Tables of common Discrete-Time Fourier Transform pairs list the transform of a sine wave:

$ \sin(\omega_0\ n) $ and its transform: $ -j\pi\ [d( \omega\ - \omega_0\ ) - d( \omega\ + \omega_0\ )] $

And the cosine:

$ \cos(\omega_0\ n) $ and its transform: $ \pi\ [d( \omega\ - \omega_0\ ) + d( \omega\ + \omega_0\ )] $

How might the results differ if the sin or cosine is causal? That is, I would like to determine the Fourier Transform of the following (sin or cosine multiplied by the unit step):

$ \sin(\omega_0\ n) u[n] $ or $ \cos(\omega_0\ n) u[n] $

So regarding the DTFT of the signals above, is the only method to realize that multiplication in the time domain is convolution in the frequency domain? Or is there a more simple rule-of-thumb, or property, of the Fourier Transform that I am missing?

(This is not homework, but it is for studying, so it would be helpful to include resources along with answers.)

  • $\begingroup$ Ive wondered what kind of support the step has in the frequency domain, because I would then imagine its just a simple multiplication in the frequency domain with your signal spectrum. (Frequency of unit step here en.wikipedia.org/wiki/Discrete-time_Fourier_transform). Either way, great question. $\endgroup$
    – Spacey
    Mar 18, 2012 at 0:42
  • $\begingroup$ Is it just multiplication in the frequency domain? I thought that since I had multiplication in the time domain that I should have convolution in the frequency domain. $\endgroup$ Mar 18, 2012 at 12:00
  • $\begingroup$ Yes, you can do it via the multiplication/correlation relationship, or you can go back to the defintion of the Fourier Transform and change the integration bounds from -infinity/+infinity to 0/+infinity. $\endgroup$
    – Jim Clay
    Mar 19, 2012 at 3:20
  • $\begingroup$ @garycomtois Oops, yes, sorry I mis-typed - since we multiply with unit-step in time, its a convolution of your signal spectrum with the spectrum of the step. $\endgroup$
    – Spacey
    Mar 19, 2012 at 20:14

1 Answer 1


Multiplication in the time domain is convolution in the frequency domain. The frequency response of a unit step has 1/frequency shape and it's pretty messy.

What you will see is that the overall frequency response is not a delta impulse anymore but that the lines widens and that the get "skirts". These will bigger and more "wiggly" for the cosine than for the sine as the step function slices right through the maximum of the cosine.

  • $\begingroup$ "... overall frequency response is not a delta impulse anymore ..." I think there might be an impulse still there in the frequency spectrum of the unit step signal. The unit step is a power signal (even though it starts at $0$), not an energy signal, which generally leads to impulses in the frequency spectrum. $\endgroup$ Mar 19, 2012 at 20:51
  • $\begingroup$ See this question and its answers for what happens in the computation of the Fourier transform of a continuous-time unit-step signal. Similar things happen for discrete-time unit-step signals as well. $\endgroup$ Mar 19, 2012 at 22:41
  • $\begingroup$ The unit step is simply the time integral of the delta impulse, hence the frequency response of then unit step is the integral over the frequency response of the delta impulse. However, the details of the integration are typically application dependent (limited to a time length, maybe windowed, maybe circular) hence my somewhat vague answer. There can definitely be impulses in there but it's more than one per frequency $\endgroup$
    – Hilmar
    Mar 20, 2012 at 13:27
  • $\begingroup$ Well, the question was quite specific and didn't mention time limitations or windowing etc. The answer to the question asked does include impulses and is not all that messy unless one has the various bells-and-whistles that you are bringing in. Perhaps you could have chosen to be a little less vague and at least mentioned impulses. $\endgroup$ Mar 20, 2012 at 13:36

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