Question 1 |
The vector function F\left ( r \right )=-x\hat{i}+y\hat{j}
is defined over a circular arc C shown in the figure.
The line integral of \int _{C} F\left ( r \right ).dr is
The line integral of \int _{C} F\left ( r \right ).dr is
\frac{1}{2} | |
\frac{1}{4} | |
\frac{1}{6} | |
\frac{1}{3} |
Question 1 Explanation:
\begin{aligned} \bar{F} &=-x i+y j \\ \int \vec{F} \cdot \overrightarrow{d r} &=\int_{c}-x d x+y d y \\ &=\int_{\theta=0}^{45^{\circ}}(-\cos \theta(-\sin \theta)+\sin \theta \cos \theta) d \theta \\ \int_{\theta=0}^{\pi / 4} \sin 2 \theta d \theta &\left.=-\frac{\cos 2 \theta}{2}\right]_{0}^{\pi / 4} \\ &=-\frac{1}{2}[0-1]=\frac{1}{2} \end{aligned}
Question 2 |
Consider the differential equation given below.
\frac{dy}{dx}+\frac{x}{1-x^{2}}y=x\sqrt{y}
The integrating factor of the differential equation is
\frac{dy}{dx}+\frac{x}{1-x^{2}}y=x\sqrt{y}
The integrating factor of the differential equation is
\left ( 1-x^{2} \right )^{-3/4} | |
\left ( 1-x^{2} \right )^{-1/4} | |
\left ( 1-x^{2} \right )^{-3/2} | |
\left ( 1-x^{2} \right )^{-1/2} |
Question 2 Explanation:
\begin{aligned} \frac{d y}{d x}+\frac{x}{1-x^{2}} y&=x \sqrt{y}, \quad \text { IF }=?\\ \text{Divided by }\sqrt{y}\\ \frac{1}{\sqrt{y}} \frac{d y}{d x}+\frac{x}{1-x^{2}} \sqrt{y}&=x \\ 2 \frac{d u}{d x}+\frac{x}{1-x^{2}} u&=x\\ \text{Let }\qquad x \sqrt{y}&=u\\ \frac{1}{2 \sqrt{v}} \frac{d y}{d x}&=\frac{d u}{d x}\\ \Rightarrow \qquad \frac{d u}{d x}+\frac{x}{2\left(1-x^{2}\right)} u&=\frac{x}{2} \rightarrow \text{ lines diff. equ.} \\ \text { I. } F&=e^{\int \frac{x}{2\left(1-x^{2}\right)} d x}=e^{-\frac{1}{4} \log \left(1-x^{2}\right)}&=e^{\log \left(1-x^{2}\right) \frac{-1}{4}} \\ \text { I.F }&=\frac{1}{\left(1-x^{2}\right)^{\frac{1}{4}}} \end{aligned}
Question 3 |
Two continuous random variables X and Y are related as
Y=2X+3
Let \sigma ^{2}_{X} and \sigma ^{2}_{Y} denote the variances of X and Y, respectively. The variances are related as
Y=2X+3
Let \sigma ^{2}_{X} and \sigma ^{2}_{Y} denote the variances of X and Y, respectively. The variances are related as
\sigma ^{2}_{Y}=2 \sigma ^{2}_{X} | |
\sigma ^{2}_{Y}=4 \sigma ^{2}_{X} | |
\sigma ^{2}_{Y}=5 \sigma ^{2}_{X} | |
\sigma ^{2}_{Y}=25 \sigma ^{2}_{X} |
Question 3 Explanation:
\begin{aligned} Y &=2 X+3 \\ \operatorname{Var}[Y] &=E\left[(Y-\bar{Y})^{2}\right] \\ E[Y] &=\bar{Y}=2 \bar{X}+3 \\ \operatorname{Var}[Y] &=E\left[(2 X+3-2 \bar{X}-3)^{2}\right] \\ &=E\left[4(X-\bar{X})^{2}\right] \\ &=4 \cdot E\left[(X-\bar{X})^{2}\right] \\ \sigma_{Y}^{2} &=4 \cdot \sigma_{X}^{2} \end{aligned}
Question 4 |
Consider a real-valued base-band signal x(t), band limited to \text{10 kHz}. The Nyquist rate for the signal y\left ( t \right )=x\left ( t \right )x\left ( 1+\dfrac{t}{2} \right ) is
\text{15 kHz} | |
\text{30 kHz} | |
\text{60 kHz} | |
\text{20 kHz} |
Question 4 Explanation:
\mathrm{NR}=2 \times f_{\mathrm{max}}=2 \times 15=30 \mathrm{kHz}
Question 5 |
Consider two 16-point sequences x[n] and h[n]. Let the linear convolution of x[n] and h[n] be denoted by y[n], while z[n] denotes the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point DFTs of x[n] and h[n]. The value(s) of k for which z[k]=y[k] is/are
k=0,1,2,,15 | |
k=0 | |
k=15 | |
k=0 and k=15 |
Question 5 Explanation:
If two' N' point signals x(n) and h(n) are convolving with each other linearly and circularly
then
y(k)=z(k) at k=N-1
where, y(n)= Linear convolution of x(n) and h(n)
z(n)= Circular convolution of x(n) and h(n)
Since, N=16 (Given)
Therefore, \quad y(k)=z(k) at k=N-1=15
then
y(k)=z(k) at k=N-1
where, y(n)= Linear convolution of x(n) and h(n)
z(n)= Circular convolution of x(n) and h(n)
Since, N=16 (Given)
Therefore, \quad y(k)=z(k) at k=N-1=15
There are 5 questions to complete.