Question 1 |
The bilateral Laplace transform of a function f(t)=\left\{\begin{matrix} 1 & if a\leq t\leq b\\ 0&otherwise \end{matrix}\right.
is
is
\frac{a-b}{s} | |
\frac{e^{x}(a-b)}{s} | |
\frac{e^{-as}-e^{-bs}}{s} | |
\frac{e^{s(a-b)}}{s} |
Question 1 Explanation:
\begin{aligned} f(t)&=\left\{\begin{array}{ll} 1 & \text { if } a \leq t \leq b \\ 0 & \text { otherwise } \end{array}\right.\\ &=U(t-\mathrm{a})-(t+b) \\ \Rightarrow \quad F(s) &=\frac{e^{-a s}-e^{-b s}}{s} \end{aligned}
Question 2 |
The value of x for which all the eigen-values of the matrix given below are real is \begin{bmatrix} 10 & 5+j&4 \\ x & 20 & 2\\ 4&2 & -10 \end{bmatrix}
5+j | |
5-j | |
1-5j | |
1+5j |
Question 2 Explanation:
For a matrix containing complex number, eigen values are real if and only if
\begin{array}{l} A=A^{0}=(\bar{A})^{T} \\ A=\left[\begin{array}{ccc} 10 & 5+j & 4 \\ x & 20 & 2 \\ 4 & 2 & -10 \end{array}\right] \\ A^{\theta}=(\bar{A})^{T}=\left[\begin{array}{ccc} 10 & \bar{x} & 4 \\ 5-j & 20 & 2 \\ 4 & 2 & -10 \end{array}\right] \end{array}
By comparing these,
x=5-j
\begin{array}{l} A=A^{0}=(\bar{A})^{T} \\ A=\left[\begin{array}{ccc} 10 & 5+j & 4 \\ x & 20 & 2 \\ 4 & 2 & -10 \end{array}\right] \\ A^{\theta}=(\bar{A})^{T}=\left[\begin{array}{ccc} 10 & \bar{x} & 4 \\ 5-j & 20 & 2 \\ 4 & 2 & -10 \end{array}\right] \end{array}
By comparing these,
x=5-j
Question 3 |
Let f(z)=\frac{az+b}{cz+d} . If f(z_{1})=f(z_{2}) \; for \; all \; z_{1}\neq z_{2} , a= 2, b = 4 and c= 5, then d should be equal to _______.
9 | |
10 | |
11 | |
12 |
Question 3 Explanation:
\begin{aligned} f\left(z_{1}\right)&=\frac{a z_{1}+b}{c z_{1}+d} \\ f\left(z_{2}\right)&=\frac{a z_{2}+b}{c z_{2}+d} \\ \frac{a z_{1}+b}{c z_{1}+d}&=\frac{a z_{2}+b}{c z_{2}+d} \\ acz_{1}z_{2}+bcz_{2}+adz_{1}&+bd=ac z_{1} z_{2}+bcz_{1}+adz_{2}+bd\\ b c\left(z_{2}-z_{1}\right)&=a d\left(z_{2}-z_{1}\right) \\ z_{2} &\neq z_{1} \\ \Rightarrow \quad b c&=a d \\ d=\frac{b c}{a}&=\frac{4 \times 5}{2}=10 \end{aligned}
Question 4 |
The general solution of the differential equation \frac{dy}{dx}=\frac{1+cos2y}{1-cos2x} is
tan y - cot x = c ( c is a constant) | |
tan x - cot y = c ( c is a constant) | |
tan y + cot x = c ( c is a constant) | |
tan x + cot y = c ( c is a constant) |
Question 4 Explanation:
\begin{aligned} \frac{d y}{1+\cos 2 y}&=\frac{d x}{1-\cos 2 x} \\ \frac{d y}{2 \cos ^{2} y}&=\frac{d x}{2 \sin ^{2} x} \\ \sec ^{2} y d y&=\text{cosec}^{2} x d x\\ \end{aligned}
Integrating both sides, we get
\begin{aligned} \tan y&=-\cot x+c\\ \tan y+\text{cot} x&=c \end{aligned}
Integrating both sides, we get
\begin{aligned} \tan y&=-\cot x+c\\ \tan y+\text{cot} x&=c \end{aligned}
Question 5 |
The magnitude and phase of the complex Fourier series coefficients a_{k} of a periodic signal x(t) are shown in the figure. Choose the correct statement from the four choices given. Notation: C is the set of complex numbers, R is the set of purely real numbers, and P is the set of purely imaginary numbers.
x(t)\in R | |
x(t)\in P | |
x(t)\in (C-R) | |
the information given is not sufficient to draw any conclusion about x(t) |
Question 5 Explanation:
\begin{aligned} \left|a_{k}\right|&-\text{ even symmetry} \\ \angle a_{k}&-\text{ Odd symmetry} &(\pi \text{ can be } -\pi) \end{aligned}
\Rightarrow x(t) is real.
\Rightarrow x(t) is real.
There are 5 questions to complete.