complex numbers and impedance

Discussion in 'General Electronics Chat' started by Metastable, Mar 31, 2012.

1. Metastable Thread Starter New Member

Feb 6, 2012
2
0
I'm having trouble understanding something, so I would really appreciate it if you could find the time to really see what will complete the gap in my understanding and not just state facts

so, impedance and reactance. Something just doesnt fit completely in my mind as to why use complex numbers to represent these. I know about complex numbers, I can see how they represent phase-shifts, I know how to solve with them, I can see that it works and I can see that it makes seemingly difficult phase shift problems very easy. But what I'm missing is a validation a proof, or a source of a proof, as to why using them is adequate representation of reality and why it still goes easily with ohm's law..

I know complex numbers came about at the first place as a way to solve problems where till now your roots of a negative number would be invalid answers. Where does the root of a negative hide in our case of reactance ? Where does this substitution to i (or j) happens ?
I mean, do we have to use complex numbers to solve these problems ? or is it merely a choice  to represent it in this manner to make our lives easier if so, what other choices do we have ? how can we prove that the complex calculations are equivalent and will hold true ?

If I came across the problem of representing a capacitor's impedance I would first go at the time domain. Impedance = voltage / current. Very quickly I would get to the expression : (sin(wt))/(cos(wt) * wc) .. right ? so this gives us (tan(wt) / wc). But now we have time varying impedance and undefined impedance when wt is half pi (where the limit is undefined aswell). So while looking at this I don't know if it's wrong, or just messy or simply unsolvable?
So let say I can define my reactance as the voltage-current ratio of the amplitude only this will give us a frequecy depended (1/wc). And then I need to express the 90 degrees phase shift somehow.
complex numbers work well here. But why ? this is what I don't understand completely why do they work? I've looked into Euler's formula to maybe find what bridges it together, but I didn't really maybe I missed something, or maybe its just difficult for me to grasp.

how can you help me ?
thanks a bunch.

2. Papabravo Expert

Feb 24, 2006
12,075
2,610
It often happens in mathematics, that two different things that follow the same rules, can arise from different sources. So it is with the square root of minus 1 and the 90 degree rotation operator. As you have already mentioned the square root of -1 arises as a necessary construction to complete the set of solutions for the roots of polynomials of degree n with real coefficients.

You can also think of j as a rotation operator that moves a voltage or a current 90 degrees counter clock wise. As you multiply things by j you get the sequence
{1, j, -1, -j, 1,...} a really nice periodic sequence.

If you understand it then fine. If you don't there is always Art History.

3. paulktreg AAC Fanatic!

Jun 2, 2008
745
163
I spent many years of my school life being told that you cannot have the square root of a negative number. Went on to do mathematics at an higher level and low and behold "j" and you can have the square root of a negative number.

I understood it once but that's over 30 years ago and it's all long forgotten.

4. Papabravo Expert

Feb 24, 2006
12,075
2,610
The technically correct way to say it is that you can't have the square root of a negative number that is a real number. You can also say that the equation
Code ( (Unknown Language)):
1.
2. x^2 +1 = 0
3.
has no real solutions.

5. crutschow Expert

Mar 14, 2008
22,027
6,371
Using imaginary numbers to represent complex impedances in a circuit and the resulting phase difference between current and voltage in such circuits is just a convenient way to represent and manipulate the values of these impedances, voltages, and currents. It does not mean that there is something "imaginary" about the values, they are indeed "real" in that they do indeed exist. Imaginary is perhaps an unfortunate term for numbers involving i or j since it implies that they somehow belong to an imaginary world, but that is the convention for such numbers and so that's what we live with.

If there were a better mathematical way deal with such values, then it would be used.

6. amilton542 Active Member

Nov 13, 2010
496
64
@ Metastable

A copy of Schaums Outline on Complex Variables would be a good place to start.

Metastable likes this.
7. steveb Senior Member

Jul 3, 2008
2,431
467
Perhaps this link is the start of what you are looking for? This shows why the mathematics of linear operations on sinusoidal waves behaves like spatial vector operations.

http://www.capphysics.ca/faculty/mfreeman/PHYS114/Notes/phasors.pdf

Perhaps this does not fully answer your question, so perhaps the next step is to understand why complex number arithmetic also obeys spatial vector operations. In other words we can treat complex numbers like they are in the complex plane.

Remember that we only apply phasors to represent sinusoidal variables in a linear system. So, the representation is limited to particular assumptions.

Last edited: Apr 1, 2012
Metastable likes this.
8. WBahn Moderator

Mar 31, 2012
24,085
7,476
There are a number of paths to answer the OP's question and, admittedly, I don't know which one will be most likely to "click" with their present understanding to let them move forward (to the degree that this is hampering them at all). Here are a couple that might help:

You do not need to use complex numbers at all. You can work strictly in the time domain in which case the voltage across an inductor is equal to the inductance multiplied by the time-derivative of the current flowing through it. Similarly, the current flowing in a capacitor is equal to the capacitance multiplied by the time-derivative of the voltage across it. The bottom line is that you are left working with differential equations. Now, there is nothing wrong with this, but working with differential equations quickly becomes very tedious and error prone.

