How Do I Solve Part C of a Low Pass Filter Exam Problem?

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Discussion Overview

The discussion revolves around solving part C of a low pass filter exam problem, focusing on the calculations involving resistances and capacitances. Participants are exploring methods to simplify circuit components and derive transfer functions, while also addressing related concepts such as Bode plots.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant describes their approach to parts A and B, simplifying resistances and capacitances to eliminate the frequency component ω.
  • Another participant suggests that if parts A and B are correct, the relationship R1C1 = R2C2 holds, and questions how changes in R1 affect the transfer function.
  • A different participant explains their method of simplifying a resistor and capacitor in parallel, leading to a complex impedance expression, and questions the correctness of their approach.
  • One participant challenges the calculation of the combined impedance, emphasizing the importance of using admittance instead of impedance for clarity in frequency-independent gain.
  • Several participants express uncertainty about drawing Bode plots and seek guidance on the process, with one providing external resources for further learning.
  • Another participant confirms their calculations for impedance and admittance, seeking validation of their approach while being advised to focus on admittance rather than impedance.

Areas of Agreement / Disagreement

Participants express varying interpretations of the calculations and methods for solving the problem, indicating that multiple competing views remain. There is no consensus on the correctness of the approaches discussed.

Contextual Notes

Participants highlight potential misunderstandings in the simplification of circuit components and the use of frequency-dependent relationships. There are unresolved mathematical steps and assumptions regarding the definitions of impedance and admittance.

Who May Find This Useful

Students preparing for exams in electrical engineering or related fields, particularly those focusing on circuit analysis and filter design, may find this discussion relevant.

Synops
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Exams are coming up and I've been doing some revision from past papers. Having issues with working out part c of the attached problem. I've already calculated part a and b, possibly correct. Basically I simplified resistances and capacitance. And to have a constant frequency I let Z equal the inverse of the simplified resistance to cancel ω the frequency component. I then used this information to determine a new resistance and capacitance that had no ω component.

Any hints or tips would be greatly appreciated.

Thanks
 

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If you did parts a and b correctly (they're really the same question) then you got R1C1 = R2C2 where 1/Z = G1 + sC1, G1 = 1/R1, and R2 = 1/G2 = 10K and C2 = 1 nF.

Now R1 → 2R1 so what is your new transfer function? And same question when R1 → R1/2. Do you know how to draw Bode plots?

Part 1 c iii does not specify the circuit so can't be answered.
 
Not sure exactly what you meant by your answer to the first two parts.. However the way i calculated it was by simplifying the resistor and capacitor in parallel. This gave me 1x10^4 + 1x10^6/jw. Then I assumed that to remove jw, since Z is now in series with this equivalent resistance, it would have to equal -1x10^6/jw.

For part b. I assumed a resistor of 1Ω in parallel with a capacitor of 0.1x10^-6. Simplifying this to obtain 1x10^6/jw + 1. I assume this answer would be negative as that's how it would cancel if it was added in series. This is the new realized Z which then adds to the result of the previous question which gave a result of 9999Ω. Which has no ω component.

Is this the correct approach?

Also, I'm not sure how to draw a bode plot. I've done them in practical classes with a simulation program called SPICE, however the book doesn't explain how to draw them very well.. Would you mind giving me an overview of the process?
 
Last edited:
Synops said:
Not sure exactly what you meant by your answer to the first two parts.. However the way i calculated it was by simplifying the resistor and capacitor in parallel. This gave me 1x10^4 + 1x10^6/jw.

What do you mean by 'simplifying'? Do you mean determining the combined impedance of the 10K resistor and 1 nF capacitor in parallel? If so, you did that wrong. Y = G + jwC so Z = 1/Y where G = 1/R. The real part of the complex impedance of the 10K and 1 nF is frequency-dependent.

BTW 1 nF = 10^(-9)F, not 10^(-6).

So you don't have the correct network for frequency-independent gain yet.

******************************************************************
In this situation I know from experience that it's much easier to deal in admittance Y, conductance G and susceptance B than impedance Z, resistance R and reactance X. The relations are simple: Y = 1/Z, B = 1/X and G = 1/R. Example: for a capacitor, B = wC and Y = jB.

You'll recall that for parallel-connected components, susceptances are added: so for your R and C, Y = G + jB = G + jwC, G = 1/R.

Given the above, what is Vout/Vin of your diagram if we replace Z with Y = 1/Z?

Also, I'm not sure how to draw a bode plot. I've done them in practical classes with a simulation program called SPICE, however the book doesn't explain how to draw them very well.. Would you mind giving me an overview of the process?

This is much too long a subject for me to discuss here. Look at

http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/Freq/Freq5.html or

www.dartmouth.edu/~sullivan/22files/Bode_plots.pdf
 
Ok thanks for all your help!

Just to get me back on track, Z is meant to equal Z0 which is: 1/(1x10^(-9)jw + 1/10000)?

I just used the relationships you were describing to find: Y = 1 x 10^(-9)jw + 1/10000

Then to find Impedance: Z = 1/Y, therefore Z = 1/(1x10^(-9)jw + 1/10000)

Is this correct so far?
 
Synops said:
Ok thanks for all your help!

Just to get me back on track, Z is meant to equal Z0 which is: 1/(1x10^(-9)jw + 1/10000)?

I just used the relationships you were describing to find: Y = 1 x 10^(-9)jw + 1/10000

Then to find Impedance: Z = 1/Y, therefore Z = 1/(1x10^(-9)jw + 1/10000)

Is this correct so far?

Fine so far! But forget Z of your network. You'll be dealing with Y's instead.

Also, don't put in numbers yet. Call the 1 nF = C2 and the 10K = R2 = 1/G2.

Now, what is Vout/Vin in terms of this Y2 = G2 + jwC2 and the Y equivalent of the Z as marked on your diagram? Defie Y1 = 1/Z.
 

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