Question 1 |
The eigen values of a skew-symmetric matrix are
always zero | |
always pure imaginary | |
either zero or pure imaginary | |
always real |
Question 1 Explanation:
A skew-synnmetric matrix is a square matrix whose
transpose is equal to its negative value, i.e.
A^{\top}=-A
The eigen values of a skew-symmetric matrix alwavs come in pairs \pm \lambda (except in odd-dimensional case where there is an additional unpaired 0 eigen value). For a real skew symmetric matrix the non-zero eigen values are all pure imaginary and thus are of the form i \lambda_{1},-i \lambda_{1}, i \lambda_{2}
-i \lambda_{2}, \ldots where each of the \lambda_{K} is real.
The eigen values of a skew-symmetric matrix alwavs come in pairs \pm \lambda (except in odd-dimensional case where there is an additional unpaired 0 eigen value). For a real skew symmetric matrix the non-zero eigen values are all pure imaginary and thus are of the form i \lambda_{1},-i \lambda_{1}, i \lambda_{2}
-i \lambda_{2}, \ldots where each of the \lambda_{K} is real.
Question 2 |
The trigonometric Fourier series for the waveform f(t) shown below contains
only cosine terms and zero values for the dc components | |
only cosine terms and a positive value for the dc components | |
only cosine terms and a negative value for the dc components | |
only sine terms and a negative value for the dc components |
Question 2 Explanation:
since f(t) is an even function, its trigonometric
Fourier series contains only cosine terms.
D.C. component,
\begin{aligned} A_{0} &=\frac{1}{T} \int_{-T / 2}^{T / 2} f(t) d t=\frac{2}{T} \int_{0}^{T / 2} f(t) d t \\ &=\frac{2}{T}\left[\int_{0}^{T / 4} A d t+\int_{T / 4}^{T / 2}(-2 A) d t\right] \\ &=\frac{2}{T}\left[\frac{A T}{4}-2 A\left(\frac{T}{2}-\frac{T}{4}\right)\right] \\ &=\frac{2}{T}\left[-\frac{A T}{4}\right]=-\frac{A}{2} \end{aligned}
Therefore, the trigonometric Fourier series for the waveform f(t) contains only cosine terms and a negative value for the dc component.
D.C. component,
\begin{aligned} A_{0} &=\frac{1}{T} \int_{-T / 2}^{T / 2} f(t) d t=\frac{2}{T} \int_{0}^{T / 2} f(t) d t \\ &=\frac{2}{T}\left[\int_{0}^{T / 4} A d t+\int_{T / 4}^{T / 2}(-2 A) d t\right] \\ &=\frac{2}{T}\left[\frac{A T}{4}-2 A\left(\frac{T}{2}-\frac{T}{4}\right)\right] \\ &=\frac{2}{T}\left[-\frac{A T}{4}\right]=-\frac{A}{2} \end{aligned}
Therefore, the trigonometric Fourier series for the waveform f(t) contains only cosine terms and a negative value for the dc component.
Question 3 |
A function n(x) satisfied the differential equation \frac{ d^{2}n(x)}{dx^{2}}-\frac{n(x)}{L^{2}}=0 where L is a constant. The boundary conditions are : n(0) = K and n(\infty ) = 0. The solution to this equation is
n(x) = Kexp(x/L) | |
n(x)=Kexp(-x/\sqrt{L}) | |
n(x)=K^{2}exp(-x/L) | |
n(x) = Kexp(- x/L) |
Question 3 Explanation:
\begin{aligned} \left(D^{2}-\frac{1}{L^{2}}\right) n(x) &=0 \\ D^{2}-\frac{1}{L^{2}} &=0 \\ D &=\pm \frac{1}{L} \end{aligned}
So, the solution is
\begin{aligned} n(x) &=C_{1} e^{x / L}+C_{2} e^{-x / L} \\ n(0) &=K \Rightarrow C_{1}+C_{2}=K \\ n(\infty) &=0 \Rightarrow C_{1}=0 \\ \Rightarrow \quad C_{2} &=K \\ \text { Therefore, } n(x) &=K \exp (-x / L) \end{aligned}
So, the solution is
\begin{aligned} n(x) &=C_{1} e^{x / L}+C_{2} e^{-x / L} \\ n(0) &=K \Rightarrow C_{1}+C_{2}=K \\ n(\infty) &=0 \Rightarrow C_{1}=0 \\ \Rightarrow \quad C_{2} &=K \\ \text { Therefore, } n(x) &=K \exp (-x / L) \end{aligned}
Question 4 |
For the two-port network shown below, the short-circuit admittance parameter matrix is
\begin{vmatrix} 4 & -2\\ -2&4 \end{vmatrix}S | |
\begin{vmatrix} 1 & -0.5\\ -0.5&1 \end{vmatrix}S | |
\begin{vmatrix} 1 & 0.5\\ 0.5&1 \end{vmatrix}S | |
\begin{vmatrix} 4 & 2\\ 2&4 \end{vmatrix}S |
Question 4 Explanation:
Short-circuit admittance parameters for a two-port \pi-network are,
\begin{array}{l} Y_{11}=Y_{a}+Y_{b} \\ Y_{12}=Y_{21}=-Y_{b} \\ Y_{22}=Y_{b}+Y_{c} \end{array}
For the given network,
\begin{aligned} Y_{a}&=Y_{b}=Y_{c}=\frac{1}{0.5}=2 \mho\\ \text{So,}Y_{11}&=2+2=4 \mho\\ Y_{12}&=Y_{21}=-2 \mho\\ Y_{22}&=2+2=4 \mho \end{aligned}
\begin{array}{l} Y_{11}=Y_{a}+Y_{b} \\ Y_{12}=Y_{21}=-Y_{b} \\ Y_{22}=Y_{b}+Y_{c} \end{array}
For the given network,
\begin{aligned} Y_{a}&=Y_{b}=Y_{c}=\frac{1}{0.5}=2 \mho\\ \text{So,}Y_{11}&=2+2=4 \mho\\ Y_{12}&=Y_{21}=-2 \mho\\ Y_{22}&=2+2=4 \mho \end{aligned}
Question 5 |
For parallel RLC circuit, which one of the following statements is NOT correct?
The bandwidth of the circuit decreases if R is increased | |
The bandwidth of the circuit remains same if L is increased | |
At resonance, input impedance is a real quantity | |
At resonance, the magnitude of input impedance attains its minimum values |
Question 5 Explanation:
Characteristic equation for a parallel RLC circuit is
s^{2}+\frac{1}{R C} s+\frac{1}{L C}=0
where, Bandwidth= \frac{1}{RC}
(i) It is clear that the bandwidth of a parallel RLC circuit is independent of L and decreases if R is increased.
(ii) At resonance, imaginary part of input impedance is zero. Hence, at resonance input impedance is a real quantity.
(iii) In parallel RLC circuit, the admittance is minimum at resonance. Hence magnitude of input impedance attains its maximum value at resonance.
s^{2}+\frac{1}{R C} s+\frac{1}{L C}=0
where, Bandwidth= \frac{1}{RC}
(i) It is clear that the bandwidth of a parallel RLC circuit is independent of L and decreases if R is increased.
(ii) At resonance, imaginary part of input impedance is zero. Hence, at resonance input impedance is a real quantity.
(iii) In parallel RLC circuit, the admittance is minimum at resonance. Hence magnitude of input impedance attains its maximum value at resonance.
There are 5 questions to complete.