tenkey,

Hmmph, I always thought the math world pretty much treated i (or EE j) as a value. I know I do.

So do I. I already explained why that works. See my last post answering Ernie within this thread.
The fact that it works well for describing electronics, to me, is remarkable.

It helps describe the phase relationships that occur whenever electrical activities are present. It is also used in other fields besides electronics.

Nice site you referenced, by the way. Ratch
Posted on 2003-05-14 00:47:36 by Ratch
Ratch,

With out all the wordy fluf, find me the roots of:

s^2 + 1

(I can build a circuit to produce a voltage in time whos characteristic equation follows this equation. So its definitely a practicle question)

Once you get the answer, tell me how you see 'j' an operator. Its not even in the starting equation!

I will give you a hint at solving it.
Use the Quadratic equation := [ -b +/- Sqrt( b^2 - 4*A*C) ] / 2*a

:NaN:
Posted on 2003-05-14 12:36:55 by NaN
Nan,

Good illustrative problem!

You gave the term s^2+1, but I assume you really mean the equation s^2+1 = 0, right?

OK, here is the solution, short 'n sweet. The quadratic formula is not needed here.

s^2+1=0----->s^2 = -1 -----> s = ?sqrt(-1); changing to polar form s = ?sqrt(1/_180?) = ?1/_90? = ?j(1) , where j(1) means 1j, or 1*j, or j, or 1 rotated by 90?, however which way you want j to be designated as a rotational operator. Terms like that appear in sinusoidal functions. Ratch

P.S. To find the sqrt of a polar number compute the square root of the magnitude and one half of the angle.
Posted on 2003-05-14 13:57:18 by Ratch
Correct, its an oscillator with a natural radial frequency of 1.

However, you jumped directly to polar notation. Im asking you to explain from quadratic how j shows up as an operator and not a mathematical term. Even how you solved it your not explaining how a number turns into an operator...

:NaN:
Posted on 2003-05-14 16:06:21 by NaN
Nan,

However, you jumped directly to polar notation. Im asking you to explain from quadratic how j shows up as an operator and not a mathematical term. Even how you solved it your not explaining how a number turns into an operator...


The first thing I did was solve the equation. Using the polar form of -1 in no way invalidates or changes the correct solution. I found the solution to be s = ?1/_90? . 'j' is defined to be a rotational operator, just like '+' is a addition operator. If they are operators, then both j and '+' have no value or meaning by themselves. They both need number(s) to operate upon. Using my cooked up notation for the functional aspects of j, I define j(1) to mean 'rotate 1 by 90?'. Substituting this value into the solution gives s = ?j(1), or in the less precise notation that most of all the world uses, s = ?j .

To iterate; the operator symbol j by itself has no meaning or value. The positive square root of -1 is j(1), not just j alone considered as a operator. The quadrature number j(1) is just called imaginary, but it really is not.

To directly answer your question, the solution turns into a j operator function because of its quadrature value. Ratch

P.S. Can you calculate the value of the term j(1)^j(1) or j^j if you want it shown that way?
Posted on 2003-05-14 20:15:57 by Ratch
You still havent convinced me of how a number turns into an operator. And to be honest, i dont care for your explainations at this point. All This petty-ness has taken a good topic way too far off its origional intentions. If this thread can be saved i would rather see this than hear more about your particular point of view on trivial inconsistancies.
Posted on 2003-05-14 21:20:34 by NaN

You still havent convinced me of how a number turns into an operator. And to be honest, i dont care for your explainations at this point. All This petty-ness has taken a good topic way too far off its origional intentions. If this thread can be saved i would rather see this than hear more about your particular point of view on trivial inconsistancies.


I agree to argue is pointless but lets put the blame where it belongs..... that is... on those who have failed to fully analyse Ratches comments.

j is a symbol that defines a quantity not on the real number line. In other words it defines a quantity in the field of complex numbers of which the real numbers are a subset.

On the real number line we can have positive or negative numbers. The symbols used to define them are - and +. But we use them as operators also. (1-1=0)

With complex numbers we have an additional dimension. It is not a number line anymore..its a plane.

We have two new directions to think of.

We can't use - or + can we? We have already used them.

Lets use j and - j.

Is j an operator?

Your damn right it is.

Why?

