December 2012 - DSP LOG

GATE-2012 ECE Q34 (signals)

Question 34 on signals from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q34. Consider the differential equation

${\frac{d^2y(t)}{dt}+2\frac{dy(t)}{dt}+y(t) = \delta(t)}$ with ${y(t)|}_{t=0^{-}}=-2$ and ${\frac{dy}{dt}\|_{t=0^{-}}=0}$

The numerical value of

${\frac{dy}{dt}|}_{t=0^{+}}$ is

(A) -2

(B) -1

(C) 0

(D) 1

Continue reading “GATE-2012 ECE Q34 (signals)”

GATE-2012 ECE Q12 (math)

Question 12 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q12. With initial condition ${x(1)=0.5}$ the solution of the differential equation,

${t\frac{dx}{dt}+x= t}$ is

(A) ${x=t-\frac{1}{2}}$

(B) ${x=t^2-\frac{1}{2}}$

(C) ${x=\frac{t^2}{2}}$

(D) ${x=\frac{t}{2}}$

Solution

From the product rule used to find the derivative of product of two or more functions,

${\frac{d}{dx}(f.g)=f\frac{dg}{dx}+g\frac{df}{dx}}$

Applying this to the above equation, we can be seen that,

$\frac{d}{dt}(t.x)=t\frac{dx}{dt}+x\frac{dt}{dt}=t\frac{dx}{dt}+x$

Plugging this in and integrating both sides,

$\begin{array}t\frac{dx}{dt}+x&=&t\\\frac{d}{dt}(t.x)&=&t\\\int\frac{d}{dt}(t.x)&=&\int t\\tx&=&\frac{t^2}{2}+C\end{array}$ .

Using the initial condition ${x(1)=0.5}$ , we can solve for the unknown $C$ , i.e.
$\begin{array}t.x&=&\frac{t^2}{2}+C\\1*0.5&=&\frac{1^2}{2}+C\\C&=&0\end{array}$ .

So the solution to the differential equation is,

${x=\frac{t}{2}}$

Based on the above, the right choice is (D) ${x=\frac{t}{2}}$ .

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[2] Wiki entry on Product rule

GATE-2012 ECE Q11 (signals)

Question 11 on signals from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q11. The unilateral Laplace transform of ${f(t)}$ is ${\frac{1}{s^2+s+1}}$ . The unilateral Laplace transform of ${tf(t)}$ is

(A) ${-\frac{s}{(s^2+s+1)^2}}$

(B) ${-\frac{2s+1}{(s^2+s+1)^2}$

(C) ${\frac{s}{(s^2+s+1)^2}}$

(D) ${\frac{2s+1}{(s^2+s+1)^2}$

GATE-2012 ECE Q2 (communication)

Question 52 on communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q2. The power spectral density of a real process ${X(t)}$ for positive frequencies is shown below. The values of ${E\[X^2(t)\]}$ and ${\|E\[X(t)\]\|}$ , respectively are

(A) ${6000/\pi, \mbox{ } 0}$

(B) ${6400/\pi, \mbox{ } 0}$

(C) ${6400/\pi, \mbox{ } 20/$\pi \sqrt{2}$}$

(D) ${6000/\pi, \mbox{ } 20/$\pi \sqrt{2}$}$

Solution

For a wide sense stationary function, the auto-correlation with delay $\tau$ is defined as,

$R$\tau$=E\[X(t+\tau)X^*(t)\]$

From the Weiner-Kinchin theorem, the auto-correlation function $R$\tau$$ and power spectral density $S(f)$ are Fourier Transform pairs, i.e.

$R$\tau$=\int_{-\infty}^{\infty}S$f$e^{j2\pi f\tau}df$

Expressing in terms of $\omega$ , where $\omega=2\pi f$

$R$\tau$=\frac{1}{2\pi}\int_{-\infty}^{\infty}S$w$e^{jw\tau}dw$ .

When the delay $\tau=0$ , the above equations simplifies to

$\begin{array}R$0$&=&E\[X(t)X^*(t)\]&=&E\[|X(t)|^2\]&=&\frac{1}{2\pi}\int_{-\infty}^{\infty}S$w$dw\end{array}$ .

