Circuit Theory/Convolution Integral/Examples/Example43

2R 1L 1C circuit to solve using the convolution integral .. for wikibook circuit analysis

Given that i_s = 1 + cos(t), find i_o using the convolution integral.

Outline:

Transfer Function

$H(s)=\frac{{𝕀}_o}{{𝕀}_s}=\frac{{𝕍}_T}{{𝕀}_s}*\frac{{𝕀}_o}{{𝕍}_T}=\frac{1}{\frac{1}{1/s}+\frac{1}{1}+\frac{1}{s+1}}*\frac{1}{s+1}$

simplify(1/((s+1)*(s + 1 + 1/(s+1))))

$H(s)=\frac{1}{s^2+2s+2}$

Homogeneous Solution

Setting the denominator to zero and find the values of s:

solve(s^2 + 2*s + 2)

$s=-1\pm i$

This means the solution has the form:

$i_{o_h}(t)=e^{-t}(A\cos t+B\sin t)+C_1$

Particular Solution

After a long time the cap opens and the inductor shorts. The current is split between the two resistors. Each will get 1/2 of the unit step function source which would be 1/2 amp:

$i_{o_p}(t)=\frac{1}{2}$

Initial Conditions

$i_o=\frac{1}{2}+e^{-t}(A\cos t+B\sin t)+C_1$

The particular solution still has to apply so at t= ∞:

$i_o(\mathrm{\infty })=\frac{1}{2}=\frac{1}{2}+C_1$

$C_1=0$

Initially the current has to be zero in this leg so:

$i_o(0)=\frac{1}{2}+A=0$

$A=-\frac{1}{2}$

The initial voltage across the cap is zero, and across the leg is zero and across the inductor is zero. So:

$V_L(t)=L*\frac{d{i}_o(t)}{dt}$

f := 1/2 + exp(-t)*((-1/2)*cos(t) + B*sin(t));
g = diff(f,t)

$V_L(t)=e^{-t}(B(cos(t)-sin(t))+\frac{1}{2}(cos(t)+sin(t))$

$V_L(0)=B+\frac{1}{2}=0$

$B=-\frac{1}{2}$

So now:

$i_o=\frac{1}{2}(1-e^{-t}(\cos t+\sin t))$

Impulse Response

Taking the derivative of the above

f := 1/2*(1-exp(-t)*(cos(t) + sin(t)));
g = diff(f,t)

get:

$i_{o_{\delta }}(t)=e^{-t}\sin t$

Convolution Integral

$i_o(t)=\underset{0}{\overset{t}{\int }}{e}^{-t+x}\sin (t-x)*(1+cos(x))dx$

f := exp(x-t)*sin(t-x)*(1 + cos(x));
S =int(f,x=0..t);

$i_o(t)=0.2\cos (t)+0.4\sin (t)-0.7\cos (t)e^{-t}-1.1\sin (t)e^{-1}+0.5$