Question:medium
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 \(s(t)\)?
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 \(s(t)\)?
Which of the following options is the correct value of equivalent Nyquist bandwidth of \(s(t)\)?
Show Hint
The Nyquist bandwidth of a baseband signal is half the symbol rate, regardless of how many amplitude levels each symbol has.
Updated On: Jul 20, 2026
- \(10\text{ kHz}\)
- \(64\text{ kHz}\)
- \(5\text{ kHz}\)
- \(20\text{ kHz}\)
Show Solution
The Correct Option is C
Solution and Explanation
Step 1: Find the time slot given to each symbol.
If $10^4$ symbols go out every second, each symbol occupies a time slot of
\[ T=\frac{1}{R_s}=\frac{1}{10^4}\text{ s}=100\ \mu\text{s} \]
Step 2: Recall how pulse duration links to bandwidth.
A pulse of duration $T$ shaped to avoid inter symbol interference needs a minimum bandwidth of
\[ B=\frac{1}{2T} \]
This comes from the idea that the smallest bandwidth channel that can still carry one independent pulse every $T$ seconds without the pulses blurring into each other is $1/(2T)$.
Step 3: Plug in the symbol period.
\[ B=\frac{1}{2\times100\ \mu\text{s}}=\frac{1}{200\ \mu\text{s}} \]
Step 4: Work out the number.
\[ B=\frac{1}{200\times10^{-6}}=5000\text{ Hz}=5\text{ kHz} \]
Step 5: Note that the four signal levels do not change this.
Whether each pulse takes one of $2$ levels or $4$ levels only changes how many bits ride on each pulse, not the physical pulse rate or its bandwidth requirement.
\[ \boxed{B=5\text{ kHz}} \]
If $10^4$ symbols go out every second, each symbol occupies a time slot of
\[ T=\frac{1}{R_s}=\frac{1}{10^4}\text{ s}=100\ \mu\text{s} \]
Step 2: Recall how pulse duration links to bandwidth.
A pulse of duration $T$ shaped to avoid inter symbol interference needs a minimum bandwidth of
\[ B=\frac{1}{2T} \]
This comes from the idea that the smallest bandwidth channel that can still carry one independent pulse every $T$ seconds without the pulses blurring into each other is $1/(2T)$.
Step 3: Plug in the symbol period.
\[ B=\frac{1}{2\times100\ \mu\text{s}}=\frac{1}{200\ \mu\text{s}} \]
Step 4: Work out the number.
\[ B=\frac{1}{200\times10^{-6}}=5000\text{ Hz}=5\text{ kHz} \]
Step 5: Note that the four signal levels do not change this.
Whether each pulse takes one of $2$ levels or $4$ levels only changes how many bits ride on each pulse, not the physical pulse rate or its bandwidth requirement.
\[ \boxed{B=5\text{ kHz}} \]
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