The unit impulse and unit step functions

This is a continuation from the previous tutorial - exponential and sinusoidal signals.

In this tutorial, we introduce several other basic signals-specifically, the unit impulse and step functions in continuous and discrete time- that are also of considerable importance in signal and system analysis.

In later tutorials, we will see how we can use unit impulse signals as basic building blocks for the construction and representation of other signals. We begin with the discrete-time case.

1. The Discrete-Time Unit Impulse and Unit Step Sequences

One of the simplest discrete-time signals is the unit impulse (or unit sample), which is defined as

\[\tag{1.63}\delta[n]=\begin{cases}0,\qquad{n\ne0}\\1,\qquad{n=0}\end{cases}\]

and which is shown in Figure 1.28. We will refer to \(\delta[n]\) interchangeably as the unit impulse or unit sample.

A second basic discrete-time signal is the discrete-time unit step, denoted by \(u[n]\) and defined as

\[\tag{1.64}u[n]=\begin{cases}0,\qquad{n\lt0}\\1,\qquad{n\ge0}\end{cases}\]

The unit step sequence is shown in Figure 1.29.

There is a close relationship between the discrete-time unit impulse nd unit step. In particular, the discrete-time unit impulse is the first difference of the discrete unit step

\[\tag{1.65}\delta[n]=u[n]-u[n-1]\]

Conversely, the discrete-time unit step is the running sum of the unit sample. That is

\[\tag{1.66}u[n]=\sum_{m=-\infty}^n\delta[m]\]

Equation (1.66) is illustrated graphically in Figure 1.30. Since the only nonzero value of the unit sample is at the point at which its argument is zero, we see from the figure that the running sum in eq. (1.66) is \(0\) for \(n\lt0\) and \(1\) for \(n\ge0\).

Furthermore, by changing the variable of summation from \(m\) to \(k=n-m\) in eq. (1.66), we find that the discrete-time unit step can also be written in terms of the unit sample as

\[u[n]=\sum_{k=\infty}^0\delta[n-k]\]

or equivalently,

\[\tag{1.67}u[n]=\sum_{k=0}^\infty\delta[n-k]\]

Equation (1.67) is illustrated in Figure 1.31. In this case the nonzero value of \(\delta[n-k]\) is at the value of \(k\) equal to \(n\), so that again we see that the summation in eq. (1.67) is \(0\) for \(n\lt0\) and \(1\) for \(n\ge0\).

An interpretation of eq. (1.67) is as a superposition of delayed impulses; i.e., we can view the equation as the sum of a unit impulse \(\delta[n]\) at \(n=0\), a unit impulse \(\delta[n-1]\) at \(n=1\), another, \(\delta[n-2]\) at \(n=2\), etc.

The unit impulse sequence can be used to sample the value of a signal at \(n=0\). In particular, since \(\delta[n]\) is nonzero (and equal to 1) only for \(n=0\), it follows that

\[\tag{1.68}x[n]\delta[n]=x[0]\delta[n]\]

More generally, if we consider a unit impulse \(\delta[n-n_0]\) at \(n=n_0\), then

\[\tag{1.69}x[n]\delta[n-n_0]=x[n_0]\delta[n-n_0]\]

This sampling property of the unit impulse will play an important role in later tutorials.

2.  The Continuous-Time Unit Step and Unit Impulse Functions

The continuous-time unit step function \(u(t)\) is defined in a manner similar to its discrete-time counterpart. Specifically,

\[\tag{1.70}u(t)=\begin{cases}0,\qquad{t\lt0}\\1,\qquad{t\gt0}\end{cases}\]

as is shown in Figure 1.32.

Note that the unit step is discontinuous at \(t=0\).

The continuous-time unit impulse function \(\delta(t)\) is related to the unit step in a manner analogous to the relationship between the discrete-time unit impulse and step functions.

In particular, the continuous-time unit step is the running integral of the unit impulse

\[\tag{1.71}u(t)=\displaystyle\int\limits_{-\infty}^t\delta(\tau)\text{d}\tau\]

This also suggests a relationship between \(\delta(t)\) and \(u(t)\) analogous to the expression for \(\delta[n]\) in eq. (1.65).

