# How condition for existence of Fourier transform is valid?

The condition for Discrete time Fourier transform to exist for function $f(n)$ is given as

$$\sum_{-\infty}^\infty |f(n)| < \infty.$$

In case of continuous Fourier transform the difference is summation is replaced by integration in the above equation

It looks something strange for me because on left side we are taking the signal within 1. infinity limits. Also,2. it is again of summation type and we are expecting right side to be less than infinity. How is it possible?

Let us take two examples

1. Suppose there is unit step signal $u(n)$ with constant unity magnitude ranging from o to $\infty$ .then we can clearly see by putting value in the above formula

L.H.S=1+1+1+......upto ${\infty}$ So R.H.S. =$\infty$

So can I say that DTFT of unit step signal doesn't exist?

1. Similarly, if I represent the power signal like $cos(wn)$ using the exponential terms and compute it's value from the formula of condition of existence ,it will come out to be $\infty$ .

So can I say that DTFT of power signal like $cos(wn)$ doesn't exist?

As mentioned in Batman's answer, the condition of the sequence being absolutely summable is only sufficient but not necessary. The Fourier transform can be extended to $\ell_2$ sequences, i.e. sequences for which

$$\sum_{n=-\infty}^{\infty}|f[n]|^2<\infty$$

is satisfied. A further generalization is possible if you allow distributions and their derivatives (such as the Dirac delta function) as Fourier transforms. Then also sequences with polynomial growth have a Fourier transform in that generalized sense. Just sequences with exponential growth have no Fourier transform. However, they can be handled by the $\mathcal{Z}$-transform. Note that everything said above also holds for continuous-time functions (if your replace "$\mathcal{Z}$-transform" by "Laplace transform").

Sequences with a constant amplitude (such as the step function or a sinusoid) will have a Dirac delta in their Fourier transform. Sequences with a linear growth will have the (generalized) derivative of a Dirac delta in their Fourier transform, etc.

And as for your question how it is possible for a sum over infinitely many elements to be finite, your should as a first basic example read up on the geometric series.

EDIT:

So if we allow a Dirac delta impulse as a Fourier transform, we can ask what the corresponding time function would be. By taking the inverse Fourier transform of a shifted Dirac delta impulse $\delta(\omega-\omega_0)$ we get

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega-\omega_0)e^{j\omega t}d\omega=\frac{1}{2\pi}e^{j\omega_0t}$$

which follows from the sifting property of the Dirac delta impulse. So we now have the Fourier transform pair

$$e^{j\omega_0t}\Longleftrightarrow2\pi\delta(\omega-\omega_0)$$

from which the transforms of $\cos(\omega_0t)$ and $\sin(\omega_0t)$ follow directly by rewriting these functions as $\cos(\omega_0t)=\frac12(e^{j\omega_0t}+e^{-j\omega_0t})$ and $\sin(\omega_0t)=\frac{1}{2j}(e^{j\omega_0t}-e^{-j\omega_0t})$.

• Matt L. sir, I want to ask you few things related to your answer. 1. I can understand that in $e^{-jwn}$ for positive values of n, as it tends to $\infty$ the whole exponential term tends to 0 . But on negative axis,for negative values of n, the exponential term will keep on increasing in exponentially. So within the range from $-\infty$ to $o$ ,how could you calculate DTFT of $cos(wn)$ function? 2. How would you calculate DTFT of unit step function using Dirac delta ? Jun 18 '15 at 9:58
• @devraj: The complex exponential will not increase in magnitude, neither for positive nor for negative $n$ because $|e^{-j\omega n}|=1$ always holds. The reason is the imaginary unit $j$ in the exponent. The computation of the Fourier transform of the unit step or of a sinusoid can probably be found on the internet, but you could also ask another question as it can't be treated properly in a comment. Jun 18 '15 at 10:02
• sir, I asked you the 1st query in the comment ,which is about DTFT formula . there is no mod of exponential term. there's exponential term . so for the negative side of DTFT formula ,it will be exponentially increasing. Jun 18 '15 at 10:22
• @devraj: You are wrong, the term $e^{-j\omega n}$ will NOT increase in magnitude for real-valued $n$ and $\omega$. It is a complex exponential. Please re-visit Euler's formula. And I'm not sure what kind of mathematical computations would aid your understanding. Jun 18 '15 at 11:40

the cosine transform can serve as an example to show that the sum is not infinite, if not rather a distribution, in the case of cosine eg are distributions deltas of Dirac spaced periodically in the frequency domain, likewise a train of Kronecker deltas (the Kronecker delta function is analogous to the Dirac delta in the discrete domain and one of its uses is to model the sampling of a signal) or sin(w0n) have transformed that are frequency distributions, I'll show you the images the case of cos(w0n) eg. On the other hand, the transform of the pulse heviside function is still a mystery to me.

• What the heck is constubre??
– jojek
Jun 18 '15 at 8:33
• @jojek: I guess it's an unsuccessful translation of a Spanish word for "customary". Jun 18 '15 at 8:48
• @Esteban sir, I didn't get satisfactory answer for my question from your answer. Jun 18 '15 at 8:51
• please wait for me, I'm new and do not know how to write formulas in this forum, but I will send a picture where we perform discrete cosine transform, what happens is that the tables showing the books are usually given for a period of frequency between - pi and pi, remember, the Fourier transform of a discrete signal is a periodic function in frecuenci, whose period is 2pi as discrete frequency signals have maxim 1/2, that is why the transformed formulas usually Formulas are given between -pi and pi as f = w / (pi) Jun 19 '15 at 11:34
• ie the minimum frequency f is -1/2 and the maximum is at the other end bone w=pi which is f = 1/2, do not give full fomula that is periodic and only give you the exprecion for a period, that is why people tend to think that the expression is finite when actually are seeing a period, in a moment upload developing transform cos (w0n) so you understand what I mean, otherwise I am Colombian and use google translator is not that much English, so please try to understand me. Jun 19 '15 at 11:34

It's sufficient for a function to be in $L^1$ (i.e. $\int_{-\infty}^\infty |f|< \infty$) or $L^2$ (i.e. $\int_{-\infty}^\infty |f|^2 < \infty$) to have the Fourier transform exist, but not necessary.

An example where a Fourier transform exists but the function is not in $L^1$ or $L^2$ is the complex sinusoids $e^{j \omega_0 t}$ (which has the Fourier transform being given as the distribution $\delta(\omega - \omega_0)$). Since you can write cosine as a sum of two complex sinusoids, you get the existence of its Fourier transform that way.

As for the unit step, it requires some finesse so see these notes.

• sir,the PDF u suggested doesn't give me proper step wise calculations for both unit step and cosine functions. Jun 18 '15 at 8:56
• Figure out the cosine case on your own by writing $\cos x = \frac{e^{j x}+e^{-j x}}{2}$, plug in $x = \omega t$ and using linearity of the FT. As for the unit step, more details are here. To make the argument more precise, you need to know some things about distributions, which would be better suited to math.SE. Jun 18 '15 at 11:23
• sir,in the DTFT formula, the summation is from $-{\infty}$ to $+{\infty}$ .so for periodic signal like $cos(wn)$ which is present for infinite interval, how could you compute its DTFT ? specially on the negative axis I.e. from 0 to $-\infty$ ? Jun 18 '15 at 12:07
• I suggest you look at the definition of the DTFT, and review your calc 1 course materials. Note that $|e^{j \omega n}| = 1$. Jun 18 '15 at 12:40