For a message signal and carrier of frequency f_{c} which of thefollowing represents a single side-band (SSB) signal ?

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The Nyquist sampling rate for the signal is given by

**A. ** 400 Hz

**B. ** 600 Hz

**C. ** 1200 Hz

**D. ** 1400 Hz

**Answer : ****Option C**

**Explaination / Solution: **

S(t) = sinc(500t) sinc(700t)

S(t) = sinc(500t) sinc(700t)

S(f ) is convolution of two signals whose spectrum covers f1 = 250 Hz and f2 = 350 Hz. So convolution extends

f = 25 + 350 = 600 Hz

Nyquist sampling rate

N = 2f = 2#600 = 1200 Hz

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X(t) is a stationary process with the power spectral density Sx(f)>0, for all f . The process is passed through a system shown below

**A. ** Sy (f ) > 0 for all f

**B. ** Sy (f ) = 0 for |f| > 1 kHz

**C. ** Sy (f ) = 0 for f = nf0, f0 = 2 kHz kHz, n any integer

**D. ** Sy (f ) = 0 for f = (2n + 1) f0 = 1 kHz, n any integer

**Answer : ****Option D**

**Explaination / Solution: **

Let Sy (f ) be the power spectral density of Y(t). Which one of the following statements is correct

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In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration(VSWR) on the 60 Ω line is

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Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is with power spectral density The lowpass filter is ideal with unity gain and cut-off frequency 1 MHz. Let Yk represent the random variable y(tk).

Yk = Nk , if transmitted bit bk = 0

Yk = a + Nk if transmitted bit bk = 1

Where Nk represents the noise sample value. The noise sample has a probability density function, (This has mean zero and variance 2/α^{2}). Assume transmitted bits to be equiprobable and threshold z is set to a/2
= 10^{-6} V.

The value of the parameter α
(in V^{-1}) is

Let response of LPF filters

Noise variance (power) is given as

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Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is with power spectral density The lowpass filter is ideal with unity gain and cut-off frequency 1 MHz. Let Yk represent the random variable y(tk).

Yk = Nk , if transmitted bit bk = 0

Yk = a + Nk if transmitted bit bk = 1

Where Nk represents the noise sample value. The noise sample has a probability density function, (This has mean zero and variance 2/α2). Assume transmitted bits to be equiprobable and threshold z is set to a/2 = 10-6 V.

The probability of bit error is

Probability of error is given by

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The block diagram of a frequency synthesizer consisting of a Phase Locked Loop (PLL) and a
divide-by-𝑁 counter (comprising ÷2, ÷4, ÷8, ÷16 outputs) is sketched below. The synthesizer is
excited with a 5 kHz signal (Input 1). The free-running frequency of the PLL is set to20 kHz.
Assume that the commutator switch makes contacts repeatedly in the order 1-2-3-4.

**A. ** 10 kHz, 20 kHz, 40 kHz, 80 kHz

**B. ** 20 kHz, 40 kHz, 80 kHz, 160 kHz

**C. ** 80 kHz, 40 kHz, 20 kHz, 10 kHz

**D. ** 160 kHz, 80 kHz, 40 kHz, 20 kHz

**Answer : ****Option A**

**Explaination / Solution: **

No Explaination.

The corresponding frequencies synthesized are:

No Explaination.

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The propagation constant of a lossy transmission line is (2 + 𝑗5) m-1 and its characteristic impedance is (50 + j0)𝜔 at 𝜔 = 106 rad s-1. The values of the line constants L, C, R, G are, respectively,

**A. ** L = 200 μH/m, C = 0.1 μH/m, R = 50 Ω/m, G = 0.02 S/m

**B. ** L = 250 μH/m, C = 0.1 μH/m, R = 100 Ω/m, G = 0.024S/m

**C. ** L = 200 μH/m, C = 0.2 μH/m, R = 100 Ω/m, G = 0.02 S/m

**D. ** L = 250 μH/m, C = 0.2 μH/m, R = 50 Ω/m, G = 0.04 S/m

**Answer : ****Option B**

**Explaination / Solution: **

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An antenna pointing in a certain direction has a noise temperature of 50K. The ambient
temperature is 290 K. The antenna is connected to a pre-amplifier that has a noise figure of 2
dB and an available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input
noise temperature T_{e} for the amplifier and the noise power P_{ao} at the output of the preamplifier,
respectively, are

**A. ** T_{e} = 169.36K and P_{ao} = 3.73 × 10^{-10}W

**B. ** T_{e} = 1.36K and P_{ao} = 3.73 × 10^{-10}W

**C. ** T_{e} = 182.5K and P_{ao} = 3.83 × 10^{-10}W

**D. ** T_{e} = 160.62K and P_{ao} = 4.6 × 10^{-10}W

**Answer : ****Option A**

**Explaination / Solution: **

No Explaination.

No Explaination.

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The electric field of a uniform plane wave travelling along the negative z direction is given by
the following equation:

**A. ** Linear, Circular (clockwise), −5dB

**B. ** Circular (clockwise), Linear, −5dB

**C. ** Circular (clockwise), Linear, −3dB

**D. ** Circular (anti clockwise), Linear, −3dB

**Answer : ****Option C**

**Explaination / Solution: **

No Explaination.

This wave is incident upon a receiving antenna placed at the origin and whose radiated electric
field towards the incident wave is given by the following equation:

The polarization of the incident wave, the polarization of the antenna and losses due to the
polarization mismatch are, respectively,

No Explaination.

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