GATE Electrical Engineering 2020

Question 1
ax^3+bx^2+cx+d is a polynomial on real x over real coefficients a, b, c, d wherein a \neq 0. Which of the following statements is true?
d can be chosen to ensure that x = 0 is a root for any given set a, b, c.
No choice of coefficients can make all roots identical.
a, b, c, d can be chosen to ensure that all roots are complex.
c alone cannot ensure that all roots are real.
Engineering Mathematics   Complex Variables
Question 1 Explanation: 
Given Polynomial ax^{3}+bx^{2}+cx+d=0;\; \; \; a\neq 0

Option (A):
If d=0, then the polynomial equation becomes
\begin{aligned} ax^3+bx^2+cx&=0\\ x(ax^2+bx+c)&=0 \\ x=0 \text{ or } ax^2+bx+c&=0 \end{aligned}
d can be choosen to ensure x=0 is a root of given polynomial.
Hence, Option (A) is correct.

Option B:
A third degree polynomial equation with all root equal is given by
(x+\alpha )^3=0
Thus, by selecting suitable values of a, b, c and d we can have all roots identical.
Hence, option (B) is incorrect.

Option (C): Complex roots always occurs in pairs,
So, the given polynomial will have maximum of 2 complex roots and 1 real root.
Hence, option (C) is incorrect.

Option (D): Nature or roots depends on other coefficients also apart from coefficient 'c'.
Hence, option (D) is correct.
Hence, the correct options are (A) and (D).
Question 2
Which of the following is true for all possible non-zero choices of integers m, n; m \neq n, or all possible non-zero choices of real numbers p, q ; p\neq q, as applicable?
\frac{1}{\pi}\int_{0}^{\pi}\sin m\theta \sin n\theta \; d\theta =0
\frac{1}{2\pi}\int_{-\pi/2}^{\pi/2}\sin p\theta \sin q\theta \; d\theta =0
\frac{1}{2\pi}\int_{-\pi}^{\pi}\sin p\theta \cos q\theta \; d\theta =0
\lim_{\alpha \to \infty }\frac{1}{2\alpha }\int_{-\alpha }^{\alpha }\sin p\theta \sin q\theta \; d\theta =0
Engineering Mathematics   Complex Variables
Question 2 Explanation: 
\begin{aligned} \because \; p& \neq q\\ &\frac{1}{2\pi}\int_{-\pi}^{\pi} \sin p\theta \cos q\theta d\theta \\ &=\frac{1}{2\pi}\cdot \frac{1}{2}\int_{-\pi}^{\pi} [\sin (p+q)\theta + \sin (p-q)\theta] d\theta \\ &=\frac{1}{4\pi}\left [ \frac{-1}{(p+q)}\cos (p+q)\theta -\frac{1}{(p-q)}\cos (p-q)\theta \right ]_{-\pi}^{\pi}\\ &=\frac{-1}{4\pi} \left \{ \frac{1}{(p+q)}(\cos (p+q) \pi -\cos (p+q)(-\pi)) \right.\\ &+\left. \frac{1}{(p-q)}(\cos (p-q) \pi -\cos (p-q)(-\pi)) \right \}\\ &=0 \end{aligned}

Question 3
Which of the following statements is true about the two sided Laplace transform?
It exists for every signal that may or may not have a Fourier transform.
It has no poles for any bounded signal that is non-zero only inside a finite time interval.
The number of finite poles and finite zeroes must be equal.
If a signal can be expressed as a weighted sum of shifted one sided exponentials, then its Laplace Transform will have no poles.
Signals and Systems   Laplace Transform
Question 3 Explanation: 
It has no poles for any bounded signal that is nonzero in a finite time interval. This is true as we know for finite amplitude finite width signal ROC is entire s plane and ROC never includes any pole.
It implies for such signals there is no poles. Hence the correct answer is option (B).
Question 4
Consider a signal x[n]=\left ( \frac{1}{2} \right )^n \; 1[n], where 1[n]=0 if n \lt 0, and 1[n]=1 if n \geq 0. The z-transform of x[n-k], k \gt 0 is \frac{z^{-k}}{1-\frac{1}{2}z^{-1}} with region of convergence being
|z| \lt 2
|z| \gt 2
|z| \lt 1/2
|z| \gt 1/2
Signals and Systems   Z-Transform
Question 4 Explanation: 
\begin{aligned}x(n)&=\left (\frac{1}{2} \right )^{n} u(n) , \; \; \; \text{ROC of }x(n):\left | z \right | \gt \frac{1}{2} \\ x(n-k)\rightleftharpoons X(z)&=\frac{z^{-k}}{1-\frac{1}{2}z^{-1}} , \; \; \; \text{ROC of }x(n-k): \left | z \right | \gt \frac{1}{2}\\ \text{For } x(n-k) \; \; \; &\text{ROC will be } \left | z \right |\gt \frac{1}{2}\end{aligned}.
Question 5
The value of the following complex integral, with C representing the unit circle centered at origin in the counterclockwise sense, is:

8 \pi i
-8 \pi i
- \pi i
\pi i
Engineering Mathematics   Complex Variables
Question 5 Explanation: 
\begin{aligned} I&=\int _C \frac{z^2+1}{z^2-2z}dz\;\;\;|z|=1 \\ \text{Using } & \text{Cauchy's integral theorem}\\ \int _C\frac{F(z)}{z-a}dz&=2 \pi i (Re_{(z=a)})\;\;\;...(i)\\ I&=\int _C \frac{z^2+1}{z(z-2)}dz \end{aligned}
Poles are at z=0 and 2 but only z=0 lies inside the unit circle.
Residue at (z=0)=\lim_{z \to 0}\frac{z^2+1}{z(z-2)}
Using equation (i)
\int _C \frac{z^2+1}{z^2-2z}dz=2 \pi i \times \left ( \frac{-1}{2} \right )=-\pi i

There are 5 questions to complete.