Help on calculating first order SD filters

[danielpasti]

Hello people again. The SD modulation has a integrator, which is the heart of the circuit. But an pratical integrator has a cut frequency which would be ideally zero. Some article I ve head, says that the filter would have a cut off freq in the audio band limit, but the quntized noise would not have and infinite atenuation in this freqs. So, my ask is, what is the influence of the cut off freq of this integrator on the circuit ? thanks

[phase_accurate]

Since no one answered I will do so.
I assume you want to dimension a ONE-BIT ds-modulator right ?

Simple answer: You can't calculate the integrator for a first-order one-bit ds modulator. The reason for this is that there is no mathematical definition of the gain of the sampling stage.
One possible assumption is that it is infinite (it is a comparator after all !!) and therefore in constant overload condition. For a multi-bit ds-loop it would be fairly easy to determine the gain of the sampling stage however.

Despite the above, one-bit modulators work quite well in practice. I'd simply try to find the best value for the integrator cap by trial and error. Use the one that gives you the best performance. Too large a cap and the loop gain will be too low. And if it is too small then the integrator might saturate too early.

Regards

Charles

[danielpasti]

Thanks for aswer. But, if I choose an small capacitor, the noise beins to enter in the audio band. So it lost its caracteristics of low pass filter for signal and high pass filter for noise. the capacitor im using is 15pf which makes it works well. But the integrator have to integrate just the noise freqeunices or the audio freq too ? i would like to know what is the cutt off freq of this filter ? idealy 0Hz or the audio band limit ? thanks

[danielpasti]

not what, but which is the cut off freq? ... sorry

[phase_accurate]

An integrator doesn't have a cutoff frequency, does it ?


H(s)=1/(sT)

Regards

Charles

[sx881663]

Charles,
You lost me, would you define your symbols please?
Roger

[phase_accurate]

This is the linear transfer functionof an (ideal) integrator. T is the integrator's time constant (usuallly R*C) and s is j*2*PI*f.
You see gain is actually depending on frequency and time-constant. For DC it is infinite and for an infinite frequency it is zero. Another interesting property is a constant phase-shift of -90 degrees.

A point that is often interesting is the unity-gain frequency given by:

f=1/(2*PI*R*C)

Regards

Charles

[sx881663]

Charles,
Thanks, makes total sense, just wasn't familiar with your terms. Not all of us have EE’s you know.
Roger

[danielpasti]

Yes I know it. But it happens only in an ideal world. All pratical lintegrator has an cut off freq. And a low passa filter with large capacitors and resistor can works like and integrator too. It makes no difference. The problem is: The modulator integrates only the high freqs or the audio band too ? Some literatures about SDM trate the integrator as an low pass filter. I know that, dependin the region of freqs it must work the same in both cases. But, in the SD has and low pass filter with cut off on, for exemple 20KHz or an integrator with ideally cut off of zero ? Or an low pass filter with cut off freq in 10 or 20Hz. I guess that Im not writing so clean, sorry but Im brazilian. i hope, that time, you undestand what im talkin about ..thanks

[phase_accurate]

I'd go fo around 100 Hz otherwise it isn't an integrator in the audio-band. I think you'd need every possible dB of SNR in the midrange.

Regards

Charles