Minimum Channel Bandwidth for PCM

Question

Please check if my answer to GATE IN 2010 question 11.20 is correct/can be improved upon.

So, every $\frac{1}{f_s}$ sec, a sample is taken. The signal amplitude is quantized into $2^n$ levels such that each sample can be encoded using $n$ bits. The question (I think) asks what is the minimum bandwidth a channel needs in order to transmit these bits.

We have $\frac{1}{f_s}$ sec time and $n$ bits to send. So the bit rate would be $nf_s$ bits per second. Since each "bit" is sent as a pulse across the channel, the pulse rate $f_p = nf_s$. So by the Shannon–Hartley theorem we can conclude,

$$nf_s \leq 2B$$ $$B \geq \frac{nf_s}{2}$$

So the answer is (c)

The question asks what the bandwidth needed for "faithful reconstruction" is, I guess they mean that if you don't have at least this large a bandwidth errors will creep in and you can't reconstruct the signal from the n bit code at the receiver end?

Answer 1

I don't think the exam question is stated properly, unless there is additional context that the student is supposed to know.

First, the bandwidth depends on the encoding. The question seems to assume binary encoding, but PCM does not specify this.

Second, the bandwidth depends on the pulse shape. Usually one assumes rectangular unless otherwise specified, but this information should bave been included in the question.

In general, the PCM signal can be written as a pulse train:

$$s_{\text{PCM}}(t) = \sum_{k=-\infty}^\infty a_k p(t-kT_p),$$

where $T_p$ is the pulse rate and $a_k$ is the encoded PCM data. In general, the bandwidth of $s_{\text{PCM}}(t)$ is equal to the bandwidth of $p(t)$.

As an example, let's say we want to transmit the bits $1010$ using a bipolar, non-return to zero binary encoding with $a_k \in \lbrace 1, -1 \rbrace$. The pulse shape is rectangular with duration $T_p = 1/(nf_s)$. Assuming $n$ starts at $n=1$, the PCM signal can be written as:

$$s_{\text{PCM}}(t) = p(t-T_p)-p(t-2T_p)+p(t-3T_p)-p(t-4T_p).$$

The problem is, what is the bandwidth of this signal? Theoretically, it's equal to the bandwidth of the rectangular pulse $p(t)$ with duration $T_p$. However, there are several ways to measure that bandwidth.

For instance, if you use the "absolute bandwidth" definition, then the bandwidth is infinite. There are ten or more engineering definitions of bandwidth, and each gives you a different number. Note, however, that transmitting $s_{\text{PCM}}(t)$ over any channel with less than infinite bandwidth will result in some amount of distortion.

Now let's say that $p(t)$ is a sinc pulse of appropriate width. The expression for the signal doesn't change. Theoretically, now the signal has infinite time duration, but its bandwidth is finite and equal to $1/(2T_p)=nf_s/2$.

Answer 2

$B_{w} = \frac{r_{b}}{2 \log_{2} \left( M \right)}$

$M$ is number of symbols, we can have per sample $0, 1 \implies 2$ symbols, so $M = 2$.

So $B_{w} = \frac{r_{b}}{ 2 \log \left( 2 \right)} = \frac{R_{b}}{2}$ and

$R_{b} = n f_{s}$