Derive Fourier transform for summation of shifted Dirac-delta function

Question

How to derive this Fourier transform:

$$ F\{ \sum_{n=-\infty}^{\infty} \delta (t- nT) \text{ }\} =\omega_o \sum_{n=-\infty}^{\infty} \delta (\omega - n \omega_o)$$

where:

$$ \omega_o = 2 \pi / T $$

Answer 1

First note that

$$f(t) = \sum_{n=-\infty}^{+\infty}\delta(t-nT)$$

is a periodic tempered distribution with period $T$. Therefore you can do a Fourier series expansion:

$$f(t) = \sum_{k=-\infty}^{+\infty}C_k\,e^{ik2\pi t/T},\;\mathrm{with}\;C_k=\frac{1}{T}\int_a^{a+T}f(t)\;e^{-ik2\pi t/T}\,dt = \frac{1}{T}\,.$$

In other words:

$$f(t) = \sum_{n=-\infty}^{+\infty}\delta(t-nT) = \frac{1}{T}\sum_{k=-\infty}^{+\infty}e^{ik2\pi t/T}\,.$$

Now, taking the Fourier transform of the last series we get the desired result:

$$F\{ f(t) \} = \frac{1}{T}F\{ \sum_{k=-\infty}^{+\infty} e^{ik2\pi t/T} \text{ }\} =\frac{2\pi}{T}\sum_{k=-\infty}^{+\infty}\delta(\omega-k\frac{2\pi}{T}).$$

Answer 2

We start with the Dirac Comb Function:

$$ f(t) = \sum\limits_{k=-\infty}^{+\infty} \delta(t - kT) $$

We know the formula for Complex Fourier Series Expansion is:

$$x(t)=\sum_{n=-a+e}^{a+e}c_n exp( j 2 \pi n t / T)$$ Where:

$$ c_n = \frac{1}{T_o}\int_{-\infty}^{\infty}x(t)exp(-j 2 \pi n t / T_o) dt$$

First, we calculate the coefficients for the series expansion of f(t):

$$ c_n = \frac{1}{T}\int\limits_{-T/2}^{T/2} \sum\limits_{k=-\infty}^{+\infty} \delta(t - kT) e^{-j 2 \pi n t /T} dt \\ $$ Since the period of f(t) is T, and we are integrating between -T/2 and T/2, it follow that only one value of k applies to the integral range, namely k=0. Thus, the equation simplifies to:

$$c_n = \frac{1}{T}\int\limits_{-T/2}^{T/2} \delta(t) e^{-j 2 \pi n t/T} dt \quad \quad\\$$

Now we use this property of dirac delta:

$$\int_{-a+e}^{a+e}f(t)\delta(t - a)dt = f(a)$$

The equation further simplifies to:

$$ \begin{align} & c_n = \frac{1}{T}e^{-j 2 \pi n 0/T} \\ & c_n = \frac{1}{T} \quad \quad \text{(for all integers of n)} \\ \end{align} $$

Substituting the coefficient value, $c_n$, into the formula for Fourier Series Expansion yields:

$$ f(t) = \sum\limits_{k=-\infty}^{+\infty} \delta(t - kT) = \sum\limits_{n=-\infty}^{+\infty} \frac{1}{T} \ e^{j 2 \pi n t/T} $$

Next we find the Fourier Transform of the Fourier Series Expansion of f(t):

we know that: $$ F \{ e^{i \omega_o t} \} = 2 \pi \delta(\omega - \omega_o) $$

Thus:

$$F\{ f(t) \} = \frac{1}{T}F\{ \sum_{k=-\infty}^{+\infty} e^{ik2\pi t/T} \text{ }\} =\frac{2\pi}{T}\sum_{k=-\infty}^{+\infty}\delta(\omega-k\frac{2\pi}{T}).$$