phasor addition

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! :)

learningphysics

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

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

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

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

anyone? :(

Gamma

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

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

A sqrt missing.

Exulus

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

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

You're quite right.

learningphysics

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

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

Amplitude = [tex] a \sqrt 2\sqrt (1 +cos\phi) [/tex]

phase = [tex] \phi + arctan(\frac{sin\phi}{1+cos \phi})[/tex]

learningphysics

Can you show how you came to the above results step by step...

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.