solution
Adding the following notations on the diagram to make it easy to do the computation
Therefore we see that \begin{equation} Y\left ( z\right ) =X\left ( z\right ) H_{0}\left ( z\right ) +E\left ( z\right ) H_{1}\left ( z\right ) \tag{1} \end{equation} So we just need to determine \(E\left ( z\right ) \). But \(E\left ( z\right ) =X\left ( z\right ) -E\left ( z\right ) H_{1}\left ( z\right ) G\left ( z\right ) \). Hence \(E\left ( z\right ) \left ( 1+H_{1}\left ( z\right ) G\left ( z\right ) \right ) =X\left ( z\right ) \) or\[ E\left ( z\right ) =\frac{X\left ( z\right ) }{1+H_{1}\left ( z\right ) G\left ( z\right ) }\] Substituting this into (1) gives\begin{align*} Y\left ( z\right ) & =X\left ( z\right ) H_{0}\left ( z\right ) +\frac{X\left ( z\right ) }{1+H_{1}\left ( z\right ) G\left ( z\right ) }H_{1}\left ( z\right ) \\ Y\left ( z\right ) & =X\left ( z\right ) \left ( H_{0}\left ( z\right ) +\frac{H_{1}\left ( z\right ) }{1+H_{1}\left ( z\right ) G\left ( z\right ) }\right ) \end{align*}
Hence\[ \frac{Y\left ( z\right ) }{X\left ( z\right ) }=H_{0}\left ( z\right ) +\frac{H_{1}\left ( z\right ) }{1+H_{1}\left ( z\right ) G\left ( z\right ) }\]
solution
Adding the following notations on the diagram to make it easy to do the computation
Therefore we see that \begin{align} E & =X-FH_{1}G_{2}\tag{1}\\ F & =EH_{2}-FH_{1}G_{1} \tag{2} \end{align}
We have 2 equations with 2 unknowns \(E,F\). Substituting first equation into the second gives\begin{align} F & =\left ( X-FH_{1}G_{2}\right ) H_{2}-FH_{1}G_{1}\nonumber \\ F & =XH_{2}-FH_{1}G_{2}H_{2}-FH_{1}G_{1}\nonumber \\ F\left ( 1+H_{1}G_{2}H_{2}+H_{1}G_{1}\right ) & =XH_{2}\nonumber \\ F & =\frac{XH_{2}}{1+H_{1}G_{2}H_{2}+H_{1}G_{1}} \tag{3} \end{align}
But \[ Y\left ( z\right ) =F\left ( z\right ) H_{1}\left ( z\right ) \] Hence using (3) into the above gives\begin{align*} Y\left ( z\right ) & =\frac{XH_{2}}{1+H_{1}G_{2}H_{2}+H_{1}G_{1}}H_{1}\\ \frac{Y\left ( z\right ) }{X\left ( z\right ) } & =\frac{H_{2}H_{1}}{1+H_{1}G_{2}H_{2}+H_{1}G_{1}} \end{align*}
solution
Figure 11.3 a is the following
Taking the Laplace transform of the ODE gives (assuming zero initial conditions)\begin{align} s^{2}Y\left ( s\right ) +sY\left ( s\right ) +Y\left ( s\right ) & =sX\left ( x\right ) \nonumber \\ \frac{Y\left ( s\right ) }{X\left ( s\right ) } & =\frac{s}{s^{2}+s+1} \tag{1} \end{align}
From the diagram, we see that \begin{equation} Y\left ( s\right ) =E\left ( s\right ) H\left ( s\right ) \tag{2} \end{equation} But \(E\left ( s\right ) =X\left ( s\right ) -R\left ( s\right ) \) and \(R\left ( s\right ) =E\left ( s\right ) H\left ( s\right ) G\left ( s\right ) \). Hence \begin{align*} E\left ( s\right ) & =X\left ( s\right ) -\left ( E\left ( s\right ) H\left ( s\right ) G\left ( s\right ) \right ) \\ E\left ( s\right ) \left ( 1+H\left ( s\right ) G\left ( s\right ) \right ) & =X\left ( s\right ) \\ E\left ( s\right ) & =\frac{X\left ( s\right ) }{1+H\left ( s\right ) G\left ( s\right ) } \end{align*}