Applying this to the problem at hand,

$\begin{array}{lll}E\[|X(t)|^2\]&=&\frac{1}{2\pi}\int_{-\infty}^{\infty}S$w$dw\\&=&\frac{2}{2\pi}\int_{0}^{\infty}S$w$dw\\&=&\frac{2}{2\pi}$\frac{1}{2}6*2*10^3 + 400$\\&=&\Large{\frac{6400}{\pi}}\end{array}$ .

Further, since the power spectral density $S(f)$ does not have any dc component, the mean of the signal $\|E\[X(t)\]\|=0$

Based on the above, the right choice is (B) ${6400/\pi, \mbox{ } 0}$

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[2] Weiner-Kinchin theorem

[3] Wide sense stationary function

[4] Auto-correlation

GATE-2012 ECE Q52 (electromagnetics)

Question 52 on electromagnetics from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

An infinitely long uniform solid wire of radius ${a}$ carries a uniform dc current of density ${\vec{j}}$ .

Q52. The magnetic field at a distance ${r}$ from the center of the wire is proportional to

(A) ${r}$ for ${r<a}$ and ${1/r^2}$ for ${r>a}$

(B) ${0}$ for ${r<a}$ and ${1/r}$ for ${r>a}$

(C) ${r}$ for ${r<a}$ and ${1/r}$ for ${r>a}$

(D) ${0}$ for ${r<a}$ and ${1/r^2}$ for ${r>a}$

Solution

To answer this question on magnetic field, we need to determine the magnetic field inside and outside the cable using Ampere’s Law.

Ampere’s Law :

The total current $I_{enc}$ inside a closed curve $C$ is the line integral of the magnetic field $B$ (in Tesla)

$\begin{array}\oint_C{B}.{dl}&=&\mu_oI_{enc}\end{array}$ ,

where

$B$ is the magnetic field (in Tesla)

$dl$ is the vector representing the infinitesimal line on the closed loop $C$ ,

$I_{enc}$ is the net current enclosed by the closed loop and

$\mu_0$ is the permeability of vacuum.

Let us use this result to find the magnetic field in a uniform solid wire of radius ${a}$ .

Magnetic field in the region ${r<a}$ :

Figure : Solid wire showing the imaginary Amperian loop inside the wire (circle with radius $r$ )

Given that the current density is ${\vec{j}}$ , the current through the area in the circle with radius $r$ is

$I = \vec{j}\pi r^2$

Applying Ampere’s law,

$\begin{array}\oint_C{B}.{dl}&=&\mu I&=&\mu\vec{j}\pi r^2\end{array}$ ,

where

$\mu=\mu_r \mu_0$ is the permeability ( $\mu_0$ is the permeability of vacuum. $\mu_r$ is the permeability of the material).

Given that the magnetic field is $B$ is parallel to the line $dl$ and is uniform across the closed loop, it can be moved out of the integral, i.e

$\begin{array}B\oint_C{dl}&=&\mu \vec{j} \pi r^2\end{array}$ .

The term $\begin{array}\oint_C{dl}\end{array}$ is the circumference of the circle with radius $r$ i.e.

$\begin{array}\oint_C{dl}&=&2\pi r\end{array}$ .

Substituting, the magnetic field in the region $\begin{array}r<a\end{array}$ is,

$\begin{array}B 2\pi r &=&{\mu \vec{j} \pi r^2}\\B&=&\frac{\mu \vec{j} r }{2} \end{array}$ .

Magnetic field in the region ${r>a}$ :

Figure : Solid wire showing the imaginary Amperian loop outside the wire (circle with radius $r$ )

Given that the current density is ${\vec{j}}$ , the current through the area in the circle with radius $r$ is determined only by the current flowing through the cable with in the radius ${a}$ , i.e.

$I = \vec{j}\pi a^2$

Applying Ampere’s law,

$\begin{array}\oint_C{B}.{dl}&=&\mu I&=&\mu\vec{j}\pi a^2\end{array}$ ,

Taking $B$ outside the intergral and substituting for the term $\begin{array}\oint_C{dl}&=&2\pi r\end{array}$ ,

The magnetic field in the region $\begin{array}r<a\end{array}$ is,

$\begin{array}B 2\pi r &=&{\mu \vec{j} \pi a^2}\\B&=&\frac{\mu \vec{j} a^2 }{2r} \end{array}$ .

