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I was trying to calculate output voltage across the inductor in the RLC circuit below. I had used the Fourier transform to convert input signal \$V_{i}(t)\$ which is a single rectangular wave pulse in time domain and I also calculated the transfer function of the circuit as $$h(jω) = \frac{V_{o}(jω)}{V_{i}(jω)}$$ and the output voltage in frequency domain as $${V_{o}(jω)} = {V_{i}(jω)} h(jω) $$ The transfer function is in terms of impedance in the circuit and the angular frequency ω. When I calculated the inverse fourier transform of \$V_{o}(jω)\$ using the equation below, I am getting complex terms in the integration result. I had used Wolfram alpha integral evaluator for this.The integral evaluation seems complex and it is not really a plausible result. Is the method of Fourier analysis really correct and how does one evaluate the complex integral from -∞ to +∞ in inverse Fourier transform? I had briefly shown the steps I obtained in calculating the fourier transform of output voltage as below.

Equation to calculate inverse fourier transform is

$$V_{o}(t) = \int_{-∞}^{+∞} v_{o}(jw) e^{jwt}dω$$

R = 1Ω, C = 500 mF and L = 800 mH in the circuit.

enter image description here

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  • \$\begingroup\$ Why not just use a circuit simulator? \$\endgroup\$
    – Andy aka
    Feb 8 at 12:43
  • \$\begingroup\$ LT spice circuit simulator works fine. I was trying to do a mathematical evaluation using the Fourier integral. \$\endgroup\$
    – Amit M
    Feb 8 at 13:20
  • \$\begingroup\$ It seems the problem has to be solved with Laplace tranforms as fourier transforms can only be used for periodic sinusoidal functions. Using fourier transform and its inverse with single pulses might cause the evaluation to be complex.. This had occurred to me but I wanted to try anyway. Thanks for your answers. \$\endgroup\$
    – Amit M
    Feb 9 at 6:15
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    \$\begingroup\$ @Amit M: Don't get discouraged so soon. Certainly the FT can be applied. The input signal is limited. I posted an answer here. \$\endgroup\$ Feb 10 at 17:48

3 Answers 3

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While the main applications of the Fourier Transform (FT) in Electrical Engineering are mainly related to the frequency response of systems and filtering, it can also be used in circuit analysis, as long as certain conditions apply to the input signals. Instead of saying that FT has disadvantages, we can say that it has limitations. The three Dirichlet conditions are sufficient for FT to exist. The main one is that the signal is absolutely integrable over an infinite interval:

$$\int_{-\infty }^{\infty }\left| x(t) \right|dt<\infty $$

But it's worth repeating, these are sufficient conditions, but not necessary. Some signals that do not satisfy the conditions can still be considered to have a FT, if impulse functions are allowed in the transform (e.g. step input, sine,...).

As well stated by Prof. B.P. Lathi:

Any signal that can be generated in practice satisfies the Dirichlet conditions and therefore has a Fourier Transform. Therefore, the physical existence of a signal is a sufficient condition for the existence of its transform.

Put another way: Has anyone ever seen an ramp voltage extending to infinite in time?

Another limitation of FT is its difficulty in dealing with initial conditions. As the Laplace transform does not have these limitations, it may be preferable in circuit analysis. But the solution via FT can be useful to the student who is learning.

THE MODEL

enter image description here

Here, instead of obtaining the expression for the output using the impedance of the components, as is traditionally done, I preferred the approach from the associated differential equation. You can ignore this if you wish.

$$v_i(t)-Ri_L(t)-\frac{1}{C}\int_{-\infty}^{t}i_L(\varsigma)d\varsigma-v_o(t)= 0$$

Deriving with repect to \$t\$:

$$\frac{dv_i(t)}{dt}-R\frac{di_L(t)}{dt}-\frac{i_L(t)}{C}-\frac{dv_o(t)}{dt}= 0$$

