Duality Property of the Fourier Transform
The duality property (or dual property) of the Fourier transform states:
If a function and its Fourier transform form a pair, then swapping them (with a sign change) also forms a valid Fourier transform pair.
1. Statement (Physics Convention)
Using the convention:
\[ \mathcal{F}\{f(x)\}(k) = \int_{-\infty}^{\infty} f(x)\, e^{-ikx}\, dx \]
If
\[ f(x) \longleftrightarrow F(k) \]
then duality says:
\[ F(x) \longleftrightarrow 2\pi\, f(-k) \]
The roles of space/time and frequency are exchanged (with a sign flip and a factor \(2\pi\)).
2. Statement (Symmetric / Engineering Convention)
Using the symmetric definition:
\[ \mathcal{F}\{f(x)\}(k) = \int_{-\infty}^{\infty} f(x)\, e^{-2\pi i k x}\, dx \]
Then duality becomes:
\[ f(x) \longleftrightarrow F(k) \quad \Rightarrow \quad F(x) \longleftrightarrow f(-k) \]
No extra constant appears.
3. Intuition
The Fourier transform is nearly symmetric in \(x\) and \(k\). The kernel \(e^{-ikx}\) treats them almost interchangeably.
- If something is narrow in time → it is wide in frequency
- The transform of that frequency function behaves similarly in the original domain
This reflects a deep symmetry between domains.
4. Classic Example (Gaussian)
\[ e^{-a x^2} \longleftrightarrow \sqrt{\frac{\pi}{a}}\, e^{-k^2/4a} \]
By duality:
\[ \sqrt{\frac{\pi}{a}}\, e^{-a x^2} \longleftrightarrow 2\pi\, e^{-k^2/4a} \]
The Gaussian essentially transforms into itself (up to scaling).
5. Example (Rect ↔ Sinc)
\[ \text{rect}(x) \longleftrightarrow \text{sinc}(k) \]
By duality:
\[ \text{sinc}(x) \longleftrightarrow \text{rect}(-k) \]
Time-domain rectangles correspond to frequency-domain sincs, and vice versa.
6. Practical Signal Processing Example
Using the engineering definition:
\[ \mathcal{F}\{f(t)\}(f) = \int_{-\infty}^{\infty} f(t)\, e^{-j2\pi f t}\, dt \]
\[ f(t) \longleftrightarrow F(f) \quad \Rightarrow \quad F(t) \longleftrightarrow f(-f) \]
Pulse Shaping Example
Known transform pair:
\[ \text{rect}\!\left(\frac{t}{T}\right) \longleftrightarrow T\,\text{sinc}(Tf) \]
\[ \text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \]
Numeric Example
\[ x(t) = \text{rect}\!\left(\frac{t}{0.01}\right) \]
\[ X(f) = 0.01\,\text{sinc}(0.01 f) \]
Suppose instead:
\[ H(f) = \text{rect}\!\left(\frac{f}{100}\right) \]
Using duality:
\[ h(t) = 100\,\text{sinc}(100 t) \]
Final Result
\[ h(t) = \frac{\sin(100\pi t)}{\pi t} \]
This is the impulse response of an ideal 100 Hz low-pass filter.
- Sharp frequency cutoff
- Long sinc in time
- Infinite duration
- Ringing (Gibbs phenomenon)