How do the complex numbers (a+bi) are related to electronics?
Posted on 2003-05-12 16:11:10 by inFinie
Geeze, do you want to open a can of worms. <grin>

First off, I'm going to use 'j' instead of 'i' for the square root of -1. In electronics, 'i' gets used for something else.

So here we have the square root of -1, or j. It's useful because it can be used to indicate not only a magnitude, but a direction as well. Here's a simple explanation:

Take a number line, and lay out a line from 0 to 5. Now imagine that line multiplied by -1, so you find its negative. You'd get a line from 0 to -5, as 0*-1=0, and 5*-1=-5. Simple enough.

Now comes the leap of faith (cause I forget the real mathematics here). I just described a way to make a 180 degree turn of a line. Call that j squared. Now imagine a turn j that when imposed twice, gives you a 180-degree turn. Simple enough you think, just turn 90 degrees twice.

That's what j does for us, its a 90 degree turn. So if I multiply my 0 to 5 line by j, I get a line from 0 to 5j. The line is 90 degrees from the number line, with the same origin (zero point).

Now instead of a number line, think of a Cartesian co-ordinate plane, the X-Y axis graph. Anything along X-axis we can denote by a 'real' number, anything along Y we use an imaginary number for.

(I really should add some graphs at this point... when time permits I'll redo this as a complete tutorial.)

Say we have a 2 dimensional line, 5 units long, at an angle of 53.13010235 degrees off the X-axis. A little trig leads you to see that the line has an X component of cos(53.13)*5 = 3 units and a Y component of sin(53.13)*5 = 4 units.

It's just a 3-4-5 right triangle, right? We want to represent the hypotenuse line in the simplest form.

One was is with polar notation, that's how I first denoted this line, a number at some angle.

But since we also know that the real part of complex number can represent length along x, while the imaginary part indicates length along Y, we can also denote our line as: 3+4j, where 3 is our length along X, and 4 is the length along Y.


WHY do this? Mostly cause it's useful in something called sinusoidal analysis. Sinusoidal analysis (or S.A. from now on) looks at a circuit at a single frequency at a time, or better, leave it in terms of a frequency variable so you can plot the response of output amplitude vs. frequency. If you've ever seen a frequency response graph, that's where it comes from (though a frequency response graph looses the phase info along the way.

In a nutshell, S.A. will allow us to use algebra to compute how a circuit responds to a sine wave input, by computing what it does at each frequency.

First off, an AC sine wave would be denoted by an amplitude and a phase. Usually, you would pick the phase of your source as zero, and then the output's phase is the difference from out to in. Or... for multi-phase power circuits such as a 3-phase power line, each line would have a phase difference of 120 from the other two, so the three phases would be:

220 at an angle of zero

220 at an angle of 120 degrees

220 at an angle of -120 degrees

That is how a three phase 220 line might be denoted. Note I've used the RMS value for the voltage here. (RMS is a whole other topic)

I'm sure everyone knows that capacitors and inductors have a different impedance over frequency. In S.A., we define this impedance as such:



j
Capacitor: Xc = ---------- where C is the capacitance, f is frequency
2*pi*f*C


Inductor: Xl = -j2*pi*f*L where L is the inductance


Impedances work mathematically just the same as resistance do. Say we have an AC voltage source driving a resistor, then a capacitor, then back to the source. If we just had two resistors here, we'd call it a voltage divider and say:



R2
Vo = --------- Vin
R1 + R2
or

Vo R2
-- = ---------
Vin R1 + R2



For the capacitor substituting for R2, just substitute Xc for R2, and you get the transfer function in terms of frequency.

NEXT time I'll even solve this and make some useful observations
Posted on 2003-05-12 19:05:42 by Ernie
Well put Ernie.. this is not an easy topic (kinda glad you broke the ice first) ;)

But i have to correct you in one spot. The reactance equations are backwards:

w = 2*pi*freq


X(l) = j.w.L = [b]j*2*pi*freq*L[/b]

X(c) = 1/ j.w.C = j/j * 1/j.w.C = -j/wC = [b]-j/2*pi*freq*C[/b]


If i may add to what you started, a way of seing this 'circle of j's' is shown rectangularly below:
                   (90 Deg - Inductive)


j

|
|
|
(180 Deg - Real) j^2 ---+--- j^4 (0 Deg - Real)
|
|
|

j^3

(-90 Deg - Capasitive)

Where j is 90 degrees phase shift (inductive), j^2 is -1 (real at 180 deg), j^3 is -1*j which is a -90 degrees phase shift (capasitive), and j^4 = -1 * -1 = 1 (real at 0 deg).