Summarizing,

${\begin{array}{llllr}B&=&\frac{\mu \vec{j} r}{2},&{ r < a}\\&=&\frac{\mu \vec{j} a^2 }{2r},&{ r > a}\end{array}}$

Based on the above, the right choice is (C) i.e. The magnetic field at a distance ${r}$ from the center of the wire is proportional to ${r}$ for ${r<a}$ and ${1/r}$ for ${r>a}$

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[2] Fields and Waves in Communication Electronics, Simon Ramo, John R. Whinnery, Theodore Van Duzer (buy from Amazon.com, buy from Flipkart.com).

[3] The youtube videos uploaded by user lasseviren1 aided me in understanding the integrals in electric field and magnetic field.

[4] Ampere’s Law

[5] Permeability

Q34. Consider the differential equation

with and

The numerical value of

is

(A) -2

(B) -1

(C) 0

(D) 1

Q12. With initial condition the solution of the differential equation,

is

(A)

(B)

(C)

(D)

Solution

References

Q11. The unilateral Laplace transform of is . The unilateral Laplace transform ofis

(A)

(B)

(C)

(D)

Q2. The power spectral density of a real process for positive frequencies is shown below. The values of and , respectively are

(A)

(B)

(C)

(D)

Solution

References

An infinitely long uniform solid wire of radius carries a uniform dc current of density .

Q52. The magnetic field at a distance from the center of the wire is proportional to

(A) for and for

(B) for and for

(C) for and for

(D) for and for

Solution

Ampere’s Law :

Magnetic field in the region :

Magnetic field in the region :

References

${\frac{d^2y(t)}{dt}+2\frac{dy(t)}{dt}+y(t) = \delta(t)}$ with ${y(t)|}_{t=0^{-}}=-2$ and ${\frac{dy}{dt}\|_{t=0^{-}}=0}$

${\frac{dy}{dt}|}_{t=0^{+}}$ is

Q12. With initial condition ${x(1)=0.5}$ the solution of the differential equation,

${t\frac{dx}{dt}+x= t}$ is

(A) ${x=t-\frac{1}{2}}$

(B) ${x=t^2-\frac{1}{2}}$

(C) ${x=\frac{t^2}{2}}$

(D) ${x=\frac{t}{2}}$

Q11. The unilateral Laplace transform of ${f(t)}$ is ${\frac{1}{s^2+s+1}}$ . The unilateral Laplace transform of ${tf(t)}$ is

(A) ${-\frac{s}{(s^2+s+1)^2}}$

(B) ${-\frac{2s+1}{(s^2+s+1)^2}$

(C) ${\frac{s}{(s^2+s+1)^2}}$

(D) ${\frac{2s+1}{(s^2+s+1)^2}$

Q2. The power spectral density of a real process ${X(t)}$ for positive frequencies is shown below. The values of ${E\[X^2(t)\]}$ and ${\|E\[X(t)\]\|}$ , respectively are

(A) ${6000/\pi, \mbox{ } 0}$

(B) ${6400/\pi, \mbox{ } 0}$

(C) ${6400/\pi, \mbox{ } 20/\(\pi \sqrt{2}\)}$

(D) ${6000/\pi, \mbox{ } 20/\(\pi \sqrt{2}\)}$

An infinitely long uniform solid wire of radius ${a}$ carries a uniform dc current of density ${\vec{j}}$ .

Q52. The magnetic field at a distance ${r}$ from the center of the wire is proportional to

(A) ${r}$ for ${r<a}$ and ${1/r^2}$ for ${r>a}$

(B) ${0}$ for ${r<a}$ and ${1/r}$ for ${r>a}$

(C) ${r}$ for ${r<a}$ and ${1/r}$ for ${r>a}$

(D) ${0}$ for ${r<a}$ and ${1/r^2}$ for ${r>a}$

Magnetic field in the region ${r<a}$ :

Magnetic field in the region ${r>a}$ :