Is my transfer function right?

[Femme_physics]

http://img803.imageshack.us/img803/7417/85681304.jpg

At least so far...

 

[I like Serena]

There seem to be a couple of loops missing.

 

[Femme_physics]

Hi ILS! Hmm you're right.

I need to add to the R line the following:

+G2H1 +G2H1

 

[I like Serena]

Yes, but that's not all (and you should only add it once and not twice ;).

 

[Femme_physics]

But there are 2! http://img710.imageshack.us/img710/4391/82552817.jpg

 

[I like Serena]

The bottom one is not a proper loop, since one of the arrows is in the wrong direction.

Actually, you have mentioned each of the loops at least once, but I haven't seen the proper expression for it yet.

There is also something missing from the C line.

 

[Femme_physics]

Before I give you the proper expression (with the missing C line stuff) I'm trying to understand why the bottom one isn't a proper loop. I start from R, but I can't go through minuses?

 

[I like Serena]

You can and should go through the minuses, but you should not go against the arrows.

 

[Femme_physics]

You can and should go through the minuses, but you should not go against the arrows.

LOL - good point.

But what's the point of this loop being there? It doesn't change anything.

 

[I like Serena]

Which loop?

 

[Femme_physics]

The bottom one, the one you told me to discount.

 

[I like Serena]

So why is the loop there?
Did someone tell you it should be there?

 

[Femme_physics]

So why is the loop there?
Did someone tell you it should be there?

Well, nobody's told me YOU going to be there! Sorry, I had to

Anyway, that loop doesn't contain anything and just goes back to R, so I say it's pointless.

As for the function:

http://img819.imageshack.us/img819/7867/gggzo.jpg

I just hope all the signs are right, too..

 

[I like Serena]

:)

Looks like you have your paths and loops almost straight now.
Where does the $-G_1$ come from in the denominator?Btw, last time you said you were taught:

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

This was for your basic exercise at that time.
However, I found it's not so simple for this circuit.
I tried to apply it. The result comes close, but is not quite what it should be.

I think you should build up your equations inside out, similar to the process of replacing resistors in series by one resistor, and replacing parallel resistors by one resistor.

 

[Femme_physics]

Where does the −G1 come from in the denominator?

http://img841.imageshack.us/img841/567/thisloop.jpg

Here.

This was for your basic exercise at that time.
However, I found it's not so simple for this circuit.
I tried to apply it. The result comes close, but is not quite what it should be.

I think you should build up your equations inside out, similar to the process of replacing resistors in series by one resistor, and replacing parallel resistors by one resistor.

We weren't really taught how to do that, we were just told to use the fast path.

 

[I like Serena]

http://img841.imageshack.us/img841/567/thisloop.jpg

Here.

That one accounts for the $+G_1$ in the denominator, but not for the $-G_1$.

We weren't really taught how to do that, we were just told to use the fast path.

Well, I'll give you a hint of what I mean.
In your circuit you could replace the loop you just marked, by a block with $G_1 \over 1 + G_1$ in it, which is the response function of just that path/loop with the G1 block in it.
This simplifies the circuit.

 

[Femme_physics]

That one accounts for the +G1 in the denominator, but not for the −G1.

You're right...

So,http://img39.imageshack.us/img39/8605/g0g0.jpg

Well, I'll give you a hint of what I mean.
In your circuit you could replace the loop you just marked, by a block with G11+G1 in it, which is the response function of just that loop with the G1 block in it.
This simplifies the circuit.

I really respect your methods and don't mind trying them but I have so much to study and relatively short period of time. Should I really try to understand this? I know it may not "sound" difficult but over the internet...u know how things can be.

 

[I like Serena]

You're right...

Yep.
That is the proper result of the application of your path/loop formula. :)

I really respect your methods and don't mind trying them but I have so much to study and relatively short period of time. Should I really try to understand this? I know it may not "sound" difficult but over the internet...u know how things can be.

No, I did not expect you would want to learn this now.
I only wanted to give you a hint that there is more to it.
Especially since your result is not quite right now.

I'll just give you the result I get:
$${C(s) \over R(s)} = {2G_1G_2 + G_2 \over 1 + G_1 + 2G_1G_2H_1 + G_2H_1}$$

As you can see it is close to what you got, except for two factors of 2 in it.
I leave it up to you what you want to do with it.

 

[Femme_physics]

I think I understand. There are two "reductors" (which I believe is the same as voltage sources) therefor your result makes sense. Thanks

Edit: That "IS" why you added that factor, yes?