To add more background, the entire buisiness of j (sqrt(-1)) comes from Euler's Identity: See Here



Any polar number is expressed as an absolute magnitude (i.e. positive only), and an angle from 0 degs. Hence a line 3m long at 45 degress would be 3<45. A line -3 at 45 degrees would be:
-3 = j^2 * 3 = 3<180 

Times the angle 1<45 makes:
3<180 * 1<45 = |3*1| < (180 + 45) = 3<225

which is the inverse of 3<45!. Now bringing this back to Euler's equation:


e^(jx) = POLAR NOTATION
cos(x) + j.sin(x) = RECTANGULAR NOTATION

So in short, we can also represent the same polor numbers as:
3<45 = 3*e^(j.45)

-3<45 = 3*e^(j.180) * 1e^(j.45) = 3*1 e^(j(180+45)) = 3e^(j.225)
Converting to rectangular format (from polar):
3*e^(j.45) = 3 * [ cos(45) + j.sin(45) ] = 3*[ j^4 *(cos(45)) + j^1*(sin(45)) ]


If you plot this out on the above rectangular "circle of j's" you will see that its:
3*cos(45) = 3*0.707 = 2.121 in the x direction (j^4 = 0 deg = x)

3*sin(45) = 3*0.707 = 2.121 in the y direction (j^1 = 90 deg = y)

So you see, all these years in school, learning that there are only positive and negative REAL numbers, means you teacher(s) has pernamently kept your brain on the REAL AXIS of the above plot! All math, and everything! When they taught you about circles (x^2 + y^2 = r^2) it can be shown that you are using this idea of "j", just you didnt realize because of one assumtion! Using trig:
x = r * cos(@)<0

y = r * sin(@)<90

Where @ is Theta (angle). y must be 90 degress from x, to be orthogonal, hence they are TRUELY defined as such above. The length in these directions are the trig you have always known. The eye opener is that you have always assumed Y is 90' from X and never shown it on paper, as math! After all its pretty obvious to see, just look at a page of graph paper ;) . But in true math, this is what your doing!

Apply our conversion:
x = r * cos(@) * e^(j.0) = r * cos(@) * e^(0) 

= r * cos(@) * [ cos(0) +j.sin(0) ] = r * cos(@)*[1]
= r * cos(@)
y = r * sin(@) * e^(j.90) = r * sin(@) * [ cos(90) + j.sin(90) ]
= r * sin(@) * [ j*1 ]
= r * j.sin(@)


Putting these two rectangular components together we get:
r * cos(@) + r * j * sin(@) = r*[cos(@) + j.sin(@)]= r*e^(j.@) !!


To get just r^2, as in our circle equation, we must multiply both sides by r*e^(-j.@) to cancle it out to e^(0) which is just 1!


r^2 = r *e^j@ * r * e^-j@
= r [cos(@) + j.sin(@)] *r [cos(-@) + j.sin(-@)]
= r^2 [cos(@)cos(-@) + j.sin(-@)cos(@) + j.sin(@)cos(-@) + j^2.sin(@)sin(@)]
= r^2 [cos^2(@) - j.sin(@)cos(@) + j.sin(@)cos(@) + j^2.sin^2(@) ]
= r^2 [(cos^2(@)) + 0 + (j^2.sin^2(@))]
= r^2 [ cos^2(@) + j^2.sin^2(@) ]
= r^2 * cos^2(@) + r^2 * j^2 * sin^2(@)
= ( r * cos(@))^2 + ( r * j * sin(@))^2

Subbing back in: x = r*cos(@)<0 and y=r*sin(@)<90 = r*j*sin(@)

r^2 == (x)^2 + (y)^2
Surprised! ;)

Anywho, this is alot of math to show you that there is *more* to the world of math than meets the eye. Electronics uses this imaginary number system alot because of something we call PHASE.

