\frac{1}{1-z^{-1}}\right ) \tag{1} \end{align}
Then (since left handed) then \(\frac{-1}{1-\frac{1}{2}z^{-1}}\longleftrightarrow \left ( \frac{1}{2}\right ) ^{n}u\left [ -n-1\right ] \). Similarly for \(\frac{1}{1-z^{-1}}\longleftrightarrow -u\left [ -n-1\right ] \) . Hence \[ x\left [ n\right ] =\left ( \frac{1}{2}\right ) ^{n}u\left [ -n-1\right ] -u\left [ -n-1\right ] \] Substituting the above in (1) gives\[ x\left [ n\right ] =2\left ( \left ( \frac{1}{2}\right ) ^{n}u\left [ -n-2\right ] -u\left [ -n-2\right ] \right ) \] Where \(u\left [ -n-1\right ] \) is changed to \(u\left [ -n-2\right ] \) because of the extra \(z\) in (1) outside, which causes extra shift and same for \(u\left [ -n-1\right ] \) changed to \(u\left [ -n-2\right ] \). Therefore the final answer is\[ x\left [ n\right ] =2\left ( \frac{1}{2}\right ) ^{n}u\left [ -n-2\right ] -2u\left [ -n-2\right ] \]
A causal LTI system is described by the difference equation\[ y\left [ n\right ] =y\left [ n-1\right ] +y\left [ n-2\right ] +x\left [ n-1\right ] \]
solution
Taking the \(Z\) transform of the difference equation gives\begin{align*} Y\left ( z\right ) & =z^{-1}Y\left ( z\right ) +z^{-2}Y\left ( z\right ) +z^{-1}X\left ( z\right ) \\ Y\left ( z\right ) \left ( 1-z^{-1}-z^{-2}\right ) & =z^{-1}X\left ( z\right ) \\ \frac{Y\left ( z\right ) }{X\left ( z\right ) } & =\frac{z^{-1}}{1-z^{-1}-z^{-2}}\\ & =\frac{z}{z^{2}-z-1}\\ & =\frac{z}{\left ( z-\left ( \frac{1}{2}\sqrt{5}+\frac{1}{2}\right ) \right ) \left ( z-\left ( \frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) \right ) } \end{align*}
Hence a pole at \(z=\frac{1}{2}\sqrt{5}+\frac{1}{2}=1.618\) and a pole at \(z=\left ( \frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) =\) \(-0.618\) and zero at \(z=0\)
Since this is a causal \(H\left ( z\right ) \) then ROC is always to the right of the right most pole. Hence ROC is \[ \left \vert z\right \vert >\frac{1}{2}\sqrt{5}+\frac{1}{2}=1.618 \] Here is a plot of the poles and zeros. The ROC is all the region to the right of \(1.618\) pole.
If the input \(x\left [ n\right ] =\delta \left [ n\right ] \) then the difference equation is now\[ y\left [ n\right ] =y\left [ n-1\right ] +y\left [ n-2\right ] +\delta \left [ n-1\right ] \] Hence taking the Z transform gives\begin{align} Y\left ( z\right ) & =z^{-1}Y\left ( z\right ) +z^{-2}Y\left ( z\right ) +z^{-1}\nonumber \\ Y\left ( z\right ) \left ( 1-z^{-2}-z^{-1}\right ) & =z^{-1}\nonumber \\ Y\left ( z\right ) & =\frac{z^{-1}}{1-z^{-1}-z^{-2}}\nonumber \\ & =\frac{-z^{-1}}{z^{-2}+z^{-1}-1}\nonumber \\ & =\frac{-z^{-1}}{\left ( z^{-1}-\left ( -\frac{1}{2}+\frac{1}{2}\sqrt{5}\right ) \right ) \left ( z^{-1}-\left ( -\frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) \right ) } \tag{1} \end{align}
