Quantization level versus bandwidth [closed]

Question

Why is it that increasing the number of quantization levels $L$ results in an increase in the signal's bandwidth?

Answer 1

Say you quantize a signal with $L=2^n$ levels. Then, every quantized sample is represented by $n$ bits. So, as you increase $L$, you need more and more bits to represent the samples. This means that transmitting the signal will require transmitting more bits.

Also, keep in mind that, in order to convey bits at a rate of $R_b$ bits per second, you need a signal with bandwidth at least $B=R_b/2$ hertz (assuming binary communications).

Now (this is important) assume that you want to transmit the signal in real time, like in a voice application, or in streaming. If you increase $L$, you will have to transmit more bits per second, which means a larger $R_b$, which in turn implies a larger $B$.

As an example, say you're transmitting telephone-quality voice with $L=256$, $n=8$, and with 8000 samples per second. Then, $R_b = 8 \times 8000 = 64000$ bits per second, which require a signal of bandwidth $B > 32000\text{ Hz}$.

Say you want to improve the signal quality and go with $L=1024$ ($n=10$). Now, you need $R_b = 80000$ and $B>40000\text{ Hz}$. An increase in $L$ produced an increase in $B$.

Note that, if there is no real-time requirement, you can instead keep a signal with the same bandwidth (so, same bit rate) and instead take longer to transmit. So, in general, increasing $L$ will cause an increase in bandwidth and/or an increase in transmission time.

Answer 2

Decreasing quantization noise might increase the signal-to-noise ratio, which according to Shannon's law can increase the potential information bit rate in a given bandwidth.