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:

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

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):

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

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

${\displaystyle i(0)=0=A+C}$

Which implies that:

${\displaystyle C=-A=1/R}$

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

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

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

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

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

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

Don't think iδ as current. It is really ${\displaystyle {d \over dt}{\frac {current}{1volt}}}$. 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:

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

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

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

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

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

This implies:

${\displaystyle C_{1}=0}$

TO DO

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

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 ${\displaystyle f(t)}$ represents ${\displaystyle i_{\delta }(t)}$.
• The red symbol ${\displaystyle g(t)}$ represents the arbitrary ${\displaystyle V_{s}(t)}$.
• 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.