All frequency based entities (current, voltage, reactance) has a relative PHASE (or shift in time).

DC circuits ( zero frequency ) have a constant PHASE: All currents and voltages may vary in amplitude 'M' but have all the same PHASE 'P' (meaning they all respond instantaneously in time). Hence resistance is cleary defined as using OHMS Law:

let:

M() < P = M().e^j.P

Be either DC Votage or Current, since P is constant.

Then:
R = V/I = [ M(v)< P ] / [ M(I) < P ]

= [ M(v) / M(I) ] < (P-P) = M(V)/M(I) < 0


Hence resistance always carrys a phase shift of ZERO (j^4). Resisters can be viewed as "Resistance * e^j(0)".

However, in AC!. A voltage and current may no longer be in phase if reactive elements are use (mentioned at the top of this post). They have constant PHASE SHIFTS of +/- 90 degrees depending if its an inductor or a capasitor (the j's in their expression). When applied to circuits, different branchs of the circuit have different constants for Phase.

All of a sudden OHM's law is no longer so simple, and the amount of PHASE shift has to be taken into account between each element (albet current, voltage or reactance)! Since they are not equal current and voltage will not be IN PHASE if reactance is used in the circuit!

The math to deal with this is EXACTLY the above eulers and 'j' notations! Depending on if your dealing with series or parallel elements, dictates when you switch to Rectangular or Polar notations for solving circuits:

Polar notation works for MUTLIPLY and DIVIDE very easily!
A<B * C<D == |A*C| < (B+D)

A<B / C<D == |A/C| < (B-D)


Rectangular notation works for ADD and SUBTRACT very easily!
(A + jB) + (C + jD) == (A+C) + j(B+D)

(A + jB) - (C + jD) == (A-C) + j(B-D)


To sum this up in general:
Resistances DONT cause current to phase shift from voltage, and vice versa. In either AC circuits or DC circuits, since their phase component is ZERO!

However REACTANCE which is complex resistance, or resistance with a phase shift component, WILL cause current to phase shift from voltage, and vice versa. Particularily in AC circuits will this be observed. However, in DC circuits, this is observed as "transients"


In particular to AC circuits, there is a quick little rhyme that I was taught to keep your head on straight and dumb all this theory down to practicle electronics:
Eliah The Ice Man! Or accuratly put:
ELI - ICE


E = Voltage (Electromotive Force)
I = Current
L = Inductance
C = Capasitance

E L I
This rhyme tells me if i have an Inductance (L) or a reactance having a positive angle +(0->180), that the resulting phase shifts will cause VOLTAGE (E) to lead CURRENT (I) in time. (i.e. at any given moment in time, the instantaneous AC voltage value will be ahead of the instantaneaous current, by an angle dictated by the angle (phase) of the reactance).

I C E
If i have an Capasitance (C) or a reactance having a negative angle -(0->180), that the resulting phase shifts will cause CURRENT (I) to lead VOLTAGE (E) in time. (i.e. at any given moment in time, the instantaneous AC voltage value will be ahead of the instantaneaous current, by an angle dictated by the angle (phase) of the reactance).

This all sounds like alot! I can understand! My advice is to read this thread over two or three times, sleep on it, and tomorrow I or Ernie will provide more solid examples using circuits and far less of this theory crap! ;)

Boy you really did open a can of worms, didn't you ;)
If you can injest all this and understand it moderately, your way ahead of the game. This is truely one of THE hardest principales in electronics to understand as a beginner. I know people in 4th year university who still didnt fully get this! So dont feel bad if your head is spinning. Just be patient, and with help from people like Ernie, myself, or others ~ you will get it!

