Question:
A binary signal is transmitted at 10kbps using NRZ signaling. Find the minimum channel bandwidth required in KHz:
A binary signal is transmitted at 10kbps using NRZ signaling. Find the minimum channel bandwidth required in KHz:
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Always distinguish between minimum theoretical bandwidth (\(R_b/2\)) and the practical bandwidth required for principal main-lobe transmission which equals \(R_b\) for standard NRZ signaling.
Updated On: Jul 4, 2026
- \( 2.5 \)
- \( 5 \)
- \( 1.5 \)
- \( 20 \)
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The Correct Option is B
Solution and Explanation
Concept:
In baseband digital communication using Non-Return-to-Zero (NRZ) line coding:
• The minimum bandwidth required to transmit a signal without Inter-Symbol Interference (ISI) is defined by the Nyquist criterion.
• For a bit rate of \( R_b \), the absolute minimum theoretical bandwidth (Nyquist bandwidth) required is: \[ B_{\min} = \frac{R_b}{2} \]
Step 1: Substitute Given Values into Formula
The given transmission rate is: \[ R_b = 10\,\text{kbps} \] Applying the minimum bandwidth formula: \[ B_{\min} = \frac{10\,\text{kbps}}{2} = 5\,\text{kHz} \] Thus, the minimum channel bandwidth required is exactly 5 kHz, corresponding directly to Option (B).
• The minimum bandwidth required to transmit a signal without Inter-Symbol Interference (ISI) is defined by the Nyquist criterion.
• For a bit rate of \( R_b \), the absolute minimum theoretical bandwidth (Nyquist bandwidth) required is: \[ B_{\min} = \frac{R_b}{2} \]
Step 1: Substitute Given Values into Formula
The given transmission rate is: \[ R_b = 10\,\text{kbps} \] Applying the minimum bandwidth formula: \[ B_{\min} = \frac{10\,\text{kbps}}{2} = 5\,\text{kHz} \] Thus, the minimum channel bandwidth required is exactly 5 kHz, corresponding directly to Option (B).
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