Submitted by:
Ashesh Nasnani
EE-32053
EEEG-313
Fourier Transform:
∞
X(jw) = 1/2 Л ∫ X(t) ejwtd t………………………………………..1
-∞
∞
X(t)= 1/2 Л ∫ X(jw) ejwt dw………………………………………..2
-∞
Equation 1 and 2 are referred to as Fourier transform pair, with the function X (jw) referred to as the Fourier transform of Fourier integral or x(t) and equation 2 referred to as inverse Fourier transform equation. The synthesis equation i.e. equation 2 plays a role for aperiodic signals since it represents a signal as a linear combination of complex exponentials. For aperiodic signals there complex exponentials occur at a continuum of frequencies and according to the synthesis equation 2 have amplitude X (jw) (dw/2Л).
The transform X(jw) of an aperiodic signal x(t) is commonly referred to as the spectrum of x(t), as it provides us with the information needed for describing x(t) as linear combination of sinusoidal signals at different frequencies.
Fourier transform of Sinw0t:
X(t)= Sinw0t:
The Fourier series coefficients for this signal are:
a1=1/2j a-1=-1/2j
ak=0; k ≠ 1 or -1
Fourier transform of Cosw0t:
X(t)= Cosw0t:
The Fourier series coefficients for this signal are:
a1=1/2j a-1=1/2j
ak=0; k ≠ 1 or -1
“:: Fourier transform of some important functions::”
………………….δ (w), δ (w-wo), δ (w+w0)………………………..
1. Find the inverse Fourier transform of δ (w).
Soln: the inverse Fourier transform is expressed as:
∞
F-1[X(jw)]= 1/2 Л ∫ X(jw) ejwt dw
-∞
Therefore we have
∞
Ø F-1[δ (w)]= 1/2Л ∫ δ (w) ejwt dw
-∞
∞
Ø F-1[δ (w)]= 1/2Л ∫ δ (w) ejwt dw
-∞
Ø F-1[δ (w)]= 1/2Л [ejwt]at w=0
Ø F-1[δ (w)]= 1/2Л *e0
Ø F-1[δ (w)] = 1/2 Л. 1
Ø F-1[δ (w)]= 1/2 Л
Ø F[1/2 Л ]= δ (w)
Hence,
1/2 Л δ (w)
1 2 Л δ
2. Find the inverse Fourier transform of δ(w-w0).
Soln: The inverse Fourier transform is expressed as:
∞
F-1[X(jw)]= 1/2 Л ∫ X(jw) ejwt dw
-∞
Therefore we have
∞
Ø F-1[δ (w-w0)]= 1/2Л ∫ δ (w-w0) ejwt dw
-∞
Using shifting and sampling property of impulse function, we get
Ø F-1[δ (w-w0)]= 1/2Л [ejwt]at w=wo
Ø F[1/2Л *ejwot]= δ (w-wo)
Hence,
Or, 1/2 ejwot δ (w-wo)
Or, ejwot 2Л δ (w-wo)
3. Find the inverse Fourier transform of δ(w+w0).
Soln: The inverse Fourier transform is expressed as:
∞
F-1[X(jw)]= 1/2 Л ∫ X(jw) ejwt dw
-∞
Therefore we have
∞
Ø F-1[δ (w+w0)]= 1/2Л ∫ δ (w+w0) ejwt dw
-∞
Using shifting and sampling property of impulse function, we get
Ø F-1[δ (w+w0)]= 1/2Л [ejwt]at w=-wo
Ø F[1/2Л *e-jwot]= δ (w+wo)
Hence,
Or, 1/2 e-jwot δ (w+wo)
Or, e-jwot 2Л δ (w+wo)
4. Find the inverse Fourier transform of gate function..
wo
F.T of gate function = 1/2Л ∫ 1. ejwt dw
-wo
=1/2Л [(ejwt- e-jwt)/j+]
=1/ Лt *Sinwct
=1/ Лt xSinwct /wct x wct
=wc /Л Sincwct
By considering impulse δ (w) in F.T. domain wd can incorporate DC signals in our spectral analysis.
5v DC 2 Л x 5 δ (w)
Adding results of questions 2 and 3
δ (w+w0) + δ (w-w0) = 1/Л [(ejwt- e-jwt )/2]
=1/ЛCoswo t
And, Cosw0 t = Л [δ (w+w0) + δ (w-w0)]
Cos is real and even function so, F.T. is real and even.
δ (w+w0) - δ (w-w0) = 1/Л x j[(ejwt- e-jwt )/2j]
=j /Л Sinwo t
Similarly,
Sinwt = Л / j [δ (w+w0)- δ (w-w0) ]
The result is imaginary, odd function.
Extension:
Signal = Real part + Imaginary Part
Any signal if real and even, then its Fourier transform is also real and even
But
Any signal if real and odd, then its Fourier transform has complex conjugate symmetry
2 Л δ (w-w0) ejwot
2 Л ak δ (w-w0) akejwot