A voice signal band limited to 3.4 kHz is sampled at 8 kHz and pulse code modulated using 64 quantization levels. Ten such signals are time division multiplexed using on 5-bit synchronising word. The minimum channel band width will be
520 kHz
Let's break down the calculation step-by-step to find the minimum channel bandwidth required for the given Pulse Code Modulated (PCM) and Time Division Multiplexed (TDM) signals.
The number of quantization levels is given as 64. The number of bits required to represent each sample is found using the formula:
\( \text{Number of levels} = 2^{\text{bits per sample}} \)
So, \( 64 = 2^n \), where \( n \) is the number of bits per sample.
Taking the logarithm base 2:
\( n = \log_2(64) = 6 \text{ bits per sample} \)
Each voice signal sample is represented by 6 bits.
The sampling rate for each voice signal is 8 kHz, which means 8000 samples are taken per second from each signal.
The bit rate generated by one voice signal is:
\( \text{Bit rate per signal} = \text{Sampling rate} \times \text{Bits per sample} \)
\( \text{Bit rate per signal} = 8000 \text{ samples/second} \times 6 \text{ bits/sample} = 48000 \text{ bits/second} \)
Ten such signals are time division multiplexed. In TDM, samples from different signals are combined into frames. Each frame contains one sample from each of the 10 signals, plus a synchronizing word.
The frame rate is equal to the sampling rate of the individual signals, which is 8 kHz (or 8000 frames per second).
Each frame contains:
Total bits per frame = (Bits from signals) + (Synchronizing word bits)
Total bits per frame = \( 60 \text{ bits} + 5 \text{ bits} = 65 \text{ bits} \)
| Component | Bits per Sample/Word | Number of Signals/Words per Frame | Total Bits per Frame |
|---|---|---|---|
| Each Voice Signal | 6 | 10 | \(10 \times 6 = 60\) |
| Synchronizing Word | 5 | 1 | \(1 \times 5 = 5\) |
| Total per Frame | - | - | 65 |
The total bit rate is the total number of bits transmitted per second. This is calculated by multiplying the bits per frame by the frame rate.
\( \text{Total bit rate} = \text{Bits per frame} \times \text{Frame rate} \)
\( \text{Total bit rate} = 65 \text{ bits/frame} \times 8000 \text{ frames/second} = 520,000 \text{ bits/second} \)
This can be written as 520 kbps (kilobits per second).
For a PCM signal, the minimum theoretical bandwidth required to transmit a bit rate of \( R_b \) bps using ideal Nyquist signaling is \( R_b/2 \) Hz. However, for common pulse shapes like Non-Return-to-Zero (NRZ), the bandwidth required up to the first null is equal to the bit rate \( R_b \). Given the options, the minimum channel bandwidth is typically considered equal to the total bit rate in such contexts, assuming a simple line coding scheme.
Minimum Channel Bandwidth \( \approx \) Total Bit Rate
Minimum Channel Bandwidth \( \approx 520,000 \text{ Hz} = 520 \text{ kHz} \)
Therefore, the minimum channel bandwidth will be 520 kHz.
| Parameter | Value | Influence on Bandwidth |
|---|---|---|
| Voice Bandwidth | 3.4 kHz | Sets the minimum sampling rate (Nyquist rate is 2 * 3.4 = 6.8 kHz, but 8 kHz is used, which is sufficient). |
| Sampling Rate | 8 kHz | Determines the frame rate and samples per second. |
| Quantization Levels | 64 | Determines the number of bits per sample (6 bits). |
| Number of Signals | 10 | Multiplies the bit rate contributed by individual signals per frame. |
| Synchronizing Word | 5 bits | Adds overhead bits to each frame. |
| Total Bit Rate | 520 kbps | Directly related to the minimum channel bandwidth. |
Pulse Code Modulation (PCM): This is a digital representation of an analog signal where the signal is sampled, quantized, and encoded. Key steps are:
Time Division Multiplexing (TDM): This technique combines multiple digital signals into one channel by dividing the transmission time into intervals, allowing each signal to use the channel sequentially for a short period. In the context of PCM, samples from different PCM signals are interleaved to form a single digital stream.
Channel Bandwidth: The range of frequencies that a communication channel can pass. For transmitting digital signals, the required bandwidth is related to the bit rate. Higher bit rates generally require wider bandwidths.