I cannot understand some part of the period signal's Fourier transform.

Here this my note's methods,

For periodic signal with period $T_0$, define as $s_{T_0}(t)$ as $$ s_{T_0} (t) = \begin{cases} s(t) for -T_0/2 <t<T_0/2\\ 0 & \text{otherwise.} \end{cases} $$ $$s(t)=\sum_{n=-\infty}^{\infty}s_{T_0}(t-n T_0) = s_{T_0}(t) * \sum_{n=-\infty}^{\infty} \delta (t-n T_0)$$ So the Fourier transform will be $$S(f)=S_{T_0}(f) \cdot f_0 \sum_{n=-\infty}^{\infty}\delta(f-n f_0) = f_0 \sum_{n=-\infty}^{\infty}S_{T_0}(n f_0) \delta(f-n f_0)$$ and the other methods is $$x(t)= s(t) \cdot \sum_{n=-\infty}^{\infty} \delta (t-n T_0)$$ $$X(f)=S(f) * f_0 \sum_{n=-\infty}^{\infty}\delta(f- n f_0) = f_0 \sum_{n=-\infty}^{\infty}S(f-n f_0)$$

I cannot understand why I have "$\cdot f_0$" after Fourier Transform and why "$\delta(f-n f_0)$" can transform to "$\delta(t-nT_0)$"? And the last questions is, for the periods signal, why the note can use Fourier transform directly with out considering the Fourier serials?

  • $\begingroup$ Your $s(t)$ is a periodic, presumably continuous or piecewise -continuous, signal taking on finite (and mostly nonzero) values for all $t$ while while your $x(t)$ is a periodic impulse train of impulses of amplitude $s(nT_0) = s(0)$ occurring every $T_0$ seconds. Thus, $x(t)$ has value $0$ almost everywhere. Why on earth do you think that the Fourier transform of $s(t)$ should be the same as the Fourier transform of $x(t)$? $\endgroup$ Feb 16, 2013 at 14:25
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    $\begingroup$ cos it is very confusing. when the Fouries transform used in periodic and non periodic signal. $\endgroup$
    – Samuel
    Feb 16, 2013 at 14:31
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    $\begingroup$ Unlike Fourier transforms of finite-energy functions, the Fourier transforms of periodic functions (which have infinite energy) are not ordinary functions but rather distributions which have a literature of their own. The salient features of this theory that are most important for Fourier analysis are simply stated in terms of impulses a.k.a. Dirac deltas and their properties. Handled properly, they do not cause difficulty for the most part, and allow treatment of periodic and non-periodic functions in a unified manner. $\endgroup$ Feb 17, 2013 at 3:41
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    $\begingroup$ To put a finer point on Dilip's answer, you can't directly calculate a Fourier transform of a periodic signal because the Fourier integral doesn't converge. The resolution to this problem is to use an alternate representation of the periodic signal as a Fourier series and then employ known properties of the Dirac delta function and Fourier transform to resolve the problem. $\endgroup$
    – user2718
    Feb 17, 2013 at 5:07
  • $\begingroup$ Except the Dirac delta, is it same to use natural log to represent the F.T. periodic signal? and both way will be same. $\endgroup$
    – Samuel
    Feb 17, 2013 at 6:09

1 Answer 1


It is not obvious what exactly OP Samuel wants to know, and so I am posting a generic answer.

Let $x(t)$ denote a finite-energy continuous-time signal by which we mean that $\displaystyle \int_{-\infty}^\infty |x(t)|^2\,\mathrm dt = \mathbb E_x < \infty$.
It is important to understand that a finite-energy signal cannot be a periodic signal, that is, there is no nonzero real number $T$ for which the following equation can hold for all time instants $t$: $$x(t+T) = x(t)~~\text{for all}~ t, -\infty < t < \infty.$$ Of course, the above might hold for some, possibly even many, values of $t$, but it cannot hold for all $t$. For example, if $x(t) =\begin{cases}\sin(200\pi t), &-1 \leq t \leq +1,\\ 0, &\text{otherwise,} \end{cases}$ is a sinusoidal pulse of duration $2$ seconds containing $200$ cycles of a $100$ Hz sine wave, then $x(t+0.01) = x(t)$ for all $t$ except for $t \in (-1.01,1)$ and $t \in(0.99,1)$. On the other hand, we do not insist that $x(t)$ be of finite duration, or that $x(t)$ be causal, etc. In these circumstances, $x(t)$ has a Fourier transform $X(f)$ given by $$X(f) = \int_{-\infty}^\infty x(t)\exp(-j2\pi ft)\,\mathrm dt, ~~ -\infty < f < \infty \tag{1}$$ where $j = \sqrt{-1}$, and $x(t)$ can be recovered from its Fourier transform using the inverse Fourier transform which gives $$x(t) = \int_{-\infty}^\infty X(f)\exp(j2\pi ft)\,\mathrm df, ~~-\infty < t < \infty.\tag{2}$$

