PCM: minimum channel bandwidth numerical

Pulse code modulation (PCM)

PCM is a method of converting an analog signal into a digital (binary) signal by three steps: sampling at or above the Nyquist rate, quantizing each sample to one of $L$ levels, and encoding each level into an $n$-bit binary code word ($L = 2^n$). The binary pulses are then transmitted.

Numerical

Given: $W = 3$ kHz, sampling at twice the Nyquist rate, step size $\Delta = 0.5\%$ of $m_p$, binary coding.

Step 1: Sampling rate

$$f_N = 2W = 6\ \text{kHz}, \qquad f_s = 2 \times 6 = 12\ \text{kHz}$$

Step 2: Number of levels. The signal swings from $-m_p$ to $+m_p$, so

$$\Delta = \frac{2m_p}{L} = 0.005\,m_p \;\Rightarrow\; L = \frac{2}{0.005} = 400$$

Step 3: Bits per sample. $n$ must be an integer with $2^n \ge 400$:

$$n = \lceil\log_2 400\rceil = \lceil 8.64\rceil = 9 \quad (2^9 = 512)$$

Step 4: Bit rate

$$R_b = n f_s = 9 \times 12000 = 108\ \text{kbps}$$

Step 5: Minimum channel bandwidth. A binary signal of rate $R_b$ needs at least $R_b/2$ Hz (Nyquist bandwidth for ISI-free transmission):

$$B_{min} = \frac{R_b}{2} = \frac{108}{2} = 54\ \text{kHz}$$

Answer: minimum channel bandwidth = 54 kHz (9 bits/sample, 108 kbps).