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:

:<math>i(t) = Ae^{-\frac{t}{\frac{L}{R}}} + C </math>

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):

:<math>v(t) = L{d i(t) \over dt}</math>:<math>v(0) = 1 = L * (-\frac{A R}{L})</math>:<math>A = -1/R</math>

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

:<math>i(0) = 0 = A + C</math>Which implies that:
:<math>C = -A = 1/R</math>So the response to a DC voltage source turning on at t=0 to one volt (called the unit response μ) is:
:<math>i_\mu (t) = \frac{1}{R}(1 - e^{-\frac{t}{\frac{L}{R}}})</math>

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

:<math>i_\delta (t) = \frac{e^{-\frac{t}{\frac{L}{R}}}}{L}</math>

Now the current due to any arbitrary V<sub>S</sub>(t) can be found using the convolution integral:

:<math>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</math>

Don't think i<sub>δ</sub> as current. It is really <math>{d \over dt}\frac{current}{1 volt}</math>. V<sub>S</sub>(τ) turns into a multiplier.
==LRC Example==
Find the time domain expression for i<sub>o</sub> given that I<sub>s</sub> = cos(t + π/2)μ(t) amp.

Earlier the step response for this problem was found:

:<math> i_{o_\mu} = \frac{1}{2}(1 - e^{-t}(\cos t + \sin t))</math>

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

:<math>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)</math>
:<math>i_{o_\delta} = \frac{1}{2}e^{-t}(\cos t + \sin t + \sin t - \cos t) = e^{-t}\sin t</math>:<math>I_s = 1 + \cos t</math>
:<math>i_o(t) = \int_0^t i_{o_\delta} (t-\tau) I_s(\tau) d\tau + C_1</math>
:<math>i_o(t) = \int_0^t e^{-(t-\tau)}\sin (t-\tau) (1 + \cos \tau) d\tau + C_1</math>
:<math>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</math>

The Mupad code to solve the integral (substituting x for τ) is:<pre>f := exp(-(t-x)) *sin(t-x) *(1 + cos(x));<br>S := int(f,x = 0..t)</pre>
==Finding the integration constant==

:<math>i_o(0_+) = 0 = \frac{1}{5}  - \frac{7}{10} + \frac{1}{2} + C_1</math>

This implies:

:<math>C_1 = 0</math>

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