1
$\begingroup$

Please help me sort this issue out.

The integration property in Fourier series is as follows:

enter image description here

So, for proving the above property, i took this approach:

enter image description here

This is where my doubt is. Some books and websites just put upper limit (ignoring the lower limit) and compare with (1) to conclude that

enter image description here

However, when the lower integral limit is substituted, we get enter image description here

This value cannot be evaluated. Though some websites say that this corresponds to the "initial value"(?) and that is zero... etc., how does one justify this mathematically?

Any help is appreciated. Thank you.

$\endgroup$
  • 1
    $\begingroup$ just a friendly suggestion: in the future, please learn to use $\LaTeX$ and express your equations with source code rather than screenshots. i want to just be able to copy them without re-expression (in $\LaTeX$). make it easier for people to help you rather than harder for them to help. $\endgroup$ – robert bristow-johnson Jun 17 at 8:14
  • 1
    $\begingroup$ Hil, using a reverse argument (what would the resulting Fourier series have to be if you differentiated it?) you could come up with all of the Fourier coefficients except for the DC coefficient. That would be undetermined unless the bottom limit of the integral was known and set to a finite value. $\endgroup$ – robert bristow-johnson Jun 17 at 8:22
  • 2
    $\begingroup$ BTW, the notation for a definite integral from whatever the book that was screen shot is horrible: $$ \int\limits_{-\infty}^{t} x(t) \, dt $$ the dummy variable of integration should not be the same as the upper limit of the integral in an definite integral. this notation is legit: $$ \int\limits_{-\infty}^{t} x(u) \, du $$ $\endgroup$ – robert bristow-johnson Jun 17 at 8:26
  • $\begingroup$ @robert bristow Johnson, I m extremely sorry for the inconvenience sir. I will make sure that I will source the equations from next post onwards. Thank you so much for your suggestion. $\endgroup$ – Nishanth A Rao Jun 17 at 8:57
  • $\begingroup$ so Nish, have you seen how to render equations with $\LaTeX$? $\endgroup$ – robert bristow-johnson Jun 17 at 18:41
2
$\begingroup$

Note that the antiderivative of a function is only defined up to a constant. Furthermore, note that if you integrate a periodic function, the result is not necessarily periodic. Let

$$x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}\tag{1}$$

Now we integrate $(1)$ with an arbitrary lower integration limit $t_0$. I'll say more about that later.

$$\begin{align}y(t)&=\int_{t_0}^tx(\tau)d\tau\\&=\sum_{k=-\infty}^{\infty}a_k\int_{t_0}^te^{jk\omega_0\tau}d\tau\\&=a_0(t-t_0)+\sum_{k=-\infty\\k\neq 0}^{\infty}\frac{a_k}{jk\omega_0}e^{jk\omega_0t}-\sum_{k=-\infty\\k\neq 0}^{\infty}\frac{a_k}{jk\omega_0}e^{jk\omega_0t_0}\tag{2}\end{align}$$

Clearly, if $a_0\neq 0$, $y(t)$ isn't periodic, so we can't hope to obtain a formula for its Fourier coefficients. So in order to end up with a periodic function $y(t)$, we require that $a_0=0$.

If $a_0=0$, we get from $(2)$

$$y(t)=\sum_{k=-\infty}^{\infty}b_ke^{jk\omega_0t}\tag{3}$$

with

$$b_k=\begin{cases}\displaystyle\frac{a_k}{jk\omega_0},&k\neq 0\\\displaystyle-\sum_{l=-\infty\\l\neq 0}^{\infty}\frac{a_l}{jl\omega_0}e^{jl\omega_0t_0},&k=0\end{cases}\tag{4}$$

So in general the antiderivative $y(t)$ has a non-zero DC component, which depends on the choice of the lower integration limit $t_0$.

The antiderivative

$$\tilde{y}(t)=\sum_{k=-\infty\\k\neq 0}^{\infty}\frac{a_k}{jk\omega_0}e^{jk\omega_0t}\tag{5}$$

is the specific antiderivative of $x(t)$ that has a zero DC Fourier coefficient, and this is the one that is meant by the formula that is usually given in textbooks.

