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# Multiple choice question for engineering

## Set 1

1. For random process X = 6 and Rxx (t, t+t) = 36 + 25 exp(|t|). Consider following statements:
(i) X(t) is first order stationary.
(ii) X(t) has total average power of 36 W.
(iii) X(t) is a wide sense stationary.
(iv) X(t) has a periodic component.
Which of the following is true?
a) 1, 2, and 4
b) 2, 3, and 4
c) 2 and 3
d) only 3

Answer: c [Reason:] X Constant and Rxx() is not a function of t, so X(t) is a wide sense stationary. So (i) is false & (iii) is true. Pxx = Rxx(0) 36+25 = 61. Thus (ii) is false if X(t) has a periodic component, then RXX(t) will have a periodic component with the same period. Thus (iv) is false.

2. White noise with power density No/2 = 6 microW/Hz is applied to an ideal filter of gain 1 and bandwidth W rad/s. If the output’s average noise power is 15 watts, the bandwidth W is
a) 2.5 x 10 (-6)
b) 2.5p x 10 (-6)
c) 5 x 10 (-6)
d) p5 x 10 (-6)

Answer: b [Reason:] Pyy = 1/2p Integral(?xx(w) |H(w)|^2 dw ) from plus infinity to minus infinity. Hence solve for W.

(Q.3-Q.4) The two-level semi-random binary process is defined by X(t) A or -A where (n 1)T < t < nt and the levels A and -A occur with equal probability. T is a positive constant and n = 0, ±1, ±2.

3. The mean value E[X(t)] is
a) 1/2
b) 1/4
c) 1
d) 0

Answer: d [Reason:] E[X(t)] = A P(A) – (-A)P(-A) which is zero.

4. The auto correlation Rxx(t1 = 0.5T, t2 = 0.7T) will be
a) 1
b) 0
c) A x A
d) 0.5 (A x A)

Answer: c [Reason:] Here Rxx is AxA if both t1 and t2 are different and zero if they are same. Hence the answer is AxA.

5. Air craft of Jet Airways at Ahmedabad airport arrive according to a Poisson process at a rate of 12 per hour. All aircraft are handled by one air traffic controller. If the controller takes a 2 – minute coffee break, what is the probability that he will miss one or more arriving aircraft?
a) 0.33
b) 0.44
c) 0.55
d) 0.66

Answer: a [Reason:] P (miss/or more aircraft) = 1 – P(miss 0) = 1 – P(0 arrive).

6. A stationary random process X(t) is applied to the input of a system for which h(t) = u(t) t2 e(-8t). If E[X(t)] = 2, the mean value of the system’s response Y(t) is
a) 1/128
b) 1/64
c) 3/128
d) 1/32

Answer: c [Reason:] The mean value of Y(t) is integral of h(t)dt over negative infinity to positive infinity which gives the value equal to 3/128.

7. A random process is defined by X(t) + A where A is continuous random variable uniformly distributed on
(0,1). The auto correlation function and mean of the process is
a) 1/2 & 1/3
b) 1/3 & 1/2
c) 1 & 1/2
d) 1/2 & 1

Answer: b [Reason:] E[X(t)X(t+t)] = 1/3 and E[X(t)] = 1/2 respectively.

(Q.8-Q.9) The auto correlation function of a stationary ergodic random process is shown below.

8. The mean value E[X(t)] is
a) 50
b) sqrt(50)
c) 20
d) sqrt(20)

Answer: d [Reason:] Lim |t| tends to infinity, Rxx(t) = 20 = X2. hence X is sqrt(20).

9. The E[X2(t)] is
a) 10
b) sqrt(10)
c) 50
d) sqrt(50)

Answer: c [Reason:] Rxx(0) = X2 = 50.

10. The variance is
a) 20
b) 50
c) 70
d) 30

Answer: d [Reason:] Here X = 0, y = 0, Rxx(0) = 5, Ryy(0) = 10. The only value that satisfies all the given conditions is 30.

## Set 2

1. Which of the following isn’t a type of rectifier?
a) Precision Half-wave Rectifier
b) Bridge Rectifier
c) Peak Rectifier
d) None of the mentioned

Answer: d [Reason:] All of the mentioned are different types of a rectifier.

