# Nyquist-Rate Converter

Joined May 19, 2013
214
In the book it says:

"We loosely define Nyquist-rate data converters as those converters that generate a series of output values in which each value has a one-to-one correspondence with a single input value.

For example, a Nyquist-rate D/A converter would generate a series of analog output levels, where each level is a result of a single N-bit input word."

But a Nyquist-rate A/D converter generates the same output value for a range of input values. So there is no one-to-one correspondence. So why the category? I mean, I see there the word "loosely", but come on.

Also, the Nyquist-rate converter is defined in opposition to the oversampling converter.
I suppose that the oversampling D/A converter generates multiple analog output values for the same bit word. But what does the oversampling A/D converter do?

It seems to me that Nyquist-rate converters are those converters which operates at a rate close to the Nyquist-rate of the signal, while the oversampling converters operate at a rate much larger than the Nyquist-rate of the signal. But why say about the one-to-one correspondence? Ok, it means the same thing in the case of the D/A, but in the case of the A/D it does not mean the same thing, and it can create confusion if one is a beginner.

So, after all this mumbling. My question. What does the oversampling A/D converter do?

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#### crutschow

Joined Mar 14, 2008
25,465
In the book it says:

"We loosely define Nyquist-rate data converters as those converters that generate a series of output values in which each value has a one-to-one correspondence with a single input value.

For example, a Nyquist-rate D/A converter would generate a series of analog output levels, where each level is a result of a single N-bit input word."

But a Nyquist-rate A/D converter generates the same output value for a range of input values. So there is no one-to-one correspondence. So why the category? I mean, I see there the word "loosely", but come on.

Also, the Nyquist-rate converter is defined in opposition to the oversampling converter.
I suppose that the oversampling D/A converter generates multiple analog output values for the same bit word. But what does the oversampling A/D converter do?

It seems to me that Nyquist-rate converters are those converters which operates at a rate close to the Nyquist-rate of the signal, while the oversampling converters operate at a rate much larger than the Nyquist-rate of the signal. But why say about the one-to-one correspondence? Ok, it means the same thing in the case of the D/A, but in the case of the A/D it does not mean the same thing, and it can create confusion if one is a beginner.

So, after all this mumbling. My question. What does the oversampling A/D converter do?
Your book definition makes little sense to me. A Nyquist-rate converter samples at a rate at least equal to the Nyquist frequency (twice the highest sine-wave Fourier signal frequency). It's the same whether it's an A/D or D/A.

An oversampling A/D samples at a rate higher than the Nyquist frequency.
One advantage of this is that you can use a digital filter at the A/D output to remove noise and signal above the desired pass-band so only a simple analog anti-alias filter is required at the A/D input.
(Do you understand about aliasing in an A/D converter?)

Joined May 19, 2013
214
Your book definition makes little sense to me. A Nyquist-rate converter samples at a rate at least equal to the Nyquist frequency (twice the highest sine-wave Fourier signal frequency). It's the same whether it's an A/D or D/A.
I agree. I will stick to this definition and ignore the one from the book. It's even in the name. Nyquist-rate => sampling close to the Nyquist rate.
But I can see why the author loosely defined them as such. But I still do not like it.

An oversampling A/D samples at a rate much higher than the Nyquist frequency.
One advantage of this is that you can use a digital filter at the A/D output to remove noise and signal above the desired pass-band so only a simple analog anti-alias filter is required at the A/D input.
(Do you understand about aliasing in an A/D converter?)
I confused things. I know that the oversampling A/D samples the analog signal at a rate higher than its Nyquist rate. This makes sense to me.

What does not make sense is actually the oversampling D/A. I do not get what it samples exactly. I know that the D/A receives a digital signal and with zero-order hold process it converts it into an analog signal. But what is with the sampling? What does it actually sample? The digital signal is already sampled in nature...the D/A receives samples.

