Circuit Analysis II Lecture 13

****THE BELOW IS INCOMPLETE, but I'm releasing it in case anyone can get some value out of what was written.****

Disclaimer: This series of posts is to serve as notes for myself as well as any others interested in the subject of Circuit Analysis II. This is the course ECE 213 at UNM.

The course will follow the book:
Electric Circuits (8th Edition)

Agenda for the day:
13.6 - AP 13.12
13.7


AP 13.12

x(t) = Acos(wt+phi) is given

Looking at the equation:
v_o/i_g = ((2+s)s * 10/s)/((2+s)+10/s)

working the algebra through to simplify you get:
10*(2+s/(s² + 2s + 10))

This is our H(s), so using that H(s) we sub it into our steady state equation:


y_ss(t) = A|H(jw)|cos(wt+phi+theta)

Now we want to find H(jw) so remember that s = jw so we can just grab w and sub j and w in for s in the H(s) equation, that will look like this:
10(2+j4)/((j4)² + 2*j4 + 10)

a little manipulation we get:
20+j40/(-16 + 8j + 10)

next we combine and multiply by the conjugate:
(20+j40)/(-6 + 8j) * (-6-8j)/(-6-8j)

finally we get:
2-j4

Converting to polar we get:
4.47e^-j63.43


Finally substituting back in our equation we get:
10*20^1/2cos(4t-63.43°)

Exercise 13.72
we have a resistor of 50 ohms and two inductors whose values are 2F and 8F in series.
v_i is attached between the resistor and the 8F inductor. the v_o is located across the 8F inductor.

Converting to the S domain and using voltage division we get:
v_o = v_i/(50+2s+8s) * 8s

Solving it out for v_o/v_i we get:
.8s/(s+5)

Now we'll sub in 75/s for v_i

so we'll get:
75*.8s/s(s+5)

Finishing the algebra we get:
60/(s+5)

Now we want to use the convolution integral on this one: