Circuit Theory/Convolution Integral

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Examples of arbitrary sources that could drive a circuit

Impulse Response[edit]

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[edit]

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 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}{1 volt}. VS(τ) turns into a multiplier.

LRC Example[edit]

RLC problem for circuit theory wikibook

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} = {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[edit]

i_o(0_+) = 0 = \frac{1}{5}  - \frac{7}{10} + \frac{1}{2} + C_1

This implies:

C_1 = 0

TO DO[edit]

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 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.