Question 1 |
The number of chords in the graph of the given circuit will be
3 | |
4 | |
5 | |
6 |
Question 1 Explanation:
Number of branches =b =6
No. of nodes = n= 4
No. of chords =b-(n-1) =6-(4-1)=3
Question 2 |
The Thevenin's equivalent of a circuit operation at \omega=5 rads/s, has V_{oc}=3.71\angle -15.9^{\circ} V and Z_0=2.38-j0.667\Omega. At this frequency, the minimal realization of the Thevenin's impedance will have a
resistor and a capacitor and an inductor | |
resistor and a capacitor | |
resistor and an inductor | |
capacitor and an inductor |
Question 2 Explanation:
Thevenin's Impedance:
Z_0=2.38-j0.667\Omega
as real part is not zero, so Z_0 has resistor
Img[Z_0]=-j0.667
CASE-I:
Z_0 has capacitor (as Img[Z_0] is negative)
CASE-II:
Z_0 has both capacitor and inductor, but inductive reactance \lt capacitive reactance.
At, \omega=5 rad/sec
For minimal realization case-I is considered. Therefore, Z_0 will have a resistor and a capacitor.
Z_0=2.38-j0.667\Omega
as real part is not zero, so Z_0 has resistor
Img[Z_0]=-j0.667
CASE-I:
Z_0 has capacitor (as Img[Z_0] is negative)
CASE-II:
Z_0 has both capacitor and inductor, but inductive reactance \lt capacitive reactance.
At, \omega=5 rad/sec
For minimal realization case-I is considered. Therefore, Z_0 will have a resistor and a capacitor.
Question 3 |
A signal e^{-\alpha t}sin(\omega t) is the input to a real Linear Time Invariant system. Given K and \phi are constants, the output of the system will be of the form Ke^{-\beta t}sin(vt+\phi ) where
\beta need not be equal to \alpha but v equal to \omega | |
v need not be equal to \omega but \beta equal to \alpha | |
\beta equal to \alpha and v equal to \omega | |
\beta need not be equal to \alpha and v need not be equal to \omega |
Question 4 |
X is a uniformly distributed random variable that takes values between 0 and 1.
The value of E{X^{3}} will be
0 | |
1/8 | |
1/4 | |
1/2 |
Question 4 Explanation:
x is uniformly distributes in [0,1]
Therefore, probability density function
\begin{aligned} f(x)&=\frac{1}{b-a} =\frac{1}{1-0}=1\\ \therefore \; f(x) &=1\;\;\;0 \lt x \lt 1 \\ &=0\;\;\;\text{elsewhere} \\ \text{Now, } E(x^3)&=\int_{0}^{1}x^3f(x)\; dx \\ &= \int_{0}^{1}x^3 \times 1 \times dx\\ &= \left [ \frac{x^4}{4} \right ]_0^1=\frac{1}{4} \end{aligned}
Therefore, probability density function
\begin{aligned} f(x)&=\frac{1}{b-a} =\frac{1}{1-0}=1\\ \therefore \; f(x) &=1\;\;\;0 \lt x \lt 1 \\ &=0\;\;\;\text{elsewhere} \\ \text{Now, } E(x^3)&=\int_{0}^{1}x^3f(x)\; dx \\ &= \int_{0}^{1}x^3 \times 1 \times dx\\ &= \left [ \frac{x^4}{4} \right ]_0^1=\frac{1}{4} \end{aligned}
Question 5 |
The characteristic equation of a (3x3) matrix P is defined as
a(\lambda )|\lambda I-P|= \lambda ^{3}+ \lambda ^{2}+ 2\lambda +1=0
If I denotes identity matrix, then the inverse of matrix P will be
a(\lambda )|\lambda I-P|= \lambda ^{3}+ \lambda ^{2}+ 2\lambda +1=0
If I denotes identity matrix, then the inverse of matrix P will be
(P^{2}+P+2I) | |
(P^{2}+P+I) | |
-(P^{2}+P+I) | |
-(P^{2}+P+2I) |
Question 5 Explanation:
If characteristic equation is
\lambda ^3+\lambda ^2+2\lambda +1=0
Then by cayley- hamilton theorem,
P^3+P^2+2P+I=0
I=-P^3-P^2-2P
Multiplying by P^{-1} on both sides,
P^{-1}=-P^2-P-2I=-(P^2+P+2I)
\lambda ^3+\lambda ^2+2\lambda +1=0
Then by cayley- hamilton theorem,
P^3+P^2+2P+I=0
I=-P^3-P^2-2P
Multiplying by P^{-1} on both sides,
P^{-1}=-P^2-P-2I=-(P^2+P+2I)
There are 5 questions to complete.