Take a hard look.

j * 1 = j

j * - j = - j ^ 2

- j ^ 2 * -1 = -1


Its an operator which changes its operation based on the sign of the terms it operates on.
Posted on 2003-05-14 22:00:08 by IwasTitan
IwasTitan,
Its an operator which changes its operation based on the sign of the terms it operates on.

I'm glad to observe you agree with me a little. However, your quote would be more correct it had said--It's an operator that changes the value of a term by rotating it by 90? CCW. Thanks again, Ratch
Posted on 2003-05-14 23:18:36 by Ratch

IwasTitan,

I'm glad to observe you agree with me a little. However, your quote would be more correct it had said--It's an operator that changes the value of a term by rotating it by 90? CCW. Thanks again, Ratch


hmmmm

Now you have me really thinking.

Its too bad this board does not have a math forum.

I put in a request moons ago for one but although many were for it..many were against. (don't know the beef)

Anyways this is an electronic forum so i guess we stick to that.

I'm still waiting for a Op Amp tutorial from NaN.

I just ripped out a power transistor from a TV that looks like it could light up New York.

Can i use it in a strobe light?
Posted on 2003-05-14 23:50:53 by IwasTitan

But CONCEPTIONALLY speaking, j is really an rotational operator.
Only has meaning in a geometric context. It is neither more true to say that j is an operator, nor less true to say that j is a number (or more precisely, a value). Whether a value represents only itself or an operation is a matter of context and point-of-view. (I will restrain myself from producing a number of examples, all math and no electronics.)
Posted on 2003-05-14 23:53:38 by tenkey
IwasTitan,

Its too bad this board does not have a math forum.

Then it would have to have a physics form also. Math and physics are the foundation sciences of electronics. But where does it end?
I'm still waiting for a Op Amp tutorial from NaN.

Until it appears, try this to hold you over. http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/opamp.html
and this http://courses.nus.edu.sg/course/mpeteocl/me2141/op%20amp.pdf
or even this http://www.eatel.net/~amptech/elecdisc/opamp.htm
I just ripped out a power transistor from a TV that looks like it could light up New York.
Can i use it in a strobe light?

Not enough information to determine that. Are you replacing a part or designing it from scratch? Try this USENET site, sci.electronics.basics . They are involved in things like that. Ratch
Posted on 2003-05-15 00:19:24 by Ratch
Afternoon, Ratch.


Great links. Thanks for supplying them:alright: .

I haven't yet looked at the pdf file, however the third link has practically all the info I'd require for when I start mucking about with opamps.

Cheers,
Scronty
Posted on 2003-05-15 03:00:16 by Scronty
IF j is an operator then on which sets (N, R, Q, C ...)it is defined?
Is it commutative?
Is it distributive?
Is it associative?
Does it have an inverse (multiplication to division, addition to substraction, logarithm to power...) ?
What is inverse of it?
Posted on 2003-05-15 06:38:54 by inFinie
inFinie,
IF j is an operator then on which sets (N, R, Q, C ...)it is defined?

What does (N, R, Q, C ...) mean?
Is it commutative?
Is it distributive?
Is it associative?

The above questions make no sense. The j operator is a unary operator, those properties you ask about are for binary operators.
Does it have an inverse (multiplication to division, addition to substraction, ...) ?
What is inverse of it?

The inverse of j is -j. For example, using my notation and starting from 4, -j(j(4)) = 4 . Using "standard" notation, starting from 4 and applying j, we get 4j. Now applying the inverse we get -j*4j = 4, the original value and showing that the inverse of j is -j . What I am trying to say is that if you rotate a number by 90? CCW, then the inverse of that operation is to rotate it 90? CW back again to where it was. You knew that, didn't you?
logarithm to power...

Did I not ask what the value of j^j, or j(1)^j(1) is in a previous post? Ratch
Posted on 2003-05-15 07:41:12 by Ratch
tenkey,
Only has meaning in a geometric context. It is neither more true to say that j is an operator, nor less true to say that j is a number (or more precisely, a value). Whether a value represents only itself or an operation is a matter of context and point-of-view.

