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)
Question 6 |
For an LTI system, the Bode plot for its gain is as illustrated in the figure shown. The number of system poles N_p and the number of system zeros N_z in the frequency range 1Hz\leq f\leq 10^7Hz is
N_p=5,N_z=2 | |
N_p=6,N_z=3 | |
N_p=7,N_z=4 | |
N_p=4,N_z=2 |
Question 6 Explanation:
Number of poles (N_{P})= 6
Number of zeros (N_{Z}) = 3
Question 7 |
A linear Hamming code is used to map 4-bit messages to 7-bit codewords. The encoder mapping is linear. If the message 0001 is mapped to the codeword 0000111, and the message 0011 is mapped to the codeword 1100110, then the message 0010 is mapped to
10011 | |
1100001 | |
1111000 | |
1111111 |
Question 7 Explanation:
Question 8 |
Which one of the following options describes correctly the equilibrium band diagram at T=300 K of a Silicon pnp^+p^{++} configuration shown in the figure?
A | |
B | |
C | |
D |
Question 9 |
The correct circuit representation of the structure shown in the figure is
A | |
B | |
C | |
D |
Question 9 Explanation:
Question 10 |
The figure shows the high-frequency C-V curve of a MOS capacitor (at T = 300 K) with \Phi _{ms}=0V and no oxide charges. The flat-band, inversion, and accumulation conditions are represented, respectively, by the points
P, Q, R | |
Q, R, P | |
R, P, Q | |
Q, P, R |
Question 10 Explanation:
Since \phi_{ms}= 0, the MOS-capacitor is ideal.
Point P Represents accumulation
Point Q Represents flat band
Point R Represents Inversion
Point P Represents accumulation
Point Q Represents flat band
Point R Represents Inversion
There are 10 questions to complete.