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Femme_physics
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Trying to get the hang of transfer function. Did I get it?
http://img24.imageshack.us/img24/5260/ssketch.jpg
http://img24.imageshack.us/img24/5260/ssketch.jpg
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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.
Femme_physics said: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... :)
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?
Femme_physics said:OK. The formula says
C/R = Sum of all the paths / 1- sum of all the loops
Femme_physics said:This is actually my NEW attempt. Pathways in black. Loops in orange.
http://img545.imageshack.us/img545/7681/laster.jpg
Femme_physics said: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)
That's the method we're being taughtOkay. That's another way to solve it.
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?
Femme_physics said:That's the method we're being taught
Femme_physics said:What do you mean by "rectify"?
Femme_physics said:I agree, I forgot the large loop.
http://img440.imageshack.us/img440/7948/gotitj.jpg
of course. It's the lower row in the image I've just made.
Femme_physics said:Oh, sorry That would be G3(+R-C)
That's the method we're being taught
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.
Let's start with the small loop.
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?
Femme_physics said:Oh, I don't mind at all, you always give me new insights.
Femme_physics said:'the hell is wrong with my English?
Femme_physics said: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.
Femme_physics said:Oh and...PIKABOO! :) Just in case u missed my cameo appearances.
http://img163.imageshack.us/img163/1453/pikabo.png
Femme_physics said: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?
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!
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).
Femme_physics said:*kissy winky*
Femme_physics said:That's exactly what all my loops are minused!
Femme_physics said:Ah, thanks. But I did get the right answer, eh?
But they aren't!
You said yourself: "C/R = Sum of all the paths / 1- sum of all the loops"
There is already a minus in your formula.
It the loop has a minus too, it becomes a plus!
Femme_physics said:You mean ALL of the minuses at the bottom row should have been written as pluses then?
A transfer function is a mathematical representation of the relationship between the input and output of a system. It describes how the system responds to different inputs, and is often used in control systems and signal processing.
A transfer function is calculated by taking the Laplace transform of the system's differential equation. This converts the differential equation into a polynomial expression, with the coefficients representing the system's parameters.
The key components of a transfer function are the numerator and denominator polynomials. The numerator represents the output of the system, while the denominator represents the input. The order of the polynomials is determined by the highest exponent of the Laplace variable, s.
The transfer function provides important information about a system's behavior, including its stability, frequency response, and time domain behavior. It also allows us to predict the output of the system for a given input.
Transfer functions have a wide range of applications, including in control systems, electrical circuits, and signal processing. They are used to design and analyze systems, predict and improve performance, and troubleshoot problems in real-world systems.