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Phasor addition

by Exulus
Tags: addition, phasor
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Exulus
#1
Feb12-05, 09:05 AM
P: 50
Hoping someone can give me a nudge in the right direction for this..ive tried searching the net and unless im putting in the wrong keywords, there doesnt seem to be much useful material out there for this. I need to add together two phasors to find the resultant amplitude and phase of each wave.

I have [tex]a(e^{j(\omega t + \phi)} + e^{j(\omega t + 2\phi)})[/tex]

The furthest i can get is:

[tex] ae^{j(\omega t)}[\cos\phi + \cos2\phi + j(\sin\phi + \sin2\phi)][/tex]

Any help on where to go next or an alternative method would be much appreciated! :)
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learningphysics
#2
Feb12-05, 10:38 AM
HW Helper
P: 4,125
Quote Quote by Exulus
Hoping someone can give me a nudge in the right direction for this..ive tried searching the net and unless im putting in the wrong keywords, there doesnt seem to be much useful material out there for this. I need to add together two phasors to find the resultant amplitude and phase of each wave.

I have [tex]a(e^{j(\omega t + \phi)} + e^{j(\omega t + 2\phi)})[/tex]

The furthest i can get is:

[tex] ae^{j(\omega t)}[\cos\phi + \cos2\phi + j(\sin\phi + \sin2\phi)][/tex]

Any help on where to go next or an alternative method would be much appreciated! :)
Ok. I'm going to divide out the [tex]e^{j\omega t}[/tex]

What is the magnitude of
[tex]a[\cos\phi + \cos2\phi + j(\sin\phi + \sin2\phi)][/tex]
?

What is the phase of
[tex]a[\cos\phi + \cos2\phi + j(\sin\phi + \sin2\phi)][/tex]
?
Galileo
#3
Feb12-05, 10:40 AM
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You could just take the absolute value (modulus) of the expression and simplify with trig rules (if that's what you're looking for).

The simplifying will be somewhat easier if you take [itex]e^{j(wt+\phi)}[/itex] outisde the brackets by using [itex]e^{j(wt+2\phi)}=e^{j(wt+\phi)}e^{j\phi}[/itex]

clive
#4
Feb12-05, 01:00 PM
P: 120
Phasor addition

The simplest way to solve this problem consists of using "phasors" (vectors). Just draw two vectors of the same magnitude a. The angles made by these two vectors with Ox axis must be [tex]\phi[/tex] and [tex]2 \phi[/tex]. Add these two vectors using your favourite method (analytical one would be a very good choice here) and find the magnitude and the angle made with Ox of the sum vector. If your results are A for the sum magnitude and [tex]\phi_{sum}[/tex] for its orientation then you can rebuild the sinusoidal function by
[tex]y_{sum}=A \cdot exp[ j(\omega t+\phi_{sum})][/tex]
Exulus
#5
Feb14-05, 04:23 AM
P: 50
Hi guys, sorry for the late reply.

learningphysics, the magnitude is [tex]a[/tex] and the phase is [tex]\cos\phi + \cos2\phi[/tex] ? I can understand that, its just i think it needs to be put into a singular cos expression. I tried using the cos addition rule (2 times cos half sum times cos half difference) but that just seems to make it more complicated as i have two cos's being multiplied then :(
Exulus
#6
Feb15-05, 06:25 AM
P: 50
anyone? :(
Gamma
#7
Feb15-05, 08:19 AM
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P: 333
Use the following to simplify your solution:

[tex] A + j B = \sqrt {A^2 + B^2} [ \frac{A}{ \sqrt (A^2 + B^2)} + j \frac{B}{ \sqrt (A^2 + B^2)} ][/tex]

= [tex] \sqrt (A^2 + B^2)[ sin\alpha + j cos\alpha] [/tex]
= [tex] \sqrt (A^2 + B^2) \ e^{j\alpha} [/tex]




where [tex] tan\alpha = A/B[/tex]
Galileo
#8
Feb15-05, 08:24 AM
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Quote Quote by Exulus
I have [tex]a(e^{j(\omega t + \phi)} + e^{j(\omega t + 2\phi)})[/tex]
If you take out [itex]e^{j(\omega t+\phi)}[/itex] you get:

[tex]a(e^{j(\omega t + \phi)} + e^{j(\omega t + 2\phi)})=ae^{j(\omega t+\phi)}(1+e^{j\phi})[/tex]

So the absolute value is:

[tex]|a||1+e^{j\phi}|=|a|\sqrt{\left((1+\cos \phi)^2+\sin^2 \phi\right)}[/tex]

simplify it further from here.
Gamma
#9
Feb15-05, 08:46 AM
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P: 333
[tex]|a||1+e^{j\phi}|=|a|\left((1+\cos \phi)^2+\sin^2 \phi\right)[/tex]

A sqrt missing.
Exulus
#10
Feb15-05, 08:48 AM
P: 50
Sorry, i dont really understand where the cos and the sin squared came from in your last line? I dont think ive been taught all the maths needed to do this but ah well :S#

Gamma, i'll try your suggestion now, cheers.
Gamma
#11
Feb15-05, 09:06 AM
Gamma's Avatar
P: 333
This is the usual way to convert a complex number in to polar form. The angle alpha is chosen such that


