# How does causality (i.e. unit step) affect the DTFT of a sine or cosine wave?

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.)

• 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. – Spacey Mar 18 '12 at 0:42
• 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. – some kind of robot Mar 18 '12 at 12:00
• 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. – Jim Clay Mar 19 '12 at 3:20
• @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. – Spacey Mar 19 '12 at 20:14

• "... 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. – Dilip Sarwate Mar 19 '12 at 20:51