# Homework help needed

#### elecstudent

Joined Nov 11, 2010
26
Am I right to represent this sine wave y=sin x; in polar form to be (1<90degrees)

or would it be in rectangualr form (2pi, 1)

Please let me know how to begin this.. as I'm unsure..

Thanks

#### Georacer

Joined Nov 25, 2009
5,182
http://en.wikipedia.org/wiki/Sine
There are many graphs for cartesian and polar (unit cycle) representation.

Do you need to use any particular software?
What do the intervals (1<90deg) and (2π,1) mean? They don't make much sense...

#### elecstudent

Joined Nov 11, 2010
26
I don't have to use any software specifically. I'm adding and subtracting functions, though thought it would be better to represent these functions as complex numbers.

So if I represent y=sine x, to rectangular form, I assumed it would be (2pi, j1).. and then work out the adding/subtracting parts in complex form.

#### Georacer

Joined Nov 25, 2009
5,182
I still don't understand what you are trying to pass across with those (x,y) pairs.

The (2π, 1) point doesn't belong to the y=sinx curve for one...

#### elecstudent

Joined Nov 11, 2010
26
ok I have the function y=sin x, in rect form I assume it is (2n/4, j1)

I'm asking would this be right?

then I can use the rect to add and subtract in complex form....

#### BillO

Joined Nov 24, 2008
990
The ordered pair for y=sin(x) in cartesian coordinates is (x, sin(x)), but I'm sure this is not what you are after. Perhaps you can give us an example or two of the problems you are trying to solve.

#### elecstudent

Joined Nov 11, 2010
26
(sin x) + (-4 cos x)

expressed as a complex number

#### Georacer

Joined Nov 25, 2009
5,182
This is a typical example of misleading from the end of the OP. Why on earth did you assume that this problem involves graphical representations of the sines? Now you lost a day's time diverting this thread out of your target.

In order to solve this kind of problems you need to take a look at these relations:
$$\cos\theta=\frac{1}{2}(e^{j\theta}+e^{-j\theta})\\ \sin\theta=\frac{1}{2j}(e^{j\theta}-e^{-j\theta})$$

#### elecstudent

Joined Nov 11, 2010
26
I'm not getting much feed back here guys.. maybe I am looking at this the wrong way..

Basically, in AC theory I can use complex numbers to represent components and how the react in the circuit, as a vector right.. so what I'm trying to do is look at the initial functions, say a sine wave and act on it, in complex form...

I have looked at the sine wave y=sin x.. this in rec form I assume it (2n/4, j1).. right??

If my above assumption is correct then I can assume -4 cos x = (2n/2,-j4)

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#### elecstudent

Joined Nov 11, 2010
26
it is yes, but I was just wondering for some feedback if my assumptions are correct aswell.

#### blah2222

Joined May 3, 2010
582
Well if you plot the phasor for y = sin(theta) on the real/imaginary plane, you only get three points all on the real axis for any theta.

Re{-1, 0, 1}

This is the same result for a cosine as well.

Where are you getting this 2n business, and what is n supposed to be?

#### elecstudent

Joined Nov 11, 2010
26
well a sine or cos wave has a period of 2n = (2pi), if I have a function -4 cos x then the amplitude is -4 with one wave being present from x=0 to x=2pi, the mid point (vector will be half this 2pi which is where i have 2pi/2) .. I have assumed that I can take a rectangular form via the cooridnates from the cos wave from x=2pi/2 and amplitude y= -4, that will give me the rect form of the cos wave.

This was what I was looking at, apologies for any confusion.

My question is my assumption correct?

#### elecstudent

Joined Nov 11, 2010
26

#### blah2222

Joined May 3, 2010
582
Can you draw out what you are trying to describe, no clue what you are trying to explain. Sorry...

#### t_n_k

Joined Mar 6, 2009
5,455
The time-domain function y(t)=|A|cos(ωt+θ)

is represented in the frequency domain by the form

Y(ω)=|A|e^(jθ)=|A|(cosθ+jsinθ)

The function z(t)=|B|sin(ωt+β)=|B|cos(90°-ωt-β)=|B|cos(ωt+β-90°)

is represented in the frequency domain by the form

Z(ω)=|B|e^(j[β-90°])=|B|(cos(β-90°)+jsin(β-90°))

If y(t)=-4cos(ωt) then Y(ω)=-4e^j0=-4

If z(t)=sin(ωt) then Z(ω)=1e^(-j90°)=1(cos(-90°)+jsin(-90°))=-j

So Y(ω)+Z(ω)=-4-j {in rectangular form}

The latter is the frequency domain equivalent of f(t)=sin(ωt)-4cos(ωt)

In your case you have the variable 'x' instead of the more conventional 'ωt'. But this ignored in the frequency domain form anyway.

#### Georacer

Joined Nov 25, 2009
5,182
@t_n_k

The whole question is full of inaccuraties. Let's set f(x)=sinx-4cosx. This is a real function $$f:\mathbb{R} \rightarrow \mathbb{R}$$. Complex numbers don't have anything to do here.

A possible interpretation would be to ask to take the function to the s-plane throug a Laplace transform, which would yield
$$F(s)=\frac{1}{s^2+1}-4\frac{s}{s^2+1}=\frac{1-4s}{s^2+1}$$
with ω0=1 (the circular frequency)

Another interpretation would be to ask to use the complex function $$g:\mathbb{R} \rightarrow \mathbb{C}$$ $$g(x)=e^{jx}$$, to express f(x), which would be, taking into account post #8
$$f(x)=\frac{1}{2j}(e^{jx}-e^{-jx})-4\frac{1}{2}(e^{jx}+e^{-jx})\\ =-2(e^{jx}+e^{-jx})-j(\frac{1}{2}(e^{jx}-e^{-jx}))\\ =(-2+\frac{j}{2}) e^{-j x}-(2+\frac{j}{2}) e^{j x}$$.
Don't be fooled, this is always a real number!

Finally, a wild guess would be to convert the graph of f(x) into a parametric equation $$h:\mathbb{R} \rightarrow \mathbb{C}, h(t)=t+jf(t)$$

As for your effort, I think you skipped a few operations there... The function f=sinx-4cosx is dependent on x and therefore is variable. However, F=4-j is constant. Clearly something is wrong here.

To conclude, the OP has repeatedly failed to ask for help in a clear and comprehensive manner.

#### t_n_k

Joined Mar 6, 2009
5,455
Since the OP had indicated in post #11 (in the affirmative) that this was a question in relation to the use of complex numbers in AC circuit theory, I assumed this was the correct path. I just took the point of view that the variable 'x' was substituted by the OP for the more commonly used 'ωt'.

Personally I don't think the OP has much idea of what he is actually asking. I was trying - like everyone else (yourself included) to interpret the meaning of the question.

Perhaps the OP may return but I'm not holding my breath.

#### tyblu

Joined Nov 29, 2010
199
Looks like phasors to me.

#### elecstudent

Joined Nov 11, 2010
26
If I took the rect form of the functions -4 cos x, and sin x, changed them to polar form, then i could draw the individual phasors.

Then I can add and subtract them, easily.

My initial question was is my assumption correct in how I took the rect form of -4 cos x, to be (2pi/2, -j4)

And sin x to be (2pi/4, j1)..