Author Topic: Complex Impedances  (Read 4246 times)

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Offline fourierpwn

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Complex Impedances
« on: May 21, 2015, 10:11:23 pm »
Hi all,

I am unsure of a problem I have been given and I'd appreciate a little guidance if possible. (I don't mean for this to be "others doing my homework" type of thing, I'm just a little unsure.)

I have attached a circuit in which I am asked to find the value of C that will satisfy Vo(t) = Vc(t)/2

My approach has been to get the complex impedance of the capacitor 0.2uF and then it is essentially a voltage divider, yeah? Though, I get a little stuck about going any further. Perhaps my approach is wrong?

I have attached the circuit and I appreciate any help that can be given.  :)
 

Offline Andy Watson

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Re: Complex Impedances
« Reply #1 on: May 21, 2015, 10:30:09 pm »
The two capacitors will act as a voltage divider, however, if you include the R you have to remember that all your calculations will need to be complex. There is much more information than you need in that circuit - have you asked the correct question? Finding the value of C that results in Vo = Vc/2 is trivially easy - you don't need a calculator.
 

Offline T3sl4co1l

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Re: Complex Impedances
« Reply #2 on: May 21, 2015, 10:35:16 pm »
Yeah, the source, resistor and current are irrelevant if you set up the system as ratios.

You can use them, to "test" the circuit, and solve everything about it (all node voltages, branch impedances, loop current), then find the parts you need, divide, and wind up with the same ratio.

The latter is a lot of busy work, but it's important to keep in mind.  Sometimes it's a good approach, like when you have a more complex circuit that doesn't yield ratios by inspection so easily.

What are you stuck on, what is your approach?  Show your steps?

Tim
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Offline fourierpwn

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Re: Complex Impedances
« Reply #3 on: May 21, 2015, 10:40:06 pm »
Thanks for the reply Andy.

Yeah, I have asked the correct question though judging from your reply I get the feeling that I am missing something simple here. There are subsequent questions too so perhaps all the info is related to those q's.

From your statement, "you don't need a calculator" then I'm starting to think that the impedance approach is not the correct way to go about this question.

I'll keep thinking  :)
 

Offline fourierpwn

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Re: Complex Impedances
« Reply #4 on: May 21, 2015, 10:48:08 pm »
What are you stuck on, what is your approach?  Show your steps?

Tim, as I mentioned my approach was to get the complex impedance of the capacitor then attempt to do some algebra to get a value for C. Though I wasn't sure about how to go about this really.

I'm not too sure I understand what you mean about setting the system up as ratios.

Honestly, I'm a little intimidated by the wealth of knowledge on these forums. Sure, I may understand this circuit eventually. Though to progress to the stage of just understanding things on sight seems like a distant wish.  :palm:

I'll keep at it though. :-+
 

Offline T3sl4co1l

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Re: Complex Impedances
« Reply #5 on: May 21, 2015, 10:52:28 pm »
Have you done resistance networks, resistors in series and parallel..?

Impedance works precisely the same way -- all the operations that are valid for real numbers also work for complex numbers.  So all the equations (like the voltage divider formula) are the same form, and work just the same.

This remains true even when you write out the complex numbers as their components, such as an impedance Z = R + jX, as long as you follow the multiplication rule for j (j * j = -1).

Tim
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Offline Andy Watson

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Re: Complex Impedances
« Reply #6 on: May 21, 2015, 10:58:41 pm »
There are subsequent questions too so perhaps all the info is related to those q's.
Ah, that would make sense.

Quote
From your statement, "you don't need a calculator" then I'm starting to think that the impedance approach is not the correct way to go about this question.
If this is part of a bigger question I suspect you need to set-out the appropriate equations as part of the preparation for the work ahead. The current is common to both capacitors - use equations that relate Vcap to i and Xc. A big hint is that Vo = Vc/2 - if it were not /2, I would be reaching for a calculator.
« Last Edit: May 21, 2015, 11:00:28 pm by Andy Watson »
 

Offline fourierpwn

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Re: Complex Impedances
« Reply #7 on: May 21, 2015, 11:01:49 pm »
Yep, I have done resistance networks. Which is why my approach was to use the impedances to go for the voltage divider. I'm a little confused though, is this the (a) correct approach?

So I have Ztotal = 50 - j5 (+ C)

then Vo(t) = Vs(t) * C/(50-j5)+C

But I'm not sure where to go from here (if this is correct)
 

Offline TimFox

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Re: Complex Impedances
« Reply #8 on: May 21, 2015, 11:10:29 pm »
If you are doing it the hard way, you also need to calculate Vc since the homework question was for the ratio between Vc and Vo, not Vs and Vo.  Starting from Vc makes the calculation extremely simple.
 

Offline fourierpwn

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Re: Complex Impedances
« Reply #9 on: May 21, 2015, 11:15:33 pm »
Starting from Vc(t), I can see that C should also equal 0.2uF for Vo(t) to be Vc(t)/2. Though I don't know how to write it out to give a proper answer.
 

