Skip to main content

PCM bit rate for 4.5 MHz TV signal with 1024 levels

Short answer

1024 levels give 10 bits per sample. Nyquist rate 9 MHz, sampling at 10.8 MHz (20% above). Bit rate = 10 * 10.8 MHz = 108 Mbps.

The bandwidth of a TV plus audio signal is 4.5 MHz. If this signal is converted to PCM with 1024 quantization levels, determine bit rate of the resulting PCM signal. Let us assume that the signal is sampled at a rate 20% above the Nyquist rate.

Given: W=4.5W = 4.5 MHz, L=1024L = 1024, sampling 20% above Nyquist rate.

  1. Bits per sample:
n=log⁡2L=log⁡21024=10 bitsn = \log_2 L = \log_2 1024 = 10\ \text{bits}
  1. Nyquist rate and actual sampling rate:
fN=2W=2×4.5=9 MHzfs=1.2×9=10.8 MHz\begin{aligned} f_N &= 2W = 2 \times 4.5 = 9\ \text{MHz} \\ f_s &= 1.2 \times 9 = 10.8\ \text{MHz} \end{aligned}
  1. Bit rate:
Rb=nfs=10×10.8×106=108×106 bpsR_b = n f_s = 10 \times 10.8 \times 10^6 = 108 \times 10^6\ \text{bps}

Answer: Bit rate = 108 Mbps.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Communication Systems chapter 4 questions →