Applying partial fractions gives \[ \frac{-z^{-1}}{\left ( z^{-1}-\left ( -\frac{1}{2}+\frac{1}{2}\sqrt{5}\right ) \right ) \left ( z^{-1}-\left ( -\frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) \right ) }=\frac{A}{z^{-1}-\left ( -\frac{1}{2}+\frac{1}{2}\sqrt{5}\right ) }+\frac{B}{z^{-1}-\left ( -\frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) }\] Hence \[ A=\left ( \frac{-z^{-1}}{\left ( z^{-1}-\left ( -\frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) \right ) }\right ) _{z^{-1}=\left ( -\frac{1}{2}+\frac{1}{2}\sqrt{5}\right ) }=\frac{-\left ( -\frac{1}{2}+\frac{1}{2}\sqrt{5}\right ) }{\left ( -\frac{1}{2}+\frac{1}{2}\sqrt{5}\right ) -\left ( -\frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) }=\frac{1}{10}\sqrt{5}-\frac{1}{2}\] And \[ B=\left ( \frac{-z^{-1}}{z^{-1}-\left ( -\frac{1}{2}+\frac{1}{2}\sqrt{5}\right ) }\right ) _{z=\left ( -\frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) }=\frac{-\left ( -\frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) }{\left ( -\frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) -\left ( -\frac{1}{2}+\frac{1}{2}\sqrt{5}\right ) }=-\frac{1}{10}\sqrt{5}-\frac{1}{2}\] Therefore (1) becomes\begin{align*} Y\left ( z\right ) & =\left ( \frac{1}{10}\sqrt{5}-\frac{1}{2}\right ) \frac{1}{z^{-1}-\left ( -\frac{1}{2}+\frac{1}{2}\sqrt{5}\right ) }-\left ( \frac{1}{10}\sqrt{5}+\frac{1}{2}\right ) \frac{1}{z^{-1}-\left ( -\frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) }\\ & =\frac{\left ( \frac{1}{10}\sqrt{5}-\frac{1}{2}\right ) }{-\frac{1}{2}+\frac{1}{2}\sqrt{5}}\frac{1}{\frac{1}{-\frac{1}{2}+\frac{1}{2}\sqrt{5}}z^{-1}-1}-\frac{\left ( \frac{1}{10}\sqrt{5}+\frac{1}{2}\right ) }{\left ( -\frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) }\frac{1}{\frac{1}{\left ( -\frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) }z^{-1}-1}\\ & =\frac{1}{5}\sqrt{5}\frac{1}{1-\left ( \frac{2}{-1+\sqrt{5}}\right ) z^{-1}}-\frac{1}{5}\sqrt{5}\frac{1}{1-\frac{2}{\left ( -1-\sqrt{5}\right ) }z^{-1}}\\ & =\frac{1}{5}\sqrt{5}\frac{1}{1-\left ( \frac{1}{2}\sqrt{5}+\frac{1}{2}\right ) z^{-1}}-\frac{1}{5}\sqrt{5}\frac{1}{1-\left ( \frac{1}{2}-\frac{1}{2}\sqrt{5}\right ) z^{-1}} \end{align*}
Now we can use the table \(\frac{1}{1-az^{-1}}\rightarrow a^{n}u\left [ n\right ] \) for \(\left \vert z\right \vert >a\). Taking the inverse Z transform of the above gives\begin{align*} y\left [ n\right ] & =-\left ( \frac{1}{5}\sqrt{5}\right ) \left ( \frac{1+\sqrt{5}}{2}\right ) ^{n}u\left [ n\right ] +\left ( \frac{1}{5}\sqrt{5}\right ) \left ( \frac{1-\sqrt{5}}{2}\right ) ^{n}u\left [ n\right ] \\ & =\left ( -\left ( 0.447\,21\right ) \left ( 1.618\right ) ^{n}+\left ( 0.447\,21\right ) \left ( -0.618\right ) ^{n}\right ) u\left [ n\right ] \end{align*}
This is unstable response \(y\left [ n\right ] \) due to the term \(\left ( 1.618\right ) ^{n}\) which grows with no limit as \(n\rightarrow \infty \).
Using the ROC where \(0.618<\left \vert z\right \vert <1.618\) instead of \(\left \vert z\right \vert >1.618\), then \begin{align*} y\left [ n\right ] & =\left ( \frac{1}{5}\sqrt{5}\right ) \left ( \frac{1+\sqrt{5}}{2}\right ) ^{n}u\left [ -n-1\right ] +\left ( \frac{1}{5}\sqrt{5}\right ) \left ( \frac{1-\sqrt{5}}{2}\right ) ^{n}u\left [ n\right ] \\ & =\left ( \left ( 0.447\,21\right ) \left ( 1.618\right ) ^{n}u\left [ -n-1\right ] +\left ( 0.447\,21\right ) \left ( -0.618\right ) ^{n}\right ) u\left [ n\right ] \end{align*}
which is now stable since the index on \(1.618^{n}\) run is negative instead of positive.