Trying to use the relation \$i_L(t)=C\frac{dv_c(t)}{dt}\$ won't help here. So, deriving one more time:

$$\frac{dv_i^2(t)}{dt^2}-R\frac{di^2_L(t)}{dt^2}-\frac{1}{C}\frac{di_L(t)}{dt}-\frac{d^2v_o(t)}{dt^2}= 0$$

Now, we can use \$\frac{di_L(t)}{dt}=\frac{v_o(t)}{L}\$:

$$\frac{dv^2_o(t)}{dt^2}+\frac{R}{L}\frac{dv_o(t)}{dt}+\frac{1}{LC}v_o(t)= \frac{dv^2_i(t)}{dt^2}$$

Taking the FT from both sides:

$$(j\omega)^2V_o(\omega)+\frac{R}{L}j\omega V_o(\omega)+\frac{1}{LC}V_o(\omega)=(j\omega)^2V_i(\omega)$$

The output, along with the Transfer Function, is:

$$V_o(\omega)=\frac{(j\omega)^2}{(j\omega)^2 + \frac{R}{L}j\omega+\frac{1}{LC}}V_i(\omega) \qquad[1]$$

THE INPUT

Consider the Fourier Transform properties, named linearity and time shifting, as shown below:

enter image description here

In our case, \$A=10 V\$, \$a=2.5 s\$ and \$\tau=1.5 s\$.

$$V_i(\omega)=20e^{-j1.5\omega}\frac{\sin(2.5\omega)}{\omega}$$

THE OUTPUT

Replacing \$V_i(\omega)\$ in \$[1]\$:

$$V_o(\omega)=\frac{20(j\omega)^2e^{-j1.5\omega}}{(j\omega)^2+1.25j\omega+2.5}\frac{\sin(2.5\omega)}{\omega} $$

Taking in account that:

$$\sin(2.5\omega)=\frac{e^{j2.5\omega}-e^{-j2.5\omega}}{2j}$$

$$V_o(\omega)=\frac{10j\omega( e^{j\omega}-e^{-4j\omega})}{(j\omega)^2+1.25j\omega+2.5}$$

In order to get the response \$v_o(t)\$ (ie the inverse Fourier Transform) it's simpler to use the form below. Start rewriting it as:

$$V_o(\omega)=F(\omega)\left( e^{j\omega}-e^{-4j\omega}\right) \qquad[2]$$

where:

$$F(\omega)=\frac{10j\omega}{(j\omega)^2+1.25j\omega+2.5}$$

That's convenient, since the exponentials in \$[2]\$ are highlighted to apply the time shifting property later.

Replacing \$j\omega\$, for another variable, like \$j\omega=v\$:

$$F(v)=\frac{10v}{v^2+1.25v+2.5}$$

Since the roots of denominator are complex conjugates, I'm going to prepare the expression to get inversion based on \$sine(.)\$ and \$cosine(.)\$ functions, along with the method completing the square:

$$F(v)=10\left[\frac{v+0.625-0.625}{(v+0.625)^2+1.452^2}\right]$$

$$F(v)=10\left[ \frac{v+0.625}{(v+0.625)^2+1.452^2}- 0.4304\frac{1.452}{(v+0.625)^2+1.452^2}\right]$$

Substituting this result into \$[2]\$ and replacing \$v=j\omega\$:

$$ \begin{align} V_o(\omega)=&\left[ 10\frac{j\omega+0.625}{(j\omega+0.625)^2+1.452^2}- 4.304\frac{1.452}{(j\omega+0.625)^2+1.452^2}\right]e^{j\omega}+\\ &\left[-10\frac{j\omega+0.625}{(j\omega+0.625)^2+1.452^2} + 4.304\frac{1.452}{(j\omega+0.625)^2+1.452^2}\right]e^{-j4\omega} \end{align} $$

Considering the the table below

enter image description here

and applying the time shifting property, we can complete the inversion to time domain:

$$ \begin{align} v_o(t)=&e^{-0.625(t+1)} \left\{10\cos\left[1.452(t+1)\right]-4.304\sin\left[1.452(t+1) \right] \right\}u(t+1)+\\ &e^{-0.625(t-4)} \left\{-10\cos\left[1.452(t-4)\right]+4.304\sin\left[1.452(t-4) \right] \right\}u(t-4) \end{align} $$

