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# AC analysis superposition

We break a sinusoidal input voltage into two complex exponentials. Using superposition, we can recover the complex output signals and reassemble them into a real sinusoidal output voltage. Created by Willy McAllister.

## Want to join the conversation?

• What needs to be true for us to be certain that the output is sinusoidal given that the input is sinusoidal? He said something about linear circuit elements. Is that why? What are those?
(1 vote)
• At if `K+` and `K-` are different, how can they still be summed into one single cosine expression? In other words, what will `B` be in terms of `K+` and `K-`?
(Or if they are equal, then why differentiate between them? Why not just call both `K` and then we would have `K=B/2`?)
• Shouldn't the Vout + and Vout - have different phases, phi + and phi -? Perhaps, I missed the answer from an earlier video. (I jumped in the middle)
• What did e to the jwt (positive or negative,it dosen't matter) actually equal?
(1 vote)
• If you select values for w (omega, radian frequency) and t (time), then e^jwt is a single point falling on the unit circle (radius = 1) in the complex plane.

If you draw a vector line from the origin to the point, the length of the vector (its magnitude) is 1, and its angle from the +real axis is wt. (positive angles are measured counterclockwise)

Example: If t = 2 seconds and w = pi/4 radians/sec, then e^(pi/4)2 = e^(pi/2).
pi/2 is an angle of 90 degrees. So this is point on the complex plane at coordinate 0,j. Straight up from the origin on the imaginary axis, a distance of 1.

If you let t run free, the point traces out a circle on the complex plane, making one revolution every 8 seconds (2pi radians per every 8 seconds = pi/4 radians/sec)