# Circuit Theory/Convolution Integral/Examples/Example43

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

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

Outline:

## Contents

### Transfer Function

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

$i_o(\infty) = \frac{1}{2} = \frac{1}{2} + C_1$
$C_1 = 0$

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:

$V_L(t) = L*{d i_o(t) \over dt}$
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

$i_o(t) = \int_0^t e^{-t+x}\sin (t-x)*(1+cos(x)) dx$
f := exp(x-t)*sin(t-x)*(1 + cos(x));
S =int(f,x=0..t);

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