Determine the system function of the 2 networks below and show that they have the same poles
For network 1),\[ y\left ( n\right ) =x(n)+2r\cos \theta y\left ( n-1\right ) -r^{2}y\left ( n-2\right ) \] Hence\begin{align*} Y\left ( z\right ) & =X(z)+2r\cos \theta z^{-1}Y\left ( z\right ) -r^{2}z^{-2}Y\left ( z\right ) \\ Y\left ( z\right ) \left [ 1-2r\cos \theta z^{-1}+r^{2}z^{-2}\right ] & =X\left ( z\right ) \\ H\left ( z\right ) & =\frac{Y\left ( z\right ) }{X\left ( z\right ) }\\ & =\frac{1}{1-2r\cos \theta z^{-1}+r^{2}z^{-2}} \end{align*}
To see the poles and zeros more easily, mutiply the above by \(z^{2}/z^{2}\)\[ H\left ( z\right ) =\frac{z^{2}}{z^{2}-2r\cos \theta z+r^{2}}\] Hence poles of system are at \(z=\frac{-b\pm \sqrt{b^{2}-4ac}}{2}=\frac{2r\cos \theta }{2}\pm \frac{1}{2}\sqrt{\left ( -2r\cos \theta \right ) ^{2}-4r^{2}}=\frac{2r\cos \theta }{2}\pm \frac{1}{2}\sqrt{4r^{2}\cos ^{2}\theta -4r^{2}}\) Hence\begin{align} z & =r\cos \theta \pm r\sqrt{\cos ^{2}\theta -1}=r\cos \theta \pm rj\sin \theta \nonumber \\ & =r\left ( \cos \theta \pm j\sin \theta \right ) \tag{1} \end{align}
For network 2, we start by labeling the corner point at \(W\)
\begin{equation} y\left ( n\right ) =r\cos \theta \ y\left ( n-1\right ) +r\sin \theta \ W\left ( n-1\right ) \tag{1} \end{equation} Now find \(W\left ( n\right ) \)\begin{equation} W\left ( n\right ) =r\cos \theta \ W\left ( n-1\right ) -r\sin \theta \ y\left ( n-1\right ) +x\left ( n\right ) \tag{2} \end{equation} Hence from the above equation (2) we need to find \(W\left ( n-1\right ) \), but we do this by delaying \(n\) one unit time, hence \[ W\left ( n-1\right ) =r\cos \theta \ W\left ( n-2\right ) -r\sin \theta \ y\left ( n-2\right ) +x\left ( n-1\right ) \] now substitute this back into (1) we obtain\begin{equation} y\left ( n\right ) =r\cos \theta \ y\left ( n-1\right ) +r\sin \theta \left [ \ r\cos \theta \ W\left ( n-2\right ) -r\sin \theta \ y\left ( n-2\right ) +x\left ( n-1\right ) \right ] \tag{3} \end{equation} We need to eliminate \(W\) from these equation to be able to obtain relation between \(y\left ( n\right ) \) and \(x\left ( n\right ) \) only. This is the tricky part.
