Question

The system shown in figure below.



can be reduced to the form with

• A. $X=c_{0}s++c_{1,}Y=1/(S^2+a_{0}s+a_1),Z=b_{o}s+b_1$
• B. $X=1,Y=(c_{o}s+c_1),/(s^2+a_{0}s+a_1),Z+b_{0}s+b_1$
• C. $X=c_{1}s+c_o,Y=(b_{1}s+b_0)(s^2+a_{1}s+a_0))Z=1$
• D. $X=c_{1}s+c_{o,Y=1/(s^2+a_{1}s-a_o),Z=b_{1}s+b_0}$

Answer 1

The correct option is D $X=c_{1}s+c_{o,Y=1/(s^2+a_{1}s-a_o),Z=b_{1}s+b_0}$

The block-diagram can be redrawn as



Signal flow graph of the block -diagram



There are two forward paths;

$P_1=1\times c_0\times \frac{1}{s}\times \frac{1}{s}\times P=\frac{c_{0}P}{s^2}$
$P_2=1\times c_1\times 1\times \frac{1}{s}\times P=\frac{c_{0}P}{s}$

There are four individual loops
$L_1=-a_1\times \frac{1}{s}=-\frac{a_1}{s}$
$L_2=\frac{1}{s}\times \frac{1}{s}\times =a_0=-\frac{a_0}{s^2}$
$L_3=\frac{1}{s}\times \frac{1}{s}\times P\times =b_0=\frac{b_{0}P}{s^2}$
$L_4=b_1\times P\times \frac{1}{s}=\frac{b_{1}P}{s}$

All the loops touch forward paths
${\Delta }_1={\Delta }_2=1$
$\Delta =1-(L_1+L_2+L_3+L_4)$
$1+\frac{a_0}{s^2}+\frac{a_1}{s}-\frac{b_{0}P}{s^2}-\frac{b_{1}P}{s}$
Using Mason's gain formula
$\frac{C\left(s\right)}{R\left(s\right)}=\frac{P_1{\Delta }_1+P_2{\Delta }_2}{s}$

$=\frac{\frac{c_{0}P}{s^2}+\frac{c_{1}P}{s}}{1+\frac{a_1}{s}+\frac{a_0}{s^2}-\frac{b_{0}P}{s^2}-\frac{b_{1}P}{s}}$

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{c_{0}P+c_1{p}_s}{s^2+(a_0-b_1)s+(a_0-b_{0}p)}$

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{(c_0+c_{1}s)/(s_2+a_{1}s+a_0)}{1-(b_0+b_{1}s)P/(s^2+a_{1}s+a_0)}$



$\frac{C\left(s\right)}{R\left(s\right)}=\frac{xyP}{1-yzP}$

Comparing equation (i) and (ii), we get
$xy=\frac{c_0+c_{1}s}{s^2+a_{1}s+a_0}$
$yz=\frac{b_0+b_{1}s}{s^2+a_{1}s+a_0}$

Hence option (d) is correct.