4.10 HW 10

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4.10.1 Problem 11.1

Figure 4.84:Problem description

solution

Adding the following notations on the diagram to make it easy to do the computation

Figure 4.85:Annotations added

Therefore we see that $$\begin{array}{}\text{(1)}& Y\left(z\right)=X\left(z\right)H_0\left(z\right)+E\left(z\right)H_1\left(z\right)\end{array}$$ 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 $E\left(z\right)(1+H_1\left(z\right)G\left(z\right))=X\left(z\right)$ 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$$\begin{array}{rl}Y\left(z\right)& =X\left(z\right)H_0\left(z\right)+\frac{X\left(z\right)}{1+H_1\left(z\right)G\left(z\right)}{H}_1\left(z\right)\\ Y\left(z\right)& =X\left(z\right)(H_0\left(z\right)+\frac{H_1\left(z\right)}{1+H_1\left(z\right)G\left(z\right)})\end{array}$$

Hence$$\frac{Y\left(z\right)}{X\left(z\right)}=H_0\left(z\right)+\frac{H_1\left(z\right)}{1+H_1\left(z\right)G\left(z\right)}$$

4.10.2 Problem 11.2

Figure 4.86:Problem description

solution

Adding the following notations on the diagram to make it easy to do the computation

Figure 4.87:Annotations added

Therefore we see that $$\begin{array}{}\text{(1)}& E& =X-F{H}_1{G}_2\text{(2)}& F& =E{H}_2-F{H}_1{G}_1\end{array}$$

We have 2 equations with 2 unknowns $E,F$. Substituting first equation into the second gives$$\begin{array}{rl}F& =(X-F{H}_1{G}_2)H_2-F{H}_1{G}_1\\ F& =X{H}_2-F{H}_1{G}_2{H}_2-F{H}_1{G}_1\\ F(1+H_1{G}_2{H}_2+H_1{G}_1)& =X{H}_2\\ \text{(3)}& F& =\frac{X{H}_2}{1+H_1{G}_2{H}_2+H_1{G}_1}\end{array}$$

But $$Y\left(z\right)=F\left(z\right)H_1\left(z\right)$$ Hence using (3) into the above gives$$\begin{array}{rl}Y\left(z\right)& =\frac{X{H}_2}{1+H_1{G}_2{H}_2+H_1{G}_1}{H}_1\\ \frac{Y\left(z\right)}{X\left(z\right)}& =\frac{H_2{H}_1}{1+H_1{G}_2{H}_2+H_1{G}_1}\end{array}$$

4.10.3 Problem 11.4

Figure 4.88:Problem description

solution

Figure 11.3 a is the following

Figure 4.89:figure from book 11.3(a)

Taking the Laplace transform of the ODE gives (assuming zero initial conditions)$$\begin{array}{rl}{s}^{2}Y\left(s\right)+sY\left(s\right)+Y\left(s\right)& =sX\left(x\right)\\ \text{(1)}& \frac{Y\left(s\right)}{X\left(s\right)}& =\frac{s}{s^2+s+1}\end{array}$$

From the diagram, we see that $$\begin{array}{}\text{(2)}& Y\left(s\right)=E\left(s\right)H\left(s\right)\end{array}$$ 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 $$\begin{array}{rl}E\left(s\right)& =X\left(s\right)-\left(E\left(s\right)H\left(s\right)G\left(s\right)\right)\\ E\left(s\right)(1+H\left(s\right)G\left(s\right))& =X\left(s\right)\\ E\left(s\right)& =\frac{X\left(s\right)}{1+H\left(s\right)G\left(s\right)}\end{array}$$

Substituting the above in (2) gives$$\begin{array}{rl}Y\left(s\right)& =\frac{X\left(s\right)}{1+H\left(s\right)G\left(s\right)}H\left(s\right)\\ \text{(3)}& \frac{Y\left(s\right)}{X\left(s\right)}& =\frac{H\left(s\right)}{1+H\left(s\right)G\left(s\right)}\end{array}$$

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)$$$\begin{array}{rl}\frac{\frac{1}{s+1}}{\frac{s+1+G\left(s\right)}{s+1}}& =\frac{s}{s^2+s+1}\\ \frac{1}{s+1+G\left(s\right)}& =\frac{s}{s^2+s+1}\\ s^2+s+sG\left(s\right)& =s^2+s+1\\ sG\left(s\right)& =s^2+s+1-s^2-s\\ G\left(s\right)& =\frac{1}{s}\end{array}$$

4.10.4 Problem 11.5

Figure 4.90:Problem description

solution

Figure 11.3 b is the following

Figure 4.91:figure from book 11.3(b)

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$$\begin{array}{rl}E\left(z\right)& =X\left(z\right)-E\left(z\right)H\left(z\right)G\left(z\right)\\ E\left(z\right)(1+H\left(z\right)G\left(z\right))& =X\left(z\right)\\ E\left(z\right)& =\frac{X\left(z\right)}{(1+H\left(z\right)G\left(z\right))}\end{array}$$

Therefore$$\begin{array}{rl}Y\left(z\right)& =E\left(z\right)H\left(z\right)\\ & =\frac{X\left(z\right)}{(1+H\left(z\right)G\left(z\right))}H\left(z\right)\\ \frac{Y\left(z\right)}{X\left(z\right)}& =\frac{H\left(z\right)}{1+H\left(z\right)G\left(z\right)}\end{array}$$

But $H\left(z\right)=\frac{1}{1-\frac{1}{2}{z}^{-1}}$ and $G\left(z\right)=1-b{z}^{-1}$. Hence the above becomes$$\begin{array}{rl}\frac{Y\left(z\right)}{X\left(z\right)}& =\frac{\frac{1}{1-\frac{1}{2}{z}^{-1}}}{1+\frac{1}{1-\frac{1}{2}{z}^{-1}}(1-b{z}^{-1})}\\ & =\frac{1}{1-\frac{1}{2}{z}^{-1}+1-b{z}^{-1}}\\ & =\frac{1}{2-\frac{1}{2}{z}^{-1}-b{z}^{-1}}\\ & =\frac{1}{2-(\frac{1}{2}+b)z^{-1}}\\ & =\frac{1}{2}\frac{1}{1-(\frac{1}{4}+\frac{b}{2})z^{-1}}\end{array}$$

The pole is $(\frac{1}{4}+\frac{b}{2})z^{-1}=1$ 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$$\begin{array}{rl}|\frac{1}{4}+\frac{b}{2}|& <1\\ -1& <\frac{1}{4}+\frac{b}{2}<1\\ -1-\frac{1}{4}& <\frac{b}{2}<1-\frac{1}{4}\\ -\frac{5}{4}& <\frac{b}{2}<\frac{3}{4}\\ -\frac{10}{4}& <b<\frac{6}{4}\\ -\frac{5}{2}& <b<\frac{3}{2}\end{array}$$

4.10.5 key solution

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