# Circuit Theory/Convolution Integral/Examples/Example43

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Given that i_{s} = 1 + cos(t), find i_{o} using the convolution integral.

Outline:

## Contents

### Transfer Function[edit]

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

### Homogeneous Solution[edit]

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

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

This means the solution has the form:

### Particular Solution[edit]

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:

### Initial Conditions[edit]

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

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

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

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

So now:

### Impulse Response[edit]

Taking the derivative of the above

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

get:

### Convolution Integral[edit]

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