Some Useful Signal Models

Sources:

1. B. P. Lathi & Roger Green. (2021). Chapter 1: Signals and Systems. Signal Processing and Linear Systems (2nd ed., pp. 64-136). Oxford University Press.
2. James McClellan, Ronald Schafer & Mark Yoder. (2015). Sinusoids. DSP First (2nd ed., pp. 9-40). Pearson.

The unit step function

The continuous-time unit step function $u(t)$

Definition: The unit step function, $u(t)$, is defined as $$u(t)=\{\begin{array}{ll}0& t<0\\ 1& t\ge 0\end{array}$$

The discrete-time unit impulse function $u[n]$

$$u[n]=\{\begin{array}{ll}0,& n<0\\ 1,& n\ge 0\end{array}.$$

The rectangular pulse

A common situation in a circuit is for a voltage to be applied at a particular time (say $t=a$ ) and removed later at $t=b$ (say). We write such a situation using unit step functions as: $$x(t)=u(t-a)-u(t-b)$$

This voltage has strength 1 , duration $(b-a)$.

The unit impulse function

The continuous-time unit mpulse function $\delta (t)$

Definition: The unit impulse function, $\delta (t)$, is defined in two parts by P. A. M. Dirac as $$\begin{array}{}\text{(1)}& \delta (t)=0t\ne 0\text{ and }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\delta (t)dt=1\end{array}$$

$\delta (t)=0$ everywhere except at $t=0$, where it is undefined.

We can visualize an impulse as a tall, narrow, rectangular pulse of unit area, as illustrated in Fig. 1.19b.

Figure 1.19

The width of this rectangular pulse is a very small value $ϵ\to 0$. Consequently, its height is a very large value $1/ϵ\to \mathrm{\infty }$. The unit impulse therefore can be regarded as a rectangular pulse with a width that has become infinitesimally small, a height that has become infinitely large, and an overall area that has been maintained at unity.

Since $\delta (t)$ is undefined at $t=0,$ a unit impulse is represented by the spearlike symbol in Fig. 1.19a.

The discrete-time unit impulse function $\delta [n]$

$$\delta [n]=\{\begin{array}{ll}0,& n\ne 0\\ 1,& n=0\end{array}$$

Multiplication of a function by an impulse

Let us now consider what happens when we multiply the unit impulse $\delta (t)$ by a function $\varphi (t)$ that is known to be continuous at $t=0$. Since the impulse has nonzero value only at $t=0$, and the value of $\varphi (t)$ at $t=0$ is $\varphi (0)$, we obtain $$\varphi (t)\delta (t)=\varphi (0)\delta (t)$$

which equals to $\varphi (0)$ .

We can generalize this result. Say that provided $\varphi (t)$ is continuous at $t=T$, $\varphi (t)$ multiplied by an impulse $\delta (t-T)$ (impulse located at $t=T)$ results in: $$\begin{array}{}\text{(2)}& \varphi (t)\delta (t-T)=\varphi (T)\delta (t-T)\end{array}$$

which equals to $(T)$.

Meanwhile, we also have $$\varphi (t)\delta (T-t)=\varphi (T)\delta (T-t)$$ The profe is rather simple. Considering $\varphi (t)\delta (T-t)$, we can use a change of variables to understand this expression. Let $u=T-t$, then $t=T-u$. The delta function $\delta (T-t)$ can be rewritten as $\delta (u)$, since $\delta (x)=$ $\delta (-x)$. Thus, we have: $$\varphi (t)\delta (T-t)=\varphi (T-u)\delta (u)=\varphi (T)\delta (u)=\varphi (T)\delta (T-t).$$

Q.E.D.

Sampling property of the unit impulse function

From $\text{(2)}$ it follows that $$\begin{array}{rl}\text{(3)}& & {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\varphi (t)\delta (t-T)dt\text{(4)}& & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\varphi (T)\delta (t-T)dt& =\varphi (T){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\delta (t)dt\\ & =\varphi (T)\end{array}$$ provided $\varphi (t)$ is continuous at $t=T$.

This property is very important and useful and is known as the sampling or sifting property of the unit impulse.

Unit impulse as a generalized function

The definition of the unit impulse function given in $\text{(1)}$ is not mathematically rigorous. It does not define a unique function, and it is undefined at $t=0$.

These difficulties are resolved by defining the impulse as a generalized function of an impulse, rather than an ordinary function. A generalized function is defined by its effect on other functions instead of by its value at every instant of time.

Because the unit step function $u(t)$ is discontinuous at $t=0$, its derivative $du/dt$ does not exist at $t=0$ in the ordinary sense. We now show that this derivative does exist in the generalized sense, and it is, in fact, $\delta (t)$. As a proof, let us evaluate the integral of $(du/dt)\varphi (t)$, using integration by parts: $$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{du(t)}{dt}\varphi (t)dt& ={u(t)\varphi (t)|}_{-\mathrm{\infty }}^{\mathrm{\infty }}-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}u(t){\varphi }^{\prime }(t)dt\\ & =\varphi (\mathrm{\infty })-0-{\int }_0^{\mathrm{\infty }}{\varphi }^{\prime }(t)dt\\ & =\varphi (\mathrm{\infty })-{\varphi (t)|}_0^{\mathrm{\infty }}=\varphi (0)\end{array}$$

This result shows that $du/dt$ satisfies the sampling property of $\delta (t)$. Therefore, it is an impulse $\delta (t)$ in the generalized sense-that is, $$\frac{du(t)}{dt}=\delta (t).$$

Consequently, # TODO $${\int }_{-\mathrm{\infty }}^t\delta (\tau )d\tau =u(t).$$

We observe that the area from $-\mathrm{\infty }$ to $t$ under the limiting form of $\delta (t)$ in Fig. $1.19\mathrm{b}$ is zero if $t<-ϵ/2$ and unity if $t\ge ϵ/2$ with $ϵ\to 0$. Consequently, $$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^t\delta (\tau )d\tau & =\{\begin{array}{ll}0& t<0\\ 1& t\ge 0\end{array}\\ & =u(t)\end{array}$$