I sent this to Rick, but thought maybe a wider reading would be beneficial. I'm designing a CIC interpolating filter (N = 3, M = 1, R = 15 if it matters) and I came to a realization I hadn't had before and haven't ever heard anyone else mention. That often means there's something badly wrong with it, but I don't see it. Let's start with the traditional implementation: Three (N) differentiators at the low sample rate A zero stuffing rate multiplier Three (N) integrators at the higher sample rate Now consider that the input to the first integrator is the output of the last differentiator for 1 out of N samples (at the higher rate) and 0 for the next N-1 samples. Then consider that the internal state and output of an integrator does not change when the input is 0. Does this not mean that an integrator after a zero stuffing rate multiplier does exactly the same thing as it would do if it was before the rate multiplier? Of course, the remaining integrators typically do NOT have zero inputs, so must operate at the higher sample rate (even though the second one adds the same value for N times). Nor should they be preceded by a zero stuffer, but rather by a sample replicator. Of course, in a synchronous system, a sample replicator would be the output of the first integrator, because it won't change for N cycles. (True statement whether the first integrator is at the low rate or the high rate, which is one reason this appears to be sound). So the refactored CIC interpolator becomes: N differentiators at the low rate an integrator at the low rate (No zero stuffer or anything special other than making sure the integrator output is available for the next N high rate clocks) N-1 integrators at the high rate Am I missing something obvious here? I think CIC interpolation just got a bit more cost effective (computationally). I can't discern any difference in the behavior, other than the power consumption of the first integrator went down by about N and there is no need for anything filling the role of a zero stuffer. Even the bit growth issues should not change. Wilton

# CIC interpolation implementation

Started by ●December 17, 2017

Reply by ●December 18, 20172017-12-18

You are correct but you can go one step further. The last slow-rate integrator cancels the previous slow-rate differentiator so you can eliminate both. You need a zero-order-hold but that comes for free as you mentioned. The industrial realizations that I’m aware of mostly use this trick, but you won’t find it in the literature for some reason. Bob

Reply by ●December 19, 20172017-12-19

Zero-order-hold is a term I'm not familiar with, and apparently isn't as widely used as some. Am I correctly inferring that it is just holding the value of the last slow rate stage for N fast rate clocks? Interesting that it is widely used and yet not in the literature. There are at least a half dozen people who have written widely on DSP topics, including CIC, with a practical emphasis on computational efficiency. You'd think it would get picked up somewhere. Once I get it implemented and tested I'll do my part to try and spread the message. Wilton

Reply by ●December 19, 20172017-12-19

Wilton Helm <wcjmhelm@gmail.com> wrote:>Interesting that it is widely used and yet not in the literature. There >are at least a half dozen people who have written widely on DSP topics, >including CIC, with a practical emphasis on computational efficiency. >You'd think it would get picked up somewhere.Perhaps because there's a difference between discussing an algorithm in the literature, where is is organized for conceptual clarity, and implementing the algorithm. S.

Reply by ●December 21, 20172017-12-21

hi Bob, Dana Massie kindly pointed me to this excellent article by Richard Lyons, which includes the optimization described. http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1657825 Reducing CIC Filter Complexity Ricardo A. Losada and Richard Lyons cheers, -Brian Clark

Reply by ●December 21, 20172017-12-21

Thanks. A related topic is the so-called noble identities, which are not widely taught but are very useful in multi-rate filter design. Bob

Reply by ●January 4, 20182018-01-04

Wilton Helm <wcjmhelm@gmail.com> writes:> Zero-order-hold is a term I'm not familiar with, and apparently isn't > as widely used as some. Am I correctly inferring that it is just > holding the value of the last slow rate stage for N fast rate clocks?A zero-order hold is a continuous-time transfer function. The impulse response of a zero-order hold is 1, 0 <= t < T_s, where T_s is the sample period. It is (was?) often used to model the operation of a DAC which held each digital sample constant until the next digital sample. I don't remember seeing it applied to a multi-rate system. See https://en.wikipedia.org/wiki/Zero-order_hold. -- Randy Yates, DSP/Embedded Firmware Developer Digital Signal Labs http://www.digitalsignallabs.com

Reply by ●January 11, 20182018-01-11

On Thursday, January 4, 2018 at 3:26:29 PM UTC-5, Randy Yates wrote:> Wilton Helm <wcjmhelm@gmail.com> writes: > > > Zero-order-hold is a term I'm not familiar with, and apparently isn't > > as widely used as some. Am I correctly inferring that it is just > > holding the value of the last slow rate stage for N fast rate clocks?it's holding it for any specified period of time. even 1 fast-rate clock.> > A zero-order hold is a continuous-time transfer function. The impulse > response of a zero-order hold is 1, 0 <= t < T_s, where T_s is the > sample period.remember our "T" problem and dimensional analysis discussion we've had in times of old here on comp.dsp. since the dimension of h(t) must be 1/time (if H(s) is dimensionless, which it is if the output and the input are the same dimension of "stuff"), i think the impulse response of a ZOH is 1/T_s when 0 <= t < T_s, despite what is popular in the textbooks. For the DC gain of a ZOH to be 1, the impulse response must have an area of 1 also.> It is (was?) often used to model the operation of a DAC > which held each digital sample constant until the next digital sample.well, if you're gonna model the sigma-delta at the high sample rate, the DAC and the analog feedback in the ADC, would still be modeled with a ZOH.

Reply by ●January 18, 20182018-01-18

robert bristow-johnson <rbj@audioimagination.com> writes:> On Thursday, January 4, 2018 at 3:26:29 PM UTC-5, Randy Yates wrote: >> Wilton Helm <wcjmhelm@gmail.com> writes: >> >> > Zero-order-hold is a term I'm not familiar with, and apparently isn't >> > as widely used as some. Am I correctly inferring that it is just >> > holding the value of the last slow rate stage for N fast rate clocks? > > it's holding it for any specified period of time. even 1 fast-rate clock. > >> >> A zero-order hold is a continuous-time transfer function. The impulse >> response of a zero-order hold is 1, 0 <= t < T_s, where T_s is the >> sample period. > > remember our "T" problem and dimensional analysis discussion we've had > in times of old here on comp.dsp. since the dimension of h(t) must be > 1/time (if H(s) is dimensionless, which it is if the output and the > input are the same dimension of "stuff"), i think the impulse response > of a ZOH is 1/T_s when 0 <= t < T_s, despite what is popular in the > textbooks. For the DC gain of a ZOH to be 1, the impulse response must > have an area of 1 also.Hey Rober, Yes, I agree. I was a bit sloppy and/or distracted by the common definition. Having a DC gain of 1 is a pretty good way to justify why it should be 1/T, IMO. -- Randy Yates, DSP/Embedded Firmware Developer Digital Signal Labs http://www.digitalsignallabs.com