1
GATE ECE 2012
MCQ (Single Correct Answer)
+2
-0.6
A BPSK scheme operating over an AWGN channel with noise power spectral density of N02, uses equi-probable signals $$${s_1}\left( t \right) = \sqrt {{{2E} \over T}\,\sin \left( {{\omega _c}t} \right)} $$$
and $$${s_2}\left( t \right) = - \sqrt {{{2E} \over T}\,\sin \left( {{\omega _c}t} \right)} $$$

over the symbol interval, $$(0, T)$$. If the local oscillator in a coherent receiver is ahead in phase by 450 with respect to the received signal, the probability of error in the resulting system is

A
$$Q\left( {\sqrt {{{2E} \over {{N_0}}}} } \right)$$
B
$$Q\left( {\sqrt {{{E} \over {{N_0}}}} } \right)$$
C
$$Q\left( {\sqrt {{{E} \over {{2N_0}}}} } \right)$$
D
$$Q\left( {\sqrt {{{E} \over {{4N_0}}}} } \right)$$
2
GATE ECE 2012
MCQ (Single Correct Answer)
+2
-0.6
A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X =0) = 9/10, then the probability of error for an optimum receiver will be
A
7/80
B
63/80
C
9/10
D
1/10
3
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
A four phase and an eight-phase signal constellation are shown in the figure below. GATE ECE 2011 Communications - Noise In Digital Communication Question 10 English

For the constraint that the minimum distance between pairs of signal points be d for both constellations, the radii r1, and r2 of the circles are

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
4
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
A four phase and an eight-phase signal constellation are shown in the figure below. GATE ECE 2011 Communications - Noise In Digital Communication Question 9 English

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

A
11.90 dB
B
8.73 dB
C
6.79 dB
D
5.33 dB
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