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

  • $\begingroup$ Hi Leolime! Quantization depth doesn't increase the bandwidth of the signal. So, not quite sure what you're asking? $\endgroup$ Aug 30, 2018 at 15:09
  • $\begingroup$ thanks for your comment :) @MarcusMüller , i have found this line in Information Transmission, Modulation And Noise by Mischa Schwartz...it says the larger quantum steps, the greater the number of bits needed to represent the signal , hence wider bandwidth needed for transmission $\endgroup$
    – Leolime
    Aug 30, 2018 at 15:14
  • 1
    $\begingroup$ @Leolime this isn’t what you asked. Conversation and transmission are different problems. $\endgroup$
    – user28715
    Aug 30, 2018 at 15:22
  • $\begingroup$ @StanleyPawlukiewicz i'm sorry for this ambiguity in my question $\endgroup$
    – Leolime
    Aug 30, 2018 at 15:25
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    $\begingroup$ The quote actually says: "larger the number of quantum steps used, the greater the number of binary digits or bits...". $\endgroup$
    – learner
    Aug 30, 2018 at 19:40

2 Answers 2


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.

  • $\begingroup$ Thanks MBaz, helped me overcome one doubt i had for a long time. $\endgroup$
    – srk_cb
    Aug 31, 2018 at 7:30
  • $\begingroup$ @srk_cb You're very welcome, and thanks for the feedback! $\endgroup$
    – MBaz
    Aug 31, 2018 at 13:46

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.


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