1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Converting phasor back into time domain

  1. Nov 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Find Amplitude of Steady-state voltage across capacitor

    2. Relevant equations


    3. The attempt at a solution
    5a96193ba4.jpg
    d57119ca1c.jpg

    I'm at the very last step of changing it back into the time domain to obtain my ampltiude. I was planning on getting it into the form a + bj, then using trigonometry and other formulae to get it into polar form from which I can get it into the form of Bsin(wt + theta), with B giving me my amplitude.

    Im stuck with 8+ 8/20 j^-1. Which isnt in the form a+bj. Any ideas how I can try and get it into that form?
     
  2. jcsd
  3. Nov 13, 2016 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    There is no specific frequency information contained in that complex number. The frequency of all voltages and currents will be that of the driving signal.

    Your complex number yields an angle and an amplitude.

    How to view ##\dfrac 1j##? Multiply both numerator and denominator by ##j## and see where that leads.

    BTW, ##\frac 8 {1+j20}## is not equal to what you say. Instead, multiply numerator and denominator by the complex conjugate of the denominator so you can end up with a real denominator.
     
  4. Nov 13, 2016 #3
    The frequency comes from the initial voltage sinusoid of 10rads/s. Phasors don't have frequencies.

    If I multiply above and below by j. I get:

    gif.latex?8%20+%20%5Cfrac%7B8%7D%7B20j%7D.gif
     
  5. Nov 13, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    Check your voltage division work. The numerator of the fraction should contain the impedance of the component that you want the voltage across and the denominator the sum of the two impedances.

    Regarding your conversion to time domain, take your rectangular form and convert it to polar form (magnitude and angle). You then have the magnitude and phase angle for the time domain.
     
  6. Nov 13, 2016 #5
    Ah.. My Voltage division is wrong. I accidently used the formula for resistors in parallel. oops
     
  7. Nov 13, 2016 #6

    NascentOxygen

    User Avatar

    Staff: Mentor

    Your expression for V right back at the start is wrong. Pretend you have resistors and write the expression, then substitute the capacitor impedance for one of the resistances.
     
  8. Nov 13, 2016 #7

    gneill

    User Avatar

    Staff: Mentor

    By the way, the following is definitely not true:
    upload_2016-11-13_11-27-17.png

    You can't split a denominator like that, even for real values. Consider an example:

    ##\frac{8}{6 + 2} \ne \frac{8}{6} + \frac{8}{2}##
     
  9. Nov 13, 2016 #8
    Okay. why am i being so stupid today. Apologies for all these stupid mistakes. Let me redo the question.
     
  10. Nov 13, 2016 #9
    Okay. So I got

    gif.latex?V%20%3D%20%5Cfrac%7B4%7D%7Bj20%20+%201%7D.gif

    Now. How do I transform that into the form a+bj.?
     
  11. Nov 13, 2016 #10

    gneill

    User Avatar

    Staff: Mentor

    That's standard complex math: multiply the numerator and denominator by the complex conjugate of the denominator.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Converting phasor back into time domain
  1. Phasor Domain Circuit (Replies: 5)

Loading...