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Question

Which of the following statements are correct :

(a) ASK gives a maximum probability of error.

(b) FSK is also known as ON-OFF keying.

(c) In DPSK, no synchronous carrier is needed at the receiver.

(d) Probability of Error in DPSK is less than BPSK.

This question was previously asked in
UGC NET 2016 Paper 3 Electronic Science Question Paper (10-Jul-2016)
The correct answer is

(a) and (c) are correct.

 Rank the schemes by error performance first — that settles two statements at once. With Eb the energy per bit and N0 the noise density,

SchemeProbability of error
BPSK (coherent)\(Q\!\left(\sqrt{2E_b/N_0}\right)\)
DPSK\(\tfrac{1}{2}e^{-E_b/N_0}\)
Coherent FSK\(Q\!\left(\sqrt{E_b/N_0}\right)\)
ASK / OOK\(Q\!\left(\sqrt{E_b/N_0}\right)\), worst in practice

Statement (a) — correct. ASK has the poorest error performance of the group. Its decision threshold sits midway between the two levels and must be set relative to an absolute amplitude, so it is vulnerable both to noise and to any fading that shifts the received level. BPSK, by contrast, places its two signals as far apart as possible (antipodal), giving a 3 dB advantage.

Statement (c) — correct. DPSK encodes information in the change of phase between successive bits, so the receiver compares each symbol with the previous one instead of with a locally regenerated carrier. No carrier recovery loop is needed, which is the whole point of the scheme — a much simpler, non-coherent receiver.

Statement (b) — wrong. On-off keying is another name for ASK, not FSK. In OOK the carrier is switched on for a 1 and off for a 0; FSK keeps the amplitude constant and switches between two frequencies.

Statement (d) — wrong, and it is the reverse of the truth. DPSK performs about 1 dB worse than coherent BPSK, because a differential detector must make its decision using a noisy previous symbol as the reference; noise therefore enters twice, and errors tend to occur in pairs. The simplicity of DPSK is bought at a small penalty in error rate.

Step — assemble. Only (a) and (c) are correct, which is option 2.

Hence, the correct statements are (a) and (c).

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Similar Questions

  1. Which of the following digital modulation systems support high bit rate ?

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    ST1 : High noise immunity
    ST2 : Large bandwidth

  3. In coherent binary FSK system the orthogonal sinusoidal signals of frequency 20 kHz and 50 kHz are used to represent '0' and '1' respectively. The maximum possible bit interval is:


Important Questions from Introduction To Digital Modulation

  1. In digital data transmission, a line code should have the following properties

    (A) Transmission bandwidth should be as small as possible

    (B) Transmitted power should be as high as possible

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  2. If the in‐phase and quadrature components in an M‐ary PSK system are permitted to be independent, then this scheme becomes a:

  3. Which of the following statements are correct?

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    B. M‐ary PSK system considers 'M' different phases in the range 'π/2'.

    C. In M‐ary modulation scheme, only coherent detection is possible.

    D. M‐ary QAM scheme uses 'm 2' carrier signals having the same frequency.

    Choose the correct answer from the options given below:

  4. In digital communication, Inter Symbol Interference (ISI) is a form of distortion where one symbol interferes with subsequent symbols. An eye diagram is used to study the extent of ISI in a communication channel. Which of the following is FALSE?

  5. The data rate of QPSK is ______ as that of BPSK for the same symbol rate.

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