# Graphing Transfer Function

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

1. ### EN0Member

Joined:
Jul 14, 2009
Messages:
476
Likes:
11
Location:
Outer Space
I saw my mistake, it's fixed now.

Thanks for the note!

2. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,057
Likes:
963
Location:
NJ
Hello again,

Oh ok that's great. If you want to work on more filters that's cool too, and now that we've done a couple like that we can look at another way to get the amplitude when we have real and imag in the numerator and real and imag in the denominator. I'll use RN and IN for numerator real and imag, and RD and ID for denominator real and imag.

For an equation thus:

(RN+IN*j)/(RD+ID*j)

we can calculate the amplitude like this:
Ampl=sqrt(RN^2+IN^2)/sqrt(RD^2+ID^2)

so that might make it a little easier for you to calculate the amplitudes of these more complex filters.

Last edited: Sep 14, 2010
3. ### EN0Member

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

I'd love to continue learning more about filters! That math gimmick you provided is very useful, I'll use that from now on.

What should I do next?

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

5. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,057
Likes:
963
Location:
NJ

Hi again,

Well, now that you have a few formulas you should probably plot a couple of them. For example, plot the low pass filter without that extra resistor R2 and then plot it with the extra resistor, making the resistor somewhat comparable to the first resistor R1. You should then find the -3db point for both and also the passband gain, then compare these findings.
This should be interesting.

6. ### EN0Member

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

That sounds great, I'll post my work when I'm done.

Thanks!

7. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,057
Likes:
963
Location:
NJ
Hi again,

You can do it however you want, but it may be more illustrative if the two plots are on the same graph, if you have the software to do that of course.

8. ### EN0Member

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

I’ll try to produce the graphs via MathCAD. If that doesn’t work, I believe I can upload the graphs I produce on my HP-50G calculator.

Thanks,

Austin

Last edited: Sep 15, 2010
9. ### EN0Member

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

I guess I'm still not quite sure how I seperate the imaginary and real parts, given a circuit? I suppose that I solve it using nodal analysis, and then try to get the terms in the order of that first equation you listed? Or is that first equation just an example equation?

Thanks!

10. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,057
Likes:
963
Location:
NJ
Hi again,

It's all in the grouping. Let me give a few examples...

A+B+C+D

Those four terms are all real, so the real part is A+B+C+D. They are all real because there is no 'j' multiplying any of them.

A+B+C*j+D

Here we have to group the terms that contain 'j' as imag and those that dont as real, so the real part is A+B+D and the imag part is C.

A+B*j+C+D*j

Here we have two terms that have 'j' multiplying them, and two that dont. The two that dont are real, and the two that do are imag, so the real is A+C and the imag is B+D.

A*j+B*j+C*j+D*j

Here everything is multiplied by j, so everything is imag and the real part is zero.

One way to look at it is we group all the 'j' terms together and that forms the imag part, and whatever is left over is real.

A+B*j+C*j+D*j

we can regroup this as A+j*(B+C+D) so A is real and B+C+D is imag.

When we have a numerator and denominator, we do the same thing except separately for each one.

(A+B*j+C*j)/(D+E+F*j+G*j)

we can regroup as:
(A+j*(B+C))/(D+E+j*(F+G))

and so real in the numerator is A, and imag numerator is B+C, and real in denom is D+E and imag is F+G.

Make sense?

Now you try one:

(A+B+C*j+D+E*j+1)/(F*j+G+H*j+3)

Find the real and imag in numerator, and real and imag in denominator.

Last edited: Sep 15, 2010
11. ### EN0Member

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

Thanks for the examples.

12. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,057
Likes:
963
Location:
NJ
Hi again,

Did that help, as in did that help you understand how to go about identifying the real parts and imag parts?

13. ### EN0Member

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

Yes, that did help.

I have produced the equations you wanted, and I've been trying to graph them but unfortunately I can't get MathCAD to graph anything! Have you had any experience with MathCAD? I'm trying to get my HP-50G to download the graphs on the PC.

Here are the equations:

Low-Pass Filter:

Simplify:

Seperate Real & Imaginary:

Calculate Amplitude:

[Error: Syntax\ error : /usr/bin/latex --interaction=nonstopmode 5227feedf0315d479379996667248f50.tex && /usr/bin/dvipng -q -D 300 -T tight -gamma 2.0 -bg Transparent -o e84e35789f06fc48d9f5335bbd49832e-2.png 5227feedf0315d479379996667248f50.dvi]

Simplify:

Seperate Real & Imaginary:

Calculate Amplitude:

Last edited: Sep 15, 2010
14. ### EN0Member

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

I got my HP-50G to show the graphs, but only individually. I don't have a large enough span in order to plot them both. The problem is that with the 50Ω load, it reduces
drastically down to the mV range. So I don't even see the latter graph, only the former one.

For some reason I still can't get MathCAD to graph anything!

Last edited: Sep 15, 2010
15. ### EN0Member

Joined:
Jul 14, 2009
Messages:
476
Likes:
11
Location:
Outer Space
Here are the individual graphs, in case you'd like to see them.

File size:
3.9 KB
Views:
110
File size:
3.8 KB
Views:
117
16. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,057
Likes:
963
Location:
NJ
Hi again,

What values are you using for R1 and C, and R2 ?

Also, did you find the -3db point for these curves?

I think you might also have to label the graph tic marks so we know what the amplitude came out to, and what the frequency is for each point.

Last edited: Sep 16, 2010
17. ### EN0Member

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

R1 and C are the same for both circuits.

There are other math softwares around for free, I'll check into them.

-3dB Point of Low-Pass:

[Error: Syntax\ error : /usr/bin/latex --interaction=nonstopmode 8db46bb2d1d002eccb9e09ed08156a19.tex && /usr/bin/dvipng -q -D 300 -T tight -gamma 2.0 -bg Transparent -o 6b1f98f545d0573fe35c576e965491b0-2.png 8db46bb2d1d002eccb9e09ed08156a19.dvi]

The -3dB point frequency is the same for both filters, since I have the same values.

Last edited: Sep 16, 2010
18. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,057
Likes:
963
Location:
NJ
Hi again,

There is something wrong with the denominator for your loaded filter. You'll have to go over that.

Also, try using 1k for R1 and 1k also for RL so that the load is comparable to R1. Once you do that, find the passband gain and -3db point for both filters and dont assume anything.

19. ### EN0Member

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

I really don't see anything wrong with the denominator, can you provide a hint?

I'll use the same values for RL and R1.

Thanks,

Austin

20. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,057
Likes:
963
Location:
NJ

THIS LINE LOOKS OK:

Seperate Real & Imaginary:
THIS LINE DOESNT:

Calculate Amplitude:
SO THIS LINE IS NOT CORRECT EITHER:

Note in the second line you lost a variable again, and in the third line it comes back but it's not correct.
I think you should take one step at a time, so instead of separating real and imag parts and squaring all in one step, do the separation first and after that is completely done THEN do the squaring or whatever else is needed.

Last edited: Sep 17, 2010
21. ### EN0Member

Joined:
Jul 14, 2009
Messages:
476
Likes:
11
Location:
Outer Space
I forgot to include the "+" into the second line, sorry.