Question 1 |
Which one of the following functions is analytic over the entire complex plane?
ln(z) | |
e^{1/z} | |
\frac{1}{1-z} | |
cos(z) |
Question 1 Explanation:
f(z) = \cos z is analytic every where.
Question 2 |
The families of curves represented by the solution of the equation
\frac{dy}{dx}=-\left (\frac{x}{y} \right )^n
for n = -1 and n = +1, respectively, are
\frac{dy}{dx}=-\left (\frac{x}{y} \right )^n
for n = -1 and n = +1, respectively, are
Parabolas and Circles | |
Circles and Hyperbolas | |
Hyperbolas and Circles | |
Hyperbolas and Parabolas |
Question 2 Explanation:
\begin{aligned} \frac{d y}{d x} &=-\left(\frac{x}{y}\right)^{n} \\ n=-1\quad\quad \frac{d y}{d x} &=-\frac{y^{\prime}}{x} \\ \frac{d y}{y} &=-\frac{d x}{x} \\ \int \frac{1}{y} d y &=-\int \frac{1}{x} d x \\ \ln y &=-\ln x+\ln c \\ \ln (y x) &=\ln c \end{aligned}
x y=c \quad (Represents rectangular hyporbola)
\begin{aligned} n=1, \quad \frac{d y}{d x}&=-\frac{x}{y} \\ y d y &=-x d x \\ y d y &=-\int x d x \\ \frac{y^{2}}{2} &=-\frac{x^{2}}{2}+c \end{aligned}
x^{2}+y^{2}=2 c \quad (Represents family of circles)
x y=c \quad (Represents rectangular hyporbola)
\begin{aligned} n=1, \quad \frac{d y}{d x}&=-\frac{x}{y} \\ y d y &=-x d x \\ y d y &=-\int x d x \\ \frac{y^{2}}{2} &=-\frac{x^{2}}{2}+c \end{aligned}
x^{2}+y^{2}=2 c \quad (Represents family of circles)
Question 3 |
Let H(z) be the z-transform of a real-valued discrete-time signal h[n]. If P(z)=H(z)H\left (\frac{1}{z} \right ) has a zero at z=\frac{1}{2}+\frac{1}{2}j, and P(z) has a total of four zeros, which one of the following plots represents all the zeros correctly?
A | |
B | |
C | |
D |
Question 3 Explanation:
P(Z)=H(Z)H\left ( \frac{1}{Z} \right )
(i) h(n) is real. Som p(n) will be also real
(ii) P(z)=P(z^{-1})
From (i) : if z_1 is a zero of P(z), then z_1^* will be also a zero of P(z).
From (ii): If z_1 is a zero of P(z), then \frac{1}{z_1} will be also a zero of P(z).
So, the 4 zeros are,
\begin{aligned} z_1&= \frac{1}{2}+\frac{1}{2}j\\ z_2&= z_1^*=\frac{1}{2}-\frac{1}{2}j\\ z_3&=\frac{1}{z_1}=\frac{1}{\frac{1}{2}-\frac{1}{2}j}=1-j \\ z_4&=\left ( \frac{1}{z_1} \right )^*=z_3^*=1+j \end{aligned}
(i) h(n) is real. Som p(n) will be also real
(ii) P(z)=P(z^{-1})
From (i) : if z_1 is a zero of P(z), then z_1^* will be also a zero of P(z).
From (ii): If z_1 is a zero of P(z), then \frac{1}{z_1} will be also a zero of P(z).
So, the 4 zeros are,
\begin{aligned} z_1&= \frac{1}{2}+\frac{1}{2}j\\ z_2&= z_1^*=\frac{1}{2}-\frac{1}{2}j\\ z_3&=\frac{1}{z_1}=\frac{1}{\frac{1}{2}-\frac{1}{2}j}=1-j \\ z_4&=\left ( \frac{1}{z_1} \right )^*=z_3^*=1+j \end{aligned}
Question 4 |
Consider the two-port resistive network shown in the figure. When an excitation of 5 V is applied across Port 1, and Port 2 is shorted, the current through the short circuit at Port 2 is measured to be 1 A (see (a) in the figure).
