Question

The Fourier transform of a function \(f(at)\) is given by

• A. \(aF(\omega)\)
• B. \(2aF(\omega)\)
• C. \frac{1}{a}F(\frac{\omega}{a})
• D. \frac{2}{a}F(\omega)

Hint:

The Time Scaling Property of Fourier Transform is \[ f(at) \;\xleftrightarrow{\mathcal F}\; \frac{1}{|a|} F\!\left(\frac{\omega}{a}\right). \] Time compression causes frequency expansion, while time expansion causes frequency compression.

Answer 1

The Correct Option is C

Concept: One of the most important properties of the Fourier Transform is the Time Scaling Property. This property explains how the spectrum of a signal changes when the time axis is compressed or expanded. If \[ f(t) \;\xleftrightarrow{\mathcal F}\; F(\omega), \] then scaling the time variable by a factor \(a\) affects both the amplitude and frequency axes of the transform.

Step 1: Write the Fourier transform definition.
The Fourier transform of \(f(t)\) is \[ F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-j\omega t}\,dt. \] Now consider the transformed signal \[ g(t)=f(at). \] We wish to determine its Fourier transform.

Step 2: Apply the Fourier transform to \(f(at)\).
Let \[ G(\omega) = \int_{-\infty}^{\infty} f(at)e^{-j\omega t}\,dt. \] Introduce the substitution \[ u=at. \] Then \[ t=\frac{u}{a}, \qquad dt=\frac{du}{a}. \] Substituting, \[ G(\omega) = \int_{-\infty}^{\infty} f(u) e^{-j\omega u/a} \frac{du}{a}. \] \[ = \frac{1}{a} \int_{-\infty}^{\infty} f(u) e^{-j(\omega/a)u} \,du. \]

Step 3: Recognize the resulting integral.
The integral \[ \int_{-\infty}^{\infty} f(u) e^{-j(\omega/a)u} \,du \] is simply \[ F\!\left(\frac{\omega}{a}\right). \] Therefore, \[ G(\omega) = \frac{1}{a} F\!\left(\frac{\omega}{a}\right). \]

Step 4: State the Time Scaling Property.
Hence, \[ \boxed{ f(at) \;\xleftrightarrow{\mathcal F}\; \frac{1}{a} F\!\left(\frac{\omega}{a}\right) } \] for \(a>0\). Thus the correct option is \[ \boxed{ \frac{1}{a} F\!\left(\frac{\omega}{a}\right) }. \]