PCM Bit Rate Calculation for Video Signal

To determine the minimum bit rate required for transmitting a video signal using Pulse Code Modulation (PCM), we need to follow a few key steps. This involves understanding the relationship between signal bandwidth, sampling rate, quantization levels, and the final bit rate.

Understanding Key Concepts in PCM

• Bandwidth (B): This refers to the range of frequencies a signal occupies. For our video signal, the bandwidth is 4.5 MHz.
• Sampling Rate (\(f_s\)): According to the Nyquist sampling theorem, to accurately reconstruct an analog signal from its sampled version, the sampling rate must be at least twice the maximum frequency (or bandwidth) of the signal. That is, \(f_s \geq 2B\). For minimum bit rate, we consider the minimum sampling rate, \(f_s = 2B\).
• Quantization Levels (L): After sampling, the analog sample values are converted into digital values by assigning them to a finite number of discrete levels. The number of such levels is \(L\). In this question, \(L = 1024\).
• Bits per Sample (n): Each quantization level needs to be represented by a certain number of bits. If there are \(L\) quantization levels, then \(L = 2^n\), where \(n\) is the number of bits required per sample.
• Bit Rate (\(R_b\)): The bit rate is the total number of bits transmitted per second. It is calculated by multiplying the sampling rate by the number of bits per sample. So, \(R_b = f_s \times n\).

Step-by-Step Calculation of Minimum Bit Rate

Let's apply these concepts to the given problem:

Given:

• Bandwidth of the video signal, \(B = 4.5 \text{ MHz}\)
• Number of quantization levels, \(L = 1024\)

1. Calculate the Minimum Sampling Rate (\(f_s\))

According to the Nyquist theorem, the minimum sampling rate is twice the bandwidth:

\[f_s = 2B\]

\[f_s = 2 \times 4.5 \text{ MHz}\]

\[f_s = 9.0 \text{ MHz}\]

2. Determine the Number of Bits per Sample (n)

The number of quantization levels \(L\) is related to the number of bits per sample \(n\) by the formula \(L = 2^n\). We need to find \(n\):

\[L = 2^n\]

\[1024 = 2^n\]

To find \(n\), we can take the logarithm base 2 of 1024, or simply recognize powers of 2:

\[2^{10} = 1024\]

Therefore,

\[n = 10 \text{ bits/sample}\]

3. Calculate the Minimum Bit Rate (\(R_b\))

The minimum bit rate is the product of the minimum sampling rate and the number of bits per sample:

\[R_b = f_s \times n\]

\[R_b = 9.0 \text{ MHz} \times 10 \text{ bits/sample}\]

\[R_b = 9.0 \times 10^6 \text{ samples/second} \times 10 \text{ bits/sample}\]

\[R_b = 90 \times 10^6 \text{ bits/second}\]

\[R_b = 90 \text{ Mbps}\]

Conclusion

The minimum bit rate required for the transmission of the video signal with a bandwidth of 4.5 MHz and 1024 quantization levels is 90 Mbps.