:alright:
NaN
:alright:
Posted on 2003-05-12 21:31:06 by NaN
I think a lot of folks don't have the correct concept of what so called "imaginary" numbers are. Too bad they have such a misrepresented name. First of all, the symbol i or j is not a ALGEBRAIC term. The term 5j does NOT mean j+j+j+... five times. The symbol j is an OPERATOR. It is rightly called that in many math and engineering books. So 5j means rotate the real number five, 90? CCW. That makes 5j a orthogonal number, or a number in quadrature. But it is not really an "imaginary" number. The term 5j^2 does not mean j to the second power. It means rotate five by 90? CCW twice, or 180?. That makes it -5 and puts it back on the real axis. The number two in j^2 should really be written as a subscript instead of a superscript to avoid confusion with exponentiation. Obviously you get correct answers by treating j as a algebraic term, but that is due to its CONFORMAL SIMILARITY. The basic concept is vastly different.

The number, x + jy, means a real number plus another number rotated by 90? CCW. This number has two parts, a real part and second part orthogonal (right angled) to the first part. This second part is called the imaginary part, but it really is not. It is just as "real" as the first part, only rotated by 90?. Combined together, x + jy, is called a complex number, but that also is a bad name for it. It implies that the number is difficult and hard to understand. So called "complex" numbers should really be called duplex numbers to emphasize their two part structure (real and quadrature), and to get away from the unfortunate connotation that they are complicated, esoteric, and originated from some mathematician's wet dream. To iterate, it can be shown that correct results can be had from j = sqrt(-1), but that is not the definition of j.

In AC circuits, the "j" designated "imaginary" rates of energy stored and released in the electrostatic fields of capacitors and electromagnetic fields of inductors, are every bit as "real" as the rate of energy dissipated in resistance as heat. Ratch
Posted on 2003-05-12 21:52:36 by Ratch
I completely disagree with the interpretation of j being an operator only. There are firm historic and mathematical reasons for its name, and j (or i) does indeed equal the square root of -1. That it also can serve as a 90-degree rotational operator is only true if you impose a graphical representation on the equation, such does not always exist.

The historic perspective comes from the solution of roots of polynomials. As soon as 'imaginary' roots were allowed, polynomial theory grew to a fully functional base.

Now for a bedtime story. I'm going from memory here, I wish I had the direct source for it, as it is a college recollection of non-other then Isaac Asimov himself. Doctor (of biochemistry I believe) Asimov is of course the author of numerous classic sci fi stories and creator of the three laws of robotics. He's also has research kudos, and loved nothing more then writing about whatever was on his mind. His monthly columns for many publications have been anthologized and are much recommended reading.

Anyway, the story he once told was he would oft drop in for the last part of a friend's philosophy class, as he had some free time before his friend's class was free. The good not yet doctor would quietly sit in the back of the class and listens in on the lectures till his friend was free to leave.

One day he noticed the prof had listed mathematicians on the blackboard (they really were once black) in the group of mystics. As the session ended, the prof asked if any questions remained. Asimov was forced to raise his hand, believing the mathematicians were listed incorrectly.

"Not at all," the prof responded. "Mathematicians believe in the square root of negative one. Since this number does not exist, they must be classified as mystics."

Rather then debate this point, Asimov asked if the prof could offer a simple proof of his mathematical prowess, and hand him half a piece of chalk. "Simple enough" agreed the prof, who promptly pulled out a fresh piece, snapped it in half, and handed one piece to Asimov.

"But professor," Asimov humbly noted. "This is not a half of a piece, it is one entire piece."

"Oh no, it is half of a standard length of chalk."

"But now you're inserting arbitrary definitions. If you are this fuzzy in your understanding of the meaning of one half, how can you hope to understand the square root of minus one?"

Asimov most charitably deleted the prof's next comment from posterity, though he did note he was offered in the future to stand in the hall till his friend was free.

Summation: Just because you cannot hold one in your hand does not mean mathematical entities do not exist.
Posted on 2003-05-12 22:42:19 by Ernie
Ernie,

The operator j stands on its own as meaning a 90? direction to a reference. As a matter of fact, one could define k as an operator that points at 90? to both a reference direction and j, like a x-y-z coordinate system. Although a graphical representation is a good way to visualize it, its validity does not depend on graphics. Most of the good texts on math and engineering call it an operator, which it is.

In my last post, I included some of the material from a quote by Emeritus Professor J. Hollingworth of the University of Manchester, England, "It would have saved a lot of trouble if it had been written j(subscript 2) " ("it" standing for j(superscript 2).* Obviously the prof thinks that j is an operator, and not something to be exponentiated.

