First of all, DSP is not my expert area. I didn't take any DSP course and don't know DSP in detail but I have a question regarding Fourier Transform confusing me.

Well, suppose that we wanted to take the fourier transform of $\cos(2\pi\cdot 3x)$ and integrated over a finite interval $[-a, a]$, in that case, we would end up with an expression like this:

$$ \dfrac{\left(f-3\right)\sin\left(2{\pi}af+6{\pi}a\right)+\left(f+3\right)\sin\left(2{\pi}af-6{\pi}a\right)}{2{\pi}\cdot\left(f^2-9\right)} $$

Considering that, why cannot we obtain the exact fourier transform of $cos(2\pi\cdot 3x)$ which is $0.5 \cdot (\delta(x + 3) + \delta(x - 3))$ as we increase the integral range $[-a, a]$? How can one explain that?

  • $\begingroup$ I lack the time but there's a satisfactory answer, your intuition is correct and nicely explains why there's any deltas at all for $a \rightarrow \infty$. $\endgroup$ May 7 at 15:11
  • $\begingroup$ I would like to see it as well. If you could share it when you have time, it would be very helpful. $\endgroup$ May 7 at 15:30

2 Answers 2


There's not much to explain here because your premise is wrong. We can approximate the Fourier transform by integrating over a finite interval and find the exact Fourier transform by taking the limit:

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt=\lim_{T\to\infty}\int_{-T}^Tx(t)e^{-j\omega t}dt\tag{1}$$

Considering your example with $x(t)=\cos(\omega_0t)$, we obtain

$$\int_{-T}^Tx(t)e^{-j\omega t}dt=\frac{\sin[(\omega+\omega_0)T]}{\omega+\omega_0}+\frac{\sin[(\omega-\omega_0)T]}{\omega-\omega_0}\tag{2}$$

For taking the limit of Eq. $(2)$ we can make use of the following result (cf. Eq. (37) here):

$$\lim_{a\to\infty}\frac{\sin(ax)}{\pi x}=\delta(x)\tag{3}$$

where $\delta(x)$ is the Dirac delta impulse. With $(3)$ we obtain from $(2)$


which is just the Fourier transform of $\cos(\omega_0t)$.


1.- Because Fourier Transforms (FT) are defined either for finite energy signals with integration interval [-Inf Inf] or for finite power signals, repetitive with cycle T with integration interval [0 1/T], or [-1/(2*T) 1/(2*T)], or any continuous time interval with span equal, neither longer, nor shorter, than T. Or amultiple of T.

2.- f=1/T where f is the carrier frequency, FT itself, as defined, is a narrow band transform.

3.- Indeed you can define any transformation similar to FT but with intgration interval [-a a] or any other variation, that may look like FT, but it is not going to be FT.

4.- If you replace a=1/6 in your expression, it becomes f*sin(2*pi*1/6*f)/(pi*(f^2-9)) and here again you can see the 2 deltas in frequency that build the sin function, windowed by the [0 T] interval, necessary to keep the transform integral finite.

Cyclic signals like cos and sin do not have finite energy, they are finite power signals over a certain interval

It is convenient to choose such interval as a multiple of T.

  • $\begingroup$ I am actually wondering whether the exact fourier transform can be obtained from the expression as we increase the interval or not. From your answer, I understand it is not possible, right? $\endgroup$ May 7 at 15:06
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    $\begingroup$ as you make a larger, the sincs at -f0 and f0 become narrower and narrower. The longer you wait, the closer that it gets to the ideal TF. But no one waits for ever; time is money. If you can do the job observing T seconds why overspend a single second more? job done, next job: to do smart Digital Signal Processing, like in any job, you need to budget wisely. $\endgroup$ May 7 at 15:13

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