Converting phasor back into time domain

You'll get a real denominator, and you know how to deal with that.That's standard complex math: multiply the numerator and denominator by the complex conjugate of the denominator. You'll get a real denominator, and you know how to deal with that.In summary, the student is trying to find the amplitude of a steady-state voltage across a capacitor. They are stuck at the last step of converting it back to the time domain and obtaining the amplitude. They plan to use trigonometry and other formulae to get the voltage in polar form and then solve for the amplitude. They also mention having difficulty with converting a complex number into the form a + bj, but are advised to multiply the numerator and denominator by the complex conjugate
  • #1
CoolDude420
198
8

Homework Statement


Find Amplitude of Steady-state voltage across capacitor

Homework Equations

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 isn't in the form a+bj. Any ideas how I can try and get it into that form?
 
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  • #2
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.
 
  • #3
NascentOxygen said:
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 ##\displaystyle {\dfrac 1j}##? Multiply both numerator and denominator by ##j## and see where that leads.

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
 
  • #4
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.
 
  • #5
gneill said:
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.

Ah.. My Voltage division is wrong. I accidently used the formula for resistors in parallel. oops
 
  • #6
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.
 
  • #7
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}##
 
  • #8
gneill said:
By the way, the following is definitely not true:
View attachment 108847

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}##

Okay. why am i being so stupid today. Apologies for all these stupid mistakes. Let me redo the question.
 
  • #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.?
 
  • #10
CoolDude420 said:
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.?
That's standard complex math: multiply the numerator and denominator by the complex conjugate of the denominator.
 

1. What is a phasor and why is it used in electrical engineering?

A phasor is a complex number that represents the magnitude and phase of a sinusoidal function. It is commonly used in electrical engineering to simplify the analysis and calculation of AC circuits, as it allows for the representation of multiple sinusoids with different frequencies and phases as a single complex number.

2. How do you convert a phasor back into the time domain?

To convert a phasor back into the time domain, we use the inverse Fourier transform. This involves taking the complex conjugate of the phasor and multiplying it by e^(jωt), where ω is the frequency of the sinusoid. The resulting function is then summed with other transformed phasors to obtain the time-domain representation.

3. Can a phasor have a negative magnitude?

Yes, a phasor can have a negative magnitude. This indicates that the sinusoid represented by the phasor is 180 degrees out of phase with the reference sinusoid. In other words, it is inverted compared to the reference.

4. How does converting a phasor back into the time domain help with circuit analysis?

Converting a phasor back into the time domain allows us to obtain the time-domain representation of a circuit, which is necessary for calculating values such as voltage and current. This simplifies the analysis of AC circuits, as it reduces the number of equations and calculations required.

5. Are there any limitations to converting a phasor back into the time domain?

Yes, there are some limitations to converting a phasor back into the time domain. One limitation is that it only works for linear circuits, as non-linear components can introduce harmonics that cannot be represented by a single phasor. Additionally, the inverse Fourier transform assumes that the phasor is constant over time, which may not be the case in some circuits.

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