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I wanted to move the summing operation (+10) that I perform before the integrator block as shown below: (MODEL-1) enter image description here

I check in this wikibook page - https://en.wikibooks.org/wiki/Control_Systems/Block_Diagrams and from the transformation "Moving a Summing Junction in front of a Block", it says to add a 1/P block, which in this case would be a differentiator. I tried adding a differentiator block like below, but the outputs (taken @ every 100th clock cycle) of the blocks does not match. (MODEL-2)

enter image description here

The output from MODEL-1 is:

enter image description here

The output from MODEL-2 is:

enter image description here

This is one part of this question.

However the major part is, when using this filter practically, the input to this filter contains high-frequency noise(now - just for example, it contains a pure sinusoid). By adding a differentiator in the front for the purpose of transformation dissolve the whole purpose of filtering?

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  • \$\begingroup\$ Where do you want to add this offset? To From3 usually for a single polarity signal source to make it bipolar, you subtract a positive offset to make the input to the discrete transform bipolar. \$\endgroup\$ – Sunnyskyguy EE75 Feb 11 at 18:41
  • \$\begingroup\$ @SunnyskyguyEE75 I want to move this offset(+10) to the beginning of the integrator block. \$\endgroup\$ – sundar Feb 11 at 18:47
  • \$\begingroup\$ I don’t see a single integrator block \$\endgroup\$ – Sunnyskyguy EE75 Feb 11 at 18:48
  • \$\begingroup\$ @SunnyskyguyEE75 \$\frac{z}{z - 1}\$ is the integration operator in the Z domain. I suppose properly it's called the "accumulation" operator due to being a discrete-time thing, but people call it integration anyway. \$\endgroup\$ – Hearth Feb 11 at 18:50
  • \$\begingroup\$ @Hearth so at DC where offset is applied to the input , is it an integrator? \$\endgroup\$ – Sunnyskyguy EE75 Feb 11 at 20:20
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You're so far off I'm not even going to attempt fixing your thinking. This. And I hope I'm not doing your homework. Note that this won't work in real life because it's causing a pole-zero cancellation of a metastable pole.

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  • \$\begingroup\$ IMO this is useless, however it is correctly moved to obtain the same result and that's what's OP is asking for (maybe? He is also confused). But unfortunatelly this won't work, because the integration goes to infinity. Integratring a constant brings you always to a dead end. \$\endgroup\$ – Marko Buršič Feb 11 at 22:24
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    \$\begingroup\$ It's not integrating a constant! It's integrating the derivative of a constant! So everything is ok! Something that I should have perhaps pointed out is that the whole block diagram manipulation thing is really just a way to do math with blocks; as soon as you start moving the blocks in the diagram away from the structure of the physical system they represent, you lose meaning. You may gain an easy path to solving for a transfer function, but you need to keep in mind that you've massacred the physical sense of the diagram. \$\endgroup\$ – TimWescott Feb 12 at 0:08

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