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why DT DSM ADC has worse SNR then CT DSM ADC??

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gggould

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Hi all,

Does anyone know why DT DSM ADC generally has worse SNR then CT DSM ADC??

Thanks
 

Well, that is not true. SNR does not depend on whether the system is implemented as discrete or continuous. All the things affecting the SNR being the same (like OSR, quantizer level, loop order, etc), SNR of a well designed DT DSM and CT DSM will be equal. In fact a CT DSM is derived from a DT DSM using impulse invariance. In other words CT DSM realizes the same Noise Transfer Function ( NTF) as DT DSM. So there is no reason why one would perform worse/better than the other in terms of SNR.
 

Ideally that is the case. But what I heard was due to switching cap nature the DT DSM has more quantization noise folding down to the passband, which degrades DT DSM SNR. Is that the case???
 

Hi
"Switching cap" by itself will not cause folding of quantization noise in baseband, as long as these operations are linear. However, if there is non-linearity in the integrator or the feedback, it will result in folding of quantization noise. But then this effect is common to both types of DSM.
 

Ideally that is the case.
But what I heard was due to switching cap nature the DT DSM has more quantization noise folding down to the passband, which degrades DT DSM SNR.
Is that the case???
For same sampling frequency, amplifier in DT-DSM require fairly large GBW than CT-DSM.
So degradation of noise folding of DT-DSM is severe than CT-DSM.

But you must not forget that a degradation due to sampling jitter exist in CT-DSM.
 

On a second thought, DT DSM do suffer from the problem of noise folding (not quantization noise). Thermal noise of the integrators and DAC around the multiples of the sampling frequency fold back into the pass band. However in CT DSM with Non Return to Zero(NRZ) pulse shape, the folding of noise around multiples of sampling frequency is not significant. In case you are interested to go in the details I can refer you to some papers.
 

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