The Fourier Transform of the Gaussian

Fourier Transform of the Gaussian

Konstantinos G. Derpanis October 20, 2005

In this note we consider the Fourier transform1 of the Gaussian. The Gaussian function, g(x), is defined as,

1        −x2[pic 1]

g(x) =    √        e 2σ2 ,        (3)

σ  2π

where ∫ ∞

---

g(x)dx = 1 (i.e., normalized). The Fourier transform of the Gaussian function is given

[pic 2]

by:        −∞

Proof:

---

tt(ω) = e−

---

2 2

2    .        (4)[pic 3]

We begin with differentiating the Gaussian function:

 dg(x)  = − x g(x)        (5)[pic 4][pic 5]

Next, applying the Fourier transform to both sides of (5) yields,

iωtt(ω) =

dG(ω)

[pic 6]

---

1

iσ2[pic 7]

---

dtt(ω) dω

---

(6)

Integrating both sides of (7) yields,

---

    dω     =        2        (7)

tt(ω)[pic 8]

ω dG(ωr)[pic 9]

[pic 10]

    dωr            j =[pic 11]

tt(ωj)

0

---

ω

ωjσ2dωj        (8)[pic 12]

0

ln tt(ω) − ln tt(0) =

---

σ2ω2

2   .        (9)[pic 13]

Since the Gaussian is normalized, the DC component tt(0) = 0, thus (9) can be rewritten as,

ln tt(ω) = −

---

σ2ω2        (10)

2[pic 14]

Finally, applying the exponent to each side yields,

ln G(ω) =

---

σ2 ω2

−

[pic 15]

---

(11)

as desired.

1The Fourier transform pair is given by: