GATE Online Mock Exam Quiz!

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.

32 KHz

14 KHz

15 KHz

7 Khz

None Of these

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.

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.

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

Flip-flop

Schmitt Trigger

Monosatable Multivibrator

Bistable Multivibrator

None Of These

The circuit in the figure consists of two CMOS-NOR gates. A flip-flop is a circuit that can store one bit of information and has two stable states. The inputs of the CMOS-NOR gates are connected in such a way that they can function as a flip-flop. Therefore, the correct answer is flip-flop.

Explanation

The circuit in the figure consists of two CMOS-NOR gates. A flip-flop is a circuit that can store one bit of information and has two stable states. The inputs of the CMOS-NOR gates are connected in such a way that they can function as a flip-flop. Therefore, the correct answer is flip-flop.

3. 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

1400 b/s

2800 b/s

5600 b/s

3600 b/s

None Of These

Explanation

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

U (t + 1) – 2u(t – 1) + 2u(t – 2)

U(t – 1) – 2u(t + 1) + 2 u(t + 2)

U(t – 1) + 2u(t + 1) – 2u(t + 2)

None Of These

The given signal x(t) can be expressed as 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).

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.

None Of These

There will no pattern E Plane Only in H Plane

Explanation

There will no pattern E Plane Only in H Plane

6. Consider a system with the transfer function G(s) =

. Its damping ratio will be 0.5 when the value of K is

1/7

2/7

3

7

None Of these

Explanation

7. The system with the open loop transfer function G(s) H(s) =

has a gain margin & cross over frequency of

0 dB, 1 rad/sec

3.5 dB, 2 rad/sec

5 dB, 5 rad/sec

7 dB, 1 rad/sec

7 dB, 2 rad/sec

Explanation

8. The root locus of the system G(s) H(s) =

has the break-away point located at

(–0.680, 0)

(–3.786, 0)

(–0.5, 0)

(–0.880, 0)

(–0.450, 0

Explanation

9. An air filled rectangular waveguide has inner dimension of 3 cm m 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)

461 Ω

400 Ω

308 Ω

435 Ω

445 Ω

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.

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.