Question 1 |
Consider a lead compensator of the form
K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta a}}, \quad \beta \gt 1, a \gt 0
The frequency at which this compensator produces maximum phase lead is 4 \mathrm{rad} / \mathrm{s}. At this frequency, the gain amplification provided by the controller, assuming asymptotic Bodemagnitude plot of K(\mathrm{~s}), is 6 \mathrm{~dB}. The values of \alpha, \beta, respectively, are
K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta a}}, \quad \beta \gt 1, a \gt 0
The frequency at which this compensator produces maximum phase lead is 4 \mathrm{rad} / \mathrm{s}. At this frequency, the gain amplification provided by the controller, assuming asymptotic Bodemagnitude plot of K(\mathrm{~s}), is 6 \mathrm{~dB}. The values of \alpha, \beta, respectively, are
1, 16 | |
2, 4 | |
3, 5 | |
2.66, 2.25 |
Question 1 Explanation:
K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta}}
Max. phase lead occur at, \omega_{m}=\sqrt{a(a \beta)} =\sqrt{\beta}
Given : \quad \omega_{\mathrm{m}}=4 \mathrm{rad} / \mathrm{sec}
a \sqrt{\beta}=4 \quad ...(1)
Now, \quad K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta a}}
Put, s=j \omega
\mathrm{K}\left(\mathrm{j} \omega_{\mathrm{n}}\right)=\frac{1+\frac{\mathrm{j} \omega_{m}}{a}}{1+\frac{j \omega_{m}}{\beta a}}=\frac{j w_{m}}{a}
Given : M=6
\begin{aligned} 20 \log \left(\frac{4}{a}\right) & =6 \\ \Rightarrow \quad a & =2 \end{aligned}
From eqn. (1), we get
\begin{aligned} 2 \sqrt{\beta} & =4 \\ \beta & =4 \end{aligned}
Max. phase lead occur at, \omega_{m}=\sqrt{a(a \beta)} =\sqrt{\beta}
Given : \quad \omega_{\mathrm{m}}=4 \mathrm{rad} / \mathrm{sec}
a \sqrt{\beta}=4 \quad ...(1)
Now, \quad K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta a}}
Put, s=j \omega
\mathrm{K}\left(\mathrm{j} \omega_{\mathrm{n}}\right)=\frac{1+\frac{\mathrm{j} \omega_{m}}{a}}{1+\frac{j \omega_{m}}{\beta a}}=\frac{j w_{m}}{a}
Given : M=6
\begin{aligned} 20 \log \left(\frac{4}{a}\right) & =6 \\ \Rightarrow \quad a & =2 \end{aligned}
From eqn. (1), we get
\begin{aligned} 2 \sqrt{\beta} & =4 \\ \beta & =4 \end{aligned}
Question 2 |
In the Nyquist plot of the open-loop transfer function
\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=\frac{3 \mathrm{~s}+5}{\mathrm{~s}-1}
corresponding to the feedback loop shown in the figure, the infinite semi-circular arc of the Nyquist contour in s-plane is mapped into a point at
\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=\frac{3 \mathrm{~s}+5}{\mathrm{~s}-1}
corresponding to the feedback loop shown in the figure, the infinite semi-circular arc of the Nyquist contour in s-plane is mapped into a point at
\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=\infty | |
\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=0 | |
\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=3 | |
\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=-5 |
Question 2 Explanation:
Nyquist Contour :
Given:
\begin{aligned} \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =\frac{3 \mathrm{~s}+5}{\mathrm{~s}-1} \\ \text { Put } \quad \mathrm{s} & =R e^{j \theta} \\ \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =\operatorname{Lim}_{R \rightarrow \infty} \frac{3 R e^{j \theta}+5}{\operatorname{Re}^{j \theta}-1} \\ \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =3 \end{aligned}
Given:
