Question 1 |
Consider a 2 x 2 square matrix
A=\begin{bmatrix} \sigma & x\\ \omega & \sigma \end{bmatrix}
where x is unknown. If the eigenvalues of the matrix A are (\sigma +j\omega ) \; and \; (\sigma -j\omega ), then x is equal to
A=\begin{bmatrix} \sigma & x\\ \omega & \sigma \end{bmatrix}
where x is unknown. If the eigenvalues of the matrix A are (\sigma +j\omega ) \; and \; (\sigma -j\omega ), then x is equal to
+j\omega | |
-j\omega | |
+\omega | |
-\omega |
Question 1 Explanation:
\begin{aligned} A &=\left[\begin{array}{ll} \sigma & x \\ \omega & \sigma \end{array}\right] \\ \text { Trace } &=\text { sum of eigen values } \\ 2 \sigma &=\sigma+j \omega+\sigma-j \omega \\ |A| &=\text { product of eigens } \\ \sigma^{2}-x \omega &=(\sigma+j \omega)(\sigma-j \omega)=\sigma^{2}+\omega^{2} \end{aligned}
which is possible only when, x=-\omega
which is possible only when, x=-\omega
Question 2 |
For f(z)=\frac{sin(z)}{z^{2}}, the residue of the pole at z=0 is __________
0.5 | |
1 | |
2 | |
3 |
Question 2 Explanation:
Residue of \frac{\sin z}{z^{2}}
= coefficient of \frac{1}{2} in \left\{\frac{z-\frac{z^{3}}{3 !}+\frac{z^{5}}{5 !}---}{z^{2}}\right\}
= coefficient of \frac{1}{z} in \left\{\frac{1}{z}-\frac{z}{3 !}+\frac{z^{3}}{5 !}---\right\}
=1
= coefficient of \frac{1}{2} in \left\{\frac{z-\frac{z^{3}}{3 !}+\frac{z^{5}}{5 !}---}{z^{2}}\right\}
= coefficient of \frac{1}{z} in \left\{\frac{1}{z}-\frac{z}{3 !}+\frac{z^{3}}{5 !}---\right\}
=1
Question 3 |
The probability of getting a "head" in a single toss of a biased coin is 0.3. The coin is tossed repeatedly till a "head" is obtained. If the tosses are independent, then the probability of getting "head" for the first time in the fifth toss is __________
0.01 | |
0.04 | |
0.07 | |
0.1 |
Question 3 Explanation:
\begin{aligned} P(H)&=0.3 \\ P(T)&=0.7 \end{aligned}
since all tosses are independent
so, probability of getting head for the first time in 5^{\text {th }} toss is
\begin{aligned} &=P(T) P(T) P(T) P(T) P(H) \\ &=0.7 \times 0.7 \times 0.7 \times 0.7 \times 0.3=0.072 \end{aligned}
since all tosses are independent
so, probability of getting head for the first time in 5^{\text {th }} toss is
\begin{aligned} &=P(T) P(T) P(T) P(T) P(H) \\ &=0.7 \times 0.7 \times 0.7 \times 0.7 \times 0.3=0.072 \end{aligned}
Question 4 |
The integral \int_{0}^{1}\frac{dx}{\sqrt{(1-x)}} is equal to __________
1 | |
2 | |
3 | |
4 |
Question 4 Explanation:
\begin{aligned} \int_{0}^{1} \frac{1}{\sqrt{1-x}} d x &=-2 \int_{0}^{1} \frac{1}{2 \sqrt{1-x}} d x \\ &=-\left.2(\sqrt{1-x})\right|_{0} ^{1} \\ &=-2(0-1)=2 \end{aligned}
Question 5 |
Consider the first order initial value problem
y'=y+2x-x^{2},y(0)=1,(0\leq x\leq \infty )
with exact solution y(x)=x^{2}+e^{x}. For x = 0.1, the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta method with step-size h = 0.1 is __________
y'=y+2x-x^{2},y(0)=1,(0\leq x\leq \infty )
with exact solution y(x)=x^{2}+e^{x}. For x = 0.1, the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta method with step-size h = 0.1 is __________
0.06 | |
0.012 | |
0.6 | |
0.12 |
Question 5 Explanation:
\begin{aligned} \frac{d y}{d x} &=y+2 x-x^{2} \\ y(0) &=1 \quad \quad 0 \leq x \leq \infty \\ f(x, y) &=y+2 x-x^{2} \\ x_{0}=0 ; y_{0}=1 ; h &=0.1 \\ k_{1} &=h f\left(x_{0}, y_{0}\right) \\ &=0.1\left(1+2 \times 0-0^{2}\right)=0.1 \\ k_{3} &=h /\left(x_{0}+h_{1} y_{0}+k_{1}\right) \\ &=0.1\left(\left(y_{0}+k_{1}\right)+2\left(x_{0}+h\right)-\left(x_{0}+h\right)^{2}\right) \\ &\left.=0.1(11+0.1)+2(0.1)-(0.1)^{2}\right) \\ &=0.129\\ y_{1} &=y_{0}+\frac{1}{2}\left(k_{1}+k_{2}\right) \\ &=1+\frac{1}{2}(0.1+0.129) \\ &=1.1145 \\ \text { Exact solution } y(x) &=x^{2}+e^{x} \\ &=(0.1)^{2}+e^{0.1} \\ &=1.1152 \\ \text { Error } &=1.1152-1.1145=0.00069 \\ \% \text { error } &=0.06 \% \end{aligned}
