Question 1 |

The order of the differential equation \frac{d^{2}y}{dt^{2}}+(\frac{dy}{dt})^{3}+y^{4}=e^{-t} is

1 | |

2 | |

3 | |

4 |

Question 1 Explanation:

Highest derivative of differential equation is 2.

Question 2 |

The Fourier series of a real periodic function has only

(P) cosine terms if it is even

(Q) sine terms if it is even

(R) cosine terms if it is odd

(S) sine terms if it is odd

Which of the above statements are correct ?

(P) cosine terms if it is even

(Q) sine terms if it is even

(R) cosine terms if it is odd

(S) sine terms if it is odd

Which of the above statements are correct ?

P and S | |

P and R | |

Q and S | |

Q and R |

Question 2 Explanation:

The Fourier series of a real periodic function has
only cosine terms if it is even and only sine terms
if it is odd.

Question 3 |

A function is given by f(t)=sin^{2}t+cos2t. Which of the following is true ?

f has frequency components at 0 and \frac{1}{2\pi } Hz | |

f has frequency components at 0 and \frac{1}{\pi } Hz | |

f has frequency components at \frac{1}{2\pi } \; and \; \frac{1}{\pi}Hz | |

f has frequency components at 0, \frac{1}{2\pi } \; and \; \frac{1}{\pi}Hz |

Question 3 Explanation:

f(t)=\frac{1}{2}(1-\cos 2 t)+\cos 2 t

frequency components are

\begin{aligned} f_{1}&=0 \\ f_{2}&=\frac{\omega_{2}}{2 \pi}=\frac{2}{2 \pi}=\frac{1}{\pi} \mathrm{Hz} \end{aligned}

frequency components are

\begin{aligned} f_{1}&=0 \\ f_{2}&=\frac{\omega_{2}}{2 \pi}=\frac{2}{2 \pi}=\frac{1}{\pi} \mathrm{Hz} \end{aligned}

Question 4 |

A fully charged mobile phone with a 12 V battery is good for a 10 minute talktime. Assume that, during the talk-time the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the figure. How much energy does the battery deliver during this talk-time?

220J | |

12kJ | |

13.2kJ | |

14.4J |

Question 4 Explanation:

\begin{aligned} P &=V I \\ \text { Energy } &=P \cdot t=V \cdot|t=(V . t)| \\ I &=2 \mathrm{A} \quad(\text { given }) \\ V \cdot t &=\text { Area under } V-t \text { curve } \\ V \cdot t &=\left(\frac{1}{2} \times 2 \times 600\right)+(10 \times 600) \\ &=600+6000 \\ V \cdot t &=6600 \\ E &=(6600) \times 2=13200=13.2 \mathrm{kJ} \end{aligned}

Question 5 |

In an n-type silicon crystal at room temperature, which of the following can have a concentration of 4 \times 10^{19}cm^{-3}?

Silicon atoms | |

Holes | |

Dopant atoms | |

Valence electrons |

Question 6 |

The full form of the abbreviations TTL and CMOS in reference to logic families are

Triple Transistor Logic and Chip Metal Oxide Semiconductor | |

Tristate Transistor Logic and Chip Metal Oxide Semiconductor | |

Transistor Transistor Logic and Complementary Metal Oxide Semiconductor | |

Tristate Transistor Logic and Complementary Metal Oxide Silicon |

Question 6 Explanation:

TTL- Transistor-transistor logic

CMOS - Complementary Metal Oxide Semiconductor

CMOS - Complementary Metal Oxide Semiconductor

Question 7 |

The ROC of z -transform of the discrete time sequence x(n)=(\frac{1}{3})^{n}u(n)-(\frac{1}{2})^{n}u(-n-1) is

\frac{1}{3} \lt |z| \lt \frac{1}{2} | |

|z|\gt \frac{1}{2} | |

|z| \lt \frac{1}{3} | |

2 \lt |z| \lt 3 |

Question 7 Explanation:

x(n)=(1 / 3)^{n} u(n)-(1 / 2)^{n} u(-n-1)

(1 / 3)^{n} u(n) is right sided signal, so ROC will be

|z| \gt 1 / 3\;\; ...(i)

-(1 / 2)^{n} u(-n-1) is left sided signal so ROC will be

|z| \lt 1 / 2 \; \; ...(ii)

from (i) and (ii) we see that ROC of the function will be

1 / 3 \lt |z| \lt 1 / 2

(1 / 3)^{n} u(n) is right sided signal, so ROC will be

|z| \gt 1 / 3\;\; ...(i)

-(1 / 2)^{n} u(-n-1) is left sided signal so ROC will be

|z| \lt 1 / 2 \; \; ...(ii)

from (i) and (ii) we see that ROC of the function will be

1 / 3 \lt |z| \lt 1 / 2

Question 8 |

The magnitude plot of a rational transfer function G(s) with real coefficients is
shown below. Which of the following compensators has such a magnitude plot ?

Lead compensator | |

Lag compensator | |

PID compensator | |

Lead-lag compensator |

Question 9 |

A white noise process X(t) with two-sided power spectral density 1 \times 10^{-10}W/Hz is input to a filter whose magnitude squared response is shown below. The power of the output process Y(t) is given by

5 \times 10^{-7}W | |

1 \times 10^{-6}W | |

2 \times 10^{-6}W | |

1 \times 10^{-5}W |

Question 9 Explanation:

PSD of white noise =1 \times 10^{-10} \mathrm{W} / \mathrm{Hz} (\equiv k)

PSD of output

\begin{aligned} G_{0}(t) &=|H(t)|^{2} \cdot G_{i}(t) \\ &=k \cdot|H(t)|^{2} \end{aligned}

output noise power

\begin{aligned} &N_{0}=\int_{-f_{0}}^{+f_{0}} G_{0}(f) d f=k \times\left(\text { areaunder }|H(f)|^{2} \text { curve }\right) \\ &=k \times 2\left(\frac{1}{2} b h\right) \\ &=k f_{0} \times 1 \\ &=1 \times 10^{-10} \times 10 \times 10^{3} \\ &=10^{-6} \mathrm{W} \end{aligned}

PSD of output

\begin{aligned} G_{0}(t) &=|H(t)|^{2} \cdot G_{i}(t) \\ &=k \cdot|H(t)|^{2} \end{aligned}

output noise power

\begin{aligned} &N_{0}=\int_{-f_{0}}^{+f_{0}} G_{0}(f) d f=k \times\left(\text { areaunder }|H(f)|^{2} \text { curve }\right) \\ &=k \times 2\left(\frac{1}{2} b h\right) \\ &=k f_{0} \times 1 \\ &=1 \times 10^{-10} \times 10 \times 10^{3} \\ &=10^{-6} \mathrm{W} \end{aligned}

Question 10 |

Which of the following statements is true regarding the fundamental mode of
the metallic waveguides shown ?

Only P has no cutoff-frequency | |

Only Q has no cutoff-frequency | |

Only R has no cutoff-frequency | |

All three have cutoff-frequencies |

Question 10 Explanation:

P is coaxial line and support TEM wave

\therefore P has no cutoff frequency

Q and R are wave-guides and cutoff frequency each depends upon their dimensions.

\therefore P has no cutoff frequency

Q and R are wave-guides and cutoff frequency each depends upon their dimensions.

There are 10 questions to complete.