Transfer functions to determine voltage of a similar circuit

Question

Goal is to solve for the voltage u₂(t)

I'm somewhat confused by this. I solved the transfer function Z(s) earlier in the top circuit, which is simply just Z. The voltage across Z in the top circuit is $$u_1(t)=5e^{(-100t)}u(t)$$ and the current source is j(t) = u(t), where u(t) is the step function. Z is a circuit system itself with no independent sources and all initial values are zero.

Using basic principles for u₂(t) you obtain e(t)·Z/(R+Z), where again e(t) = u(t), so it's u(t)·Z/(R+Z).

Obviously, just from that equation you can't solve for voltage. I'm thinking this has something to do with the frequency response, but I don't know how I'm supposed to use it for this problem.

Z(s) is the same in both circuits.

Answer 1

Lets assume that the source voltage \$e(t)\$ is given by $$ e(t) = u(t) V \leftrightarrow E(s) = \frac{1}{s} V$$ where \$E(s)\$ is the laplace transform of \$e(t)\$ and \$u(t)\$ is a unit step function. The voltage \$u_2(t)\$ with laplace transform \$U_2(s)\$ then follows to $$ U_2(s) = \frac{Z(s)}{Z(s) + R} \cdot E(s)$$ by making use of kirchhoffs current and voltage laws. In the following i will assume that \$Z(s)\$ takes the form $$ Z(s) = K\frac{s + a}{s + b}$$ based on your description of the response of the first circuit. It is left as an exercise for you to determine the correct parameters \$K, \,a,\,b\$. This leaves us with the following expression for the desired voltage in the laplace domain $$ U_2(s) = \frac{Ks + Ka}{Ks + Ka + R(s + b)} \frac{1}{s} V.$$ Finally we need to perform the inverse laplace transform of \$U_2(s)\$ to get \$u_2(t)\$ in time domain. First we identify the poles by setting the characteristic polynomial in the denominator equal to zero and calculating the roots as done below $$ s((K+R)s + (Ka + Rb)) = 0 $$ $$\rightarrow s = 0 =:s_1 \,\, or \,\, s = - \frac{(Ka + Rb)}{(K+R)} =: s_2.$$ In a second step we perform a partial fraction decomposition $$ U_2(s)/V = \frac{[U_2(s)\cdot (s - s_1)](s \rightarrow s_1)}{s - s_1} + \frac{[U_2(s)\cdot (s-s_2)](s \rightarrow s_2)}{s - s_2}$$ where $$[U_2(s)\cdot (s - s_1)](s \rightarrow s_1) = \lim_{s\rightarrow s_1} U_2(s)\cdot (s - s_1).$$ (Warning: This direct method of calculating the coefficients in the denominator work only for poles with multiplicity of 1. The method has to be adjusted for poles with greater multiplicity and for cases where the numerator polynomial is of higher or equal order than the denominator polynomial.)

The third step of attaining \$u_2(t)\$ in time domain is done by applying the inverse laplace transform to each term of the partial fraction sum (i.e. looking up the inverse transform in corresponding tables). In our case we will get $$u_2(t)/V = ([U_2(s)\cdot (s - s_1)](s \rightarrow s_1) + [U_2(s)\cdot (s - s_2)](s \rightarrow s_2)\cdot e^{s_2 \cdot t } ) \cdot u(t) .$$ To summarize, first determine all the poles (and relative degree). In a second step perform a partial fraction decomposition. And third, apply the inverse laplace transform to each term. The final result will be the response signal in the time domain.