2. For a half wave or full wave rectifier the Peak Inverse Voltage of the rectifier is always
a) Greater than the input voltage
b) Smaller than the input voltage
c) Equal to the input voltage
d) Greater than the input voltage for full wave rectifier and smaller for the half wave rectifier

Answer: b [Reason:] The peak input voltage is smaller than the input voltage due to the presence of diode(s). A single diode reduces the output voltage by approximately 0.7V.

3. For a half-wave rectifier having diode voltage VD and supply input of VI, the diode conducts for π – 2Θ, where Θ is given by
a) tan -1 VD/VI
b) tan-1 VD/VI – VI
c) sin-1 VD/VI
d) sin-1 VD/VI – VI

Answer: c [Reason:] The diode doesn’t conducts when VD ≥VI . Hence Θ = sin-1 (D/VI).

4. Bridge rectifier is an alternative for
a) Full wave rectifier
b) Peak rectifier
c) Half wave rectifier
d) None of the mentioned

Answer: a [Reason:] Bridge rectifier is a better alternative for a full wave rectifier.

5. Which of the following is true for a bridge rectifier?
a) The peak inverse voltage or PIV for the bridge rectifier is lower when compared to an identical center tapped rectifier
b) The output voltage for the center tapped rectifier is lower than the identical bridge rectifier
c) A transistor of higher number of coil is required for center tapped rectifier than the identical bridge rectifier
d) All of the mentioned

Answer: d [Reason:] All of the given statements are true for a bridge rectifier.

6. The diode rectifier works well enough if the supply voltage is much than greater than 0.7V. For smaller voltage (of few hundreds of millivolt) input which of the following can be used?
a) Superdiode
b) Peak rectifier
c) Precision rectifier
d) None of the mentioned

Answer: a [Reason:] For the supply voltages less than 0.7V super diodes are used.

7. A simple diode rectifier has ‘ripples’ in the output wave which makes it unsuitable as a DC source. To overcome this one can use
a) A capacitor in series with a the load resistance
b) A capacitor in parallel to the load resistance
c) Both of the mentioned situations will work
d) None of the mentioned situations will work

Answer: b [Reason:] A capacitor is parallel with a resistor can only makes ripples go away. Series connection will become equal to an open circuit once the capacitor is fully charged.

8. Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value Vp = 100 V. Let the load resistance R = 10 kΩ. Calculate the fraction of the cycle during which the diode is conducting
a) 1.06 %
b) 2.12 %
c) 3.18%
d) 4.24%

Answer: c [Reason:] w Δt ~ √(2Vr/Vp) Θ = √(2 X 2/100) Θ = 0.2 rad or 3.18% of the cycle

(Q.9-Q.10) The op amp in the precision rectifier circuit is ideal with output saturation levels of ±12 V. Assume that when conducting the diode exhibits a constant voltage drop of 0.7 V.

9. Find V when VI is -1V.
a) 0V
b) 0.7V
c) 1V
d) 1.7V

Answer: a [Reason:] VI = -1v Vo = 1v VA = 1.7v V = 0v Virtual gnd as negative feedback is closed through R. VI > 0 D1 conducts D2 cutoff. VI < 0 D2 conducts D1 cutoff. V0 VI = -1.

10. Find V0 when VI is 2V.
a) 0V
b) 0.7V
c) 1V
d) 1.7V

Answer: a [Reason:] VI = 2v Vo = 0v VA = -0.7v V = 0v Virtual gnd as negative feedback is closed through R. VI > 0 D1 conducts D2 cutoff. VI < 0 D2 conducts D1 cutoff. V0 VI = -1.

## Set 3

In the problems assume the parameter given in following table. Use the temperature T= 300 K unless otherwise stated.