We can get a smaller form by replacing each pair of \$sine(.) \pm cosine(.)\$ of same frequency) for just one \$cosine\$ with a phase shift:

$$ \begin{align} v_o(t)=&\left[10.89e^{-0,625(t+1)}\cos(1.452t+0.592\pi)\right]u(t+1)+\\ &\left[10.89e^{-0,625(t-4)}\cos(1.452t-0.719\pi)\right]u(t-4) \end{align} $$

A plot showing the pulse \$v_i(t)\$ and the corresponding response \$v_o(t)\$:

enter image description here

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  • \$\begingroup\$ There may be different ways to solve this problem. Your approach seemed best. Thanks! \$\endgroup\$
    – Amit M
    Feb 11 at 3:25
  • \$\begingroup\$ Thank you Amit. Although the Laplace Transform has a wider range of applications, I tried to give an answer that involved the Fourier Transform, as you requested in question. As you can notice, much of the solution method ends up being common between both. \$\endgroup\$ Feb 11 at 12:55
  • \$\begingroup\$ It appears that fourier integral evaluation is easy for symmetrical waveforms which can be converted to exponential and sinusoidal expressions using time shifting property as in this method. Laplace transform is applicable for a wide variety of waveforms and its inverse evaluation from look up tables of laplace transform is much easier. \$\endgroup\$
    – Amit M
    Feb 14 at 16:28
  • \$\begingroup\$ You are correct. Only note that my answer made use of "completing the squares" method applied with complex conjugate pairs of roots and also, the use of an inversion table. Both of these procedures would be very simiar in the Laplace Transform case. But I used \$F(v)\$ rather than \$F(s)\$ just for avoid a premature association with the Laplace T. As I have mentioned in answer, the Laplace variant can deal with initial conditions, too. Finally, I took the opportunity of your comment to put the phase angles in radians on last expression (more consistent). \$\endgroup\$ Feb 14 at 19:59
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Well, notice that your transfer function is given by:

$$\mathscr{H}\left(\text{s}\right):=\frac{\displaystyle\text{V}_\text{o}\left(\text{s}\right)}{\displaystyle\text{V}_\text{i}\left(\text{s}\right)}=\frac{\displaystyle\text{sL}}{\displaystyle\text{sL}+\frac{\displaystyle1}{\displaystyle\text{sC}}+\text{R}}=\frac{\displaystyle\text{CLs}^2}{\displaystyle\text{CLs}^2+\text{CRs}+1}\tag1$$

So, using the convolution theorem of the Laplace transform, we can write:

$$\text{v}_\text{o}\left(t\right)=\int\limits_0^t\text{v}_\text{i}\left(\tau\right)\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle\text{CLs}^2}{\displaystyle\text{CLs}^2+\text{CRs}+1}\right]_{\left(t-\tau\right)}\space\text{d}\tau\tag2$$

Using Mathematica I found that the inverse Laplace transform is given by:

enter image description here

And \$\displaystyle\text{v}_\text{i}\left(\tau\right)\$ is given by:

$$\text{v}_\text{i}\left(\tau\right)=\begin{cases} 10&\space\text{if}\space-1\le\tau\le4\\ \\ 0&\space\text{if}\space\tau>4\\ \\ 0&\space\text{if}\space\tau<-1 \end{cases}\tag3$$

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Calculation of the voltage transient across the inductance. Generally the Fourier transform is used when the system is in a sinusoidal regime, in which case the Vi is omitted i.e. we work with the phasors.

enter image description here

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  • \$\begingroup\$ Timing of the pulse is different in your solution. The actual timing is t= -1 to t= 4 seconds. Voltage of the pulse is 10V unlike 18V above. \$\endgroup\$
    – Amit M
    Feb 10 at 17:00

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