Offline TimFox

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Re: Complex Impedances
« Reply #10 on: May 21, 2015, 11:18:00 pm »
If both capacitors have the same value, then at all frequencies they have the same reactance and the same impedance.  Continue from there.
 

Offline fourierpwn

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Re: Complex Impedances
« Reply #11 on: May 21, 2015, 11:21:48 pm »
If both capacitors have the same value, then at all frequencies they have the same reactance and the same impedance.  Continue from there.

I can see this, though I arrived at the value for C by the simple voltage divider rule considering that you're treating the caps. as impedances. My problem now is, I don't know how to derive this value (0.2uF)

Thank you for your input though.
 

Offline German_EE

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Re: Complex Impedances
« Reply #12 on: May 22, 2015, 06:34:27 am »
Do they still teach engineers how to use a Smith Chart these days or is it now all based around computers?
Should you find yourself in a chronically leaking boat, energy devoted to changing vessels is likely to be more productive than energy devoted to patching leaks.

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Offline fourierpwn

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Re: Complex Impedances
« Reply #13 on: May 22, 2015, 10:03:23 am »
Do they still teach engineers how to use a Smith Chart these days or is it now all based around computers?

Well I've never come across Smith charts before, and I wouldn't say that it's all computers now. Only a little bit for simulation etc.
 

Offline Asmyldof

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Re: Complex Impedances
« Reply #14 on: May 22, 2015, 10:05:54 am »
Do they still teach engineers how to use a Smith Chart these days or is it now all based around computers?
I wish I could actually +1 that. Although, that might be too computerry ^-^
If it's a puzzle, I want to solve it.
If it's a problem, I need to solve it.
If it's an equation... mjeh, I've got Matlab
...
...
(not really though, Matlab annoys me).
 

Offline T3sl4co1l

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Re: Complex Impedances
« Reply #15 on: May 22, 2015, 10:57:52 am »
Ehh, Smith charts are good for transmission lines and RF stuff, not so much for general circuits.  You're better off working the algebra.

Yep, I have done resistance networks. Which is why my approach was to use the impedances to go for the voltage divider. I'm a little confused though, is this the (a) correct approach?

So I have Ztotal = 50 - j5 (+ C)

then Vo(t) = Vs(t) * C/(50-j5)+C

But I'm not sure where to go from here (if this is correct)

What is C?  -- I think you forgot to multiply by j*w (or s).  First step (easy to forget!): convert component values to impedances (Zc = 1 / (j*w*C), etc.).  Then substitute those into the node/loop equations and continue.

Generally, frequency is w (lowercase omega, but good luck getting that to work on this forum..), so that time-domain voltage is written as A * sin(w*t).  So here, w = 10^6 rad/s.

Let's label the resistor R, the "C" capacitor C1 and the "0.2uF" capacitor C2.

Total impedance Zt = Zr + Zc1 + Zc2
= R + 1 / (j*w*C1) + 1 / (j*w*C2)
= R - (j / w) * (1/C1 + 1/C2)

(Note the use of 1/j == -j, an identity of the multiplicative rule.)

Current = I = Vs / Z
= Vs / [R - (j / w) * (1/C1 + 1/C2)]
(and so on, however far you wish to simplify it)

Voltage on C2 = Vo = Zc2 * I
= Vs * Zc2 / (Zr + Zc1 + Zc2)

Voltage on C1 + C2 = Vc = (Zc1 + Zc2) * I
= Vs * (Zc1 + Zc2) / (Zr + Zc1 + Zc2)

Vo / Vc = 1/2 = capacitor divider gain (as requested) =
Zc2 * I / ((Zc1 + Zc2) * I)
= Zc2 / (Zc1 + Zc2)

The 1/(j*w) stuff is common to all terms, so we can continue and write...
= (1/C2) / (1/C1 + 1/C2)
= 1 / [C2 * (C2 + C1)/(C1*C2)]
= C1*C2 / [C2 * (C2 + C1)]
= C1 / (C1 + C2)

Which is, by value, the opposite of the voltage divider equation, because capacitance is inversely proportional to impedance.  An important thing to keep in mind when you're doing capacitors in series (uses resistors in parallel formula) and parallel (they add) and such!

Since we want 1/2 = C1 / (C1 + C2), then
2*C1 = (C1 + C2)
C1 = C2

Which is what we intuitively expect. :)

I skipped writing out all the currents and voltages exactly, because it's a lot of bother for no value, but again, you're welcome to expand any of the equations along the way by substituting in the relevant bits.

By the way, it's mildly interesting to note what they didn't say: when asking for a ratio of node voltages, note that that ratio itself is, in general, a complex number.  When analyzing a filter, for example, usually one is interested in Vo/Vi == H(w), the transfer function (gain) of the system.  The resulting form is a rational polynomial in j*w.  By...manifest destiny, rather than sheer luck, we are given H(w) = 1/2 -- a function that is constant with w.  But, that is, of course, the only function we can have here, because the frequency response of two capacitors in series is exactly proportional, and anything else nearby is a red herring (or busywork).

