Fourier Transform of Pulse Train
We want to prove:
∞
X
∞
2π X
δ(t − nT ) ←→
δ(Ω − nΩs )
T n=−∞
n=−∞
Start by doing the inverse Fourier transform of right-hand side (RHS):
#
Z ∞"
∞
2π X
1
x(t) =
δ(Ω − nΩs ) ejΩt dt
2π −∞ T n=−∞
We can change the order of summation and integration:
∞ Z ∞
1 X
δ(Ω − nΩs )ejΩt dt
x(t) =
T n=−∞ −∞
(1)
(2)
(3)
Due to the sifting property of the delta function, we can write above equation as
∞
1 X jnΩs t
e
T n=−∞
(4)
N
1 X jnΩs t
x(t) =
e
T n=−N
(5)
x(t) =
We modify the limits of above integration to
where later we shall have N → ∞. After lots of algebra, the above power series reduces to
1 ej(N +1)Ωs t − e−jN Ωs t
T
ejΩs t − 1
1 sin[(2N + 1)Ωs t/2]
=
T
sin(Ωs t/2)
x(t) =
Figure 1 shows the effect of N on the shape of the x(t).
The function shows its peaks at times t = − · · · − 2T , −T , 0, T , 2T · · · . The peak value is
x(0) =
2N + 1
T
We can clearly see that from the plots of x(t).
As N → ∞ the peak value will reach ∞. Therefore, Eq. 6 can be expressed as
x(t)N →∞ = sum∞
n=−∞ δ(t − nT )
This proves Eq. 1.
1
(6)
Figure 1: Plot of x(t) vs. t for T = 1. Green line is for N = 10 and blue line is for N = 2.
2