1
\$\begingroup\$

What will be the Fourier transform of ramp function? Actually I got some results regarding this on the internet, but I was wondering how it could be possible because Fourier Transform (as I have read) is defined only for those functions which are converging or in other words which follow Dirichlet conditions.

There is one more question: Fourier transform is defined for signals with finite energy (converging functions), or in other words energy signals (which means finite energy and zero power) but when it comes to power signals (which means infinite energy and finite power), how power signals can also have their Fourier transform though they have infinite energy (e.g sinusoidal signals, unit step, DC signal etc.)?

\$\endgroup\$
1
\$\begingroup\$

You don't say what you mean by a "ramp function" so I'll just show you how pictorially, analytically you can get your function that you need.

remember that multiplication in the time domain is the same as convolving in the frequency domain and the converse is true too. Convolving in the time domain is the same as multiplying in the frequency domain.

I know the relationships between several shapes in both the time and frequency domains. One I've chosen is the rect function (rectangle) which has a sinc(f) frequency function.

I then think about how to generate a ramp function that increases and then decreases.

enter image description here

Here is shown two rect functions one red and one black (for ease of understanding). Figure 1 is before they intercept. 2 is when they start to intercept (with the black rectangle being the over lap area) 3 is at maximal overlap (black area is maximal) 4. is when the overlap is decreasing as the red passes through the black rect and 5. is when it is over.

Figure 6 shows the result of this convolution with a ramp up and then a ramp down.

Now the fun part.

A convolution of two rect functions in the time domain is a multiplication in the frequency domain. Since the two rect's are the same we simply get \$sinc^2\$ in the frequency domain.

I'll leave it to you to fill in the details of the actual mathematic steps. the combining and use of fundamental functions is the key insight you need.

\$\endgroup\$
2
  • \$\begingroup\$ Don't have to think of this as a convolution of square waves. Think of it as the integral of a 50% duty cycle square wave (either between +/-A, or just a positive square wave with a constant subtracted), and use integration rules. \$\endgroup\$ – Scott Seidman Sep 23 '14 at 22:14
  • \$\begingroup\$ @ScottSeidman of course you can, but I was trying to show them an intuitive way of assembling/understanding the mechanisms. Use the power of the transform! (tm) \$\endgroup\$ – placeholder Sep 23 '14 at 22:40
0
\$\begingroup\$

Any periodic function does have a Fourier transform. Even though the Fourier integral does not converge in the conventional sense, it can be interpreted in terms of the Dirac delta distribution. A very important transform pair is the following:

$$e^{j\omega_0t}\Longleftrightarrow 2\pi\delta(\omega-\omega_0)\tag{1}$$

where \$\delta(\omega)\$ is the Dirac delta impulse. Using (1) the Fourier transform of any periodic function \$f(t)=f(t+T)\$ can be directly obtained from its Fourier series:

$$f(t)=\sum_{k=-\infty}^{\infty}c_ke^{2\pi j kt/T}\Longleftrightarrow 2\pi\sum_{k=-\infty}^{\infty}c_k\delta(\omega-2\pi k/T)$$

where \$c_k\$ are the complex Fourier coefficients of \$f(t)\$.

All the functions you mentioned in your question do have a Fourier transform which contains Dirac delta impulses (cf. Fourier transform relations with distributions).

\$\endgroup\$
0
\$\begingroup\$

Periodic signals, like a sinus function, can be represented in the frequency domain by expanding it into its Fourier series, therefore you don't need to worry about Fourier transform.

The latter is defined for aperiodic signals. The basic idea behind Fourier transform is, that an aperiodic signal can be represented as a periodic signal with a periodic time approaching infinity. As you said, it is only applicable for signals with finite energy, more formally, for every \$x(t)\$, for which $$\int_{-∞}^{∞}|x(t)|^2 dt < ∞$$

Functions with non-finite energy can have their Fourier transforms, but in most cases you will have to use some kind of mathematical "trick". Refer to this paper about the Fourier transform of a unit step: https://www.cs.uaf.edu/~bueler/M611heaviside.pdf

However, you can use the unilateral Laplace transform for anything else, but be careful, as it will ignore everything that happens before \$t=0\$.

And lastly, there is no need to transform a DC signal, as it is clear to have only a 0 Hz component in the frequency domain.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ Finite energy signals satisfy \$\int_{-\infty}^{\infty}|x(t)|^2dt<\infty\$ (i.e. with a square). Furthermore, in general the Laplace transform does not ignore everything for \$t<0\$, only if you specifically refer to the unilateral Laplace transform. There is, however, also the bilateral Laplace transform. \$\endgroup\$ – Matt L. Aug 14 '14 at 19:27
  • \$\begingroup\$ Thank you for your comments and help! I'm going to fix those in my answer. Didn't know about the bilateral Laplace, I was only taught the unilateral. As for the finite energy formula, I mixed it up with another condition for the Fourier transform. \$\endgroup\$ – hryghr Aug 14 '14 at 20:20
  • \$\begingroup\$ Yes, the condition that the function be absolutely integrable (as given in your original answer) is sufficient (but not necessary) for the Fourier integral to exist. Unfortunately, a lot of useful functions don't satisfy this condition (sinc, sinusoids, step function, periodic functions, etc.). \$\endgroup\$ – Matt L. Aug 14 '14 at 20:30
  • \$\begingroup\$ It was time I read through my old notes again, thank you for the corrections! :) \$\endgroup\$ – hryghr Aug 14 '14 at 20:36
  • \$\begingroup\$ that's really nice of you all. but i am still confused that i could i get the Fourier transform of ramp with that i mean the process of computing its Fourier. \$\endgroup\$ – Pankaj Kumar Aug 15 '14 at 7:01
0
\$\begingroup\$

If you're talking about a ramp (y=0, t<0; y=t, t>=0), as opposed to some sort of sawtooth periodic wave, think of the ramp as the integral of a step function.

The Fourier Xform of the step function is (1/jw). The Xform of the integral of x(t) is (1/jw)X(jw). When you apply both of these rules, the Fourier Transform of the ramp is (1/jw)^2

\$\endgroup\$

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.