[tex] sin\alpha = \frac{A}{ \sqrt (A^2 + B^2)} [/tex]

You will see then, the following is correct

[tex] cos\alpha = \frac{B}{ \sqrt (A^2 + B^2)} [/tex]

So you get the desired format [tex] sin\alpha +i cos\alpha [/tex]

Hope this helps.
Galileo
#12
Feb15-05, 10:08 AM
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Quote Quote by Gamma
A sqrt missing.
You're quite right.
learningphysics
#13
Feb15-05, 11:11 AM
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P: 4,125
Quote Quote by Exulus
Hi guys, sorry for the late reply.

learningphysics, the magnitude is [tex]a[/tex] and the phase is [tex]\cos\phi + \cos2\phi[/tex] ? I can understand that, its just i think it needs to be put into a singular cos expression. I tried using the cos addition rule (2 times cos half sum times cos half difference) but that just seems to make it more complicated as i have two cos's being multiplied then :(
No the magnitude and phase you've given aren't correct. Have a look through the posts in the thread.

[tex]a[\cos\phi + \cos2\phi + j(\sin\phi + \sin2\phi)]=[/tex]

[tex]a[\cos\phi + \cos2\phi] + j[a(\sin\phi + \sin2\phi)][/tex]

Convert this complex number into polar form. The magnitude isn't a. If you're studying phasors, I'm certain your text has a section explaining how to convert to polar form. Review it, and you'll see the answer.

Galileo gave a nice way to make the math a lot easier.
Exulus
#14
Feb15-05, 12:59 PM
P: 50
Is the magnitude [tex]2a[/tex] and the phase [tex]\cos3/2\phi\cos1/2\phi[/tex] ? I hope that comes out alright, my internet is being incredibly slow at the moment and i cant see any of the equations above. I reached my answer using cos addition rules.
Gamma
#15
Feb15-05, 02:33 PM
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P: 333
Amplitude = [tex] a \sqrt 2\sqrt (1 +cos\phi) [/tex]

phase = [tex] \phi + arctan(\frac{sin\phi}{1+cos \phi})[/tex]
learningphysics
#16
Feb15-05, 03:03 PM
HW Helper
P: 4,125
Quote Quote by Exulus
Is the magnitude [tex]2a[/tex] and the phase [tex]\cos3/2\phi\cos1/2\phi[/tex] ? I hope that comes out alright, my internet is being incredibly slow at the moment and i cant see any of the equations above. I reached my answer using cos addition rules.
Can you show how you came to the above results step by step...
Exulus
#17
Feb15-05, 04:20 PM
P: 50
ok :)

[tex]a(e^{j(\omega t + \phi)} + e^{j(\omega t + 2\phi)})[/tex]
[tex]ae^{j\omega t}(e^{j\phi} + e^{j2\phi})[/tex] (1)

Taking the bit in brackets:

[tex]cos\phi + j\sin\phi + \cos2\phi + j\sin2\phi[/tex]

Taking the real part and using: [tex]\cos{u} + \cos{v} = 2\cos{(u + v)/2}\cos{(u - v)/2}[/tex]

[tex]\cos\phi + \cos2\phi = 2\cos{3/2}\phi\cos{1/2}\phi[/tex] (2)

Subbing (2) into (1):

[tex]2a\cos{3/2}\phi\cos{1/2}\phi\cos\omegat[/tex]

And then you can read off the amplitude and phase.
learningphysics
#18
Feb15-05, 07:32 PM
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P: 4,125
Quote Quote by Exulus
ok :)

[tex]a(e^{j(\omega t + \phi)} + e^{j(\omega t + 2\phi)})[/tex]
[tex]ae^{j\omega t}(e^{j\phi} + e^{j2\phi})[/tex] (1)

Taking the bit in brackets:

[tex]cos\phi + j\sin\phi + \cos2\phi + j\sin2\phi[/tex]

Taking the real part and using: [tex]\cos{u} + \cos{v} = 2\cos{(u + v)/2}\cos{(u - v)/2}[/tex]

[tex]\cos\phi + \cos2\phi = 2\cos{3/2}\phi\cos{1/2}\phi[/tex] (2)

Subbing (2) into (1):

[tex]2a\cos{3/2}\phi\cos{1/2}\phi\cos\omegat[/tex]

And then you can read off the amplitude and phase.
Ok. What you've done above is just written the real part in a different form. That technique is useful when multiply two cosines in the time domain. But it has nothing to do with getting the magnitude or phase of a phasor.

When you have a complex number in rectangular form:
a+jb

The magnitude is [tex]M=\sqrt{a^2+b^2}[/tex]
and the phase is [tex]\theta=tan^{-1}(b/a)[/tex]

So:
[tex]a+jb = Me^{j\theta}[/tex]
in polar form, or you can use the notation [tex]M\angle \theta[/tex]

Once you've found the phasor in polar form you can then immediately write the time domain solutions:
[tex]Mcos(wt + \theta)[/tex]

Please review your text for dealing with complex numbers. It'll probably go through all this in more detail.


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