All this, of course, is old hat to most readers, so now let us construct a periodic signal from $x(t)$. Let $T_0$ denote some positive number, and define

$$s(t) = \sum_{n=-\infty}^\infty x(t - nT_0).\tag{3}$$

Then, we claim that $s(t)$ is a signal of period $T_0$. Note that for every choice of real number $t$, we have that $$\begin{align} s(t+T_0) &= \sum_{n=-\infty}^\infty x(t+T_0-nT_0)\\ &= \sum_{n=-\infty}^\infty x(t-(n-1)T_0) &\text{and now replace}~ n-1~\text{by}~m\\ &= \sum_{m=-\infty}^\infty x(t-mT_0)\\ &= s(t). \end{align}$$ But, what does one period of $s(t)$ look like? Well, for $-T_0/2 \leq t \leq T_0/2$, $s(t)$ is not the same as $x(t)$ except in the special case when the support of $x(t)$ is a subset of $[-T_0/2,T_0/2]$. Instead, one period (what OP Samuel calls $s_{T_0}(t)$) can be found by slicing up the graph of $x(t)$ into pieces of duration $T_0$ centered at integer multiples of $T_0$, moving all the pieces to be centered at $0$, and then adding them up. This sum, one period of the periodic signal that we have created, is now replicated at intervals of $T_0$ seconds all along the time axis. Thus, we have another representation of $s(t)$, namely,

$$s(t) = \sum_{n=-\infty}^\infty s_{T_0}(t - nT_0). \tag{4}$$

Let us at this point introduce Dirac deltas or impulses $\delta(t)$ which have the property that for any $g(t)$ that is continuous at $t = a$, $$\int_{-\infty}^\infty g(t)\delta(t-a)\,\mathrm dt = \int_{-\infty}^\infty g(t)\delta(a-t)\,\mathrm dt = g(a).$$ Now, the convolution of $g(t)$ and $h(t)$ is a signal $(g\star h)$ whose value $(g\star h)(\tau)$ at particular time instant $\tau$ is given by $$(g\star h)(\tau) = \int_{-\infty}^\infty g(t)h(\tau-t)\,\mathrm dt$$ and so if $h_n(t) = \delta(t-nT_0)$, we have that $$\begin{align} (x\star h_n)(\tau) &= \int_{-\infty}^\infty x(t)h_n(\tau-t)\,\mathrm dt\\ &= \int_{-\infty}^\infty x(t)\delta(\tau-t-nT_0)\,\mathrm dt\\ &= \int_{-\infty}^\infty x(t)\delta((\tau-nT_0) - t)\,\mathrm dt\\ &= x(\tau - nT_0). \end{align}$$ This gives us yet another representation of the periodic signal $s(t)$, namely,

$$s(t)= x(t) \star \sum_{n=-\infty}^\infty \delta(t-nT_0) = \sum_{n=-\infty}^\infty x(t-nT_0). \tag{5}$$

It is worth reminding ourselves at this point that $s(t)$ is a continuous-time finite amplitude signal that, despite what it looks like in the middle sum in $(5)$ does not contain any impulses.