A couple centuries ago, a number of people were working on methods to solve differential equations that would be easier and less error prone and one of the tools they developed was the Laplace Transform. A mathematical transform is nothing more than a mapping from one set of variables to another. Changing from rectangular to polar coordinates is one such transform and is handy because some things are much easier to do in one set than the other; so we transform from the set that would be tedious to the set where it is easy, then do whatever it is we want to do the easy way, then transform back to the original set. So it is with the Laplace Transform. With it, we can transform a differential equation from the time domain to what is called the "complex frequency" domain. It is simply a mathematical mapping, but in this new domain differential equations become algebraic ones and we have lots of tools and techniques to solve algebraic equations quickly, cleanly, and with significantly lower chances of making mistakes. Once we have the solution in the complex frequency domain, we can then transform the result back into the time domain and get it into a form that makes sense to us.

Now, believe it or not, the transformation between the time domain and the complex frequency domain is so easy and simple that we learn to do it by inspection. We then show others how to do it by inspection without bothering to ever tell them what they are actually doing. Whenever you replace an inductor having inductance L with an element having an impedance of jwL, you have just taken the Fourier Transform of the differential equation describing how an inductor behaves when driven by a steady state sinusoidal signal. The same is true when you replace a capacitance C with -j/(wC). Notice that I said "Fourier" instead of "Laplace". This is because the Laplace Transform is a very general transform and allows us to solve for the transient response, as well as the steady state response after all of the transients have died out. The Fourier Transform is merely a simplified subset of the Laplace Transform that only lets us get at the steady state solution, which is frequently good enough.

So that is one answer. Here is another.

Let's assume that I have a linear circuit (resistors, inductors, and capacitors along with some simple voltage and current sources) and all of my sources are putting out steady sinusoidal voltages/currents and have been on for a long time (long enough for all of the elements to settle into their steady-state response). Let's focus on one such voltage signal, v(t), which has the form:

v(t) = V*cos(wt+phi)

While it may not be obvious why we would do this, we could express this signal as:

v(t) = V*cos(wt+phi) = Re{[V*e^(j*phi)]*e^(jwt)}

The factor V*e^(j*phi) is what you are used to working with as the "phasor" for this voltage when working with complex impedances. While very hand-wavy, this expression represents the transformation between the time domain and the complex frequency domain and back. In the complex frequency domain, everything has the frequency exponential factor, e^(jwt), multiplying it, so we simply divide it out and work with the phasors to get an answer and then multiply by the frequency exponential factor again and take the real part to get our final time-domain answer. Again, the steps involved in going from the time domain equation to the phasor representation is so simple that we usually do it by inspection.

None of this explains why we can use all of our normal techniques that we learned when working with nothing but resistors (and batteries and constant current sources). That answer lies in the fact that the solutions to the algebraic equations, after transforming the underlying differential equations to the complex frequency domain using either the Laplace or the Fourier Transform, have the exact same form as the time-domain equations used to solve a circuit having nothing but resistors (and batteries, and constant current sources).

In fact, if this weren't the case, people would have continued devising other transforms (and there are many of them out there with new ones being devised all the time for different purposes) until they came up with one that did. Engineers are inherently lazy; we spend huge amounts of time and effort trying to find easier ways to do our job.

Last edited: Apr 1, 2012
screen1988, Metastable and MrChips like this.
9. Metastable Thread Starter New Member

Feb 6, 2012
2
0
@crutschow yeah, I really do get the fact that imaginary or complex numbers exist like any other number. unfortunately, thats not what hinders my understanding of its use.

@amilton542 & @steveb - thank you for your suggestions ! I will check them out.

@WBahn - thank you very much for this explanation, and sending me in the right direction.
.. I study everything I'm interested in by myself and can't afford otherwise (making this site, khan academy, etc - a goldmine) , so one of the biggest problems I face is I have no higher guidance as to where to go.. and so little resources take the time to help you connect the dots, or like you said "tell them what they are actually doing".

If anyone has any other resources, suggestions or explanations to add - it will only help. thanks !

10. MrChips Moderator

Oct 2, 2009
18,463
5,855
I like WBahn's explanation of treating it as a mathematical transformation.

We transform intensity and power levels into decibels using a log function. This way multiplication gets transformed into simple addition.

With the Laplace transform, we transform differential equations into linear equations.

Fourier transform turns convolution into multiplication.

Transforming rectangular coordinates to polar coordinates gives us a different perspective.

Replacing L with jwL and C with -j/(wC) allows us to use phasor or vector analysis.

Oct 2, 2009
18,463
5,855