Another thing, perhaps less "mildly interesting" and more "pedantic": note that the question is phrased entirely in the time domain.  Lowercase script v and i, functions of time.  In that case, you shouldn't write the impedances at all (which are a frequency domain, steady state form), but the differential equations of the components (I = C * dV/dt, etc.).  And, in that case, the transfer function "vo(t) / vi(t)" == h(t) could be tremendously more complex than the mere ratio 1/2, or of a polynomial in frequency; in general, one cannot divide functions, one must perform a convolution (written "vo(t) = vi(t) star h(t)") instead!  It seems all the more bizarrely providential that, out of the space of possible functions used with convolution, the function "1/2" should be chosen!

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 

Online bson

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Re: Complex Impedances
« Reply #16 on: May 22, 2015, 12:24:20 pm »
Here's a solution to illustrate how to use sympy, but pretty much any complex calculator could be used.  It shows how you can easily walk around the circuit using Kirchoff's laws, only using complex arithmetic instead of real.  (Sympy is my go-to calculator these days, along with numpy for graphing.)  Only the frequency of the input (Vs) matters, not phase, so cos vs sin is unimportant, and w is simply 10MHz*2*pi.  The 2.0V of Vs @ w is also unimportant for this particular problem since you're only looking for a relationship... but I included it anyway.

Code: [Select]
import sympy, numpy
from sympy import I as j
from numpy import pi

# Define symbols
R,C,Co,w,Vs = sympy.symbols('R C Co w Vs')

# Impedances across the two caps
Zc = 1./(j*w*C)
Zco = 1./(j*w*Co)

# Total impedance
Z = R+Zc+Zco

# Current
Iz = Vs/Z

# Voltage across Zco
Vo = Zco*Iz

# Vc
Vc = Vs-Iz*R


# Solve for C at Vo = Vc/2  |  Co=0.2e-6; R=50; w=10MHz; Vs=2.0
solution = sympy.solve(sympy.Eq(Vo, Vc/2.0).subs({Co: 0.2e-6, Vs: 2.0, R: 50.0, w: 10e6*2*numpy.pi}), {C})

print solution

Running it:

Code: [Select]
[2.00000000000000e-7]

Since it's a symbolic solver you can experiment for other relationships, for instance obtain C as a function of w.
« Last Edit: May 22, 2015, 12:27:30 pm by bson »
 

Offline smjcuk

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Re: Complex Impedances
« Reply #17 on: May 22, 2015, 04:52:01 pm »
This is a really interesting thread. Nice to see some maths :)

Never used a Smith Chart here (EE deg 1995-1998) and would solve this algebraically too.
 

Offline TimFox

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Re: Complex Impedances
« Reply #18 on: May 22, 2015, 10:54:50 pm »
Do they still teach engineers how to use a Smith Chart these days or is it now all based around computers?
Back in grad school, an old-line German professor was discussing elementary electron optics.  He wrote an equation of motion on the board, which would lead to a solution in the form of a sinusoidal dependence on the co-ordinate z.  He asked the class how to solve it, expecting the answer "force a sine function and find the parameters".  A theoretician in the front row suggested finding the Green's function.  The professor replied "you could also pull on your pants with tongs".
The original problem has the obvious simple solution, since it only asks for the ratio of Vout to Vc.  Recognizing how to simplify a problem is often more educational than brute-force calculations.
 

Offline German_EE

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Re: Complex Impedances
« Reply #19 on: May 23, 2015, 04:09:56 am »
Well, I have heard a Smith Chart described as an analog computer and it does come pretty close. If you have a complex impedance and need to convert it to (normally) fifty ohms which is the center of the circle then you have a number of solutions which can be worked out by drawing curves on the chart. You then work out your capacitor or inductor values using standard formula as the chart is frequency independent.

Plugging the values into a web page then getting a result to two decimal places may get you the result you needed but you won't know WHY it is the correct result. This is why the Smith Chart is still of use.
Should you find yourself in a chronically leaking boat, energy devoted to changing vessels is likely to be more productive than energy devoted to patching leaks.

Warren Buffett
 

Offline IanB

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Re: Complex Impedances
« Reply #20 on: May 23, 2015, 04:16:00 am »
Question:

If you have an AC network with a single frequency you can solve it for steady state conditions using Kirchhoff's laws with complex voltages, currents and impedances.

Now, suppose you have sources with a mixture of frequencies. Is there an analogous approach to finding steady state solutions with (simple) algebra? (For example, could you use superposition to solve for each frequency independently and then sum the solutions?)
I'm not an EE--what am I doing here?
 

Offline T3sl4co1l

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Re: Complex Impedances
« Reply #21 on: May 23, 2015, 08:19:43 am »
As long as the system is linear, it will obey superposition, and can be analyzed the same for any frequency.

A simple example is an integrator: doing the analysis for each frequency in a square wave gets you a different phase shift (cos goes to sin, or whatever), and a different gain term for each one.  The most useful part comes when you keep the format as a geometric series, and spot the pattern: whereas the amplitude of harmonics of a square wave go as 1/N, the integrated result (a triangle wave) goes as 1/N^2.

You use Parseval's theorem to convert the sum into RMS measure, or power.

Tim
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Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 


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