PCM rate for digitizing a colour TV baseband A colour TV video signal has 4.6 MHz bandwidth and is digitized using 10 bits per sample. Using basic PCM (Nyquist) sampling, what is the required bit rate in kilobits per second (kb/s)?

Introduction / Context:
Pulse Code Modulation (PCM) digitizes an analog baseband by sampling at or above twice its highest frequency component and quantizing each sample to a fixed number of bits. Determining the raw bitrate helps appreciate why video compression is necessary in practice.

Given Data / Assumptions:

• Video baseband bandwidth B = 4.6 MHz.
• Quantization = 10 bits per sample.
• Use Nyquist sampling: f_s = 2 * B for a low-pass baseband.
• No compaction, coding overhead, or compression considered.

Concept / Approach:
For baseband PCM, the minimum sampling rate is f_s = 2 * B. The raw bit rate R is then R = f_s * bits_per_sample. Unit conversion from bits per second to kilobits per second is straightforward (divide by 10^3).

Step-by-Step Solution:
Compute sampling rate: f_s = 2 * 4.6 * 10^6 = 9.2 * 10^6 samples/s.Multiply by quantization: R = 9.2 * 10^6 * 10 = 92 * 10^6 bits/s.Convert to kb/s: 92 * 10^6 / 10^3 = 92,000 kb/s.

Verification / Alternative check:
Order-of-magnitude sanity check: several tens of megabits per second are expected for raw uncompressed video; 92 Mb/s is consistent.

Why Other Options Are Wrong:

• 920 or 230 kb/s: far too low for raw 4.6 MHz video at 10 bits/sample.
• 23,000 or 46,000 kb/s: underestimates because they imply sampling below 2B or fewer bits per sample.

Common Pitfalls:

• Using f_s = B instead of 2B for baseband signals.
• Forgetting to multiply by bits per sample.
• Ignoring overheads; here the question asks for basic PCM only.

Final Answer:
92,000 kb/s