Question

Consider a discrete memoryless source with an alphabet of four source symbols. \(s(t)\) is a multi-level \((-1,0,+1,+2)\) signal representing a long sequence of random symbols from the above source which is generating \(10^4\) symbols per second.
Which of the following options is the correct value of equivalent Nyquist bandwidth of \(s(t)\)?

• A. \(10\text{ kHz}\)
• B. \(64\text{ kHz}\)
• C. \(5\text{ kHz}\)
• D. \(20\text{ kHz}\)

Hint:

The Nyquist bandwidth of a baseband signal is half the symbol rate, regardless of how many amplitude levels each symbol has.

Answer 1

The Correct Option is C

Step 1: Recall the Nyquist minimum bandwidth theorem.
For an M-ary baseband pulse signal transmitted at a symbol (baud) rate of \(R_s\) symbols per second, the minimum bandwidth needed to send it without inter-symbol interference is
\[ B_{\text{Nyquist}}=\frac{R_s}{2} \]
This result depends only on the symbol rate, not on how many levels \(M\) each symbol can take. The number of amplitude levels changes how much information each symbol carries, but it does not change the minimum pulse bandwidth needed at a fixed symbol rate.

Step 2: Identify the symbol rate from the question.
The source produces \(10^4\) symbols per second, so
\[ R_s=10^4\text{ symbols/s} \]
The fact that each symbol is one of four levels \((-1,0,+1,+2)\) tells us this is a 4-ary signal, but as noted in Step 1, this does not enter the Nyquist bandwidth formula directly.

Step 3: Apply the formula.
\[ B_{\text{Nyquist}}=\frac{R_s}{2}=\frac{10^4}{2}=5\times10^3\text{ Hz}=5\text{ kHz} \]

Step 4: Rule out the other options.
\(10\text{ kHz}\) is simply the symbol rate itself, mistaking the bandwidth for \(R_s\) rather than \(R_s/2\). \(20\text{ kHz}\) doubles the symbol rate rather than halving it. \(64\text{ kHz}\) does not follow from either the symbol rate or the number of levels in this problem at all.

Final Answer:
\[ \boxed{B_{\text{Nyquist}}=5\text{ kHz}} \]