 

[I like Serena]

I think I understand. There are two "reductors" (which I believe is the same as voltage sources) therefor your result makes sense. Thanks

Edit: That "IS" why you added that factor, yes?

Sorry, but I do not know what you mean by "reductor".

I calculated the result of the circuit with 2 sure-fire methods that are different from yours, which appears to be a rule of thumb.
Both came up with the extra factors 2.

 

[Femme_physics]

By "reductors" I mean the circly-thingies with the plus and minus signs, like the one that appears right after "R". If they are not reductors, what are they?

 

[I like Serena]

By "reductors" I mean the circly-thingies with the plus and minus signs, like the one that appears right after "R".

Ah okay, I looked them up before and came up with "summing junctions".
Are they also called reductors?

Either way, apparently it depends on how the reductors are placed whether they introduce a factor 2 or not.
Normally they wouldn't, but in this particular circuit they do.

 

[Femme_physics]

I'll go with "summing junctions", then! Whatever you say. I just thought they were said to be reductors from some reason, but summing junctions sounds more right.

Normally they wouldn't, but in this particular circuit they do.

Hmm...and how do you know when they do, and when they do not?

EDIT: Off to bed..will check again tomorrow :)

 

[I like Serena]

Hmm...and how do you know when they do, and when they do not?

When my 2 sure-fire methods say they do or do not. ;)

Your formula of paths/loops was new to me and it is pretty slick.
However, I was not ready to rely on it, so I checked...

 

[Femme_physics]

How is your system called?Also,...

In this case

http://img20.imageshack.us/img20/3033/clarify.jpg
(used a pencil to clarify text)
Should I use a factor of 3 on the routes because I have 3 summing junctions at the beginning?

 

[I like Serena]

My methods are:

• the replace-subcircuit-by-block-and-repeat method,
• the trace-signal-with-formula-and-solve method.

The first is similar to finding an equivalent impedance for an electrical circuit.
The second is similar to analyzing a circuit with KVL/KCL.

I'll get back to you about your new circuit later.

 

[I like Serena]

Also,...

In this case

http://img20.imageshack.us/img20/3033/clarify.jpg
(used a pencil to clarify text)
Should I use a factor of 3 on the routes because I have 3 summing junctions at the beginning?

No, you do not get a factor of 3.
Your rule C/R=(sum paths)/(1 - sum loops) works for this circuit.

In your original circuit, the factor of 2 seems to be caused by the extra forward path that bypasses a loop.

 

[Femme_physics]

I finally got it, and resolved that exercise and others and got the right result. So just wanted to say a belated thanks! Test is this tuesday :)

 

[I like Serena]

I finally got it, and resolved that exercise and others and got the right result. So just wanted to say a belated thanks! Test is this tuesday :)

You're welcome!

Will you let me know what the result of the test was?

 

[Femme_physics]

Of course :)

My hydraulics test wasn't so good, but we had a terrible teacher. I had no resolved all the question from the test successfully and will go for second term this friday.

The engineering test is this tuesday. I still have a question about Mason's Gain formula!

In this case - did I get it right?

http://img163.imageshack.us/img163/9947/tfun.jpg

 

[I like Serena]

I see you're notching it up!
This is looking challenging! :)
And you have a name for the formula, Mason's gain.
That is something I can look up.

Now let's see... the sign of loop G3H3 seems wrong.

And your non-touching loops do not appear to be correct.
If I select the first 2 non-touching loops, I find G1G2H1 x G3H3.
But you have G3G4 after that??

Furthermore there are 4 combinations of 2 non-touching loops, but you have only 2 combinations...

 

[Femme_physics]

And your non-touching loops do not appear to be correct.
If I select the first 2 non-touching loops, I find G1G2H1 x G3H3.
But you have G3G4 after that??

Furthermore there are 4 combinations of 2 non-touching loops, but you have only 2 combinations...

Ooohhhhhhh... I understand how it works. Wait :)

http://img24.imageshack.us/img24/444/songut.jpg

Plus fixing the sign of G3H3 to minus before the non-touching loops - agreed

 

[I like Serena]

Looking at you first combination, G1G2H1 is a loop, but G4 is not a loop.
You should use G3G4 instead.

Same thing in your 3rd term.

That's all! You have the signs correct this time round! :)

 

[Femme_physics]

Looking at you first combination, G1G2H1 is a loop, but G4 is not a loop.
You should use G3G4 instead.

Same thing in your 3rd term.

That's all! You have the signs correct this time round! :)

You're right, I got it, I was rushing because I thought I had it. Thank you! :)