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I am trying to solve the inductor current for the circuit below by using Fourier transform instead of Laplace. The purpose is to see if Fourier transform also works for problems with initial conditions like this. The voltage source is a DC voltage source.

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By KVL: \$\ -1 + L\frac{\mathrm{di} }{\mathrm{d} x} + R i = 0\$

Fourier transform the equation above:

\$\ -2\pi \delta (\omega ) + j\omega L *I(\omega ) + R*I(\omega ) = 0\$

However, I don't see anywhere that the condition i(0) is used here. How can I include the initial condition and solve for the current?

PS. The circuit is taken from this site.

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  • \$\begingroup\$ Is the voltage source a unit step? If so, the Fourier Transform is wrong. Also the KVL equation has a wrong sign. \$\endgroup\$ – Chu Jul 19 '17 at 0:01
  • \$\begingroup\$ @Chu: voltage source is a DC source, yes, the sign is mistake I will correct it now. \$\endgroup\$ – anhnha Jul 19 '17 at 0:03
  • \$\begingroup\$ The Fourier Spectrum of current in the voltage in the resistor should be \$\dfrac{V_{in}}{R}-I(0)=\Delta I\$ with an exponential attenuation of continuous spectrum, same as a LPF starting from \$ \Delta V=\Delta I*R\$ with a breakpoint @ ω=L/R then attenuating 20dB/decade. Start with the exponential time equation. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 19 '17 at 0:04
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First of all, the Laplace Transform and the Fourier Transform spring forth from the same body of water. They are sorta the same thing, and if you have common conditions using either, they are the same thing with \$ s=j\omega \$. the unilateral Laplace transform is better for circuit problems that are defined only for \$t \ge 0\$ and have initial conditions because there is a convenient mechanism for dealing with those initial conditions at \$t=0\$ with the unilateral Laplace transform.

The bilateral Laplace transform is most like the Fourier transform (which always has a bilateral definition) with the substitution of \$s=j\omega=j2\pi f\$.

That said, the problem above can only be dealt with with the Fourier transform if the circuit, which is completely "relaxed" (having all initial conditions equal to zero) at \$t=0\$, is modeled as driven by a unit step of 1 volt, rather than a constant input of 1 volt. If the above circuit was connected to 1 volt for all time, there is no way that initial current, \$i(0)\$ can be anything other than \$i(0) = \frac{1 \text{V}}{R}\$. If \$i(0)\$ is equal to anything other than that, you must represent the input to be a step function.

Then the other problem with the Fourier Transform that the Laplace doesn't really have, is that the Fourier Transform does not converge nicely for the unit step. Through an indirect method, the F.T. of a unit step can be inferred, but it's natural with the Laplace. So with the F.T. you would have to represent the unit step as a limiting case of a function that does have a legitimate F.T.:

$$ u(t) = \lim_{\tau \to +\infty} \begin{cases} e^{-t/\tau} \qquad & \text{ if } t \ge 0 \\ 0 & \text{ if } t < 0 \\ \end{cases}$$

for a finite \$\tau\$, that has a legit F.T. and you can solve this system using the F.T. for a finite \$\tau\$, get an answer, and then let \$\tau\$ go to \$\infty\$.

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  • \$\begingroup\$ I had considered adding that Laplace transforms work with known initial conditions much more easily. But you've said that, and probably better than I would have. Good stuff. +1. \$\endgroup\$ – jonk Jul 19 '17 at 4:29
  • \$\begingroup\$ Thanks. I have two questions relating to this. 1. You said that if you have common conditions, use either. What is meant by common conditions here? Is it convergence condition? 2. That said, the problem above can only be dealt with with the Fourier transform if the circuit, which is completely "relaxed" (having all initial conditions equal to zero) at t=0, is modeled as driven by a unit step of 1 volt, rather than a constant input of 1 volt. What is the reason why Fourier only works with all initial conditions equal to zero? \$\endgroup\$ – anhnha Jul 19 '17 at 11:11
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    \$\begingroup\$ 1. a simple example: $$ x(t) = \begin{cases} e^{-t/\tau} \qquad & \text{ if } t \ge 0 \\ 0 & \text{ if } t < 0 \\ \end{cases}$$ has, with \$ s = \sigma + j \omega \$ the same one-sided Laplace Transform and two-sided Laplace Transform and the same Fourier Transform with \$ \sigma=0 \$. it is clear that the definitions of the three are exactly the same if a) the argument \$x(t)=0\$ for all \$t<0\$ and if the integrals converge in all three cases. (for the unit step where \$x(t) = u(t)\$, the Fourier integral does not converge without some goofy handwaving.) \$\endgroup\$ – robert bristow-johnson Jul 19 '17 at 17:03
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    \$\begingroup\$ 2. the two-sided Laplace Transform can be expressed with two integrals: $$ X(s) = \lim_{\epsilon \to 0} \int\limits_{-\infty}^{-\epsilon} x(t) \, e^{-st} \, dt + \int\limits_{-\epsilon}^{\infty} x(t) \, e^{-st} \, dt $$ as \$\epsilon>0\$ approaches zero, the integral on the right is simply the one-sided Laplace Transform and the integral on the left encompasses all of the activity of \$x(t)\$ for \$ t<0 \$. it is the left-hand term that sets up the initial conditions and if that term can be replaced with equivalent values, that is how the one-sided Laplace incorporates initial conditions. \$\endgroup\$ – robert bristow-johnson Jul 19 '17 at 17:11
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    \$\begingroup\$ in order for the initial current to be \$i(0) \ne 0\$, then the applied voltage from long ago until \$t=0\$ would have to be the constant \$i(0)R\$. then if the input voltage suddenly changes from \$i(0)R\$ to \$1 \text{ V}\$ at time \$t=0\$, then the input voltage is $$ v(t) = i(0)R + (1\text{V} - i(0)R) u(t) $$ where \$u(t)\$ is the unit step function. but, since it's Fourier and not Laplace, you have to represent your unit step as a limit, like i have shown above in the answer. \$\endgroup\$ – robert bristow-johnson Jul 20 '17 at 19:31

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