Consider the linear, discrete-time, shift-invariant system with input \(x[n]\) and output \(y[n]\) for which\[ y\left [ n-1\right ] -\frac{10}{3}y\left [ n\right ] +y\left [ n+1\right ] =x\left [ n\right ] \] is stable. Determine the unit sample response.
solution
Taking the Z transform of the difference equation gives\begin{align*} z^{-1}Y\left ( z\right ) -\frac{10}{3}Y\left ( z\right ) +zY\left ( z\right ) & =X\left ( z\right ) \\ Y\left ( z\right ) \left ( z^{-1}-\frac{10}{3}+z\right ) & =X\left ( z\right ) \end{align*}
Hence the unit sample is when \(x\left [ n\right ] =\delta \left [ n\right ] \). Hence \(X\left ( z\right ) =1\). Therefore the impulse response is\begin{align*} H\left ( z\right ) & =\frac{1}{z^{-1}-\frac{10}{3}+z}\\ & =\frac{z^{-1}}{z^{-2}-\frac{10}{3}z^{-1}+1}\\ & =\frac{z^{-1}}{\left ( z^{-1}-3\right ) \left ( z^{-1}-\frac{1}{3}\right ) } \end{align*}
Applying partial fractions\[ H\left ( z\right ) =\frac{A}{\left ( z^{-1}-3\right ) }+\frac{B}{\left ( z^{-1}-\frac{1}{3}\right ) }\] Hence \(A=\left ( \frac{z^{-1}}{\left ( z^{-1}-\frac{1}{3}\right ) }\right ) _{z^{-1}=3}=\frac{3}{\left ( 3-\frac{1}{3}\right ) }=\frac{9}{8}\) and \(B=\left ( \frac{z^{-1}}{\left ( z^{-1}-3\right ) }\right ) _{z^{-1}=\frac{1}{3}}=\frac{\frac{1}{3}}{\left ( \frac{1}{3}-3\right ) }=-\frac{1}{8}\). Therefore\begin{align} H\left ( z\right ) & =\frac{9}{8}\frac{1}{\left ( z^{-1}-3\right ) }-\frac{1}{8}\frac{1}{\left ( z^{-1}-\frac{1}{3}\right ) }\nonumber \\ & =\frac{3}{8}\frac{1}{\left ( \frac{1}{3}z^{-1}-1\right ) }-\frac{3}{8}\frac{1}{\left ( 3z^{-1}-1\right ) }\nonumber \\ & =\frac{3}{8}\frac{1}{1-3z^{-1}}-\frac{3}{8}\frac{1}{1-\frac{1}{3}z^{-1}} \tag{1} \end{align}
We see a pole at \(z=3\) and a pole at \(z=\frac{1}{3}\).
For \(\frac{1}{1-3z^{-1}}\), this is stable only for a left sided signal, this is because \(a\) which is \(3\) here is larger than \(1\). Hence its inverse Z transform is of this is \(x_{1}\left [ n\right ] =-\frac{3}{8}3^{n}u\left [ -n-1\right ] \) and for the second term \(\frac{1}{1-\frac{1}{3}z^{-1}}\) is stable for right sided signal, since \(\frac{1}{3}<1\). Hence its inverse Z transform is \(-\frac{3}{8}\left ( \frac{1}{3}\right ) ^{n}u\left [ n\right ] \). Therefore\[ h\left [ n\right ] =-\frac{3}{8}\left ( 3\right ) ^{n}u\left [ -n-1\right ] -\frac{3}{8}\left ( \frac{1}{3}\right ) ^{n}u\left [ n\right ] \]
solution
Let the value at the branch just to the right of \(x\left [ n\right ] \) summation sign be called \(A\left [ z\right ] \).
