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Consider a discrete memoryless source with an alphabet of four source symbols. s(t) is a multi-level (-1, 0, +1, +2) ...

 

Q. Consider a discrete memoryless source with an alphabet of four source symbols. The signal s(t) has four levels (-1, 0, +1, +2) and generates 104 symbols per second. Find the equivalent Nyquist bandwidth.

(A) 10 kHz (B) 64 kHz (C) 5 kHz (D) 20 kHz 

Given

  • Number of signal levels: M = 4
  • Symbol rate: Rs = 104 symbols/s
  • Signal levels: -1, 0, +1, +2

Calculate Nyquist bandwidth

For ideal Nyquist signaling, the minimum bandwidth can be expressed in terms of the symbol rate:

B = Rs / 2

B = 10,000 / 2 = 5,000 Hz

Correct answer

Option (C) — 5 kHz

Using the symbol-rate definition of Nyquist bandwidth, B = Rs/2 gives 5 kHz.

Don't confuse symbol rate with bit rate

The bit rate is Rb = Rs*log2(M) = 20 kbps, while the symbol rate is 10 kbaud. These are different quantities because each symbol represents 2 bits.

Symbol-rate form: B = Rs/2 = 5 kHz

Bit-rate form: B = Rb/2 = 10 kHz

The formulas use different rate conventions. For this question, use the symbol-rate convention if that is what your course intends by equivalent Nyquist bandwidth.

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