3 Inversion of the Fourier transform

Formal inversion of the Fourier transform, i.e. finding f ( t ) for a given F ( ω ) , is sometimes possible using the inversion integral (4). However, in elementary cases, we can use a Table of standard Fourier transforms together, if necessary, with the appropriate properties of the Fourier transform.

The following Examples and Tasks involve such inversion.

Example 3

Find the inverse Fourier transform of   F ( ω ) = 20 sin 5 ω 5 ω .

Solution

The appearance of the sine function implies that f ( t ) is a symmetric rectangular pulse.

We know the standard form   F { p a ( t ) } = 2 a sin ω a ω a  or   F − 1 { 2 a sin ω a ω a } = p a ( t ) .

Putting a = 5   F − 1 { 10 sin 5 ω 5 ω } = p 5 ( t ) .  Thus, by the linearity property

f ( t ) = F − 1 { 20 sin 5 ω 5 ω } = 2 p 5 ( t )

Figure 4

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Example 4

Find the inverse Fourier transform of  G ( ω ) = 20 sin 5 ω 5 ω exp ( − 3 i ω ) .

Solution

The occurrence of the complex exponential factor in the Fourier transform suggests the time-shift property with the time shift t 0 = + 3 (i.e. a right shift).

From Example 3

F − 1 { 20 sin 5 ω 5 ω } = 2 p 5 ( t ) so g ( t ) = F − 1 { 20 sin 5 ω 5 ω e − 3 i ω } = 2 p 5 ( t − 3 )

Figure 5

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Task!

Find the inverse Fourier transform of

H ( ω ) = 6 sin 2 ω ω e − 4 i ω .

Firstly ignore the exponential factor and find the inverse Fourier transform of the remaining terms:

We use the result: F − 1 { 2 a sin ω a ω a } = p a ( t )

Putting  a = 2 gives F − 1 { 2 sin 2 ω ω } = p 2 ( t ) ∴ F − 1 { 6 sin 2 ω ω } = 3 p 2 ( t )

Now take account of the exponential factor:

Using the time-shift theorem for t 0 = 4

h ( t ) = F − 1 { 6 sin 2 ω ω e − 4 i ω } = 3 p 2 ( t − 4 )

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Example 5

Find the inverse Fourier transform of

K ( ω ) = 2 1 + 2 ( ω − 1 ) i

Solution

The presence of the term ( ω − 1 ) instead of ω suggests the frequency shift property.

Hence, we consider first

K ̂ ( ω ) = 2 1 + 2 i ω .

The relevant standard form is

F { e − α t u ( t ) } = 1 α + i ω or F − 1 { 1 α + i ω } = e − α t u ( t ) .

Hence, writing K ̂ ( ω ) = 1 1 2 + i ω k ̂ ( t ) = e − 1 2 t u ( t ) .

Then, by the frequency shift property with ω 0 = 1 k ( t ) = F − 1 { 2 1 + 2 ( ω − 1 ) i } = e − 1 2 t e i t u ( t ) .

Here k ( t ) is a complex time-domain signal.

Task!

Find the inverse Fourier transforms of

  1. L ( ω ) = 2 sin 3 ( ω − 2 π ) ( ω − 2 π )
  2. M ( ω ) = e i ω 1 + i ω
  1. Using the frequency shift property with ω 0 = 2 π

    l ( t ) = F − 1 { L ( ω ) } = p 3 ( t ) e i 2 π t

  2. Using the time shift property with t 0 = − 1

    m ( t ) = e − ( t + 1 ) u ( t + 1 )

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