Exam Questions Papers - Exam Paper 1 (6)

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Exam Questions Papers - Exam Paper 1

41.

What is the propagation constant for air filled wave guide with dimensions a = 1.59" and b = 0.795" at 4.95 GHz?

A.	0.6698
B.	0.7698
C.	0.503
D.	0.6598

Answer: Option A

Explanation:

12-05-37

Here, 12-05-37-1

a = 1.59'' = 40.386 mm = 4.04 cm

12-05-37-2

42.

A MOSFET has a threshold voltage of 1 V and oxide thickness of 500 x 10^-8 [ε_r = 3.9; ε₀ = 8.85 x 10^-14 F/cm, q = 1.6 x 10^-19 c]. The region under the gate is ion implanted for threshold voltage tailoring. The base and type of impant required to shift threshold voltage to - 1 V are __________ .

A.	8.6 x 10 11/cm2, p-type
B.	8.6 x 10 11/cm2, n-type
C.	0.86 x 109/cm2, p-type
D.	1.02 x 10 12/cm2, n-type

Answer: Option A

Explanation:

V_T(new) = V_T(odd) + 4-796-57-1

4-796-57-2

4-796-57-3 = 6.903 x 10^-8

4-796-57-4

4-796-57-5

f_B = - 8.6 x 10¹¹

^{The threshold voltage is always negative for p-channel and hence implant is of p-type.}

43.

Consider the system 6-02-18 with 6-02-18-1 and 6-02-18-2 where p and q are arbitrary real numbers. Which of the following statements about the controllability of the system is true?

A.	The system is completely state controllable for any non zero values p and q
B.	Only p = 0 and q = 0 result in controllability
C.	The system is uncontrollable for all values of p and q
D.	We cannot conclude about controllability from the given data

Answer: Option C

Explanation:

Use the condition of controllability.

44.

In a semiconductor material. The hole concentration is found to be 2 x 2.5 x 10¹⁵ cm^-3. If mobility of carriers is 0.13 m²/ v-s. Then find the current density if electric field intensity is 3.62 x 10^-19

A.	7.6237 x 10-4 A/cm2
B.	7.6237 x 10-5 A/cm2
C.	7.6237 x 10-3 A/cm2
D.	none of these

Answer: Option A

Explanation:

Current density J = σE

Where σ = conductivity

Given -> μ = 0.13 m²/v-s = 0.13 x 10⁴ cm²/V sec

P = 2.25 x 10¹⁵/cm³

We have, ∴ n_i = 1.5 x 10¹⁰

^{Also n.p. =}

^{∴ n = /p}

^{= (1.6 x 10-19 x 0.13 x 104 x 2.25 x 1015) x}

= (0.468) (4.5 x 10¹⁵)

σ = 2.106 x 10¹⁵ μ/cm

J = σE

∴ Current density = 2.106 x 10¹⁵ x 3.620 x 10^-19

= 7.6237 x 10^-4 A/m².

45.

In the circuit shown, in switch S is open for a long time and is closed at t = 0. The current i(t) for t ≥ 0+ is

7-04-32

A.	i(t) = 0.5 - 0.125e-1000t A
B.	i(t) = 1.5 - 0.125e-1000t A
C.	i(t) = 0.5 - 0.5e-1000t A
D.	i(t) = 0.375e-1000t A

Answer: Option A

Explanation:

i(f) = 0.5, i(i) = 0.75

i(t) = V_r + (i_i - i_j)e^-1/p = 0.5 - 0.125e^-1000t .

46.

A radar receiver has a noise figure of 10 dB at 300 K having a bandwidth of 2.5 MHz. The minimum power it can receive is

A.	3.45 x 10-15 W
B.	1.38 x 10-15 W
C.	7.5 x 10-15 W
D.	93.15 x 10-15 W

Answer: Option D

Explanation:

P_m = kT₀Δ_f(F - 1) = 93.15 x 10^-15 W.

47.

A rectangular waveguide, in dominant TE mode, has dimensions 10 cm x 15 cm. The cut off frequency is

A.	10 GHz
B.	1 GHz
C.	15 GHz
D.	25 GHz

Answer: Option B

Explanation:

For TE10 mode

2-767-73-1

48.

A certain 8 bit uniform quantization PCM system can accommodate a signal ranging from - 1 V to + V. The rms value of the signal is 16-227-2 V . The signal to quantization noise ratio is:

A.	30 dB
B.	46.91 dB
C.	40 dB
D.	49.92 dB

Answer: Option D

Explanation:

SNR)_dB = 1.76 + 6.02n

n = 8

Thus (SNR)_dB = 1.76 + 6.02 x 8 = 49.92 dB.

Exam Questions Papers - Exam Paper 1 (6)

Exam Questions Papers - Exam Paper 1

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