# GATE Online Mock Exam Quiz!

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Appliedjobz
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Quizzes Created: 23 | Total Attempts: 17,040
Questions: 9 | Attempts: 784  Settings  • 1.

### A Carrier is a frequency modulated with a sinusoidal signal of 2 kHz, resulting in a maximum frequency deviation of 5 kHz, the Bandwidth of the modulated signal.

• A.

32 KHz

• B.

14 KHz

• C.

15 KHz

• D.

7 Khz

• E.

None Of these

B. 14 KHz
Explanation
When a carrier signal is frequency modulated with a sinusoidal signal, the bandwidth of the modulated signal can be calculated using Carson's rule. According to Carson's rule, the bandwidth is equal to twice the sum of the maximum frequency deviation and the highest frequency component of the modulating signal. In this case, the maximum frequency deviation is 5 kHz and the highest frequency component of the modulating signal is 2 kHz. Therefore, the bandwidth of the modulated signal is 2 * (5 kHz + 2 kHz) = 14 kHz.

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• 2.

### A Certain telephone line bandwidth is 3.5 kHz, the data rate can be transmitted if we use binary signaling with the raised cosine pulses and a roll-off factor α = 0.25

• A.

1400 b/s

• B.

2800 b/s

• C.

5600 b/s

• D.

3600 b/s

• E.

None Of These

C. 5600 b/s
• 3.

### Consider a system with the transfer function G(s) = . Its damping ratio will be 0.5 when the value of K is

• A.

1/7

• B.

2/7

• C.

3

• D.

7

• E.

None Of these

A. 1/7
• 4.

### The root locus of the system G(s) H(s) =  has the break-away point located at

• A.

(–0.680, 0)

• B.

(–3.786, 0)

• C.

(–0.5, 0)

• D.

(–0.880, 0)

• E.

(–0.450, 0

D. (–0.880, 0)
• 5.

### Two identical and parallel dipole antenna are kept apart by a distance of λ/4 in the H-plane. They are fed with equal currents but the right most antenna has a phase shift of +90º.

• A.
• B.
• C.
• D.
• E.

None Of These

A.
Explanation
There will no pattern E Plane Only in H Plane

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• 6.

• A.

• B.

• C.

• D.

• E.

• 7.

### The signal x(t) shown in figure can be expressed as

• A.

U (t + 1) – 2u(t – 1) + 2u(t – 2)

• B.

U(t – 1) – 2u(t + 1) + 2 u(t + 2)

• C.

U(t – 1) – 2u(t + 1) + 2 u(t + 2)

• D.

U(t – 1) + 2u(t + 1) – 2u(t + 2)

• E.

None Of These

A. U (t + 1) – 2u(t – 1) + 2u(t – 2)
Explanation
The given signal x(t) can be expressed as u (t + 1) – 2u(t – 1) + 2u(t – 2).

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• 8.

### An air filled rectangular waveguide has inner dimension of 3 cm × 2 cm. The wave impedance of the TE20 mode of propagation in the waveguide at a frequency of 20 GHz is (free space impedance  η0 = 377Ω)

• A.

461 Ω

• B.

400 Ω

• C.

308 Ω

• D.

435 Ω

• E.

445 Ω

A. 461 Ω
Explanation
The wave impedance of the TE20 mode of propagation in a rectangular waveguide can be calculated using the formula Z = (2π/λ) * sqrt(1 - (λ/λc)^2), where λ is the wavelength and λc is the cutoff wavelength. The cutoff wavelength for the TE20 mode can be calculated using the formula λc = 2a/sqrt(4 - m^2), where a is the width of the waveguide and m is the mode number. Substituting the given values into the formulas, we can calculate the wave impedance to be approximately 461 Ω.

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• 9.

### The circuit in the figure has two CMOS-NOR gates. This circuit functions as a

• A.

Flip-flop

• B.

Schmitt Trigger

• C.

Monosatable Multivibrator

• D.

Bistable Multivibrator

• E.

None Of These Back to top