Home » Uncategorized » GATE 2015 EE - SET 1 Question 25

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 ]}

Translate »