For the signal flow graph show in figure, the transfer function is C(s)/R(s)=

\frac{G_{1} G_{2} G_{3}}{1 + G_{1} G_{2} H_{1} + G_{2} G_{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{G_{1} G_{2} G_{3}}{1 + G_{1} G_{2} H_{1} + G_{2} G_{3} H_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{G_{1} G_{2} G_{3}}{1 + G_{1} G_{2} H_{1} + G_{1} + G_{2} G_{3} H_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{G_{1} G_{2}}{1 + G_{1} G_{2} + G_{2} G_{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{G_{1} G_{2} G_{3}}{1 + G_{1} G_{2} H_{1} + G_{2} G_{3} H_{2}}$
There is one forward path and two loops,
$P_{1} = G_{1} G_{2} G_{3}$
$L_{1} = - H_{1} G_{1} G_{2}$ ,
$L_{2} = - H_{2} G_{2} G_{3}$
$T . F . = \frac{P_{1} Δ_{1}}{Δ}$
$= \frac{G_{1} G_{2} G_{3} [1 - 0]}{1 + H_{1} G_{1} G_{2} + H_{2} G_{2} G_{3}}$
$= \frac{G_{1} G_{2} G_{3}}{1 + G_{1} G_{2} H_{1} + G_{2} G_{3} H_{2}}$

Suggest Corrections

Similar questions

\frac{C (s)}{R (s)}

\frac{C (s)}{R (s)}

U_{1} (s)

U_{2} (s)

Y (s) / U_{1} (s)

\frac{C (s)}{R (s)}

Join BYJU'S Learning Program