# RC circuit Transfer Function/bode plot

Discussion in 'Electronics Homework Help' started by Mitchy190, Apr 19, 2012.

1. ### Mitchy190

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Apr 19, 2012
1. The problem statement, all variables and given/known data

I have for the attached circuit measured the amplitude and phase response, but the question I have to answer is too theoretically, derive, similar results and then compare the two methods. It is suggested to figure out the transfer function and plot the magnitude and phase response.

:
The component values are know, and are:

R1 = 47k
R2 = 4K7
C = 10nF

I am having difficulty driving the transfer function for this circuit and some much needed help would be appreciated!

2. Relevant equations

H(jω) = R2/(R2+(R1*1/jωC)/(R1+1/jωC))

3. The attempt at a solution

So far I have converted the circuit into 'standard form':

Z1 = R1, Z2 = R2, Zc = 1/jωC, Vin = Vip and Vout = Vop (p = Phasor)

and then I know that H(jω) = Vop/Vip

And knowing this I worked out that

Vop = (R2/(R2+(R1*1/jωC)/(R1+1/jωC)))*Vip

H(jω) = R2/(R2+(R1*1/jωC)/(R1+1/jωC))

And this is where I am stuck. I'm not sure if this is the correct method to derive the transfer function, or even right at all, but this is just my thinking and attempt . So much help is needed

Thank you

Last edited: Apr 19, 2012
2. ### Harald KappModeratorModerator

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Nov 17, 2011
The transfer function is Vout/Vin=H(jw) as you have correctly noted,
Also you equation (2) for H8jw) seems entirely correct to me.

So you have the correct solution. Where exactly is the problem?

Harald

3. ### Mitchy190

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Apr 19, 2012
As I recall from my lectures the transfer function has to be in a specific form, where the break frequency, frequency response and phase response can be approximated (bode plot). The problem is I do not recall how to get my derived transfer function into this form, or to be honest, I'm not entirely sure why it has to be in this form?

Mitch

4. ### Harald KappModeratorModerator

11,824
2,756
Nov 17, 2011
What you're after is, i think, simplifying the equation such that no complex terms are in the denominator.

H(jω) = R2/(R2+(R*1/jωC)/(R+1/jωC)) (note that R1 is only named R in the picture)

pick the term (R*1/jωC)/(R+1/jωC) multiply nom. and denom. by jωC to get
(R*1/jωC)/(R+1/jωC) = R/(1+jωRC)

You now have H(jω) = R2/(R2+(R/(1+jωRC))

multiply nom. and denom. by (1+jωRC) and you'll get
H(jω) = (R2*(1+jωRC))/(R2*(1+jωRC)+R)

multiply the parentheses in the denom into the long form and you'll get
H(jω) = (R2+jωRR2C))/((R+R2)+jωRR2C)

now divide nom. and denom. by (R+R2) and get
H(jω) = (R2/(R+R2) * (1+jωRC)) / (1+(R2/(R+R2))*jωRC)

simplify by substituting (R2/(R+R2) = A to get
H(jω) = A *(1+jωRC) / (1+A*jωRC)

now you need the conjugate of (1+A*jωRC) which is (1-A*jωRC). Multiply nom. and denom. by the conjugate and get
H(jω) = A *(1+jωRC)*(1-A*jωRC) / ((1+A*jωRC)*(1-A*jωRC))

Since (1+A*jωRC)*(1-A*jωRC) = 1^2-(A*jωRC)^2 = 1+(AωRC)^2 the complex expression in the denominator has now been eliminated:
H(jω) = A*(1+jωRC)*(1-A*jωRC) / (1+(AωRC)^2)

Now re-subsitute A and you're almost finished.

Harald

P.S.: I hope i did get all those parentheses right

5. ### Mitchy190

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Apr 19, 2012
Thank you that did help!

So now I have my transfer function in the form:

I understand that the function has one pole at

-1/R1*C

And one zero at

-1/R3*C

I wish to approximate the magnitude and phase response through a bode plot, but I am unsure on the next step, and some advice would help. Thanks

Last edited: Apr 20, 2012

11,824
2,756
Nov 17, 2011
7. ### Laplace

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Apr 4, 2010
I took your transfer function and had Mathcad display it as a Bode plot. If anyone does it the long way, I'm wondering if Mathcad provides an accurate graph.

File size:
29.6 KB
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442
8. ### Mitchy190

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Apr 19, 2012
Thats really cool! I have actually measured the real life frequency response of the circuit, and your plot looks accurate. But a bode plot is only an a approximation so your plot is more accurate than a bode plot.

below are the results I obtained from measuring the circuits response

View attachment Frequency1.zip

9. ### Mitchy190

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Apr 19, 2012
Actually after inspection, the frequency axis seem to be a factor of x10 out.

And the magnitude a factor of x2 out.

Do you have any idea why?

Last edited: Apr 21, 2012
10. ### Mitchy190

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Apr 19, 2012
I am unsure that my transfer function is right now, I have plotted a bode for the magnitude and it does not compare to my real life results at all, which i am assuming are right.

I worked out the bode plot by

|H(Jω)|=20Log(|A|)+20Log(|1+jωRC|)−20Log(|1+jωR_3C|)

And I got these results

View attachment 20log.zip

The real measured results I obtained are:

View attachment Frequency1.zip

As you can see they look completely different? Is my transfer function wrong, or is it the method in which I have plotted the bode?

Thanks.

11. ### Mitchy190

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Apr 19, 2012
Hey Laplace

I am just wondering are using mathcad prime 1.0? or 2.0?

And how did you plot transfer functions in them?

thanks

12. ### Laplace

1,252
184
Apr 4, 2010
@Mitchy190:

I'm still using Mathcad 14, running on Windows 2003, running on a virtual machine in CentOS Security-Enhanced Linux 6 with a host-only network. (I would never expose a working copy of Windows to the internet.)

As for how to plot a trasnfer function, I made a pdf copy of the Mathcad window contents. You see exactly what I see. I did it that way.