Now, if an excitation of 5 V is applied across Port 2, and Port 1 is shorted (see(b) in the figure), what is the current through the short circuit at Port 1?
Now, if an excitation of 5 V is applied across Port 2, and Port 1 is shorted (see(b) in the figure), what is the current through the short circuit at Port 1?
0.5 A | |
1.0 A | |
2.0 A | |
2.5 A |
Question 4 Explanation:
According to reciprocity theorem,
In a linear bilateral single source network the ratio of response to excitation remains the same even after their positions get interchanged.
\therefore \quad \frac{I}{5}=\frac{1}{5} \Rightarrow I=1 \mathrm{A}
In a linear bilateral single source network the ratio of response to excitation remains the same even after their positions get interchanged.
\therefore \quad \frac{I}{5}=\frac{1}{5} \Rightarrow I=1 \mathrm{A}
Question 5 |
Let Y(s) be the unit-step response of a causal system having a transfer function
G(s)=\frac{3-s}{(s+1)(s+3)}
that is, Y(s)=\frac{G(s)}{s}. The forced response of the system is
G(s)=\frac{3-s}{(s+1)(s+3)}
that is, Y(s)=\frac{G(s)}{s}. The forced response of the system is
u(t)-2e^{-t}u(t)+e^{-3t}u(t) | |
2u(t)-2e^{-t}u(t)+e^{-3t}u(t) | |
2u(t) | |
u(t) |
Question 5 Explanation:
Given, \quad G(s)=\frac{3-s}{(s+1)(s+3)}
\therefore \quad Y(s)=\frac{G(s)}{s}=\frac{3-s}{s(s+1)(s+3)}
Using partial fractions, we get,
\begin{aligned} Y(s)&=\frac{A}{s}+\frac{B}{(s+1)}+\frac{C}{(s+3)} \\ A\left(s^{2}+4 s+3\right)&+B\left(s^{2}+3 s\right)+C\left(s^{2}+s\right)=3-s \\ A+B+C&=0\\ 4 A+3 B+C&=-1 \\ \text{and }3 A&=3 \\ \text{Therefore, }&\text{we get,}\\ A=1, B&=-2 \text { and } C=1\\ \text{So, }\quad Y(s)&=\frac{1}{s}-\frac{2}{(s+1)}+\frac{1}{(s+3)} \\ \text{and}\quad \mathrm{y}(t)&=u(t)-2 e^{-t} u(t)+e^{-3 t} u(t) \\ \end{aligned}
Forced response,
y_{t}(t)=u(t) \Rightarrow \text { option }(D)
\therefore \quad Y(s)=\frac{G(s)}{s}=\frac{3-s}{s(s+1)(s+3)}
Using partial fractions, we get,
\begin{aligned} Y(s)&=\frac{A}{s}+\frac{B}{(s+1)}+\frac{C}{(s+3)} \\ A\left(s^{2}+4 s+3\right)&+B\left(s^{2}+3 s\right)+C\left(s^{2}+s\right)=3-s \\ A+B+C&=0\\ 4 A+3 B+C&=-1 \\ \text{and }3 A&=3 \\ \text{Therefore, }&\text{we get,}\\ A=1, B&=-2 \text { and } C=1\\ \text{So, }\quad Y(s)&=\frac{1}{s}-\frac{2}{(s+1)}+\frac{1}{(s+3)} \\ \text{and}\quad \mathrm{y}(t)&=u(t)-2 e^{-t} u(t)+e^{-3 t} u(t) \\ \end{aligned}
Forced response,
y_{t}(t)=u(t) \Rightarrow \text { option }(D)
There are 5 questions to complete.