# Circuit Theory/Convolution Integral/Examples/Example42

Given that V_{s} = (1 + 2t)μ(t), find i_{o} using the convolution integral.

## Contents

### Transfer function[edit]

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

solve(s^2 + 3*s + 1,s)

### Homogeneous Solution[edit]

Two real, different roots:

So I_{o} has this form:

### Particular Solution[edit]

After a very long time, inductors are shorts, the voltage across both 1 ohm resistors is 1 volt, so i_{o} is 1 amp:

### Initial Conditions[edit]

Adding the particular and homogeneous solutions together, have this form:

After a long period of time (using the particular initial condition again), the current is going to be 1:

Initially, the two inductors are going to be opens, thus i_{o} has to be 0:

Initially, all the drop is going to be across the first inductor, leaving the voltage across the second zero. The current i_{o} flows through both the second inductor and it's serial resistor, so an equation for this voltage can be obtained by taking the derivative:

solve([1+A+B,s1*A + s2*B],[A,B])

So:

### Impulse Solution[edit]

Taking the derivative of the above get:

### Convolution Integral[edit]

syms t x s1 = (-3 + sqrt(5))/2; s2 = (-3 - sqrt(5))/2; f = (s1*s2/(s1-s2)*(exp(s1*(t-x))-exp(s2*(t-x)))*(1+2*x)); S =int(f,0,t); vpa(S, 3)

At t=0:

Which means that C_1 = 0 so finally: