A transmission line of characteristic impedance 50 W is terminated in a load
impedance Z_{L}. The VSWR of the line is measured as 5 and the first of the
voltage maxima in the line is observed at a distance of λ/4 from the load. The
value of Z_{L} is

**A. ** 1Ω

**B. ** 250Ω

**C. ** (19.23 + j46.15)Ω

**D. ** (19.23 - j46.15)Ω

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

**Explaination / Solution: **

Since voltage maxima is observed at a distance of λ/4 from the load and we know that the separation between one maxima and minima equals to λ/4 so voltage minima will be observed at the load, Therefore load can not be complex it must be pure resistive.

Since voltage maxima is observed at a distance of λ/4 from the load and we know that the separation between one maxima and minima equals to λ/4 so voltage minima will be observed at the load, Therefore load can not be complex it must be pure resistive.

also R_{L} = R_{0}/s (since voltage maxima is formed at the load)

R_{L }= (50/5)Ω

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x(t) is a stationary random process with auto-correlation function. This process is passed through the system shown below. The
power spectral density of the output process y(t) is

**A. ** (4π^{2}f^{2} + 1) exp (-πf^{2})

**B. ** (4π^{2}f^{2} - 1) exp (-πf^{2})

**C. ** (4π^{2}f^{2} + 1) exp (-πf)

**D. ** (4π^{2}f^{2} - 1) exp (-πf)

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

**Explaination / Solution: **

The given circuit can be simplified as

Power spectral density of output is

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Suppose that the modulating signal is and the carrier signal is , which one of the following is a conventional AM signal without over-modulation
**A. ** x(t) = A_{C }m(t)
cos (2𝜋f_{C} t)

**B. ** x(t) = A_{C }[1 + m(t)] cos (2𝜋f_{C} t)

**C. ** x(t) = A_{C }cos (2𝜋f_{C} t) +AC /4 m(t) cos (2𝜋fC t)

**D. ** x(t) = AC cos (2𝜋fC t) cos (2𝜋f_{m} t)+AC sin (2𝜋fm t) sin (2𝜋fC t)

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

**Explaination / Solution: **

No Explaination.

No Explaination.

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Consider an angle modulated signal The average power of x(t) is

**A. ** 10 W

**B. ** 18 W

**C. ** 20 W

**D. ** 28 W

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

**Explaination / Solution: **

No Explaination.

No Explaination.

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If the scattering matrix [S ] of a two port network is then the network is

**A. ** lossless and reciprocal

**B. ** lossless but not reciprocal

**C. ** not lossless but reciprocal

**D. ** neither lossless nor reciprocal

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

**Explaination / Solution: **

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A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1Ω/m . If the line is distortion less, the attenuation constant(in Np/m) is

**A. ** 500

**B. ** 5

**C. ** 0.014

**D. ** 0.002

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

**Explaination / Solution: **

For distortion less transmission line characteristics impedance

For distortion less transmission line characteristics impedance

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A message signal m(t) = cos 200πt + 4cosπt modulates the carrier
c(t) = coe 2πf_{c}t where f_{c} = 1MHz to produce an AM signal. For demodulating
the generated AM signal using an envelope detector, the time constant RC of
the detector circuit should satisfy

**A. ** 0.5 ms < RC < 1 ms

**B. ** 1 μs << RC < 0.5 ms

**C. ** RC << μs

**D. ** RC >> 0.5 ms

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

**Explaination / Solution: **

Highest frequency component in m(t) is

Highest frequency component in m(t) is

Carrier frequency fc = 1 MHz

For Envelope detector condition

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In the circuit shown below, the network N is described by the following Y matrix: the voltage gain (V_{2}/V_{1})is

**A. ** 1/90

**B. ** -1/90

**C. ** -1/99

**D. ** -1/11

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

**Explaination / Solution: **

From given admittance matrix we get

From given admittance matrix we get

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A four-phase and an eight-phase signal constellation are shown in the figure
below.

**A. ** r1 = 0.707d, r2 = 2.782d

**B. ** r1 = 0.707d, r2 = 1.932d

**C. ** r1 = 0.707d, r2 = 1.545d

**D. ** r1 = 0.707d, r2 = 1.307d

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

**Explaination / Solution: **

Four phase signal constellation is shown below

For the constraint that the minimum distance between pairs of signal points be
d for both constellations, the radii r_{1}, and r_{2} of the circles are

Four phase signal constellation is shown below

d^{2} = r_{1}^{2} + r_{1}^{2}

d^{2} = 2r_{1}^{2}

r_{1} = d/√2 = 0.707d

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A four-phase and an eight-phase signal constellation are shown in the figure
below.

**A. ** 11.90 dB

**B. ** 8.73 dB

**C. ** 6.79 dB

**D. ** 5.33 dB

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

**Explaination / Solution: **

Here P_{e} for 4 PSK and 8 PSK is same because P_{e} depends on d . Since P_{e} is same,
d is same for 4 PSK and 8 PSK.

Assuming high SNR and that all signals are equally probable, the additional
average transmitted signal energy required by the 8-PSK signal to achieve the
same error probability as the 4-PSK signal is

Here P

Additional Power SNR

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