Transfer Function - Basic

1. Dec 1, 2011

Femme_physics

Last edited by a moderator: May 5, 2017
2. Dec 1, 2011

I like Serena

I don't think so...
Do you have some work to go with that?

3. Dec 2, 2011

Femme_physics

What do you mean some work to go with that?

4. Dec 2, 2011

I like Serena

Well, at each point between the blocks you should have a formula representing the signal there.
Solving the entire formula should give you C/R.

5. Dec 2, 2011

Femme_physics

Hmm..I still don't get where exactly my mistake is in the C/R I created?

If you can help me see that, please... :)

6. Dec 3, 2011

I like Serena

I don't know how you derived it.
If you could say how you got it...?

Let's start on the left, where you have the input signal R, which is combined with the output signal C.
The circle signifies how they are combined, which will result in (+R-C) between the circle and block G1.

Next step is between blocks G1 and G2.
A block multiplies the signal, meaning the resulting signal is G1(+R-C).

Can you say what the next step is?
And the resulting signal?

7. Dec 3, 2011

Femme_physics

OK. The formula says

C/R = Sum of all the paths / 1- sum of all the loops

This is actually my NEW attempt. Pathways in black. Loops in orange.

http://img545.imageshack.us/img545/7681/laster.jpg [Broken]

G1(+R-C)+G2(+R-C)+G3(+R-C)

(IMO that's only 1 pathway, there are 2 pathways since there's a pathway that skips over G1)

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8. Dec 3, 2011

I like Serena

Okay. That's another way to solve it.

This is a different diagram than your problem states...

You have the sum of the paths right though.

For the loops you should first rectify you diagram.
After that you should consider that you also have a large loop.

Perhaps you can list the loops that you see with their corresponding formulas?

You're moving to fast with this method.
Let's make a stop immediately to the right of G2.
So left of G2 you have G1(+R-C).
What would you have immediately to the right of G2?

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9. Dec 3, 2011

Femme_physics

That's the method we're being taught

What do you mean by "rectify"?

I agree, I forgot the large loop.

http://img440.imageshack.us/img440/7948/gotitj.jpg [Broken]

of course. It's the lower row in the image I've just made.

Oh, sorry That would be G3(+R-C)

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10. Dec 3, 2011

I like Serena

Okay, so that's what we'll do. ;)

If you'd like we can also do it my way, if only as a verification.

You just did. You have the proper diagram now.

You have the right diagram now, but not the right loops.

It contains G3 and H, but that does not match either of the loops you got.

No, not quite.
We had G1(+R-C) and we apply block G2.
How did you get G3?
And where did G1 go?

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11. Dec 4, 2011

Femme_physics

Oh, I don't mind at all, you always give me new insights.
'the hell is wrong with my English?

Oh, I thought loops can only start with the "R". I guess that's not the case. So, here, I coloured all the R (loops) only.

Oh and.........PIKABOO! :) Just in case u missed my cameo appearances.

http://img163.imageshack.us/img163/1453/pikabo.png [Broken]

Well, it didn't go anywhere, but you asked me what's the the right of G2, and G1 is not at the right of G2.

To the right of G2 is what appears to be a voltage source?

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12. Dec 4, 2011

I like Serena

rec·ti·fy (rkt-f)
tr.v. rec·ti·fied, rec·ti·fy·ing, rec·ti·fies
1. To set right; correct.
2. To correct by calculation or adjustment. See Synonyms at correct.
3. Chemistry To refine or purify, especially by distillation.
4. Electronics To convert (alternating current) into direct current.
5. To adjust (the proof of alcoholic beverages) by adding water or other liquids.

It turns out there's even a specific electronics meaning, but that's not the one I intended. ;)

Yes, I did miss your cameo appearances.
Thanks!

There!
You have the right loops now:
1. G1G2G3
2. G3
3. G3H

Only one thing... the loops go through summing junctions.
If a loop goes through the minus of a summing junction, it becomes negative...

No, not a voltage, but a summing junction.
It sums the signals that come in (or subtracts them if there is a minus).

I asked what the signal would be to the right of G2.
To the left of G2 the signal is G1(+R-C).
(Note the + and - signs that have resulted from the summing junction.)
Since a block multiplies the signal, the resulting signal is G2G1(+R-C).

13. Dec 5, 2011

Femme_physics

I know! You didn't have to go through the full explanation! It's the electronics definition that confused me because I was aware of it. I was worried it has a system engineering definition. Heh. Nevermind.

*kissy winky*

That's exactly what all my loops are minused!

Gotcha.

Ah, thanks. But I did get the right answer, eh?

14. Dec 5, 2011

I like Serena

But they aren't!
You said yourself: "C/R = Sum of all the paths / 1- sum of all the loops"
It the loop has a minus too, it becomes a plus!

Nope.

15. Dec 6, 2011

Femme_physics

You mean ALL of the minuses at the bottom row should have been written as pluses then?

16. Dec 6, 2011

I like Serena

Yep!

17. Dec 10, 2011

Femme_physics

Thanks ILS, we solved it in class and you were of course all right