# Difference between revisions of "Impulse Response"

 Revision as of 17:38, 12 July 2018 (Tag: 2017 source edit)← Older edit Latest revision as of 19:51, 29 October 2018 (Multilanguage link)(Tag: 2017 source edit)

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

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

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));<br>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: $C_1 = 0$
==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):

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

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

: $i(0) = 0 = A + C$<br />Which implies that:

: $C = -A = 1/R$<br />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<sub>S</sub>(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<sub>δ</sub> as current. It is really ${d \over dt}\frac{current}{1 volt}$. 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:

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

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

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

: $i_{o_\delta} = \frac{1}{2}e^{-t}(\cos t + \sin t + \sin t - \cos t) = e^{-t}\sin t$<br />:$I_s = 1 + \cos t$

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

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

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

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

==Finding the integration constant==

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

This implies:

: $C_1 = 0$

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