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https://designers-guide.org/forum/YaBB.pl Design >> Mixed-Signal Design >> Another difficult fundamental question: about delta sigma modulator https://designers-guide.org/forum/YaBB.pl?num=1253702751 Message started by tony_taoyh on Sep 23rd, 2009, 3:45am |
Title: Another difficult fundamental question: about delta sigma modulator Post by tony_taoyh on Sep 23rd, 2009, 3:45am Hi, All, I have another difficult fundamental question for the first order delta sigma modulator. ff | x -----(+) ----- {z^-1/(1-z^-1)} ------(+) ------> [8-bit ADC] ------> y | | ----------------------------------------------------------------- The feedback is a negative one, with 8-bit ideal DAC. The ADC is also a ideal one. ff is white noise. x is a DC input. According to the linear model, the TF is as: y = x*z^-1 + (ff+qq) * (1-z^-1), where qq is the quantization noise. According to this model, the ff will also be first-order shaped. Right? However, from the Matlab behavioral model, ff is NOT shaped. Is there anybody who can explain this well? Thanks a lot. |
Title: Re: Another difficult fundamental question: about delta sigma modulator Post by tony_taoyh on Sep 23rd, 2009, 6:12pm Some additional information. In the matlab program, the integrator output will be: v1(n) = v1(n-1) + 1*(x(n-1) - y(n-1)) + ff(n-1) It is equivalent to: v1(n) = v1(n-1) + 1*({x(n-1) + ff(n-1) } - y(n-1)) Assuming the gain of the integrator is 1. Then, the x + ff can be considered as the input of the modulator. Both x and ff will NOT be shaped inside the behavioral simulation. There shoud be some breaking point somewhere, to explain the contradition between behavioral simulation and linear model. Thanks. |
Title: Re: Another difficult fundamental question: about delta sigma modulator Post by Berti on Sep 23rd, 2009, 11:01pm I think your conclusion is wrong. The output y will contain ff and x. You basically consider the modulator open-loop. When you close the loop you get y(n) = x(n-1) + (ff(n) - ff(n-1)). How do you simulate this modulator? Cheers |
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