Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

RC circuit Transfer Function/bode plot

Status
Not open for further replies.

Mitchy190

Newbie level 4
Joined
Apr 19, 2012
Messages
6
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,333
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.

circuit d.png

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
 

The preferred forms to express the transfer function are a quotient of polynomials or zeros and poles.

If you work algebraically a little bit more the expression of H:

H(jω) = R2/(R2+(R1*1/jωC)/(R1+1/jωC)) = R2/(R+R2) (1+jωCR)/(1+jωCRR2/(R+R2))​

or more readable:

H(jω) = Ao (1+jωCR)/(1+jωCR3)

with:
Ao = R2/(R+R2) ; The DC gain
R3 = RR2/(R+R2) = R//R2 ; parallel R,R2​
 

The preferred forms to express the transfer function are a quotient of polynomials or zeros and poles.

If you work algebraically a little bit more the expression of H:

H(jω) = R2/(R2+(R1*1/jωC)/(R1+1/jωC)) = R2/(R+R2) (1+jωCR)/(1+jωCRR2/(R+R2))​

or more readable:

H(jω) = Ao (1+jωCR)/(1+jωCR3)

with:
Ao = R2/(R+R2) ; The DC gain
R3 = RR2/(R+R2) = R//R2 ; parallel R,R2​

Thank you for you reply! :) I understand this is the form I need to get my equation into, but I don't fully understand how you did? If you would be as kind to elaborate, I would much appreciate it? :D

Thanks
 

Thank you for you reply! :) I understand this is the form I need to get my equation into, but I don't fully understand how you did? If you would be as kind to elaborate, I would much appreciate it? :D
Thanks

\[Z_s = \frac{1}{\frac{1}{R}+j {\omega}C}\]

\[H(j{\omega}) = \frac{R2}{R2 + Zs} = \frac{R2}{R2 + \frac{1}{\frac{1}{R}+j{\omega}C}} = \frac{R2\,(\frac{1}{R}+j{\omega}C)}{R2 \,(\frac{1}{R}+j{\omega}C) + 1 } \\

= \frac{R2 \,(1+j{\omega}CR)}{(R2+j{\omega}CRR2 + R)} = \left(\frac{R2}{R+R2}\right) \;\frac{1+j{\omega}CR}{1+j{\omega}C \frac{RR2}{R+R2}}\]

This is a step by step, but if you have some skill in algebra are only one intermediate step.
 
Thank you that did help!

So now I have my transfer function in the form:

Transfer function.png

And 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:

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top