Then we see that \[ Y\left ( z\right ) =A\left ( z\right ) -\frac{k}{4}z^{-1}A\left ( z\right ) \] We just need to find \(A\left ( z\right ) \). We see that \(A\left ( z\right ) =X\left ( z\right ) -\frac{k}{3}z^{-1}A\left ( z\right ) \). Hence \(A\left ( z\right ) \left ( 1+\frac{k}{3}z^{-1}\right ) =X\left ( z\right ) \) or \(A\left ( z\right ) =\frac{X\left ( z\right ) }{1+\frac{k}{3}z^{-1}}\). Therefore the above becomes\begin{align*} Y\left ( z\right ) & =\frac{X\left ( z\right ) }{1+\frac{k}{3}z^{-1}}-\frac{k}{4}z^{-1}\frac{X\left ( z\right ) }{1+\frac{k}{3}z^{-1}}\\ & =X\left ( z\right ) \left ( \frac{1}{1+\frac{k}{3}z^{-1}}-\frac{k}{4}\frac{z^{-1}}{1+\frac{k}{3}z^{-1}}\right ) \end{align*}
Hence \begin{align*} H\left ( z\right ) & =\frac{Y\left ( z\right ) }{X\left ( z\right ) }\\ & =\frac{1}{1+\frac{k}{3}z^{-1}}-\frac{k}{4}\frac{z^{-1}}{1+\frac{k}{3}z^{-1}}\\ & =\frac{1-\frac{k}{4}z^{-1}}{1+\frac{k}{3}z^{-1}} \end{align*}
The pole is when \(\frac{k}{3}z^{-1}=-1\) or \(z=-\frac{k}{3}\). Zero is when \(1-kz^{-1}=0\) or \(kz^{-1}=1\) or \(z=k\). Since this causal system, then the ROC is to the right of the most right pole. Hence \(\left \vert z\right \vert >\frac{\left \vert k\right \vert }{3}\) is the ROC.
System is stable if it has a Discrete time Fourier transform. This implies the ROC must include the unit circle. Hence \(\frac{\left \vert k\right \vert }{3}<1\) or \(\left \vert k\right \vert <3\).
From part (a), the unit sample response is \(H\left ( z\right ) =\frac{1-\frac{k}{4}z^{-1}}{1+\frac{k}{3}z^{-1}}\). When \(k=1\) this becomes \(H\left ( z\right ) =\frac{1-\frac{1}{4}z^{-1}}{1+\frac{1}{3}z^{-1}}\)
Since \(x\left [ n\right ] =\left ( \frac{2}{3}\right ) ^{n}\) for all \(n\) and this is casual system, then this means \(x\left [ n\right ] =\left ( \frac{2}{3}\right ) ^{n}u\left [ n\right ] \). Therefore \[ X\left ( z\right ) =\frac{1}{1-\frac{2}{3}z^{-1}}\] Hence from part (a)\begin{align*} Y\left ( z\right ) & =H\left ( z\right ) X\left ( z\right ) \\ & =\frac{1-\frac{1}{4}z^{-1}}{1+\frac{1}{3}z^{-1}}\frac{1}{1-\frac{2}{3}z^{-1}}\\ & =\frac{1-\frac{1}{4}z^{-1}}{\left ( 1+\frac{1}{3}z^{-1}\right ) \left ( 1-\frac{2}{3}z^{-1}\right ) }\\ & =\frac{A}{1+\frac{1}{3}z^{-1}}+\frac{B}{1-\frac{2}{3}z^{-1}} \end{align*}
Therefore \(A=\left ( \frac{1-\frac{1}{4}z^{-1}}{\left ( 1-\frac{2}{3}z^{-1}\right ) }\right ) _{z^{-1}=-3}=\frac{1-\frac{1}{4}\left ( -3\right ) }{\left ( 1-\frac{2}{3}\left ( -3\right ) \right ) }=\frac{7}{12}\) and \(B=\left ( \frac{1-\frac{1}{4}z^{-1}}{\left ( 1+\frac{1}{3}z^{-1}\right ) }\right ) _{z^{-1}=\frac{3}{2}}=\left ( \frac{1-\frac{1}{4}\left ( \frac{3}{2}\right ) }{\left ( 1+\frac{1}{3}\left ( \frac{3}{2}\right ) \right ) }\right ) =\frac{5}{12}\). Hence \[ Y\left ( z\right ) =\frac{7}{12}\frac{1}{1+\frac{1}{3}z^{-1}}+\frac{5}{12}\frac{1}{1-\frac{2}{3}z^{-1}}\] Therefore\[ y\left [ n\right ] =\frac{7}{12}\left ( -\frac{1}{3}\right ) ^{n}u\left [ n\right ] +\frac{5}{12}\left ( \frac{2}{3}\right ) ^{n}u\left [ n\right ] \]
The following is a plot of the solution