I and other folks of my ilk believe that conceiving j as an operator helps to better understand the underlying principles of duplex numbers. Just about any measurement or value in math and physics can be displayed in a geometric context. Ratch
Posted on 2003-05-15 07:56:09 by Ratch
a ^ b = e ^ ( b * ln( a ) )

logarithm in complex numbers:

ln( r * cis( t ) ) = ln(r) + ( t + 2kP ) * i , k is an element of Integers

j^j= e ^ ( j * ln( j ) )

j= 1 * cis(P/2) = cos(P/2) + j * sin(P/2) = 0 + j * 1

ln( j ) = ln ( cis(P/2) ) = ln1 + (P/2 + 2kP) * j

j ^ j = e ^ (j * j * (P/2 + 2kP)) = e ^ ( -P/2 - 2kP )
For k=0
j^j = e^(-P/2) = 0,207879576350761908546955619834979

PS cis(t) is a shorthand for cos(t) + i * sin(t)
P=pi


N, Q, R, C ...
Natural, Rational, Real, Complex ... (there are many others)

I can not make a connection between

logarithm to power...

AND

Did I not ask what the value of j^j, or j(1)^j(1) is in a previous post?

Regards
Posted on 2003-05-15 12:47:55 by inFinie
That's an irony that you are asking j^j Ratch.

Can you think such a thing:
2 (*^2) 2 = ...
two, multiplication squared, two

No, i don't think so.

You can not raise an operator to a power.
Posted on 2003-05-15 12:51:39 by inFinie
inFinie,

a ^ b = e ^ ( b * ln( a ) )

logarithm in complex numbers:

ln( r * cis( t ) ) = ln(r) + ( t + 2kP ) * i , k is an element of Integers

j^j= e ^ ( j * ln( j ) )

j= 1 * cis(P/2) = cos(P/2) + j * sin(P/2) = 0 + j * 1

ln( j ) = ln ( cis(P/2) ) = ln1 + (P/2 + 2kP) * j

j ^ j = e ^ (j * j * (P/2 + 2kP)) = e ^ ( -P/2 - 2kP )
For k=0
j^j = e^(-P/2) = 0,207879576350761908546955619834979

PS cis(t) is a shorthand for cos(t) + i * sin(t)
P=pi


Congratulations inFinie! Your answer is correct. It appears that in Turkey they use commas for decimal points and periods for triples separators. That is the way the Europeans do it too, which is opposite of the U.S.

A more concise method is to use Euler's formula, e^jx=cos(x)+jsin(x). substituting x = pi/2, we get j = e^[(pi/2)*j]. Raising both sides to the j power gives j^j = e^[(pi/2)*j*j] = e^-pi/2 = 0.20787.....


I can not make a connection between logrithm to power....


Powers and logs are inverses of each other, once the log base is established. See http://mathworld.wolfram.com/EulerFormula.html, and notice how Euler's formula is changed from the power form into logarithmic form. I gave my problem to illustrate that quadrature values raised to a quadrature value power can be defined, and in this case even have a non-quadrature value solution.

That's an irony that you are asking j^j Ratch.

Can you think such a thing:
2 (*^2) 2 = ...
two, multiplication squared, two

No, i don't think so.

You can not raise an operator to a power.


It is not a irony, inFinie, it is a misinterpretation on your part. Look back at my previous posts. When I gave the problem, I said j^j, or j(1)^j(1) . The term j^j is the way most of the math and engineering world write the problem. A better and more precise way is j(1)^j(1) where j(1) means 'rotate 1 by 90? CCW'. In the more ambiguous form, j by itself is taken to mean rotate 1 by 90?. 30j is taken to mean rotate 30 by 90?. In the more precise form j(30), j is an operator only and has no dual meaning. Yes, you are right, an operator cannot be taken to a power, but that is not what I was doing. Ratch
Posted on 2003-05-15 21:18:52 by Ratch
Yes commas for decimal points.

Now your misinterpretation:

I can not make a connection between logrithm to power....

Did I not ask what the value of j^j, or j(1)^j(1) is in a previous post?

You claim that j is an operator. Then j^j is raising an operator to a power like i exampled in my previous post.

It is a Complex Number which's square is -1, named specially imaginary number
Posted on 2003-05-16 02:33:58 by inFinie
inFinie,

You claim that j is an operator. Then j^j is raising an operator to a power like i exampled in my previous post.

It is a Complex Number which's square is -1, named specially imaginary number


I believe I addressed your question and concerns in the last paragraph of my last post. Did your read and understand it? Ratch
Posted on 2003-05-16 06:37:18 by Ratch