Fourier Sine and Cosine Transforms

FOURIER SINE & COSINE TRANSFORMS:

FOURIER COSINE TRANSFORM:

The infinite Fourier cosine transform of f(x) is defined by

The inverse Fourier cosine transform Fc [f(x)] is defined by

INVERSION FORMULA FOR FOURIER COSINE TRANSFORM

Let F_C(s) denote the F.C.T of f(x). Then

Proof: By the definition of F.C.T,

Here, f(x) is defined for all x ≥0

Now define g(x) by g(x) = 

Clearly,

g(‒x) = g(x) for all x and hence g is an even function.

To prove that, the Fourier transform of g(x) is the F.C.T of f(x).

= Fc [g(x)]

= Fc [f(x)]               [ g(x) = f(x) for all x ≥ 0]

Hence, by inversion formula for F.T, we have

FOURIER SINE TRANSFORM:

The infinite Fourier sine transform of f(x) is defined by

The inverse Fourier sine transform of Fs [f(x)] is defined by

INVERSION FORMULA FOR FOURIER SINE TRANSFORM

Let F(s) denote the F.S.T of f(x). Then

Proof: By the definition of F.S.T

Properties of Fourier sine transform and Fourier cosine

1. Linear property

 (i) Fs [af(x) + bg (x)] = a Fs [f(x)] + b Fs [g (x) ]

 (ii) Fc [af(x) + bg (x)] = a Fc [f(x)] + bFc[g (x)]

Proof: (i) We know that,

= a Fs [f(x)] + b Fs (g(x)]

 (ii) We know that,

2. Modulation property:

(i) F_S[f (x) sin ax ] = 1⁄2 [ Fc(s − a) − Fc(s + a) ]

(ii) F_S [f (x) cos ax] = 1/2 [Fc(s + a) + Fc(s − a) ]

 (iii) F_C [f(x) sin ax] = 1⁄2 [Fs(a + s) + Fs(a − s) ]

 (iv) F_C [f (x) cos ax] = 1/2 [Fc(s + a) + Fc(s − a) ]

Proof:

3. F_S [f(ax) ] = 1/a Fs [s/a] [Change of scale property]

Proof:

F_S [f(ax) ] = √(2/π) ₀ʃ^∞ f(ax) sin sx dx

4. Fs [f ‘(x)] = −s Fc(s), if f(x) → 0 as x→ ∞.

 [Transform of derivative]

Proof:

5. F_C [f ' (x)] = − √(2/π) f(0) + sF_S(s)    if f(x) → 0 as x→ ∞.

 [Transform of derivative]

Proof:

6. Fs [xf(x)] = ‒ d/ds [F_C(s)]

[Derivatives of transform]

Proof: We know that,

7. Fc [xf(x)] = d/ds F_S(s)

Proof: We know that,