Question 6 |
Consider the signal x(t)=cos(6\pi t)+sin(8\pi t), where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y(t)=x(2t+ 5) is
8 | |
12 | |
16 | |
32 |
Question 6 Explanation:
\begin{aligned} X(t) & =\cos (6 \pi t)+\sin (8 \pi t) \\ Y(t) & =x(2 t+5) \\ Y(t) & =\cos [6 \pi(2 t+5)+\sin (8 \pi(2 t+5)] \\ & =\cos (12 \pi t+30 \pi)+\sin (16 \pi t+40 \pi) \\ f_{m_{1}} & =6 \mathrm{Hz}_{i} \qquad f_{m 2}=8 \mathrm{Hz}\\ \text{Nyquist }&\text{sampling rate},\\ f_{s} &=2 f_{\max } \\ &=16 \text { samples/second } \end{aligned}
Question 7 |
If the signal x(t)=\frac{sin(t)}{\pi t}*\frac{sin(t)}{\pi t} with * denoting the convolution operation, then x(t) is equal to
\frac{sin(t)}{\pi t} | |
\frac{sin(2t)}{2\pi t} | |
\frac{2sin(t)}{\pi t} | |
(\frac{sin(t)}{\pi t})^{2} |
Question 7 Explanation:
\begin{aligned} x(t) &=x_{1}(t) * x_{1}(t) \\ X(\omega) &=X_{1}(\omega) \cdot X_{1}(\omega) \end{aligned}
\therefore \quad x(t)=\frac{\sin t}{\pi t}
Question 8 |
A discrete-time signal x[n]=\delta [n-3]+2\delta [n-5] has z-transform X(z). If X(z)=X(-z) is the z-transform of another signal y[n], then
y[n]=x[n] | |
y[n]=x[-n] | |
y[n]=-x[n] | |
y[n]=-x[-n] |
Question 8 Explanation:
Given,
\begin{aligned} x[n] &=\delta[n-3]+2 \delta[n-5] \\ X(z) &=z^{3}+2 z^{5} \\ X(-z) &=(-z)^{-3}+2(-z)^{-5} \\ Y(z) &=X(-z)=-z^{-3}-2 z^{-5} \\ &=-\left[z^{-3}+2 z^{-5}\right] \\ y[n] &=-x[n] \end{aligned}
\begin{aligned} x[n] &=\delta[n-3]+2 \delta[n-5] \\ X(z) &=z^{3}+2 z^{5} \\ X(-z) &=(-z)^{-3}+2(-z)^{-5} \\ Y(z) &=X(-z)=-z^{-3}-2 z^{-5} \\ &=-\left[z^{-3}+2 z^{-5}\right] \\ y[n] &=-x[n] \end{aligned}
Question 9 |
the RLC circuit shown in the figure, the input voltage is given by
v_{i}(t)=2cos(200t)+4sin(500t)
The output voltage v_{0}(t) is
v_{i}(t)=2cos(200t)+4sin(500t)
The output voltage v_{0}(t) is
cos(200t) + 2 sin(500t) | |
2cos(200t) + 4 sin(500t) | |
sin(200t) + 2 cos(500t) | |
2sin(200t) + 4 cos(500t) |
Question 9 Explanation:
Where, V_{i}(t)=2 \cos (200 t)+4 \sin (500 t)
As different frequencies are operating, using superposition theorem, we get
for \omega=200 \mathrm{rad} / \mathrm{sec}
\begin{array}{l} x_{L}=\omega L=(200)(0.25)=50 \Omega \\ x_{C}=\frac{1}{\omega C}=\frac{1}{200 \times 100 \times 10^{-6}}=50 \Omega \end{array}
\begin{aligned} V_{0}(t) &=V_{i}(t) \\ X_{L} &=0.4 \times 500=200 \Omega \\ X_{C} &=\frac{1}{10 \times 10^{-6} \times 500}=200 \Omega \end{aligned}
\begin{aligned} \therefore \quad V_{0}(t)&=V_{i}(t) \\ \text { Therefore } \quad V_{0}(t)&=2 \cos (200 t)+4 \sin (500 t) \end{aligned}
Question 10 |
The I-V characteristics of three types of diodes at the room temperature, made of semiconductors X, Y and Z, are shown in the figure. Assume that the diodes are uniformly doped and identical in all respects except their materials. If E_{gX},E_{gY} \; and \; E_{gZ} are the band gaps of X, Y and Z, respectively,
then
E_{gX} \gt E_{gY}\gt E_{gZ} | |
E_{gX}=E_{gY} = E_{gZ} | |
E_{gX} \lt E_{gY}\lt E_{gZ} | |
no relationship among these band gaps exists. |
Question 10 Explanation:
\begin{array}{l} X \rightarrow \text { Ge Diode } \ldots \ldots \ldots \ldots \ldots E_{G}=0.2 \mathrm{V} \\ Y \rightarrow \text { Si Diode } \ldots \ldots \ldots \ldots E_{G}=0.7 \mathrm{V} \\ Z \rightarrow \text { GaAs Diode }(\text { or }) \mathrm{LED} \ldots \ldots \ldots . E_{G}=1.3 \mathrm{V} \end{array}
There are 10 questions to complete.