exponential Fsereis?

Question

express periodic impulse delta T = sigma of (delta (t-nT) for n from - to + infinity in terms of exponential Fourier series Cn using exponential Fourier series.

Answer 1

The crucial point of this problem, is the evaluation of an integral with the delta function:
For any function f(t) and a < T < b it is

b
∫ δ(t - Τ) · f(t) dt = f(T)
a

The rest is easy . The coefficients of the exponential Fourier series are:
Cn · T =
t₂
∫ δ(t - n·Τ) · exp(-i·n·ω·t) dt =
t₁
(where ω = 2·π/T)
t₂
∫ δ(t - n·Τ) · exp(-i·n·2·π· t/T ) dt =
t₁
exp(i·n²·2·π) =
apply eulers formula
cos(-n²·2·π) + i·sin(-n²·2·π) =
cos(n²·2·π) - i·sin(n²·2·π) =
1 + i·0 = 1


The exponential fourier series is of periodic pulse with period T is:
∞
Σ exp(i·n·2·π/T)
-∞

Answer 2

Adapting Equations (30) & (31) from the reference below to your notation,

"For a function periodic in [-T/2, T/2], these become:

f(t) = sum (-inf, +inf) Cn e^ i[ (2 pi n t) / T ]

Cn = 1/T int (-T/2, T/2) f(x) e^ -i[ (2 pi n t) / T ] dt "

Now, the very important thing to realize is that if we redefine your periodic impulse function as a single impulse at t=0 over the interval [-T/2,T/2], the mechanics of the Fourier Transform assumes that it keeps repeating itself!

So, "sigma of (delta (t-nT) for n from - to + infinity" simply becomes delta(t - 0) over the interval [-T/2,T/2] (which repeats!)

Working the integral in the second equation:

Cn = 1/T int (-T/2, T/2) delta(t-0) e^ -i[ (2 pi n t) / T ] dt

Employing the sifting property of integrals that contain the delta function as the kernel:

Cn = 1/T e^ -i[ (2 pi n 0) / T ]

Cn = 1/T

Substituting this back into the first equation;

f(t) = sum (-inf, +inf) 1/T e^ i[ (2 pi n t) / T ]

the 1/T comes outside:

f(t) = 1/T sum (-inf, +inf) e^ i[ (2 pi n t) / T ]