Pulse Signal and Its Representation

We are introducing pulse signal, which is a common function seen in the future.

Definition

A pulse signal at time $0$ is written as,

$$\delta(t)$$

It is defined as a limit of any distribution, that has a mean value of zero, and the chance of being non-zero is one.

That is to say,

$$\delta(t) = \lim_{\epsilon \to 0} \text{distribution}_\epsilon(t)$$

Where $\epsilon$ is the chance of $t$ being non-zero.

For example,

$$\delta(t) := \lim_{\epsilon \to 0} \text{rect}_\epsilon(t)$$

Where,

$$\text{rect}_\epsilon(t) := \begin{cases} \frac{1}{2\epsilon} \quad ||t|| \leq 2\epsilon \\ 0 \; \text{otherwise} \end{cases} \quad \; s.t. \; \epsilon >0$$

tip

$\text{rect}_\epsilon$ is also called the gate function. It is a uniform distribution from $-\epsilon$ to $\epsilon$. Where $\epsilon > 0$.

Or the gaussian distribution,

$$\delta(t) := \lim_{\epsilon \to 0} \frac{1}{\sqrt{2\pi\epsilon^2}}e^{-\frac{t^2}{2\epsilon^2}}$$

tip

Whenever you need to prove some properties of a pulse signal, you must keep in mind that, $\delta$ is the limit of a probability distribution.

note

There are more than one ways to define a $\delta$ signal. We use the probability distribution as the definition. It can also be defined via test function.

Obviously, $\delta(0) = +\infty$, and $\delta(t) = 0$ for $t \neq 0$.

We often draw $\delta(t)$ signal as an arrow pointing upwards, with a height of 1.

tip

In a physical sense, the dimension is,

$$[ \delta(t) ] = \frac{1}{[t]}$$

Properties

Even

This is obvious because $\delta(t) = \delta(-t) = 0$ for $t \neq 0$.

Unit Area

$$\int_{-\infty}^{+\infty} \delta(t) \mathrm{d}t = 1$$

Because it is the limit of a distribution, the area under the curve is 1.

Selective

$$\int_{-\infty}^{+\infty} \delta(t-t_0) f(t) \mathrm{d}t = f(t_0)$$

Consider,

$$\int_{-\infty}^{+\infty} \delta(t) f(t) \mathrm{d}t = \mathbb{E}_{t \sim \delta(t)}(f(t)) = f(0)$$

So the selectivity can be proved via substitution.

Derivative

$$\int_{-\infty}^{+\infty} f(t) (\frac{\mathrm{d}}{\mathrm{d}t})^n \delta(t - t_0) \mathrm{d}t = (-1)^{n}((\frac{\mathrm{d}}{\mathrm{d}t})^n f(t))|_{t=t_0}$$

This is easy from integration by parts. Consider,

$$\int_{-\infty}^{+\infty} f(t) \frac{\mathrm{d}}{\mathrm{d}t} \delta(t - t_0) \mathrm{d}t \\ = \int_{-\infty}^{+\infty} f(t) \mathrm{d} \delta(t - t_0) \\ = (f(t) \delta(t - t_0))|^{+\infty}_{-\infty} - \int_{-\infty}^{+\infty} \frac{\mathrm{d}}{\mathrm{d}t} f(t) \delta(t - t_0) \mathrm{d}t \\ = - (\frac{\mathrm{d}}{\mathrm{d}t} f(t))|_{t=t_0}$$

This can be done recursively for the final result.

Scale

$$\delta(\alpha t) = \frac{1}{||\alpha||} \delta(t)$$

This is done by definition. Assume $\alpha > 0$,

We first calculate the CDF for the distribution $\alpha t$,

$$F(t')=P(\alpha t \leq t') = P(t \leq \frac{t'}{\alpha}) \\ = \int_{-\infty}^{\frac{t'}{\alpha}} \text{distribution}_{\epsilon}(t) \mathrm{d}t$$

Because the derivative of CDF is a PDF, so,

$$p(t') = \frac{1}{\alpha}\text{distribution}_{\epsilon}(t)$$

We enforce $\epsilon \to 0$, so,

$$\delta(\alpha t) = \frac{1}{\alpha}\delta(t)$$

If $\alpha < 0$, we can utilize the evenness of the delta function.

Decomposition

If a polynomial $P(x)$ can be decomposed into product of linear terms $p_i = x_i - \zeta_i$.

That is,

$$P(x) = \prod_{i=0}^{i=n} p_i$$

Then,

$$\delta(P(x)) = \sum_{i=0}^{i=n} \frac{\delta(x_i - \zeta_i)}{||P'(\zeta_i)||}$$

This is hard to prove directly, we use an indirect proof.

Consider,

$$\int_{-\infty}^{+\infty} \delta(P(x)) f(x) dx$$

Because in most of the space, $\delta(P(x))$ is zero. We can make that into,

$$\sum_{i=0}^{i=n}\int_{\zeta_i-\epsilon_i}^{\zeta_i+\epsilon_i} \delta(P(x)) f(x) dx$$

We can use substitution if we assume $\epsilon_i \to 0$. That is to say, we can perceive $P(x)$ as locally linear, so,

$$dx = \frac{1}{||P'(\zeta_i)||}dP(x)$$

So,

$$\int_{-\infty}^{+\infty} \delta(P(x)) f(x) dx \\ = \sum_{i=0}^{i=n}\int_{\zeta_i-\epsilon_i}^{\zeta_i+\epsilon_i} \delta(P(x)) f(x) dx\\ = \sum_{i=0}^{i=n}\int_{\zeta_i-\epsilon_i}^{\zeta_i+\epsilon_i} \frac{1}{||P'(\zeta_i)||} \delta(P(x)) f(x) dP(x) \\ = \sum_{i=0}^{i=n} \frac{f(\zeta_i)}{||P'(\zeta_i)||}$$

Which is equivalent to,

$$\delta(P(x)) = \sum_{i=0}^{i=n} \frac{\delta(x_i - \zeta_i)}{||P'(\zeta_i)||}$$

Integration

If we integrate $\delta(t)$, we get a step function,

info

By integrate, we will always mean a definite integral from $-\infty$ to a variable. So that integrating a function yields another function.

$$u(t) := \int_{-\infty}^{t} \delta(\tau) d\tau$$

Or, alternatively,

$$u(t) = \begin{cases} 0 & \text{if } t < 0 \\ 1 & \text{if } t \geq 0 \end{cases}$$

The step function is a common function seen in the future.

info

It is obvious that,

$$\text{rect}_\epsilon(t) = u(\epsilon + t) - u(\epsilon - t)$$

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More precisely,

$$u(t) = \begin{cases} 0 & \text{if } t < 0 \\ 1 & \text{if } t > 0 \\ \text{undefined} & \text{if } t =0 \end{cases}$$

The value at $0$ depends on whether we are integrating on $\delta(t-0)$ or $\delta(t+0)$.