# Circuit Theory/Convolution Integral

!Examples of arbitrary sources that could drive a
circuit

## 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. !Series LR
circuit with impulse δ function as voltage
source

Choose a DC source of 1 volt (the real Vs then can scale off this).
!Series LR circuit with unit μ step function voltage source


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):

$$v(t) = L{d i(t) \over dt}$$

$$v(0) = 1 = L * (-\frac{A R}{L})$$

$$A = -1/R$$ 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 V~S~(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 i~δ~ as current. It is really
${d \over dt}\frac{current}{1 volt}$. V~S~(τ) turns into a multiplier.

## LRC Example

!RLC problem for circuit theory
wikibook Find
the time domain expression for i~o~ given that I~s~ = 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:

$$i_{o_\delta} = {d i_{o_\mu} \over dt} = 0 + \frac{1}{2}e^{-t}(\cos t + \sin t) - \frac{1}{2}e^{-t}(-\sin t + \cos t)$$

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

$$I_s = 1 + \cos t$$

$$i_o(t) = \int_0^t i_{o_\delta} (t-\tau) I_s(\tau) d\tau + C_1$$

$$i_o(t) = \int_0^t e^{-(t-\tau)}\sin (t-\tau) (1 + \cos \tau) d\tau + C_1$$

$$i_o(t) = \frac{\cos t}{5} + \frac{2 \sin t}{5} - \frac{7 e^{-t}\cos t}{10} - \frac{11 e^{-t}\sin t}{10} + \frac{1}{2} + C_1$$
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

$$i(t) = \int_0^t i_\delta (t-\tau) V_s(\tau) d\tau = \int_0^t f(t-\tau)g(\tau)d\tau$$
!Convolution of spiky function with
box2.gif

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 V~S~ 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.

