If the Fourier transform of f(t) is F(ω) then Fourier transform o

SOLUTION

Concept:

The time scaling property of Fourier transform states that:

If \(x\left( t \right)\mathop \leftrightarrow \limits^{\;FT\;} X\left( \omega \right)\)

Then:

\(x\left( {at} \right)\mathop \leftrightarrow \limits^{\;FT\;} \frac{1}{{\left| a \right|}}X\left( {\frac{\omega }{a}} \right)\)

‘a’ any real constant.

Proof:

For a > 0:

\(FT\left[ {x\left( {at} \right)} \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( {at} \right){e^{ - j\omega t}}dt\)

Let at = τ. Thus adt = dτ

The above expression can now be written as:

\( = \frac{1}{a}\mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( \tau \right){e^{ - j\left( {\frac{\omega }{a}} \right)\tau }}d\tau = \frac{1}{a}X\left( {\frac{\omega }{a}} \right)\)

For a < 0:

\(FT\left[ {x\left( { - at} \right)} \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( { - at} \right){e^{ - j\omega t}}dt\)

Let -at = ui – adt = du

\( = - \frac{1}{a}\mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( u \right){e^{ - j\left( { - \frac{\omega }{a}} \right)u}}du\)

\(= \frac{1}{2}\mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( u \right){e^{ - j\left( { - \frac{\omega }{a}} \right)u}}du = \frac{1}{a}X\left( { - \frac{\omega }{a}} \right)\)

\(\therefore FT\left[ {x\left( {at} \right)} \right] = \frac{1}{{\left| a \right|}}X\left( {\frac{\omega }{a}} \right)\)

Note:

The scaling property states that time compression of signal results in its spectral expansion and vice-versa.