Returning to Fourier transforms (bet you thought I forgot about them, didn't you?), we come up against the uncomfortable fact that $s(t)$ does not have a Fourier transform in the usual sense of the word because it is not a finite-energy signal. What it does have is a Fourier series representation, as pointed out by Fourier himself almost two hundred years ago. We can incorporate Fourier series into Fourier transforms if we use impulses in the frequency domain. So, what is the Fourier series for $s(t)$? We write $$s(t) = \sum_{k=-\infty}^\infty c_k \exp(j2\pi kt/T_0), -\infty < t < \infty$$ where $$\begin{align} c_k &= \frac{1}{T_0}\int_{-T_0/2}^{T_0/2}s(t)\exp(-j2\pi kt/T_0)\,\mathrm dt\\ &= \frac{1}{T_0}\int_{-T_0/2}^{T_0/2}\sum_{n=-\infty}^\infty x(t-nT_0) \exp(-j2\pi kt/T_0)\,\mathrm dt &\text{using}~ (3)\\ &= \frac{1}{T_0}\sum_{n=-\infty}^\infty \int_{-T_0/2}^{T_0/2}x(t-nT_0) \exp(-j2\pi kt/T_0)\,\mathrm dt &\text{now change variables}~ \tau = t-nT_0\\ &= \frac{1}{T_0}\sum_{n=-\infty}^\infty \int_{nT_0-T_0/2}^{nT_0+T_0/2} x(\tau) \exp(-j2\pi k(\tau+nT_0)/T_0)\,\mathrm d\tau \\ &= \frac{1}{T_0}\sum_{n=-\infty}^\infty \int_{nT_0-T_0/2}^{nT_0+T_0/2} x(\tau) \exp(-j2\pi k\tau/T_0)\,\mathrm d\tau &\text{now combine the integrals into one}\\ &=\frac{1}{T_0}\int_{-\infty}^\infty x(\tau)\exp(-j2\pi k\tau/T_0)\,\mathrm d\tau\\ &= \frac{1}{T_0}X\left(\frac{k}{T_0}\right) &\text{using the definition}~ (2) \end{align}$$

So we have the Fourier series representation $$s(t) = \frac{1}{T_0}\sum_{k=-\infty}^{\infty} X\left(\frac{k}{T_0}\right)\exp(j2\pi kt/T_0).\tag{6}$$ What should $S(f)$ be so that when it is plugged into the inverse Fourier transform $(2)$ will give us $$s(t) = \int_{-\infty}^\infty S(f)\exp(j2\pi ft)\,\mathrm df = \frac{1}{T_0}\sum_{k=-\infty}^{\infty} X\left(\frac{k}{T_0}\right)\exp(j2\pi kt/T_0)??$$ The answer is readily worked out to be

$$\begin{align} S(f) &= \frac{1}{T_0}\sum_{k=-\infty}^{\infty} X(f) \delta\left(f-\frac{k}{T_0}\right) = \frac{1}{T_0}\sum_{k=-\infty}^{\infty} X\left(\frac{k}{T_0}\right) \delta\left(f-\frac{k}{T_0}\right)\\ &= f_0 \sum_{k=-\infty}^{\infty} X(kf_0)\delta(f-kf_0) \end{align}$$

I will not say much about OP Samuel's "other method" since it is comparing apples and oranges. The signal $$x(t)\cdot \sum_{n=-\infty}^\infty \delta(t-nT_0) = \sum_{n=-\infty}^\infty x(nT_0)\delta(t-nT_0)\tag{7}$$ is a impulse train in the time domain (one way of representing a sampled-data or discrete-time signal with sampling interval $T_0$), and is very different from the signal $$x(t) \star \sum_{n=-\infty}^\infty \delta(t-nT_0) = \sum_{n=-\infty}^\infty x(t-nT_0)$$ in $(5)$ which is a continuous-time signal containing no impulses. There is no reason for the two to have similar or identical Fourier transforms, and the OP's Procrustean efforts to equate both is leading him into a great deal of confusion. In fact, duality suggests that, just as the Fourier transform of a periodic signal is a set of equally-spaced impulses (of different amplitudes) in the frequency domain, the Fourier transform of a set of equally-spaced impulses (of different amplitudes) in the time domain is a periodic function in the frequency domain. For details, see this answer to a different question in which the Fourier transform of $(7)$ has been described in more depth than the hand-waving in the previous sentence.


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