Question 1 |
In the following block diagram, R(s) and D(s) are two inputs. The output Y(s) is expressed as Y(s)=G_{1}(s) R(s)+G_{2}(s) D(s).
G_{1}(s) and G_{2}(s) are given by
G_{1}(s) and G_{2}(s) are given by
G_{1}(s)=\frac{G(s)}{1+G(s)+G(s) H(s)} and G_{2}(s)=\frac{G(s)}{1+G(s)+G(s) H(s)} | |
G_{1}(s)=\frac{G(s)}{1+G(s)+H(s)} and G_{2}(s)=\frac{G(s)}{1+G(s)+H(s)} | |
G_{1}(s)=\frac{G(s)}{1+G(s)+H(s)} and G_{2}(s)=\frac{G(s)}{1+G(s)+G(s) H(s)} | |
G_{1}(s)=\frac{G(s)}{1+G(s)+G(s) H(s)} and G_{2}(s)=\frac{G(s)}{1+G(s)+H(s)} |
Question 1 Explanation:
Y(s)=\underbrace{G_{1}(s) R(s)}_{Y_{1}(s)}+\underbrace{G_{2}(s) D(s)}_{Y_{2}(s)} Considering first R(s) only, then Y(s) is Y_{1}(s)
\begin{aligned} \frac{Y_{1}(s)}{R(s)}&=\frac{\frac{G(s)}{1+G(s) H(s)}}{1+\frac{G(s)}{1+G(s) H(s)}} \\ \frac{Y_{1}(s)}{R(s)}&=\frac{G(s)}{1+G(s) H(s)+G(s)} \\ Y_{1}(s)&=\left[\frac{G(s)}{1+G(s)+G(s) H(s)}\right] R(s) \\ G_{1}(s)&=\frac{G(s)}{1+G(s)+G(s) H(s)} \end{aligned}
Now considering D(s) only, then Y(s) is Y_{2}(s)
\begin{aligned} \frac{Y_{2}(s)}{D(s)} & =\frac{G(s)}{1+G(s)[1+H(s)]} \\ Y_{2}(s) & =\left[\frac{G(s)}{1+G(s)+G(s) H(s)}\right] D(s) \\ G_{2}(s) & =\frac{G(s)}{1+G(s)+G(s) H(s)} \end{aligned}
Hence, G_{1}(s) and G_{2}(s) both are equal.
Question 2 |
The open loop transfer function of a unity negative feedback system is G(s)=\frac{k}{s\left(1+s T_{1}\right)\left(1+s T_{2}\right)}, where k, T_{1} and T_{2} are positive constants. The phase crossover frequency, in rad/s, is
\frac{1}{\sqrt{T_{1} T_{2}}} | |
\frac{1}{T_{1} T_{2}} | |
\frac{1}{T_{1} \sqrt{T_{2}}} | |
\frac{1}{T_{2} \sqrt{T_{1}}} |
Question 2 Explanation:
We know phase crossover frequency is that frequency at which phase of the open loop transfer function is -180^{\circ}.
