# Graphing Transfer Function

Discussion in 'Mathematics and Physics' started by EN0, Aug 18, 2010.

1. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,070
Likes:
963
Location:
NJ
Hi again,

On the graphs make sure to find the pass band gains and the -3db points as that is what this is all about.

Last edited: Sep 21, 2010
2. ### EN0Member

Joined:
Jul 14, 2009
Messages:
476
Likes:
11
Location:
Outer Space
I apologize for not coming on sooner! I fixed my error in the other post, so it should look ok now.

I’ve found some math software’s, which I’m looking into. I’ll let you know when the graphs are produced.

3. ### EN0Member

Joined:
Jul 14, 2009
Messages:
476
Likes:
11
Location:
Outer Space
Hey MrAl,

I am very sorry this took so long to generate, but I had a lot of school work and then I needed to learn the syntax of Freemat. Anyway, here it is:

-3dB Point of Ordinary LPF: 15915.49 Hz
-3dB Point of LPF With Load: 15915.49 Hz

Given:

R1 = 1kΩ
C = 10nF
R2 = 1kΩ
Vin = 5V

Is that correct?

#### Attached Files:

• ###### TransferFunctionCombined1.PNG
File size:
8 KB
Views:
124
Last edited: Sep 28, 2010

Joined:
Jan 12, 1997
Messages:
-
Likes:
0

5. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,070
Likes:
963
Location:
NJ

Hi again,

Something doesnt look right in the graphs. How does the amplitude get as high as 5v in the top trace for example?

6. ### EN0Member

Joined:
Jul 14, 2009
Messages:
476
Likes:
11
Location:
Outer Space
Well the top graph is without a load, so it should start from 5V, shouldn't it? The lower graph is with the load so it begins at 2.5V. I should have told you beforehand which was which, sorry.

7. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,070
Likes:
963
Location:
NJ
Hi again,

So you are using an input voltage of 5v then? Usually you would use 1v to graph it. Of course it helps to graph in db gain too rather than voltage so you can easily spot the -3db point.

8. ### EN0Member

Joined:
Jul 14, 2009
Messages:
476
Likes:
11
Location:
Outer Space
After all that, I need to change the graphs again?

Alright, I'll fix it up and put it in terms of dB. One volt for the input voltage is a good idea, I'll implement that as well.

Till Then,

Austin

Last edited: Oct 1, 2010
9. ### EN0Member

Joined:
Jul 14, 2009
Messages:
476
Likes:
11
Location:
Outer Space
Hey MrAl,

Apparently there is a mistake in our mathematics? Hayato provided the graph that is attached and his equations are different.

File size:
51.5 KB
Views:
125
10. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,070
Likes:
963
Location:
NJ
Hello,

I'll double check and get back here.

11. ### EN0Member

Joined:
Jul 14, 2009
Messages:
476
Likes:
11
Location:
Outer Space
Hi MrAl,

I believe our first transfer function is correct, but the second one isn't. I went over the math myself and will show you my results.

Conventional Low-Pass Filter

Final Equation

[Error: Syntax\ error : /usr/bin/latex --interaction=nonstopmode 3ea467687f52918902822f9177f893ab.tex && /usr/bin/dvipng -q -D 300 -T tight -gamma 2.0 -bg Transparent -o 5e0bfaa73c137e992b3eb5d28e591195-2.png 3ea467687f52918902822f9177f893ab.dvi]

Final Equation

I believe that's correct?

Last edited: Oct 23, 2010
12. ### HayatoMember

Joined:
Jun 21, 2006
Messages:
694
Likes:
14
Location:
BR
Plot the graph again, if they are similar to the ones I've plotted, they are ok.

13. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,070
Likes:
963
Location:
NJ

Hello again,

Ok since you only have a problem with the circuit with the load resistor we'll concentrate on that one.

With R1 being the input resistor and R2 being the load resistor and C the only capacitor, the transfer function is:
Vout=(Vin*R2)/(s*C*R1*R2+R1+R2)

and the amplitude (after subst s=j*w) is:
VoutAmpl=(Vin*R2)/sqrt(w^2*C^2*R1^2*R2^2+R2^2+2*R1*R2+R1^2)

Now this last result has been verified in a circuit simulator, which by the way you should be doing too. You can compare your result to that VoutAmpl and if you dont get that or an equivalent expression then it can not be correct so you'll have to see what went wrong.
You might note that both R1 and R2 appear in the denominator too. I also think that using Latex just complicates matters rather than making them simpler. You should probably avoid that until you get the final correct solution.

Last edited: Oct 24, 2010