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)

Question 6 |

If the rank of a (5x6) matrix Q is 4, then which one of the following statement
is correct ?

Q will have four linearly independent rows and four linearly independent columns | |

Q will have four linearly independent rows and five linearly independent columns | |

QQ^{T}will be invertible | |

Q^{T}Q will be invertible |

Question 6 Explanation:

If rank of (5 x 6) matrix is 4, then surely it must have exactly 4 linearly independent rows as will as 4 linearly independent columns.

Question 7 |

A function y(t) satisfies the following differential equation :

\frac{dy(t)}{dt}+y(t)=\delta (t)

where \delta(t) is the delta function. Assuming zero initial condition, and denoting the unit step function by u(t), y(t) can be of the form

\frac{dy(t)}{dt}+y(t)=\delta (t)

where \delta(t) is the delta function. Assuming zero initial condition, and denoting the unit step function by u(t), y(t) can be of the form

e^{t} | |

e^{-t} | |

e^{t} u(t) | |

e^{-t} u(t) |

Question 7 Explanation:

Taking (L.T.) on both sides

Y(s)(s+1)=1

\therefore \;\; Y(s)=\frac{1}{s+1}

Taking inverse laplace transform

y(t)=e^{-t}u(t)

Y(s)(s+1)=1

\therefore \;\; Y(s)=\frac{1}{s+1}

Taking inverse laplace transform

y(t)=e^{-t}u(t)

Question 8 |

The equivalent circuits of a diode, during forward biased and reverse biased
conditions, are shown in the figure.

If such a diode is used in clipper circuit of figure given above, the output voltage v_0 of the circuit will be

If such a diode is used in clipper circuit of figure given above, the output voltage v_0 of the circuit will be

A | |

B | |

C | |

D |

Question 8 Explanation:

V_P=\frac{10}{10+10}\times 10sin\omega t

\;\; =5sin\omega t

Since maximum voltage across at the point P may be 5 V, hence voltage across the diode always will be less than or equal to zero. So it will be reversed always.

\therefore V_0=\frac{10}{10+10}\times 10sin\omega t=5sin\omega t

Question 9 |

Two 8-bit ADCs, one of single slope integrating type and other of successive
approximate type, take T_A \; and \; T_B times to convert 5 V analog input signal to
equivalent digital output. If the input analog signal is reduced to 2.5 V, the
approximate time taken by the two ADCs will respectively, be

T_{A},T_{B} | |

T_{A}/2,T_{B} | |

T_{A},T_{B}/2 | |

T_{A}/2,T_{B}/2 |

Question 9 Explanation:

Single slope integrating type ADC utilize digital counter techniques to measure time required for a voltage ramp to rise from zero to the input voltage.

If conversion time for input voltage 5V=T_A

So, conversion time for input voltage 2.5V=T_A/2

Conversion time in successive type ADC does not depend on input voltage. So, conversion time for input voltage 2.5V is also T_B.

If conversion time for input voltage 5V=T_A

So, conversion time for input voltage 2.5V=T_A/2

Conversion time in successive type ADC does not depend on input voltage. So, conversion time for input voltage 2.5V is also T_B.

Question 10 |

An input device is interfaced with Intel 8085A microprocessor as memory mapped
I/O. The address of the device is 2500H. In order to input data from the device
to accumulator, the sequence of instructions will be

LXI H, 2500H MOV A, M | |

LXI H, 2500H MOV M, A | |

LHLD 2500H MOV A, M | |

LHLD 2500H MOV M, A |

There are 10 questions to complete.