Vout=-1.302*10^-3 - j7.86*10^-3 = 7.97 < 260.59 mV

 

Vout=-1.302*10^-3 - j7.86*10^-3 = 7.97 < 260.59 mV

This is with 40 volts applied to the input, right?

Are you satisfied that this is correct? Is it the answer your book gives?

I get something substantially larger.

 

This is with 4 mV applied to the input, so I'm satisfied with an answer of 7 mV. I wrote a mistake on the link in my very first attempt which I think I fixed in a link I posted in a later post.

 

Do you have some independent means to verify your solution, such as an answer from the book?

I get Vout = (-1.29730 -j7.78378)mV

 

No but I will in a few days, I'll update this thread when I do.

 

The answer is 7.89 < -9.46 degrees mV

Electrician, the answer you gave is correct for the magnitude, but your values give an angle of 181.4 degrees.

 

The answer is 7.89 < -9.46 degrees mV

Electrician, the answer you gave is correct for the magnitude, but your values give an angle of 181.4 degrees.

The reason I got that is because back in post #41, your answer had been derived assuming that the input had a -90° phase shift. I think you have since realized that you don't need to do that. But, I followed your lead so that I would get the same numbers you did.

What I get if I don't assume any phase shift associated with the input signal is:

(7.78378 - j1.297297)mV = 7.89115 mV < -9.46232°