Delta function

The Fourier transform of the Dirac delta function $δ (t - t_{0})$ is given by

F {δ (t - t_{0})} = \int_{- \infty}^{\infty} δ (t - t_{0}) exp (- jω t) d t = exp (- jω t_{0})

That means, the Fourier transform of $δ (t - t_{0})$ has a fixed norm $1$ and $t_{0}$ only affects its phase.

If we express the result by $f = \frac{ω}{2 π}$ , we will have $F {δ (t - t_{0})} = exp (- j 2 π f t_{0})$ .

To generalize, we have

F {x (t)} = X (ω) ⟹ F {x (t - t_{0})} = exp (- jω t_{0}) X (ω)

and

F {x (t)} = X (ω) ⟹ F {exp (j ω_{0} t) x (t - t_{0})} = X (ω - ω_{0})

Square Waves

Consider a square wave signal with the following representation

x (t + NT) = ⎩ ⎨ ⎧ 1, 0, ∣ t ∣ < T_{1} T_{1} < ∣ t ∣ < \frac{T}{2}, N = 0, 1, 2, \dots

We assume the Fourier Series is $x (t) = k = - \infty \sum \infty a_{k} exp (jk ω_{0} t)$ where $ω_{0} = \frac{2 π}{T}$ , then we have

a_{0} = \frac{1}{T} \int_{- T /2}^{T /2} x (t) d t = \frac{1}{T} \int_{- T_{1}}^{T_{1}} 1 d t = \frac{2 T _{1}}{T}

and

a_{k} = \frac{1}{T} \int_{- T /2}^{T /2} x (t) exp (jk ω_{0} t) d t = \frac{1}{T} \int_{- T_{1}}^{T_{1}} exp (jk ω_{0} t) d t = - \frac{1}{jk ω _{0} T} [exp (- jk ω_{0} T_{1}) - exp (jk ω_{0} T_{1})] = \frac{2 sin ( k ω _{0} T _{1} )}{k ω _{0} T} = \frac{sin ( k ω _{0} T _{1} )}{kπ}

Using Sinc function, we can rewrite $a_{k}$ as

a_{k} = \frac{sin ( k ω _{0} T _{1} )}{kπ} = \frac{2 T _{1}}{T} sinc (k ω_{0} T_{1}) = \frac{2 T _{1}}{T} sinc (2 kπ \frac{T _{1}}{T})

If we let $λ = \frac{2 T _{1}}{T}$ be valid length of the square signal, we can just have $a_{k} = λ sinc (λkπ)$