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Homework Help: Phasor addition

  1. Feb 12, 2005 #1
    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! :)
  2. jcsd
  3. Feb 12, 2005 #2


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    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]
  4. Feb 12, 2005 #3


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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]
  5. Feb 12, 2005 #4
    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]
    Last edited: Feb 12, 2005
  6. Feb 14, 2005 #5
    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 :(
  7. Feb 15, 2005 #6
    anyone? :(
  8. Feb 15, 2005 #7
    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]
  9. Feb 15, 2005 #8


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    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.
    Last edited: Feb 15, 2005
  10. Feb 15, 2005 #9

    A sqrt missing.
  11. Feb 15, 2005 #10
    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.
  12. Feb 15, 2005 #11
    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.
  13. Feb 15, 2005 #12


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    You're quite right. :eek:
  14. Feb 15, 2005 #13


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    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.
  15. Feb 15, 2005 #14
    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.
  16. Feb 15, 2005 #15
    Amplitude = [tex] a \sqrt 2\sqrt (1 +cos\phi) [/tex]

    phase = [tex] \phi + arctan(\frac{sin\phi}{1+cos \phi})[/tex]
  17. Feb 15, 2005 #16


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    Can you show how you came to the above results step by step...
  18. Feb 15, 2005 #17
    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):


    And then you can read off the amplitude and phase.
    Last edited: Feb 15, 2005
  19. Feb 15, 2005 #18


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    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:

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

    [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.
    Last edited: Feb 15, 2005
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