1. In germanium semiconductor material at T 400 K the intrinsic concentration is (x 10^14 per cc)
a) 26.8
b) 18.4
c) 8.5
d) 3.6

2. The intrinsic carrier concentration in silicon is to be no greater than ni = 1 x 10^12 cc. The maximum temperature allowed for the silicon is ( Eg = 1.12 eV)
a) 300 K
b) 360 K
c) 382 K
d) 364 K

3. Two semiconductor material have exactly the same properties except that material A has a bandgap of 1.0
eV and material B has a bandgap energy of 1.2 eV. The ratio of intrinsic concentration of material A to that of
material B is
a) 2016
b) 47.5
c) 58.23
d) 1048

4. In silicon at T = 300 K the thermal-equilibrium concentration of electron is n0 = 5 x 10^4 cc. The hole concentration is
a) 4.5 x 1015 cc
b) 4.5 x 1015 m3
c) 0.3 x 10-6 cc
d) 0.3 x 10-6 m3

Answer: a [Reason:] ni x ni = no x po.

5. In silicon at T = 300 K if the Fermi energy is 0.22 eV above the valence band energy, the value of p0 is
a) 2 x 1015 cm3
b) 1015 cm3
c) 3 x 1015 cm3
d) 4 x 1015 cm3

6. The thermal-equilibrium concentration of hole p0 in silicon at T = 300 K is 1015 cm3. The value of n0 is
a) 3.8 x 108 cm3
b) 4.4 x 104 cm3
c) 2.6 x 104 cm3
d) 4.3 x 108 cm3

7. In germanium semiconductor at T 300 K, the acceptor concentrations is Na 1013 cm3 and donor concentration is Nd 0. The thermal equilibrium concentration p0 is
a) 2.97 x 109 cm3
b) 2.68 x 1012 cm3
c) 2.95 x 1013 cm3
d) 2.4 cm3

8. A silicon sample doped n type at 10^18 cm3 have a resistance of 10 ohm. The sample has an area of 10^(-6) cm2 and a length of 10 µm . The doping efficiency of the sample is (µn = 800 cm2/V-s )
a) 43.2%
b) 78.1%
c) 96.3%
d) 54.3%

9. Six volts is applied across a 2 cm long semiconductor bar. The average drift velocity is 104 cms. The electron mobility is
a) 4396 cm2/V-s 2
b) 3 x 104 cm2/V-s
c) 6 x 104 cm2V-s
d) 3333 cm2/V-s

10. A particular intrinsic semiconductor has a resistivity of 50 (ohm-cm) at T = 300 K and 5 (ohm-cm) at T = 330 K. If change in mobility with temperature is neglected, the bandgap energy of the semiconductor is
a) 1.9 eV
b) 1.3 eV
c) 2.6 eV
d) 0.64 eV

## Set 4

1. A cosine wave voltage signal has a 10V RMS value and 60Hz frequency. Also at time, t=0, the value of the voltage signal is equal to its RMS value. Which of the following is the correct mathematical representation of the voltage signal?
a) 10 cos(60t)
b) 10 cos (120πt)
c) 14.14 cos(60t + π/4)
d) 14.14 cos(120πt + π/4)

Answer: d [Reason:] Only equation 14.14 cos(120πt + π/4) satisfies all the given parameters.

2. Which of the following is a characteristic of digital signal?
a) It takes quantized value
b) Its waveform is a continuous function
c) The maximum number of signals that can be produced by N bits is 2N-1
d) There is no loss of value after converting an analog signal to digital signal

Answer: a [Reason:] Digital signal is non continuous and has discrete sets of possible value which it can take.

3. Consider an N-bits ADC (Analog to Digital Converter) whose analog input varies from 0 to Vmax, then which of the following is not true?
a) The least significant bit correspond to a change of Vmax/2N -1 in the analog signal
b) The resolution of the ADC is Vmax/2N -1
c) The maximum error in the conversion (or quantization error) is Vmax/2(2N -1)
d) None of the mentioned

Answer: d [Reason:] None of the statements are true.

4. In compact disc (CD) audio technology, the signal is sampled at 44kHz. Each sample is represented by 16bits. What is the speed of the system in bits/second?
a) 1.34 bits/second
b) 2,750 bits/second
c) 704,000 bits/second
d) 1,441,792,000 bits/second

Answer: c [Reason:] 44000 X 16 = 704000 bits/s

5. An electrical signal can be represented in either Thevenin form or Norton From. However, Thevenin representation is preferred when
a) The load resistance is very large as compared to the source resistance
b) The load resistance is very low compared to the source resistance
c) There is no preferred case.
d) Both of the cases mentioned are the preferred case

