Going from phasor form to instantaneous time sinusoidal functions

[Lolsauce]

So I'm currently reviewing for my electromagnetism class, but I do not remember the techniques from going to phasor to instantaneous time. Maybe someone can explain to me what is going on.

Homework Statement

Find the instantaneous time sinusoidal functions corresponding to the following phasors:My question is specifically on Part A and Part B

Homework Equations

Relevant equations includes Eulers:

Which let you go from polar to rectangular.

The Attempt at a Solution

For Part A, we are given V~ = -5exp(j*pi/3)

I just went straight from phasor to instantaneous time:

V~ = v(t) -> -5exp(j*pi/3) = Re{V~ * exp(jwt) = -5exp(jwt + pi/3)

But for some strange reason, in the solution manual they subtract pi from the exponential to remove the negative sign. I'm not exactly sure why they do this. It is the same for Part B, why do they subtract pi/2. Can anyone explain this to me? I've also attached the solution to part A and B below.

 

[rude man]

First, you need to multiply all your coefficients (voltages and currents) by √2. The transformed voltages and currents represent rms values. The time-domain ones obviously represent peak voltages.

OK, now realize that -1 = +1*exp(jπ) = +1*exp(-jπ). You don't want a negative sign in front of your time-domain voltage expression, it's meaningless. What counts is phase. -1V = +1V shifted 180 degrees.

Use Euler to convince yourself that -1 = exp(jπ) = exp(-jπ). Also that j = exp(jπ/2).

Subtracting π/2 from the exponent is subtracting 90 degrees from the phase shift. Multiplying any transformed voltage by j is equivalent to adding (or subtracting) π from the exponent. So you need to subtract π/2, then add π, to figure out part b.

 

[Lolsauce]

First, you need to multiply all your coefficients (voltages and currents) by √2. The transformed voltages and currents represent rms values. The time-domain ones obviously represent peak voltages.

OK, now realize that -1 = +1*exp(jπ) = +1*exp(-jπ). You don't want a negative sign in front of your time-domain voltage expression, it's meaningless. What counts is phase. -1V = +1V shifted 180 degrees.

Use Euler to convince yourself that -1 = exp(jπ) = exp(-jπ). Also that j = exp(jπ/2).

Subtracting π/2 from the exponent is subtracting 90 degrees from the phase shift. Multiplying any transformed voltage by j is equivalent to adding (or subtracting) π from the exponent. So you need to subtract π/2, then add π, to figure out part b.

Thanks for the reply, so you are saying that the the subtraction of the ∏ phase in part is to get rid of the negative. Meaning if I add ±∏, I should get the same answer since it shifts the negative to positive phase either way. That means:

5exp(j(4∏/3 + wt)) should also be correct? Yes?

 

[rude man]

Yes, but convention is to never exceed pi/2 in magnitude.

So exp(-2pi/3) is more conventionally expressed as exp(pi/3) etc.

 

[DeeNos]

I can offer some insight into why the solution manual subtracts pi or pi/2 from the exponential in the phasor form to instantaneous time conversion.

First, let's review the concept of phasors and instantaneous time sinusoidal functions. Phasors are complex numbers that represent the amplitude and phase of a sinusoidal function. They are useful in analyzing AC circuits because they simplify the calculation of complex quantities such as voltage and current. On the other hand, instantaneous time sinusoidal functions represent the actual time-varying behavior of the AC signals.

Now, in order to go from phasor form to instantaneous time, we use the Euler's formula, which relates the exponential function to trigonometric functions. In this case, we are using the complex exponential form, which is e^(jwt). This form represents a sinusoidal function with an amplitude of 1 and a phase of wt.

In the solution manual, they subtract pi or pi/2 from the exponential in order to adjust for the initial phase angle given in the phasor form. This is because in the phasor form, the phase angle is measured with respect to the positive real axis, whereas in the instantaneous time form, the phase angle is measured with respect to the positive imaginary axis. Subtracting pi or pi/2 shifts the phase angle by the appropriate amount to align it with the positive imaginary axis.

In Part A, the initial phase angle is pi/3, so subtracting pi from the exponential aligns the phase angle with the positive imaginary axis. Similarly, in Part B, the initial phase angle is pi/2, so subtracting pi/2 from the exponential aligns the phase angle with the positive imaginary axis.

I hope this explanation helps you understand the technique of going from phasor form to instantaneous time sinusoidal functions. Remember to always keep track of the initial phase angle and adjust the exponential accordingly. Good luck with your studies in electromagnetism!