Consequently, the expression

$$y(t)=\int_{-\infty}^tx(\tau)d\tau\tag{6}$$

for the antiderivative of $x(t)$ with Fourier coefficients $b_k=a_k/(jk\omega_0)$, $k\neq 0$, and $b_0=0$, is at least inaccurate; it is in fact a sloppy way of expressing the specific antiderivative $\tilde{y}(t)$ given by $(5)$, i.e., the one with a zero DC coefficient.

$\endgroup$
  • $\begingroup$ i like Eq. (4). $\endgroup$ – robert bristow-johnson Jun 17 at 19:51
  • $\begingroup$ @MattL. Thank you very much for your insight sir. However, this doesn't answer one of the queries i have asked in the post i.e, what happens when $$ t_{o} = \infty ? $$ The complex exponential in equation 4 still doesn't converge to any value right? $\endgroup$ – Nishanth A Rao Jul 6 at 16:09
  • $\begingroup$ @NishanthARao: As you've already noticed, that limit doesn't exist. $\endgroup$ – Matt L. Jul 6 at 18:19
1
$\begingroup$

It appears to me that the "initial value" issue would affect only the DC value, $a_0$, of the Fourier series. In my opinion, the author should have left the integrals as indefinite integrals (which are the same as the "anti-derivative") or have expressed this relationship only in terms of differentiation, and not integration.

Actually, now that I think of it, the original $x(t)$ would have to have $a_0 = 0$ in order for the integration theorem to make any sense. And the resulting Fourier series can have any finite DC value you want. You would have to determine the resulting DC value by other means.

To do this theorem properly, you must first do it for the derivative of $x(t)$ first and then apply the results in reverse.

I dunno what book you're using but there are both some notational and even mathematical problems with the expression of this problem.

$\endgroup$
  • 1
    $\begingroup$ i'm gonna let someone else (maybe @MattL. ) make the appropriate and rigorous answer to this question. as it is posed, it's sorta a flawed problem in definition. (and i do not blame the OP for this, it could be the fault of the book.) $\endgroup$ – robert bristow-johnson Jun 17 at 8:34
1
$\begingroup$

Oh, hell...

Let's say you have two periodic functions, $x(t)$ and $y(t)$ having exactly the same period (and fundamental frequency):

$$ x(t) \triangleq \sum\limits_{k=-\infty}^{\infty} a_k \, e^{j k \omega_0 t} $$

$$ y(t) \triangleq \sum\limits_{k=-\infty}^{\infty} b_k \, e^{j k \omega_0 t} $$

where the period common to both is $\frac{2 \pi}{\omega_0}$.

Suppose you were to differentiate $y(t)$:

$$ y'(t) = \sum\limits_{k=-\infty}^{\infty} j k \omega_0 b_k \, e^{j k \omega_0 t} $$

you can see right away that the DC term of $y'(t)$ is zero (as it should be):

$$ j k \omega_0 b_k \bigg|_{k=0} = 0 $$

Now let's say that you set that differentiated periodic function to $x(t)$:

$$ x(t) = y'(t) $$

That means $y(t)$ is the indefinite integral of $x(t)$. Then you see that all of the coefficients of $y(t)$ are well-defined in terms of the coefficients of $x(t)$ except for the DC coefficient, $b_0$, which could be any finite value.

$$ a_k = j k \omega_0 b_k $$

or

$$ b_k = \frac{a_k}{j k \omega_0} \qquad \qquad \forall k \in \mathbb{Z} \ne 0 $$.

That is, in my opinion, the only legit way to look at the integration of a Fourier series.

$\endgroup$
  • $\begingroup$ This is a wonderful way to look at the property. Extremely thankful for your insight sir. With due respect however, I just want to know where I am going wrong, and if at all, why? I am pretty sure there must be some way to continue from that step. $\endgroup$ – Nishanth A Rao Jun 17 at 8:59
  • $\begingroup$ @NishanthARao: rbj's answer is probably the most straightforward way to look at the problem. If you're interested in how to figure out the solution by integration, have a look at my answer. $\endgroup$ – Matt L. Jun 17 at 10:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.