Complex Numbers

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
darth sidious wrote,..
Ooh yes, good stuff!

Some older books use cis(theta), I can't remember if it was/is mathematically correct to use degrees with cis. It's not mathematically correct to use degrees with e^j{theta}, but many engineers DO! (Myself included
Click to expand...

The above quote coming from the thread "Who Loves an Ice Cream Cone"

Hello Darth,

cis(theta) comes from a Taylor Series although I think form memory it is actually a Maclaurin Series expansion of the e^x. Replace x with j(theta). Group real and imaginary terms and factor out j. Now expand using Taylor Series (Maclaurin Series) cos(theta) and sin(theta). The even terms correspond with cos while the odd terms correspond with sin...that is to say that they are identical to those obtained by expanding e^x, substituting in j(theta), simplifying and grouping terms, so e^j(theta) = cos(theta) + jsin(theta).

The angles can be in either radians or degrees, same applies when expressed in the format...A cos(wt + x) where A is a real number and x is an angle in either radians or degrees.

In the case of Complex Fourier Series, the limits of the integral being a measure of the period of g(t) and the exponent within the integrand g(t)e^jwnt are all in radians. Sometimes it is easier to always stay with radians, but I prefered using degrees when solving circuit problems containing reactive components.

Ron.
 
Same here, I much prefer to think in degrees!

Well, you are right in saying it's a Maclaurin expansion... and also saying Taylor's Expansion; Maclaurin's is really a special case of Taylor's by expanding about the point x = 0 (or t = 0, or y = 0, or whatever = 0 :p) The main man with maths at my university always referred to it as a Taylor's Expansion, even though it was actually Maclaurin's!

Taylor's Expansion : f(x+h) = f(x) + h.f'(x) + h^2.f''(x)/2! + h^3.f'''(x)/3! + h^4.f''''(x)/4! + .....

Macalaurin's Expansion: f(x) = f(0) + x.f'(0) + x^2.f''(0)/2! + x^3.f'''(0)/3! + x^4.f''''(0)/4! + .....

If we 'reverse' the Taylor's Expansion i.e. f(x+h) =f(h+x) = f(h) + x.f'(h) + x^2.f''(h)/2! + x^3.f'''(h)/3! + x^4.f''''(h)/4! + ..... (the x becomes h, and vice versa)

set h = 0, f(h+x) = f(x) = f(0) + x.f'(0) + x^2.f''(0)/2! + x^3.f'''(0)/3! + x^4.f''''(0)/4! + ..... i.e. it becomes Maclaurin's Expansion!

There's another way of deriving e^j{theta}:-

H = cos(x) + j.sin(x) so,

dH/dx = -sin(x) +j.cos(x) = j.(cos(x) + j.sin(x)) = j.H

(1/H).dH/dx = j,

Addendum, I forgot a rather important part here! :oops: The end result is the same, though!

by integrating, ln (H) + C = jx (C being a constant of integration, There is really a constant on both sides of the equation, but can be 'lumped' into one side i.e.

ln (H) + Const1 = jx + Const2, so ln (H) + (Const1 - Const2) = jx, setting C = (Const1 - Const2), we have ln (H) + C = jx )

It's customary to place the 'lumped' constant on the side of the independent variable, but is mathematically correct to place it on the side of the dependent variable, and is convenient to do so at times, as I shall show...

C is a constant, therefore let C = ln(k), where k = e^C, and is therefore another constant.

ln (H) + ln (k) = ln (k.H)

k.H = e^jx. At x = 0, k.H = e^j0 = e^0 = 1, so k = 1/H

Recall that originally we had H =cos (x) + j.sin (x), at x = 0, H = 1 + j0 = 1

therefore, k = 1/H = 1/1 = 1, so we have H = e^j(x)

put y=-x, so x = -y

H = cos(-y) + j.sin(-y) = cos(y) -j.sin(y) = e^j(-y) = e^-j(y) [since cos(-y) = cos(y) and sin(-y) = -sin(y)]

So... cos(x) ±j.sin(x) = e^±j(x)

Good stuff!
 
Hello Darth,

Yes indeed,....and I like your alternative method for deriving e^j(theta).. :D

Wonderful stuff!

Ron.
 
SydneyRoverP6B said:
Hello Darth,

Yes indeed,....and I like your alternative method for deriving e^j(theta).. :D

Wonderful stuff!

Ron.
Click to expand...

Not really mine, I saw it in a textbook many moons back! Glad ya like it though! :)
 
Oh my lord, I've just had a flash back to university!

I've disengaged my brain somewhat since then so I won't pretend to understand it fully but I do admire people who do this sort of thing on a daily basis and try to encourage others to join in.
 
Complex numbers are of the form a + jb where 'a' and 'b' are real numbers and 'j' or equally 'i' has the property of being equal to the square root of negative 1. This format is know as the rectangular form, and equally it can be expressed in polar (where it displays magnitude and an angle in radians or degrees) or in exponential form where it does the same as polar but in a more formal way.

Complex numbers first appeared in the 16th century as a means of solving third order polynomials. From that point on their applications grew to encompass many areas of mathematics including geometry, trigonometry, set theory, sequences and series, mappings and curves, functions, limits and continuity, differential operators and equations, integrals and more. As a consequence they have a direct application in engineering and physics including areas of electrical circuit analysis that involves alternating values, signal analysis, electromagnetics, heat flow, elasticity, hydrodynamics and aerodynamics.

Ron.
 
darth sidious said:
quattro said:
I am completely lost with it all :shock:
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LMAO!!!!!!!!!!!!!!! Says the one who solves all the puzzles before anyone else! :p
Click to expand...

I never learnt any advanced maths, so am lost with this lot. :?

I am just quite good at solving puzzles and problems. 8)

Richard
 
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