From (1), we solve for \(W\left ( n-1\right ) \)\[ W\left ( n-1\right ) =\frac{y\left ( n\right ) -r\cos \theta \ y\left ( n-1\right ) }{r\sin \theta }\] Hence delay it one more time to find \(W\left ( n-2\right ) \)\[ W\left ( n-2\right ) =\frac{y\left ( n-1\right ) -r\cos \theta \ y\left ( n-2\right ) }{r\sin \theta }\] Substitute the above into (3) to finally remove \(W\) from the equation, we obtain\begin{align*} y\left ( n\right ) & =r\cos \theta \ y\left ( n-1\right ) +r\sin \theta \left [ \ r\cos \theta \overset{W\left ( n-2\right ) }{\ \overbrace{\left [ \frac{y\left ( n-1\right ) -r\cos \theta \ y\left ( n-2\right ) }{r\sin \theta }\right ] }}-r\sin \theta \ y\left ( n-2\right ) +x\left ( n-1\right ) \right ] \\ & =r\cos \theta \ y\left ( n-1\right ) +\ r\cos \theta \ \left [ y\left ( n-1\right ) -r\cos \theta \ y\left ( n-2\right ) \right ] -r^{2}\sin ^{2}\theta \ y\left ( n-2\right ) +r\sin \theta x\left ( n-1\right ) \\ & =r\cos \theta \ y\left ( n-1\right ) +r\cos \theta \ y\left ( n-1\right ) -r^{2}\cos ^{2}\theta \ y\left ( n-2\right ) -r^{2}\sin ^{2}\theta \ y\left ( n-2\right ) +r\sin \theta x\left ( n-1\right ) \\ & =2r\cos \theta \ y\left ( n-1\right ) -r^{2}\ y\left ( n-2\right ) \overset{1}{\overbrace{\left ( \cos ^{2}\theta +\sin ^{2}\theta \right ) }}+r\sin \theta x\left ( n-1\right ) \end{align*}
Take Z transform\begin{align*} Y\left ( z\right ) & =2r\cos \theta z^{-1}Y\left ( z\right ) -r^{2}z^{-2}Y\left ( z\right ) +r\sin \theta z^{-1}X\left ( z\right ) \\ Y\left ( z\right ) \left [ 1-2r\cos \theta z^{-1}+r^{2}z^{-2}\right ] & =r\sin \theta z^{-1}X\left ( z\right ) \end{align*}
Hence \begin{align*} H\left ( z\right ) & =\frac{Y\left ( z\right ) }{X\left ( z\right ) }=\frac{r\sin \theta z^{-1}}{1-2r\cos \theta z^{-1}+r^{2}z^{-2}}\\ & =\frac{r\sin \theta z}{z^{2}-2r\cos \theta z+r^{2}} \end{align*}
Compare the above to network (1) which was\[ H\left ( z\right ) =\frac{z^{2}}{z^{2}-2r\cos \theta z+r^{2}}\] Same denomator, hence same poles.
\begin{align*} y\left ( n\right ) -\frac{3}{4}y\left ( n-1\right ) +\frac{1}{8}y\left ( n-2\right ) & =x\left ( n\right ) +\frac{1}{3}x\left ( n-1\right ) \\ y\left ( n\right ) & =\frac{3}{4}y\left ( n-1\right ) -\frac{1}{8}y\left ( n-2\right ) +x\left ( n\right ) +\frac{1}{3}x\left ( n-1\right ) \\ y\left ( n\right ) & =a_{1}y\left ( n-1\right ) +a_{2}y\left ( n-2\right ) +a_{0}x\left ( n\right ) +a_{1}x\left ( n-1\right ) \end{align*}
Direct form I and II:
To find cascade form:\begin{align*} Y\left ( z\right ) & =\frac{3}{4}z^{-1}Y\left ( z\right ) -\frac{1}{8}z^{-2}Y\left ( z\right ) +X\left ( z\right ) +\frac{1}{3}z^{-1}X\left ( z\right ) \\ Y\left ( z\right ) \left ( 1-\frac{3}{4}z^{-1}+\frac{1}{8}z^{-2}\right ) & =X\left ( z\right ) \left ( 1+\frac{1}{3}z^{-1}\right ) \\ H\left ( z\right ) & =\frac{Y\left ( z\right ) }{X\left ( z\right ) }=\frac{1+\frac{1}{3}z^{-1}}{1-\frac{3}{4}z^{-1}+\frac{1}{8}z^{-2}}=\frac{1+\frac{1}{3}z^{-1}}{\left ( 1-\frac{1}{4}z^{-1}\right ) \left ( 1-\frac{1}{2}z^{-1}\right ) }\\ & =\frac{1+\frac{1}{3}z^{-1}}{\left ( 1-\frac{1}{4}z^{-1}\right ) }\frac{1}{\left ( 1-\frac{1}{2}z^{-1}\right ) } \end{align*}