In the book it says:
"
Oversampling converters are those converters that operate much faster than the input signal’s Nyquist rate (typically 10 to 512 times faster) and increase the output’s signal-to-noise ratio (SNR) by filtering out quantization noise that is not in the signal’s bandwidth. In A/D converters, this filtering is performed digitally, whereas in D/A converters, analog filtering is used. Most often, oversampling converters use noise shaping to place much of the quantization noise outside the input signal’s bandwidth."

So there are oversampling D/A converters. But what do they sample?

PS: Aliasing is when the signal is sampled below its Nyquist rate, I think.

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#### nsaspook

Joined Aug 27, 2009
7,612
What does not make sense is actually the oversampling D/A. I do not get what it samples exactly. I know that the D/A receives a digital signal and with zero-order hold process it converts it into an analog signal. But what is with the sampling? What does it actually sample? The digital signal is already sampled in nature...the D/A receives samples.
Oversampling a DAC is easy, you just add zeros to the stream (in fixed-rate audio up-sampling) between the actual samples. Your original signal stays the same but spectrum of the energy that needs to be filtered from gets pushed further way from the needed energy. This increases the CPU load so you don't get something for nothing.

#### crutschow

Joined Mar 14, 2008
25,465
The meaning of the word "sample"is more general than just "to sample".
It also can be the sample taken by the A/D and converted to a digital word which is now being regenerated by the D/A back into analog form.
It's an output "sample".

Joined May 19, 2013
214
The meaning of the word "sample"is more general than just "to sample".
It also can be the sample taken by the A/D and converted to a digital word which is now being regenerated by the D/A back into analog form.
It's an output "sample".
Ok. I see.

If the voltage reference used by the D/A is larger than the voltage reference used by the A/D, the signal will obviously be amplified on the y axis. But would the sampling frequency need to be decreased also (so the period be increased) so that the proportion between the y axis and x axis stay the same?
The sampling frequency being the time a sample is hold with the zero-order hold process.

(both converters have the same resolution)

I do not think so. If we would want to amplify a sine wave, then we would increase only its amplitude. And not look at the frequency also. So only the y axis. Even if the proportion is not there anymore.

#### crutschow

Joined Mar 14, 2008
25,465
If the voltage reference used by the D/A is larger than the voltage reference used by the A/D, the signal will obviously be amplified on the y axis. But would the sampling frequency need to be decreased also (so the period be increased) so that the proportion between the y axis and x axis stay the same?
The sampling frequency being the time a sample is hold with the zero-order hold process.
The x axis and y axis are independent of each other.
Amplifying in the vertical axis is the same as changing the gain on a amplifier. Obviously that has no effect on the horizontal axis.
If you changed the sampling frequency then you would be changing the Fourier spectrum of the output, which you normally don't want.

• Joined May 19, 2013
214

Suppose an analog signal is converted into a 10-bit digital signal by a 10-bit A/D.
Then if the 10-bit digital signal is converted into an analog signal by a 10-bit D/A using the zero-order hold process, I can see what happens.

But what if it is converted into an analog signal by a 6-bit D/A using the zero-order hold process? How does it do it?

We have samples going from 0 to 1023, and these samples must be relocated to {0,...,63}.

I suppose it divides the {0,..,1023} in 64 intervals and does quantization. Right? And quantization error appears. And depending on the chosen voltage reference, it amplifies or not this distorted 6-bit signal.

#### crutschow

Joined Mar 14, 2008
25,465

Suppose an analog signal is converted into a 10-bit digital signal by a 10-bit A/D.
Then if the 10-bit digital signal is converted into an analog signal by a 10-bit D/A using the zero-order hold process, I can see what happens.

But what if it is converted into an analog signal by a 6-bit D/A using the zero-order hold process? How does it do it?
You simply use the 6 most-significant bits from the 10-bit word and ignore the 4 least-significant.
The result is essentially the same as if the original signal was converted using a 6-bit A/D.