The bedtime story was interesting, but neither the story or the summation proved your point.

*Hollingworth, J. "The symbolic Method" Bulletin of Electrical Engineering Education Number 10 P 22, England(June-1953)
Posted on 2003-05-13 00:12:15 by Ratch
Nan,

In your above post, you state that reactance is "complex resistance". Although both resistance and reactance have the same units of measurements (ohms), there are more profound differences than just phase. Resistance dissipates energy as heat whenever current is present, but "pure" reactance stores and releases energy to/from the circuit without any loss. The effect of this behavior difference is huge, so looking at reactance as an out of phase resistance is misleading. Ratch
Posted on 2003-05-13 00:28:16 by Ratch
Originally posted by Ratch
The bedtime story was interesting, but neither the story or the summation proved your point.


To be excessivly pedantic, j is defined as the square root of -1.

A definition is either usefull or not.

A definition (by definition) needs no proof.

Nor will any be offered.
Posted on 2003-05-13 06:34:31 by Ernie
Nan,

Of course you have it correct. Its been years since I actually needed to use this stuff (Bodie plots usually being sufficient).

Me bad, thanks for the correction.
Posted on 2003-05-13 06:46:10 by Ernie
Ernie,

Quote by Ernie
A definition (by definition) needs no proof.


In this case you luck out. The definition gives you correct answers due to its conformal similarity. But most good texts on engineering "define" j to be a operator, i.e. the quadrature part of a duplex number. Ratch
Posted on 2003-05-13 07:10:39 by Ratch
Can you post the results of your survey please?

How many engineering texts did you review?

By what criterion did you deem them 'good'?

And finally, do you feel lucky?



I'll find it acceptable if you can provide one half of an answers.
Posted on 2003-05-13 11:44:19 by Ernie
i or j or k you can call operators if you like.

The Argand plane was created for a graphical representation of complex numbers.

"i" alone means a purely imaginary number with a magnitude of 1 and a direction of 90 degrees.

Although 1 + 1i does represents two components, the result is what is important , that is... a radius vector of sqr root(2) at 45 degrees in the first quadrant.

Yes there are two components but (1 + i) is only one number with magnitude and direction. It represents a single point in the Argand plane disk or field of complex numbers if you like.
Posted on 2003-05-13 14:18:14 by IwasTitan
Ernie,


Quotes by Ernie
Can you post the results of your survey please?


No I cannot. I was not doing a survey. I just perused many texts over the years. If you will fund it, I will be happy to do a survey. It is not hard to find the term "j operator", however. In no time at all I found this on Google http://www.ycc.ac.uk/yc/ENGINEER/Freelance%20Presentations/index13.htm . Take my word for it, a lot of people agree with me that j is an operator. Consensus alone is not proof, but I gave my reasons why I think it is so. Can you give your reasons if you think it is not? Can you find anything wrong with the logic that I used to present my point?

By what criterion did you deem them 'good'?

The way the material is organized and presented. Good and plentiful use of examples. Answers to all the exercises. Completeness of the subject coverage. And most of all, absence of cutesy ways of explaining basic subjects and ideas.

And finally, do you feel lucky?

Not particularly, should I? Ratch
Posted on 2003-05-13 19:08:02 by Ratch
IwasTitan,


Quotes by IwasTitan
"i" alone means a purely imaginary number with a magnitude of 1 and a direction of 90 degrees.

My point is that the i or j symbol is not a number, imaginary or otherwise, but a rotational operator. And you get correct answers by treating it as a number because of its property of conformal similiarity. I gave reasons for my statement, can you refute them?
Although 1 + 1i does represents two components, the result is what is important , that is... a radius vector of sqr root(2) at 45 degrees in the first quadrant.

And is it not just as important that the polar form can be changed into rectangular coordinates? What is your point in the above?
Yes there are two components but (1 + i) is only one number with magnitude and direction. It represents a single point in the Argand plane disk or field of complex numbers if you like.