Solution: To better understand this representation, this is a small example. Assume \(H\left ( z\right ) =\frac{3+4z^{-1}}{1-5z^{-1}-6z^{-2}}\,\), where \(N=2,M=1,\) start by multiplying numerator and denominator by \(z^{2}\) we obtain \[ H\left ( z\right ) =\frac{3z^{2}+4z}{z^{2}-5z-6}=\overset{A_{0}}{\overbrace{3}}+\overset{G_{0}\left ( z\right ) }{\overbrace{\frac{19z-18}{z^{2}-5z-6}}}\] Now, for \(G_{0}\left ( z\right ) ,\) divide numerator and denominator by \(19z-18\), we obtain\[ H\left ( z\right ) =3+\frac{1}{\frac{z^{2}-5z-6}{19z-18}}=3+\overset{G_{0}\left ( z\right ) }{\overbrace{\frac{1}{\underset{A_{1}}{\underbrace{\frac{1}{3}}}+\underset{B_{1}}{\underbrace{\frac{1}{9}}}z+\underset{G_{1}\left ( z\right ) }{\underbrace{\frac{-12}{9z-18}}}}}}\] Now, for \(G_{1}\left ( z\right ) ,\) divide numerator and denominator by \(-12\), we obtain\[ H\left ( z\right ) =3+\overset{G_{0}\left ( z\right ) }{\overbrace{\frac{1}{\frac{1}{3}+\frac{1}{9}z+\overset{G_{1}\left ( z\right ) }{\overbrace{\frac{1}{\underset{A_{2}}{\underbrace{\frac{18}{12}}}-\underset{B_{2}}{\underbrace{\frac{9}{12}z}}}}}}}}\] Now we need to draw a network diagram when \(N\) is odd. For example, if \(N=3\), then equation P4.10-1 will be\[ H\left ( z\right ) =A_{0}+\frac{1}{A_{1}+B_{1}z+\frac{1}{A_{2}+B_{2}z+\frac{1}{A_{3}+B_{3}z}}}\] or\[ H\left ( z\right ) =A_{0}+\frac{\frac{1}{B_{1}}z^{-1}}{1+\frac{A_{1}}{B_{1}}z^{-1}+\frac{\frac{1}{B_{2}}z^{-1}}{1+\frac{A_{2}}{B_{2}}z^{-1}\frac{\frac{1}{B_{3}}z^{-1}}{1+\frac{A_{3}}{B_{3}}z^{-1}}}}\] So, network will look like
After doing the above, I started having doubts about it. Here is another interpretation of the network:
\begin{align*} H\left ( z\right ) & =\frac{1}{1-2r\cos \theta z^{-1}+r^{2}z^{-2}}\\ & =\frac{z^{2}}{z^{2}-2r\cos \theta z+r^{2}} \end{align*}
Divide polynomials, we obtain\begin{align*} H\left ( z\right ) & =1+\frac{2r\cos \theta z-r^{2}}{z^{2}-2r\cos \theta z+r^{2}}\\ & =1+\frac{1}{\frac{z^{2}-2r\cos \theta z+r^{2}}{2r\cos \theta z-r^{2}}} \end{align*}
Divide the polynomials, we obtain\begin{align*} H\left ( z\right ) & =1+\frac{1}{\frac{1-4\cos ^{2}\theta }{4\cos ^{2}\theta }+\frac{z}{2r\cos \theta }+\frac{r^{2}\left ( 1+\frac{1-4\cos ^{2}\theta }{4\cos ^{2}\theta }\right ) }{2r\cos \theta z-r^{2}}}\\ & =1+\frac{1}{\frac{1-4\cos ^{2}\theta }{4\cos ^{2}\theta }+\frac{z}{2r\cos \theta }+\frac{1}{\frac{2r\cos \theta z-r^{2}}{r^{2}\left ( 1+\frac{1-4\cos ^{2}\theta }{4\cos ^{2}\theta }\right ) }}}\\ & =1+\frac{1}{\frac{1-4\cos ^{2}\theta }{4\cos ^{2}\theta }+\frac{z}{2r\cos \theta }+\frac{1}{-\frac{1}{\left ( 1+\frac{1-4\cos ^{2}\theta }{4\cos ^{2}\theta }\right ) }+\frac{2\cos \theta z}{r\left ( 1+\frac{1-4\cos ^{2}\theta }{4\cos ^{2}\theta }\right ) }}} \end{align*}
Let \(\frac{1-4\cos ^{2}\theta }{4\cos ^{2}\theta }=\beta \), then above can be written as\[ H\left ( z\right ) =1+\frac{1}{\beta +\frac{z}{2r\cos \theta }+\frac{1}{-\frac{1}{\beta }+\frac{2\cos \theta z}{r\left ( 1+\beta \right ) }}}\] Part(c): Network diagram using part(a) for the above is the following:
