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Dirac delta function

Dirac delta function

A dirac delta “function” is a function which puts area 1 on top of a single point, and area zero on top of the rest of the points. We have written the word function within quotes, as dirac delta function is in reality not a function. In mathematical terminology, it is called a “distribution” or a “generalized function”. We will see why this is so later.

In the usual functions, area on top of a single point x is always zero, assuming that the function has a finite value there. Consider the function $f(x)$.

Periodic Dirac Delta Functions

Theorem:
$$
\sum_{k=-\infty}^{\infty}e^{{2 \pi i} kt}=\sum_{n=-\infty}^{\infty}\delta(t-n)$$
Proof:
$$\begin{eqnarray}
\sum_{k=-N}^{N}e^{{2 \pi i} kt}&=& e^{-{2 \pi i} Nt}+e^{-{2 \pi i} (N-1)t}+\ldots+e^{{2 \pi i} (N-1)t}+e^{{2 \pi i} Nt} \\
&=&e^{-{2 \pi i} Nt} (1+e^{2 \pi i t} + \ldots +e^{\pi i (4N-1)t}+e^{4 \pi i Nt})
\end{eqnarray}$$
The part within the parantheses is a truncated geomertic series. After summing it, and performing some algebraic manipulations, we arrive at
$$\begin{eqnarray}
\sum_{k=-N}^{N}e^{2 \pi i kt}&=&e^{-2 \pi i Nt}\frac{1-e^{2 \pi i (2N+1)t}}{1-e^{2 \pi i t}}\\
&=&e^{-2 \pi i Nt}\frac{e^{\pi i (2N+1)t}(e^{- \pi i (2N+1)t}-e^{ \pi i (2N+1)t})}{e^{\pi i t}(e^{-\pi i t}-e^{ \pi i t})}\\
&=&\frac{e^{- \pi i (2N+1)t}-e^{ \pi i (2N+1)t}}{e^{-\pi i t}-e^{ \pi i t}}\\
\end{eqnarray}$$
Remembering the definition of sin(.), we arrive at
$$\begin{eqnarray}
\sum_{k=-N}^{N}e^{2 \pi i kt}=\frac{\sin[(2N+1)\pi t]}{\sin(\pi t)}
\end{eqnarray}$$
Now, let us take the limit $N \rightarrow \infty$, ie, consider the expression
$$\begin{eqnarray}
\lim_{N \rightarrow \infty}\frac{\sin[(2N+1)\pi t]}{\sin(\pi t)}
\end{eqnarray}$$
As this expression is periodic with period 1, it is sufficient to analyze it only on a single period. Hence, let us limit our analysis to the interval $[-1/2,1/2]$. This interval can be decomposed into three subintervals:
$$\begin{eqnarray}
[-\frac{1}{2},\frac{1}{2}]=[-\frac{1}{2},-\delta] \cup (-\delta,\delta) \cup [\delta,\frac{1}{2}].
\end{eqnarray}$$
where $\delta$ is a small positive number.

  • In $[-\frac{1}{2},-\delta]$, $\sin(\pi t)$ is not zero and the function is a bounded and infinitely fast oscillation.
  • In $(-\delta,\delta)$, $t$ is very small hence $\sin(\pi t) \approx \pi t$. Therefore
    $$\begin{eqnarray}
    \lim_{N \rightarrow \infty}\frac{\sin[(2N+1)\pi t]}{\sin(\pi t)} \approx \lim_{N \rightarrow \infty}\frac{\sin[(2N+1)\pi t]}{\pi t}=\delta(t)
    \end{eqnarray}$$
  • In $[\delta,\frac{1}{2}]$, $\sin(\pi t)$ is not zero and the function is a bounded and infinitely fast oscillation.

Therefore,
$$\begin{eqnarray}
\lim_{N \rightarrow \infty}\frac{\sin[(2N+1)\pi t]}{\sin(\pi t)} =\delta(t)
\end{eqnarray} \qquad,\qquad t \in [-\delta,\delta] $$
If we consider all the periods, we obtain the theorem..

Scaled version of the Periodic Dirac Delta Functions

Let us scale time in the above expression
$$ \begin{eqnarray}
t \rightarrow \frac{t}{T}
\end{eqnarray}$$
where $T$ is the scaling factor. Then
$$
\sum_{k=-\infty}^{\infty}e^{{2 \pi i} k\frac{t}{T}}=\sum_{n=-\infty}^{\infty}\delta(\frac{t}{T}-n)$$
Or, using the modulation property of dirac deltas,
$$
\delta(at)=\frac{1}{|a|}\delta(t)
$$
we get
$$
\sum_{k=-\infty}^{\infty}e^{{2 \pi i} k\frac{t}{T}}=T\sum_{n=-\infty}^{\infty}\delta(t-nT)$$

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