# Circuit Theory/Convolution Integral

## Impulse Response

So far circuits have been driven by a DC source, an AC source and an exponential source. If we can find the current of a circuit generated by a Dirac delta function or impulse voltage source δ, then the convolution integral can be used to find the current to any given voltage source!

## Example Impulse Response

The current is found by taking the derivative of the current found due to a DC voltage source! Say the goal is to find the δ current of a series LR circuit .. so that in the future the convolution integral can be used to find the current given any arbitrary source.

Choose a DC source of 1 volt (the real Vs then can scale off this).

The particular homogeneous solution (steady state) is 0. The homogeneous solution to the non-homogeneous equation has the form:

$i(t)=Ae^{-{\frac {t}{\frac {L}{R}}}}+C$

Assume the current initially in the inductor is zero. The initial voltage is going to be 1 and is going to be across the inductor (since no current is flowing):

$v(t)=L{di(t) \over dt}$
$v(0)=1=L*(-{\frac {AR}{L}})$
$A=-1/R$

If the current in the inductor is initially zero, then:

$i(0)=0=A+C$

Which implies that:

$C=-A=1/R$

So the response to a DC voltage source turning on at t=0 to one volt (called the unit response μ) is:

$i_{\mu }(t)={\frac {1}{R}}(1-e^{-{\frac {t}{\frac {L}{R}}}})$

Taking the derivative of this, get the impulse (δ) current is:

$i_{\delta }(t)={\frac {e^{-{\frac {t}{\frac {L}{R}}}}}{L}}$

Now the current due to any arbitrary VS(t) can be found using the convolution integral:

$i(t)=\int _{0}^{t}i_{\delta }(t-\tau )V_{s}(\tau )d\tau =\int _{0}^{t}f(t-\tau )g(\tau )d\tau +C_{1}$

Don't think iδ as current. It is really ${d \over dt}{\frac {current}{1volt}}$ . VS(τ) turns into a multiplier.

## LRC Example

Find the time domain expression for io given that Is = cos(t + π/2)μ(t) amp.

Earlier the step response for this problem was found:

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

The impulse response is going to be the derivative of this:

$i_{o_{\delta }}={di_{o_{\mu }} \over dt}=0+{\frac {1}{2}}e^{-t}(\cos t+\sin t)-{\frac {1}{2}}e^{-t}(-\sin t+\cos t)$
$i_{o_{\delta }}={\frac {1}{2}}e^{-t}(\cos t+\sin t+\sin t-\cos t)=e^{-t}\sin t$
$I_{s}=1+\cos t$
$i_{o}(t)=\int _{0}^{t}i_{o_{\delta }}(t-\tau )I_{s}(\tau )d\tau +C_{1}$
$i_{o}(t)=\int _{0}^{t}e^{-(t-\tau )}\sin(t-\tau )(1+\cos \tau )d\tau +C_{1}$
$i_{o}(t)={\frac {\cos t}{5}}+{\frac {2\sin t}{5}}-{\frac {7e^{-t}\cos t}{10}}-{\frac {11e^{-t}\sin t}{10}}+{\frac {1}{2}}+C_{1}$

The Mupad code to solve the integral (substituting x for τ) is:

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


## Finding the integration constant

$i_{o}(0_{+})=0={\frac {1}{5}}-{\frac {7}{10}}+{\frac {1}{2}}+C_{1}$

This implies:

$C_{1}=0$

## TO DO

$i(t)=\int _{0}^{t}i_{\delta }(t-\tau )V_{s}(\tau )d\tau =\int _{0}^{t}f(t-\tau )g(\tau )d\tau$

This was created with matlab, turned into a gif with ImageMagick, cropped with a photo editor and then released into the public domain.

Several others have created an alternative animation.

• The blue symbol $f(t)$  represents $i_{\delta }(t)$ .
• The red symbol $g(t)$  represents the arbitrary $V_{s}(t)$ .
• The current due to the VS black (on top of the yellow).
• The turn on event occurs at t = 5 seconds, not 0.
• The voltage of the source is not on indefinitely. It turns on at zero and off at 5 time constants.