# Block reduction how to move summer

Discussion in 'Homework Help' started by Watershadow, Jul 9, 2011.

Jul 6, 2011
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I so far reduced the blocks to the current sketch. I am not sure what to do with the feedback loop of A and C. How do you move the two summing points A to B. Or is there another easier way to reduce the block in its current form?

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2. ### Georacer Moderator

Nov 25, 2009
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Think about it: You have three summing points in a row. Isn't addition associative? You can substitute all of these summing points with one.
I also suggest you move the origin of the H1G2 feedback loop from C to the output, making it (H1G2)/(G2G3+G4).
That way you 'll end up with three parallel feedback loops, waiting for you to simplify them.

Is that clear?

3. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
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Here is a link to an application note that discussion block manipulation early on in the text. It reinforces what georacer has stated.

hgmjr

Jul 6, 2011
9
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Georacer

What does addition associative mean? Does it mean that I can move the summing points around?

5. ### Georacer Moderator

Nov 25, 2009
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It means that a+b+c=(a+b)+c=a+(b+c).

In other words you can make a single summing point out of the three smaller ones just by summing all of the components.

Jul 6, 2011
9
0
Georacer

I dont know how to add summing points. But like you said if its a+b+c then I can move the order around like a+c+b, that way I can reduce the feedback loops slowly 1 by 1.

7. ### Georacer Moderator

Nov 25, 2009
5,181
1,289
A summing point doesn't have to add only two signals. You can lead all three feedback loops to a single summing point.

That said, you will end up with a system whit G1 in series with G2G3+G4 in the forward branch, and three parallel branches for feedback with TFs H1G2, H2/G1 and 1, all of them giving negative feedback.

Isn't that the same as F=G1*(G2G3+G4), G=(H1G2/(G2G3+G4)+H2/G1+1) and the overall TF being H=F/(1+FG)?