# Fourier transform of the impulse unit function

Discussion in 'Homework Help' started by nyasha, Nov 17, 2011.

1. ### nyasha Thread Starter Active Member

Mar 23, 2009
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1
Can someone explain to me how do they use the sampling property to end up having the fourier transform being one. In my attempt to get the answer, am in the correct direction ?

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2. ### steveb Senior Member

Jul 3, 2008
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No, you made an invalid step there. The delta function is not one over the whole range of integration. Ask yourself where the delta function is zero and where is it nonzero. Then ask yourself what the value of exp(-jwt) is in the places where the delta function is not zero.

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3. ### nyasha Thread Starter Active Member

Mar 23, 2009
90
1
The dirac function is one when x=0 of which the exponential function is also one when x=0. Thanks for the help.

4. ### steveb Senior Member

Jul 3, 2008
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469
That's not quite correct, but you are homing in on the correct viewpoint. Remember that the Dirac delta function is not one at zero, but goes to infinity. It is the area under the delta function that is one, as shown by your integral equation for the sifting property.

Often the integral (or area under) the delta function is derived as a limiting function. For example a rectangular pulse with finite width and finite height (with an area of one) is used. Then the limit is taken as the width gets smaller and the height gets taller, but the area is held constant as one. In the limit of the width going to zero and the height going to infinity (still with area of one) you obtain the delta function.

However, you don't have to use a rectangular pulse. You could use a triangular pulse, or a Gaussian pulse. You can then derive the delta function by taking the integral of the pulse times another function that equals one at zero. Since the pulse is getting narrower and narrower in the limit, the multiplication with this other function has less and less effect, until it has no effect once the limit is taken.