Answer: a [Reason:] a is preferred in Thevenin’s case and b in Norton’s case

6. An electrical signal can be represented in either Thevenin form or Norton From. However, Norton representation is preferred when
a) The load resistance is very large as compared to the source resistance
b) The load resistance is very low as compared to the source resistance
c) There is no preferred case
d) Both of the cases mentioned are the preferred case

Answer: b [Reason:] a is preferred in Thevenin’s case and b in Norton’s case

Consider the DAC (digital to Analog Converter) shown below. Where b1, b2….bn can be either 0 or 1. (Q.7 to Q.9)

7. What will be the value of the output current io, if the digital signal is in the form of b1b2b3…bn
a) Vref/2R (b1/2+ b2/4 + b3/8….+bn/(2N))
b) 2Vref/R (b1/2+ b2/4 + b3/8….+bn/(2N))
c) Vref/R (b1/2+ b2/4 + b3/8….+bn/(2N))
d) none of the above

Answer: c [Reason:] It is the total current that will flow individual resistances.

8. Which is the Most Significant Bit and Least Significant Bit in this case?
a) MSB: bN, LSB: bN
b) MSB: bN, LSB: b1
c) MSB: b1, LSB: bN
d) MSB: b1, LSB: b1

Answer: b [Reason:] LSB will have maximum resistance (hence least contribution to the total current) and MSB will have least resistance (Hence maximum contribution to the total current).

9. For vref = 10V, n = 6, and R= 5000 ohm, which of the following is true?
a) The maximum value of the output current is 1.9375 mA
b) The change in the output current if the LSB is changed from 0 to 1 is 0.03125 mA
c) The change in the output current is the MSB is change from 0 to 1 is 0.5 mA
d) None of the mentioned

Answer: b [Reason:] According to the figure, the LSB is given by 10 / 50006 mA.

10. What is the binary representation of 57?
a) 0011 1101
b) 0101 1001
c) 0111 1001
d) 0011 1001

Answer: d [Reason:] The binary equivalent of 57 is 00111001.

## Set 5

1. i(t) = ?

a) 20 cos (300t + 68.2) A
b) 20 cos(300t – 68.2) A
c) 2.48 cos(300t + 68.2) A
d) 2.48 cos(300t – 68.2) A

2. Vc(t) = ?

a) 0.89 cos (1000t – 63.43) V
b) 0.89 cos (1000t + 63.43) V
c) 0.45 cos (1000t + 26.57) V
d) 0.45 cos (1000t – 26.57) V

3. Vc(t) = ?

a) 2.25 cos (5t + 150) V
b) 2.25 cos (5t – 150) V
c) 2.25 cos (5t + 140.71) V
d) 2.25 cos (5t – 140.71) V

4. i(t) = ?

a) 2 sin (2t 5.77) A
b) cos (2t 84.23) A
c) 2 sin (2t 5.77) A
d) cos (2t 84.23) A

5. In the bridge shown, Z1 = 300 ohm, Z2 = 400 – j300 ohm, Z3 = 200 + j100 ohm. The Z4 at balance is

a) 400 + j300 ohm
b) 400 – j300 ohm
c) j100 ohm
d) -j900 ohm

Answer: b [Reason:] Use Z1 x Z4 = Z2 x Z3.

Circuit for Q.6 and Q.7

6. i1(t) = ?
a) 2.36 cos (4t 41.07) A
b) 2.36 cos (4t 41.07) A
c) 1.37 cos (4t 41.07) A
d) 2.36 cos (4t 41.07) A

7. i2(t) = ?
a) 2.04 sin (4t 92.13) A
b) 2.04 sin (4t 2.13) A
c) 2.04 cos (4t 2.13) A
d) 2.04 cos (4t 92.13) A

8. In a two element series circuit, the applied voltage and the resulting current are v(t) = 60 + 66 sin (1000t) V, i(t) = 2.3sin (1000t + 68.3) 3 A. The nature of the elements would be
a) R C
b) L C
c) R L
d) R R

Answer: a [Reason:] RC circuit causes a positive shift in the circuit.

9. P = 269 W, Q = 150 VAR (capacitive). The power in the complex form is
a) 150 – j269 VA
b) 150 + j269 VA
c) 269 – j150 VA
d) 269 + j150 VA