# How does delta modulation work practically?

How does delta modulation work in practice? And where is it used? I find it confusing that, in 1-bit quantization we use only 2 level; if we apply this quantization, there will be loss in information. So, why do we use this technique?

One key point to note about delta modulation schemes is that they quantize the change between the current value and the previously quantized value.

The other key point to note is that, in general, delta modulation schemes are run at sampling rates MUCH higher than the Nyquist rate would suggest. Early portable CD players used a scheme called MASH (multi-stage noise shaping) on their DACs to allow cheaper, more accurate reproduction of sound.

I can't find a reference quickly, but I seem to recall that for the 44.1kHz CD sampling rate, MASH 1-bit DACs ran at sampling rates in the order of 5-6 MHz.

• this looks an awful lot like $\Sigma \Delta$ modulation. at one time i was very familiar with the difference between $\Sigma \Delta$ and $\Delta$. now i need to look it up. – robert bristow-johnson Dec 11 '15 at 18:40
• @robertbristow-johnson : Feel free to correct / update. – Peter K. Dec 11 '15 at 18:44
• okay, i looked it up. $\Sigma \Delta$ has the integrator appearing after the summer, before the quantizer. to be equivalent to $\Delta$ would imply a differentiator in the input. actually $\Sigma \Delta$ has a more sophisticated LPF before the quantizer. and MASH is even more complicated. – robert bristow-johnson Dec 11 '15 at 18:45

There won't be any loss of information if all the information desired to be transmitted is already quantized and band-limited well below the cut-off of the low-pass filter network that removes anything near the (delta) modulation rate (or its noise-shaped side-bands). The high frequency 1-bit quantization gets "averaged out" by a low pass filter over a long enough period of time to transmit lots of bits (or signal levels) of information at a much lower rate.

• well, there is always some loss of information in any quantization. – robert bristow-johnson Dec 11 '15 at 18:37
• I was assuming the information was already pre-quantized, just not at the delta modulation rate. – hotpaw2 Dec 11 '15 at 19:07
• for practical purposes if you have 24 bits or equivalent to 8 bit u-Law logarithmic compander, there is no perceivable noise or difference – Sunnyskyguy EE75 Apr 18 '18 at 16:02