4 Some special Fourier transform pairs

As mentioned in the previous subsection it is possible to obtain Fourier transforms for some important functions that violate the Dirichlet conditions. To discuss this situation we must introduce the unit impulse function , also known as the Dirac delta function . We shall study this topic in an inituitive, rather than rigorous, fashion.

Recall that a symmetrical rectangular pulse

p a ( t ) = 1 − a < t < a 0 otherwise

has a Fourier transform

P a ( ω ) = 2 ω sin ω a .

If we consider a pulse whose height is 1 2 a rather than 1 (so that the pulse encloses unit area), then we have, by the linearity property of Fourier transforms,

F 1 2 a p a ( t ) = sin ω a ω a .

As the value of a becomes smaller, the rectangular pulse becomes narrower and taller but still has unit area.

Figure 7

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We define the unit impulse function δ ( t ) as

δ ( t ) = lim a → 0 1 2 a p a ( t )

and show it graphically as follows:

Figure 8

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Then,

F { δ ( t ) } = F lim a → 0 1 2 a p a ( t ) = lim a → 0 F 1 2 a p a ( t ) = lim a → 0 sin ω a ω a = 1.

Here we have assumed that interchanging the order of taking the Fourier transform with the limit operation is valid.

Now consider a shifted unit impulse δ ( t − t 0 ) :

Figure 9

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We have, by the time shift property

F { δ ( t − t 0 ) } = e − i ω t 0 ( 1 ) = e − i ω t 0 .

These results are summarized in the following Key Point:

Key Point 4

The Fourier transform of a Unit Impulse

F { δ ( t − t 0 ) } = e − i ω t 0 .
If t 0 = 0 then F { δ ( t ) } = 1.
Task!

Apply the duality property to the result

F { δ ( t ) } = 1.

(From the way we have introduced the unit impluse function it must clearly be treated as an even function.)

We have F { δ ( t ) } = 1. Therefore by the duality property

F { 1 } = 2 π δ ( − ω ) = 2 π δ ( ω ) .

We see that the signal

f ( t ) = 1 , − ∞ < t < ∞

which is infinitely wide, has Fourier transform   F ( ω ) = 2 π δ ( ω )  which is infinitesimally narrow. This reciprocal effect is characteristic of Fourier transforms.

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This result is intuitively plausible since a constant signal would be expected to have a frequency representation which had only a component at zero frequency ( ω = 0 ).

Task!

Use the result F { 1 } = 2 π δ ( ω ) and the frequency shift property to obtain

F { e i ω 0 t } .

F { e i ω 0 t } = F { e i ω 0 t f ( t ) }   where f ( t ) = 1 , − ∞ < t < ∞ .

But   F { f ( t ) } = 2 π δ ( ω ) ,  therefore, by the frequency shift property   F { e i ω 0 t } = 2 π δ ( ω − ω 0 ) .  

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Task!

Obtain the Fourier transform of a pure cosine wave

f ( t ) = cos ω 0 t − ∞ < t < ∞

by writing f ( t ) in terms of complex exponentials and using the result of the previous Task.

We have f ( t ) = cos ω 0 t = 1 2 e i ω 0 t + e − i ω 0 t

so

F { cos ω 0 t } = 1 2 F { e i ω 0 t } + 1 2 F { e − i ω 0 t } = π δ ( ω − ω 0 ) + π δ ( ω + ω 0 )

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Note that because ∫ − ∞ ∞ cos ω 0 t d t diverges, one of the Dirichlet conditions is violated. Nevertheless, as we can see via the use of the unit impulse functions, the Fourier transform of cos ω 0 t exists.

By similar reasoning we can readily show

F { sin ω 0 t } = π i δ ( ω − ω 0 ) − π i δ ( ω + ω 0 ) .

Note that the usual results for Fourier transforms of even and odd functions still hold.