Converting Complex Exponential Functions to Trigonometric Functions

[brollysan]

"Reduce" a trig.function

Homework Statement

Z(t) = the real part of :
(-5.8 + 2.2i)exp(iwt)

1. Reform it into: Acos(wt + ø)
2. Then reform it into Bsin(wt)+Dcos(wt)

Homework Equations

The Attempt at a Solution

I found the 2nd step to be much easier as I just have to use eulers formula to remove the exponential then just multiply it with the complex number that is before it and take the real part of it, leaving me with Z(t) = -5.8cos(wt) -2.2sin(wt)

The first step should be equally easy but for some odd reason (my lack of knowledge in trig) I get an angle that is wrong.

Acos(wt +ø ) = Asin(wt)cos(ø) + Acos(wt)sin(ø) =>

-5.8cos(wt) - 2.2sin(wt) = Asin(wt)cos(ø) + Acos(wt)sin(ø) =>
(cause cos and sin are lin.independant?)

-5.8cos(wt) = Acos(wt)sin(ø) => A = -5.8/sin(ø)
-Same logic- =>A = -2.2/cos(ø) =>

-2.2/cos(ø) = -5.8/sin(ø) or: -2.2tan(ø) = -5.8
=> ø = 1.21
Which means A = -5.8/sin(1.21) = -6.2
A = -2.2/cos(1.21) = +.35

??

That is the inconsistency and I do not know why, I am quite sure everything in the math is correct :S

 

[dacruick]

just split it up into -5.8exp(iwt) +2.2exp(iwt).
exp(iwt) = cos(wt) + isin(wt).
the real part of this formula is the cos(wt)
so you will have -5.8cos(wt) + 2.2cos(wt)
that gives -3.6cos(wt), and your answer I believe

 

[brollysan]

But

(-5.8 + 2.2i)exp(iwt) = (-5.8 + 2.2i)(cos(wt) + isin(wt)
= -5.8cos(wt) -5.8isin(wt) +2.2icos(wt) + 2.2(i*i)sin(wt)
= (-5.8cos(wt) -2.2sin(wt)) + i(2.2cos(wt) -5.8sin(wt))

Where the first part is the real part, leaving me with both a sine and cosine and unfortunantly not a sum of cosine ;(

 

[dacruick]

gah i did make a mistake. the two imaginary terms multilpy and become a real term.

 

[dacruick]

so what we have here now is -5.8cos(wt) -2.2sin(wt) and your question is how can you put that in the form of Acos(wt + phi)??

 

[dacruick]

sine and cosine are out of phase by pi/2, so you could technically just add pi/2 or subtract p/2 and you would have either both sines or both cosines. But then you would have a phase on one and no phase on the other, making them unable to be summed. so i don't know, check out some trig identities

 

[brollysan]

Actually plotting the function let's me "see" by how much it is out of phase so numerically I can approximate the Cos(wt + ø) but analytically? (There is an answer, so says the book)

How do you find the amplitude of 2 signals being added? -5.8cos(wt) -2.2sin(wt)

Still not able to find the phase.

 

[l'Hôpital]

Consider the following:
$$\cos(x - \theta) = \cos x \cos\theta + \sin x \sin\theta$$

Therefore, let's multiple by a constant R.
$$\begin{array}{1} R \cos(x - \theta) = R\cos(x)\cos(\theta) + R\sin(x)sin(\theta) \\ R\cos(x - \theta) = (R\cos(\theta))\cos x + (R\sin(\theta))\sin x \\ R\cos(x - \theta) = a\cos x + b\sin x$$
Where
$$a = R\cos\theta \ , \ b = R\sin\theta$$
Obviously, $$R = \sqrt{a^2 + b^2}$$

So, you have a and b. Solve for theta.

 

[Dick]

Actually plotting the function let's me "see" by how much it is out of phase so numerically I can approximate the Cos(wt + ø) but analytically? (There is an answer, so says the book)

How do you find the amplitude of 2 signals being added? -5.8cos(wt) -2.2sin(wt)

Still not able to find the phase.

You started out with the right general idea. But cos(wt+phi)=cos(wt)*cos(phi)-sin(wt)*sin(phi). It looks like you used the sin addition formula instead of cos. And your inconsistency at the end is just a mistake. -5.8/sin(1.21)=(-6.20) and -2.2/cos(1.21)=(-6.23). NOT +.35. The reason the two values of 'A' are slightly different is that you rounded off the angle to 1.21.

 

[brollysan]

Ah yes thank you it all makes sense, I used that formula for the sum of an angle because the book insists that the following are equivalent, which does not agree with the formula for the cosine of sum of angles.

(1) Z(t) = Acos(wt +ø)
(2) Z(t) = Asin(wt)cos(ø) + Acos(wt)sin(ø)
(3) Z(t) = Bsin(wt) + Dcos(wt)

 

[Dick]

(1) is only the same as (2) if you pick different phi's for each one.