An analog speech signal contains signals from \(200\) Hz to \(2400\) Hz. It is sampled at \(6\) kHz and quantized with \(512\) levels for pulse code modulation (PCM). The bit rate will be _____ kbps.
Show Hint
Find the number of bits needed to represent \(512\) quantization levels, then multiply by the sampling rate given in the question.
Step 1: Express the level count as a power of two.
The number of bits per sample $n$ is found from the number of quantization levels $L$ using $L = 2^n$. Here $L=512$.
Step 2: Find $n$ by repeated halving.
Divide $512$ by $2$ repeatedly and count the steps until reaching $1$:
$512 \to 256 \to 128 \to 64 \to 32 \to 16 \to 8 \to 4 \to 2 \to 1$.
That is $9$ divisions, so $n=9$ bits per sample, confirming $2^9=512$.
Step 3: Multiply by the sampling rate to get the data rate.
Each second, $f_s = 6000$ samples are taken, and each sample carries $9$ bits, so the total number of bits transmitted per second is
\[ R_b = 6000 \times 9 = 54000 \text{ bps} \]
This matches option (D). The frequency band of the speech signal ($200$ Hz to $2400$ Hz) is only context for why a $6$ kHz sampling rate was chosen (comfortably above the Nyquist rate of $4800$ Hz); it is not itself used in the bit rate formula once the actual sampling rate is stated.
\[ \boxed{R_b = 54 \text{ kbps}} \]