Didn't I previously say that x+iy was a duplex number? And yes, it does represent a single point that needs two coordinates to describe its position. Again, what is the point? Ratch
Posted on 2003-05-13 19:44:16 by Ratch

IwasTitan,


My point is that the i or j symbol is not a number, imaginary or otherwise, but a rotational operator. And you get correct answers by treating it as a number because of its property of conformal similiarity. I gave reasons for my statement, can you refute them?

And is it not just as important that the polar form can be changed into rectangular coordinates? What is your point in the above?

Didn't I previously say that x+iy was a duplex number? And yes, it does represent a single point that needs two coordinates to describe its position. Again, what is the point? Ratch


Ratch:

I was just stating what i know about complex numbers.

I wasn't denying anything you said nor was i looking for an argument about such.
Posted on 2003-05-13 20:48:34 by IwasTitan
Funny, in no time at all, clicking ahead a few panels from a previous post,
I found the j is be equated to the square root of -1.

Thank you for making my point for me.

This page is an excellent introduction to phasor analysis. I strongly advise going there and reading it all. For those of you new to this, work out the samples too. Then we can all have another long thread on the correct answers (with an extra bonus side digression about how Alexander Bell didn't invent the first working telephone, but another man with the same exact name did).
Posted on 2003-05-13 22:24:57 by Ernie
Ernie,

Quotes by Ernie
Funny, in no time at all, clicking ahead a few panels from a previous post, I found the j is be equated to the square root of -1.

Thank you for making my point for me.


I thought you would say that. But no, it does not make your point. As I said before, correct answers are obtained by treating j as a exponentiated number. I gave the reason why that works. I even use that method to churn out an answer. But CONCEPTIONALLY speaking, j is really an rotational operator.

I also thought you would notice the educational value of the site. That is why I am of the opinion that it is a waste of time to write tutorials when so much material is easily available. Unless what is wanted is not available elsewhere. Ratch
Posted on 2003-05-13 23:36:43 by Ratch
Afternoon, Ratch.

If you haven't realized it yet, this Electronics forum is attached to the Win32Asm board.

Members are allowed, even encouraged, to post anything which doesn't contravene the rules. :alright:

You may find it a waste of time, but others do not. And it is not your place to mention that it's a waste of time with every one of your posts.:mad:

Instead of being a "talking head", why don't you supply a list of topics and their associated page links for members to instantly obtain relevant information?

The idea here is to have a "one-stop-shop" so that information, and links to information, can be found on everyones favourite board.:grin:

Cheers,
Scronty
Posted on 2003-05-13 23:53:02 by Scronty
Hmmph, I always thought the math world pretty much treated i (or EE j) as a value. I know I do.

The complex numbers can be used without reference to coordinate geometry. (I know, the geometry gives meaning to complex number arithmetic.) The fact that it works well for describing electronics, to me, is remarkable.

See http://mathworld.wolfram.com/ComplexNumber.html
Posted on 2003-05-14 00:27:23 by tenkey
Good Morning Scronty,

If you haven't realized it yet, this Electronics forum is attached to the Win32Asm board.

Certainly I am aware of that. I can see that everytime I sign in. And the point is?
Members are allowed, even encouraged, to post anything which doesn't contravene the rules.

Sounds like a good policy. And the point is?
You may find it a waste of time, but others do not. And it is not your place to mention that it's a waste of time with every one of your posts

You did not say what the waste of time is about, but I assume it is about writing tutorials, right? If I can "post anything which doesn't contravene the rules" and am encouraged to do so, what is the problem? Anyway, I don't mention it in every one of my posts, only the ones that pertain to the subject. Everyone is free to ignore my statements and write as many tutorials as they want.
Instead of being a "talking head", why don't you supply a list of topics and their associated page links for members to instantly obtain relevant information?

Because everyone has different circumstances. It it up to the individual to seek out and ask for what s/he needs and wants. To assist, there are many search engines like my favorite, Google.
The idea here is to have a "one-stop-shop" so that information, and links to information, can be found on everyones favourite board.
When you come across that board, let me know. I usually always have to scrounge from several sources to find complete information about something. Is there anything within reason that you cannot find out about? Maybe I can help you. Cheerio, Ratch
Posted on 2003-05-14 00:28:12 by Ratch