GATE 2015 EE – SET 1 Question 25

Solution :

P$$\Rightarrow$$  Forward paths $$\Rightarrow (1)(G_{1})(G_{2}])$$

Individual loop gains

$$P_{11}$$ = – $$G_{1}$$
$$P_{21}$$ = – $$G_{1}G_{2}$$

TF = $$\frac{1}{\Delta }\sum_{k}\Delta_{k}P_{k}$$

TF = $$TF = \frac{P_{1}\Delta_{1} }{\Delta }$$

Where $$\Delta _{1}$$ is the sum of gain loops not touching the forward path  and $$\Delta$$ is 1 – (sum of gain of all single loops)+(sum of gain of all two non touching loops)

TF =$$ \frac{G_{1}G_{2}(1)}{1-\left [ -G_{1} – G_{1}G_{2}\right ]}$$

TF = $$\frac{y(s)}{X_{2}(s)}$$ = $$\frac{G_{2}}{1+G_{1}+G_{1}G_{2}}$$ = $$\frac{G_{2}}{1+G_{1}(1+G_{2})}$$

Forward path = $$P_{1}$$ = (1)$$(G_{1}(G_{2} = G_{1} G_{2}$$

Individual loop gains =$$ P_{11}$$= (1)$$(G_{1})$$(-1) = – $$G_{1}$$

$$P_{22} $$=$$ (G_{1}G_{2})$$(-1) =$$ G_{1}G_{2}$$

TF =$$ \frac{1}{\Delta }\left [ \Delta _{1}P_{1} \right ]$$ = $$ \frac{\Delta _{1}P_{1}}{1-\left [ P_{11}+P_{22} \right ]}$$

TF = $$\frac{G_{1}G_{2}}{1-\left [ -G_{1}-G_{1}G_{2} \right ]}$$

TF = $$\frac{G_{1}G_{2}}{1+\left [ G_{1} + G_{1}G_{2} \right ]}$$

$$\because$$ TF =$$ \frac{Y_{s}}{X_{2}} $$

$$\therefore$$  $$ \frac{Y_{s}}{X_{2}} $$ = $$\frac{G_{1}G_{2}}{1+\left [ G_{1} + G_{1}G_{2} \right ]}$$

since the input is X_{2} , we can omit G_{1} ,then we can have

$$ \frac{Y_{s}}{X_{2}} $$ = $$\frac{G_{2}}{1+\left [ G_{1} + G_{1}G_{2} \right ]}$$