Question

The signal flow graph of a system is shown below. U(s) is the input and C(s) is the output.



Assuming, $h_1=b_1$ and $h_o=b_0-b_1{a}_1$, the input- output transfer function, $G\left(s\right)=\frac{C\left(s\right)}{U\left(s\right)}$ of the system is given by

• A. $G\left(s\right)=\frac{b_{0}s+b_1}{s^2+a_{0}s+a_1}$
• B. $G\left(s\right)=\frac{a_{1}s+a_0}{s^2+b_{1}s+b_0}$
• C. $G\left(s\right)=\frac{b_{1}s+b_0}{s^2+a_{1}s+a_0}$
• D. $G\left(s\right)=\frac{a_{0}s+a_1}{s^2+b_{0}s+b_1}$

Answer 1

The correct option is C $G\left(s\right)=\frac{b_{1}s+b_0}{s^2+a_{1}s+a_0}$

Using Mason gain formula,
Transfer function,

$G\left(s\right)=\frac{C\left(s\right)}{U\left(s\right)}=\frac{\sum P_K{\Delta }_k}{\Delta }$

Forward paths are:
$P_1=\frac{h_0}{s^2},P_2=\frac{h_1}{s}$

Individual loops are :
$L_1=\frac{-a_1}{s},L_2=\frac{-a_0}{s^2}$

Non-touching loops=zero
$\therefore {\Delta }_1=1,$
${\Delta }_2=1-\left(\frac{-a_1}{s}\right)=(1+\frac{a_1}{s})=\left(\frac{s+a_1}{s}\right)$

$\Delta =1-(\frac{-a_1}{s}-\frac{a_0}{s^2})=\frac{s^2+a_{1}s+a_0}{s^2}$

$\therefore G\left(s\right)=\frac{C\left(s\right)}{U\left(s\right)}=\frac{P_1{\Delta }_1+P_2{\Delta }_2}{\Delta }$

$=\frac{\left(\frac{h_0}{s^2}\right)\times \left(\frac{h_1}{s}\right)\times \left(\frac{s+a_1}{s}\right)}{\left(\frac{s^2+a_{1}s+a_0}{s^2}\right)}$

$\left[\frac{h_0+h_1(s+a_1)}{S^2+a_{1}s+a_0}\right]$

$Given,h_1=b_{1}and{h}_0=b_0-b_1{a}_1$

$Thus,G\left(s\right)=\frac{(b_0-b_1{a}_1)+b_1(s+a_1)}{(s^2+a_{1}s+a_0)}$

$=\left(\frac{b_{1}s+b_0}{s^2+a_{1}s+a_0}\right)$

$G\left(s\right)=\left(\frac{b_{1}s+b_0}{s^2+a_{1}s+a_0}\right)$