GATE ECE (Electronics) Test: Modem Digital Modulation & Detection Techniques

BPSK is the most efficient of the three digital modulation, i.e. ASK, FSK and PSK because BPSK is used for high bit rates. Reason is also a true statement but not the correct explanation of assertion.

In BPSK, baud rate is half the bit rate due to which there is more effective utilization of the available bandwidth of the transmission channel. Given, bit rate = 34 Mbps

• Statement-1 is correct.
• BPSK has the best performance of all the three digital modulation techniques in presence of noise, it yields the minimum value of probability of error compared to ASK, FSK and PSK. Hence, statement-2 is false.
• Statement-3 is correct since binary FSK has highest system complexity while binary ASK has least system complexity.
• Binary ASK is demodulated using envelope detection while binary FSK and PSK are demodulated using coherent detection.

Hence, statement-4 is fals

More energy (or power) is contained in the main lobe of MSK (about 99% of the total energy or power) as compared to the QPSK (about 90% of the total energy or power).

The bit period of NRZ data stream is

In BPSK, each binary bit is a symbol, therefore symbol duration T_s = T_b = 1 μs.
∴ The bandwidth required for BPSK system is

In QPSK, we group two successive bits to form one symbol, therefore symbol duration,

The error probability of QPSK is same as BPSK.
The bit error probability is determined when the signal interference and Gaussian noise are applied to the input of the QPSK receiver. The interference is approximated by a sum of sine waves with constant amplitudes and uniformly distributed phases.

Minimum bandwidth required in BPSK = 2f_b.
Minimum bandwidth required in BFSK = 4f_b.
Hence, bandwidth of BPSK is lower than that of BFSK.

The error probability of QPSK is better (or lower) than that of BPSK due to which QPSK is used for very high bit rate data transmission.