Substituting the above in (2) gives\begin{align} Y\left ( s\right ) & =\frac{X\left ( s\right ) }{1+H\left ( s\right ) G\left ( s\right ) }H\left ( s\right ) \nonumber \\ \frac{Y\left ( s\right ) }{X\left ( s\right ) } & =\frac{H\left ( s\right ) }{1+H\left ( s\right ) G\left ( s\right ) } \tag{3} \end{align}
Comparing (3) and (1) shows that\[ \frac{H\left ( s\right ) }{1+H\left ( s\right ) G\left ( s\right ) }=\frac{s}{s^{2}+s+1}\] But we are given that \(H\left ( s\right ) =\frac{1}{s+1}\). Hence the above becomes\[ \frac{\frac{1}{s+1}}{1+\frac{1}{s+1}G\left ( s\right ) }=\frac{s}{s^{2}+s+1}\] Now we solve for \(G\left ( s\right ) \)\begin{align*} \frac{\frac{1}{s+1}}{\frac{s+1+G\left ( s\right ) }{s+1}} & =\frac{s}{s^{2}+s+1}\\ \frac{1}{s+1+G\left ( s\right ) } & =\frac{s}{s^{2}+s+1}\\ s^{2}+s+sG\left ( s\right ) & =s^{2}+s+1\\ sG\left ( s\right ) & =s^{2}+s+1-s^{2}-s\\ G\left ( s\right ) & =\frac{1}{s} \end{align*}
solution
Figure 11.3 b is the following
From the diagram \(Y\left ( z\right ) =E\left ( z\right ) H\left ( z\right ) \) but \(E\left ( z\right ) =X\left ( z\right ) -R\left ( z\right ) \) and \(R\left ( z\right ) =E\left ( z\right ) H\left ( z\right ) G\left ( z\right ) \), hence\begin{align*} E\left ( z\right ) & =X\left ( z\right ) -E\left ( z\right ) H\left ( z\right ) G\left ( z\right ) \\ E\left ( z\right ) \left ( 1+H\left ( z\right ) G\left ( z\right ) \right ) & =X\left ( z\right ) \\ E\left ( z\right ) & =\frac{X\left ( z\right ) }{\left ( 1+H\left ( z\right ) G\left ( z\right ) \right ) } \end{align*}
Therefore\begin{align*} Y\left ( z\right ) & =E\left ( z\right ) H\left ( z\right ) \\ & =\frac{X\left ( z\right ) }{\left ( 1+H\left ( z\right ) G\left ( z\right ) \right ) }H\left ( z\right ) \\ \frac{Y\left ( z\right ) }{X\left ( z\right ) } & =\frac{H\left ( z\right ) }{1+H\left ( z\right ) G\left ( z\right ) } \end{align*}
But \(H\left ( z\right ) =\frac{1}{1-\frac{1}{2}z^{-1}}\) and \(G\left ( z\right ) =1-bz^{-1}\). Hence the above becomes\begin{align*} \frac{Y\left ( z\right ) }{X\left ( z\right ) } & =\frac{\frac{1}{1-\frac{1}{2}z^{-1}}}{1+\frac{1}{1-\frac{1}{2}z^{-1}}\left ( 1-bz^{-1}\right ) }\\ & =\frac{1}{1-\frac{1}{2}z^{-1}+1-bz^{-1}}\\ & =\frac{1}{2-\frac{1}{2}z^{-1}-bz^{-1}}\\ & =\frac{1}{2-\left ( \frac{1}{2}+b\right ) z^{-1}}\\ & =\frac{1}{2}\frac{1}{1-\left ( \frac{1}{4}+\frac{b}{2}\right ) z^{-1}} \end{align*}
The pole is \(\left ( \frac{1}{4}+\frac{b}{2}\right ) z^{-1}=1\) or \(z=\frac{1}{4}+\frac{b}{2}\). For causal system the pole should be inside the unit circle for stable system (so that it has a DFT). Therefore\begin{align*} \left \vert \frac{1}{4}+\frac{b}{2}\right \vert & <1\\ -1 & <\frac{1}{4}+\frac{b}{2}<1\\ -1-\frac{1}{4} & <\frac{b}{2}<1-\frac{1}{4}\\ -\frac{5}{4} & <\frac{b}{2}<\frac{3}{4}\\ -\frac{10}{4} & <b<\frac{6}{4}\\ -\frac{5}{2} & <b<\frac{3}{2} \end{align*}