so I am a little bit confused about the formula of the step size in a uniform quantizer.

from what I have studied:

StepSize = $\displaystyle\frac{maxValue-minValue}{2^n}$, where n is the number of bits.

so for example, if we have a signal that changes from (-6 to 6) and the number of bits is 3 bits (also suppose it's midrise).

so according to the formula:

StepSize = $\displaystyle\frac{6-(-6)}{2^3}$ = 1.5

so the possible values for a signal that it could take are:

-6, -4.5, -3, -1.5, 0, 1.5, 3, 4.5, 6

which are 9 levels not 8 so it needs 4 bits, not 3, also if it's a midrise signal then it shouldn't have a value at zero. so is there anything that I misses here?

  • $\begingroup$ Hi! You can read about Lloyd-Max quantizer to get a better appreciation of this topic... $\endgroup$
    – Fat32
    Mar 10, 2023 at 9:20

1 Answer 1


The step size is correctly computed as the OP has shown. The difference between mid-tread and mid-riser is the range within those steps that are used to map to the quantized values.

With reference to the convenient graphic below from this Georgia Tech course pdf by Drs. Butera and Williams (Intro to ECE Design):

mistep and misrise

The OP has computed $\Delta$ to be 1.5 and is using a midrise quantizer. As we see in the graphic, $V_{min}$ is at $-4\Delta = -6$ and $V_{max}$ is at $+4\Delta = +6$. Where to be clear $V_{min}$ and $V_{max}$ are ranges on the input that are within the maximum quantization error; extending the range further would result in "clipping error". So far so good. What we also see, is that the output binary values represent values symmetrically about $0$ extending to $\pm 7\Delta/2$. With that all input values $v(t) \in [-4\Delta, -3\Delta) = [-6, -4.5)$ map to $-7\Delta/2 = -5.25$, and so on over the range of input values. The the possible values that the signal can take once quantized are:

$$-5.25, -3.75, -2.25, -.75, 0.75, 2.25, 3.75, 5.25$$

The midrise quantizer is conveniently symmetric about zero, but since it does not include zero it will be more sensitive to noise at low signal levels. The midstep quantizer, although conveniently including zero for minimizing the small signal noise, will have an offset: the signed quantization extends more negative than positive (as is commonly done for signed fixed point representations). The offset becomes insignificant for larger quantization levels, so the noise reduction benefit is generally preferred.

  • $\begingroup$ It's just the fencepost error, Dan. Nothing more. $\endgroup$ Mar 10, 2023 at 3:47
  • $\begingroup$ @robertbristow-johnson Are you referring to my [ ) instead of [ ]? $\endgroup$ Mar 10, 2023 at 3:49
  • $\begingroup$ No, the OP question. $\endgroup$ Mar 10, 2023 at 3:50
  • $\begingroup$ If he wasn't midrise that would be the case, if you are referring to the extra 6 $\endgroup$ Mar 10, 2023 at 3:51
  • 1
    $\begingroup$ doesn't matter if the OP is midrise or midtread. The step size for a uniform quantizer is always: $$ \Delta = \frac{V_\text{max} - V_\text{min}}{2^n - 1} $$ if $n$ is the number of bits. $\endgroup$ Mar 10, 2023 at 3:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.