# Circuits in the Frequency Domain

Discussion in 'Homework Help' started by jbriaris, May 7, 2013.

1. ### jbriaris Thread Starter New Member

Apr 11, 2013
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Hi All

I'm trying to get a better understanding of analysing circuits in the frequency domain. Attached are two circuits consisting of resistors and capacitors, and powered by a constant current source.

Figure 1: Here the resistor R1 and capacitor C1 are in parallel, so we can calculate the equivalent impedance $Z_{eq}$ using

$Z_{eq} = \frac{Z_{R1} Z_{C1}}{Z_{R1} + Z_{C1}}$

where

$Z_{R1} = R$ and $Z_{C1} = \frac{1}{j\omega C}$

which after substitution and rearranging gives

$Z_{eq} = \frac{R}{1+j\omega RC}$

Then, given that I know the frequency of the current source and its magnitude, I can simply apply Ohm's law to get $V_1$, i.e.,

$V_1 = I_1 Z_{eq}$

I understand this. However, in Figure 2, I'm a little lost as to how to calculate the equivalent impedances if I want to calculate a value for $V_2$. My initial attempt was to say R2 and C2 are in parallel and use the above method to calculate $Z_{eq2}$, then to say R3 and C3 are in parallel to calculate $Z_{eq3}$. This then forms a simple potential divider circuit where

$V_2 = I_2\times \frac{Z_{eq3}}{Z_{eq2}+Z_{eq3}}$

but this doesn't seem to work, i.e., the spice simulation doesn't match my calcs. Am I doing something fundamentally wrong by calculating the impedances in this manner?

Thanks for any tips.

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Would you be able to calculate V2 voltage if you replace C1 and C2 with a resistors?

3. ### WBahn Moderator

Mar 31, 2012
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Your problem is the claim that, in the second circuit, R2 and C2 are in parallel. What makes you think that? Remember, to be in parallel, both components have to have exactly the same voltage across them at all times. A tast for this is that you must be able to put your left index finger on a node that is connected to one side of both components and your right index finger on a node that is connected to the other side of both components. You can do that with R3 and C3, so they are in parallel. But you can't do that with R2 and C2, so they are not.

Once you have your equivalent impedance for R3 and C3, what can you say about that equivalent impedance and R2? Once you combine those into a new equivalent impedance, THEN you have something that is in parallel with C2.

Another way to approach it is to use Nodal or Mesh analysis. Mesh is probably the easier since you have a current source in an outer branch and, hence, only have two non-trivial equations to solve.

4. ### jbriaris Thread Starter New Member

Apr 11, 2013
19
0
Okay. So if I replace C2 and C3 with resistors R4 and R5, I can work out the total impedance of the circuit by working back from R3, i.e., R3 & R5 are in parallel, thus

$R_{eq1} = \frac{R_3R_5}{R_3+R_5}$

R2 is then in series with the above such that

$R_{eq2} = R_2 + R_{eq1}$

and the above is then in parallel with R4, so that

$R_{eq3} = \frac{R_4+R_{eq2}}{R_4R_{eq2}}$

The above will be the total impedance of the circuit. I can use that to calculate V1, I was then hoping to calculate V2 using

$V_2 = V_1\frac{R_{eq1}}{R_{eq3}}$

but I'm not getting the right answer. I realise you would usually use Kirchoff's laws to set up a system of equations, but I was hoping to do it like this as I have quite a large multi-branched system and want to work back from the terminal branches to the source calculating total impedances at the nodes. Then work back out from the source to the terminals calculating currents and voltages.

5. ### WBahn Moderator

Mar 31, 2012
20,064
5,665
For the most part you are on the right track, but you are botching some of the details.

Always, always, ALWAYS check your units!

Explain carefully, as if to a classmate WHY you think this will work and you will probably spot why it won't and what you need to do differently.

It's hard for us to provide feedback about work that is not shown. We are not mind readers.

6. ### jbriaris Thread Starter New Member

Apr 11, 2013
19
0
Yes, that was a silly typo.

Sorry. I find it difficult to visualise these circuits. If I use Kirchoff's current law at node V2, i.e.,

$I_2 - I_3 - I_5 = 0$

where

$I_2 = \frac{V_1-V_2}{R_2}, I_3 = \frac{V_2}{R_3}, I_5 = \frac{V_2}{R_5}$

then

$-V_2(\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_5}) = -\frac{V_1}{R_2}$

which after rearranging gives

$V_2 = \frac{V_1}{1+\frac{R_2}{R_3}+\frac{R_2}{R_5}}$

This (I think) provides the right answer for V2??

So I think it is plausible then to work through a multi-branched network as previously suggested. When I replace resistors R4 and R5 with capacitors, I simply use their reactance - making the circuit a function of frequency.

Apologies. I was just testing the formulae by setting resistances to 1 and input current to 1 and modeling spice.

7. ### WBahn Moderator

Mar 31, 2012
20,064
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That's okay. It takes practice. But it's also all the more reason to take things carefully and step by step.

Correct.

Inline with your prior work, you have an expression for V1 and you have V2 which liies along a path from V1 to ground. V2 can therefore be found by treating it as a voltage divider, or (equivalently) finding the current down that path and multiplying that by the impedance between V2 and ground.

So you had V1 going through Req2, thus the current is

V1/Req2

The impedance that V2 is developed across is Req1, so you have:

V2 = (V1/Req2)*Req1 = V1*(Req1/Req2)

Do you see why using Req3 instead of Req2 is incorrect?

jbriaris likes this.
8. ### jbriaris Thread Starter New Member

Apr 11, 2013
19
0
Great! Thanks. .

Req3 has a path to ground via R4, so its nonsense to try and use that to find V2. The way you describe it makes sense, i.e., thinking about the loads/impedances in relation to their path to ground.

Will take-on the multi-branch network tomorrow with my new-found knowledge (head needs some rest!). Thank you so much.

9. ### WBahn Moderator

Mar 31, 2012
20,064
5,665
Very good. You are doing just fine.