\begin{aligned} \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =\frac{3 \mathrm{~s}+5}{\mathrm{~s}-1} \\ \text { Put } \quad \mathrm{s} & =R e^{j \theta} \\ \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =\operatorname{Lim}_{R \rightarrow \infty} \frac{3 R e^{j \theta}+5}{\operatorname{Re}^{j \theta}-1} \\ \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =3 \end{aligned}
Question 3 |
The open loop transfer function of a unity gain negative feedback system is given
as
G(s)=\frac{1}{s(s+1)}
The Nyquist contour in the s-plane encloses the entire right half plane and a small neighbourhood around the origin in the left half plane, as shown in the figure below. The number of encirclements of the point (-1+j0) by the Nyquist plot of G(s), corresponding to the Nyquist contour, is denoted as N. Then N equals to
G(s)=\frac{1}{s(s+1)}
The Nyquist contour in the s-plane encloses the entire right half plane and a small neighbourhood around the origin in the left half plane, as shown in the figure below. The number of encirclements of the point (-1+j0) by the Nyquist plot of G(s), corresponding to the Nyquist contour, is denoted as N. Then N equals to
0 | |
1 | |
2 | |
3 |
Question 3 Explanation:
Given: P = 1 (Because, Nyquist contour encircle one
pole i.e s = 0)
We have, N = P - Z
N = 1-Z
Characteristic equation
\begin{aligned} 1+G(s)H(s) &=0 \\ 1+\frac{k}{s(s+1)}&=0 \\ s^2+s+k&= 0 \end{aligned}
R-H criteria:
\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k\\ 1 & 0\\ k & \end{matrix}
Hence, z = 0 (because no sign change in first column of R-H criteria)
(where, z = closed loop pole on RHS side of s-plane)
Therefore, N = 1
We have, N = P - Z
N = 1-Z
Characteristic equation
\begin{aligned} 1+G(s)H(s) &=0 \\ 1+\frac{k}{s(s+1)}&=0 \\ s^2+s+k&= 0 \end{aligned}
R-H criteria:
\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k\\ 1 & 0\\ k & \end{matrix}
Hence, z = 0 (because no sign change in first column of R-H criteria)
(where, z = closed loop pole on RHS side of s-plane)
Therefore, N = 1
Question 4 |
An LTI system is shown in the figure where
G(s)=\frac{100}{s^2+0.1s+10}
The steady state output of the system, to the input r(t) , is given as y(t)=a+b\sin (10t+\theta ). The values of a and b will be
G(s)=\frac{100}{s^2+0.1s+10}
The steady state output of the system, to the input r(t) , is given as y(t)=a+b\sin (10t+\theta ). The values of a and b will be
a=1, b=10 | |
a=10, b=1 | |
a=1, b=100 | |
a=100, b=1 |
Question 4 Explanation:
We knaow, y(t)=A|G(j\omega ) \sin (\omega t+\phi )
Here, G(j\omega )=\frac{100}{-\omega ^2+j0.1\omega +100}
Put \omega =0
|G(j\omega )|=1
Now, put \omega =10 rad/sec
|G(j\omega )|=\left | \frac{100}{-100+i1+100} \right |=100
Therefore, y(t)=1+0.1 \times 100 \sin(10t+\theta ) =1+10 \sin (10t+\theta )
On comparision : a=1 ,b=10
Here, G(j\omega )=\frac{100}{-\omega ^2+j0.1\omega +100}
Put \omega =0
|G(j\omega )|=1
Now, put \omega =10 rad/sec
|G(j\omega )|=\left | \frac{100}{-100+i1+100} \right |=100
Therefore, y(t)=1+0.1 \times 100 \sin(10t+\theta ) =1+10 \sin (10t+\theta )
On comparision : a=1 ,b=10
Question 5 |
The Bode magnitude plot of a first order stable system is constant with frequency.
The asymptotic value of the high frequency phase, for the system, is -180^{\circ}. This
system has
one LHP pole and one RHP zero at the same frequency | |
one LHP pole and one LHP zero at the same frequency | |
two LHP poles and one RHP zero | |
two RHP poles and one LHP zero. |
Question 5 Explanation:
The given system is non-minimum phase system
Therefore, transfer function, T.F=\frac{s-1}{s+1}
Hence, one LHP pole and one RHP zero at the same frequency.
Hence, one LHP pole and one RHP zero at the same frequency.
There are 5 questions to complete.