\begin{aligned} \therefore \quad G(s) & =\frac{K}{s\left(1+s T_{1}\right)\left(1+s T_{2}\right)} \\ G(j \omega) & =\frac{K}{(j \omega)\left(1+j \omega T_{1}\right)\left(1+j \omega T_{2}\right)} \\ \text{Phase of }G(j \omega) & =\phi =-90-\tan ^{-1}\left(\omega T_{1}\right) - \tan ^{-1}\left(\omega T_{2}\right) \\ \therefore \quad \text { At } \omega &=\omega_{p c}, \phi =-180 \\ -180 & =-90-\tan ^{-1}\left(\omega_{p c} T_{1}\right)-\tan ^{-1}\left(\omega_{p c} T_{2}\right) \\ 9 \quad 90 & =\tan ^{-1}\left(\omega_{p c} T_{1}\right)+\tan ^{-1}\left(\omega_{p c} T_{2}\right) \\ \tan ^{-1}\left(\frac{\omega_{p c} T_{1}+\omega_{p c} T_{2}}{1-\omega_{p c}^{2} T_{1} T_{2}}\right) & =90 \\ 1-\omega_{p c}^{2} T_{1} T_{2} & =0 \\ \omega_{p c} & =\frac{1}{\sqrt{T_{1} T_{2}}} \end{aligned}
\begin{aligned} \therefore \quad G(s) & =\frac{K}{s\left(1+s T_{1}\right)\left(1+s T_{2}\right)} \\ G(j \omega) & =\frac{K}{(j \omega)\left(1+j \omega T_{1}\right)\left(1+j \omega T_{2}\right)} \\ \text{Phase of }G(j \omega) & =\phi =-90-\tan ^{-1}\left(\omega T_{1}\right) - \tan ^{-1}\left(\omega T_{2}\right) \\ \therefore \quad \text { At } \omega &=\omega_{p c}, \phi =-180 \\ -180 & =-90-\tan ^{-1}\left(\omega_{p c} T_{1}\right)-\tan ^{-1}\left(\omega_{p c} T_{2}\right) \\ 9 \quad 90 & =\tan ^{-1}\left(\omega_{p c} T_{1}\right)+\tan ^{-1}\left(\omega_{p c} T_{2}\right) \\ \tan ^{-1}\left(\frac{\omega_{p c} T_{1}+\omega_{p c} T_{2}}{1-\omega_{p c}^{2} T_{1} T_{2}}\right) & =90 \\ 1-\omega_{p c}^{2} T_{1} T_{2} & =0 \\ \omega_{p c} & =\frac{1}{\sqrt{T_{1} T_{2}}} \end{aligned}
Question 3 |
The block diagram of a feedback control system is shown in the figure
The transfer function \dfrac{Y{\left ( s \right )}}{X {\left ( s \right )}} of the system is
The transfer function \dfrac{Y{\left ( s \right )}}{X {\left ( s \right )}} of the system is
\frac{G_{1}+G_{2}+G_{1}G_{2}H}{1+G_{1}H} | |
\frac{G_{1}+G_{2}}{1+G_{1}H+G_{2}H} | |
\frac{G_{1}+G_{2}}{1+G_{1}H} | |
\frac{G_{1}+G_{2}+G_{1}G_{2}H}{1+G_{1}H+G_{2}H} |
Question 3 Explanation:
\frac{Y}{R}=\frac{G_{1}(1-0)+G_{2}(1-0)}{1-\left[-G_{1} H\right]}=\frac{G_{1}+G_{2}}{1+G_{1} H}
Question 4 |
The block diagram of a system is illustrated in the figure shown, where X(s) is the input and Y(s) is the output. The transfer function H(s)=\frac{Y(s)}{X(s)} is
H(s)=\frac{s^2+1}{s^3+s^2+s+1} | |
H(s)=\frac{s^2+1}{s^3+2s^2+s+1} | |
H(s)=\frac{s+1}{s^2+s+1} | |
H(s)=\frac{s^2+1}{2s^2+s+1} |
Question 4 Explanation:
Using block diagram reduction, we get,
\frac{Y(s)}{X(s)}=H(s)=\frac{s^{2}+1}{s^{3}+2 s^{2}+s+1}
\frac{Y(s)}{X(s)}=H(s)=\frac{s^{2}+1}{s^{3}+2 s^{2}+s+1}
Question 5 |
For the system shown in the figure, Y(s)/X(s)= __________.
1 | |
2 | |
3 | |
4 |
Question 5 Explanation:
\begin{aligned} Y(s) &=[X(s)-Y(s)] G(s)+X(s) \\ [1+G(s)] Y(s) &=[G(s)+1] X(s) \\ \frac{Y(s)}{X(s)} &=1 \end{aligned}
There are 5 questions to complete.