In particular, it follows from eq. (1.71) that the continuous-time unit impulse can be thought of as the first derivative of the continuous-time unit step:

\[\tag{1.72}\delta(t)=\frac{\text{d}u(t)}{\text{d}t}\]

In contrast to the discrete-time case, there is some formal difficulty with this equation as a representation of the unit impulse function, since \(u(t)\) is discontinuous at \(t=0\) and consequently is formally not differentiable.

We can, however, interpret eq. (1.72) by considering an approximation to the unit step \(u_\Delta(t)\), as illustrated in Figure 1.33, which rises from the value \(0\) to the value \(1\) in a short time interval of length \(\Delta\).

The unit step, of course, changes values instantaneously and thus can be thought of as an idealization of \(u_\Delta(t)\) for \(\Delta\) so short that its duration doesn't matter for any practical purpose. Formally, \(u(t)\) is the limit of \(u_\Delta(t)\) as \(\Delta\rightarrow0\).

Let us now consider the derivative

\[\tag{1.73}\delta_\Delta(t)=\frac{\text{d}u_\Delta(t)}{\text{d}t}\]

as shown in Figure 1.34.

Note that \(\delta_\Delta(t)\) is a short pulse, of duration \(\Delta\) and with unit area for any value of \(\Delta\). As \(\Delta\rightarrow0\), \(\delta_\Delta(t)\) becomes narrower and higher, maintaining its unit area.

Its limiting form,

\[\tag{1.74}\delta(t)=\lim_{\Delta\rightarrow0}\delta_\Delta(t)\]

can then be thought of as an idealization of the short pulse \(\delta_\Delta(t)\) as the duration \(\Delta\) becomes insignificant.

Since \(\delta(t)\) has, in effect, no duration but unit area, we adopt the graphical notation for it shown in Figure 1.35, where the arrow at \(t=0\) indicates that the area of the pulse is concentrated at \(t=0\) and the height of the arrow and the "1" next to the arrow are used to represent the area of the impulse.

More generally, a scaled impulse \(k\delta(t)\) will have an area \(k\), and thus

\[\displaystyle\int\limits_{-\infty}^tk\delta(\tau)\text{d}\tau=ku(t)\]

A scaled impulse with area \(k\) is shown in Figure 1.36, where the height of the arrow used to depict the scaled impulse is chosen to be proportional to the area of the impulse.

As with discrete time, we can provide a simple graphical interpretation of the running integral of eq. (1.71); this is shown in Figure 1.37.

Since the area of the continuous-time unit impulse \(\delta(\tau)\) is concentrated at \(\tau=0\), we see that the running integral is \(0\) for \(t\lt0\) and \(1\) for \(t\gt0\).

Also we note that the relationship in eq. (1.71) between the continuous-time unit step and impulse can be rewritten in a different form, analogous to the discrete-time form in eq. (1.67), by changing the variable of integration from \(\tau\) to \(\sigma=t-\tau\):

\[u(t)=\displaystyle\int\limits_{-\infty}^t\delta(\tau)\text{d}\tau=\int\limits_\infty^0\delta(t-\sigma)(-\text{d}\sigma)\]

or equivalently,

\[\tag{1.75}u(t)=\displaystyle\int\limits_0^\infty\delta(t-\sigma)\text{d}\sigma\]

The graphical interpretation of this form of the relationship between \(u(t)\) and \(\delta(t)\) is given in Figure 1.38. Since in this case the area of \(\delta(t-\sigma)\) is concentrated at the point \(\sigma=t\), we again see that the intergral in eq. (1.75) is \(0\) for \(t\lt0\) and \(1\) for \(t\gt0\).

This type of graphical interpretation of the behavior of the unit impulse under integration will be extremely useful in later tutorials.

As with the discrete-time impulse, the continuous-time impulse has a very important sampling property.

In particular, for a number of reasons it will be important to consider the product of an impulse and more well-behaved continuous-time functions \(x(t)\). The interpretation of this quantity is most readily developed using the definition of \(\delta(t)\) according to eq. (1.74). Specifically, consider

\[x_\text{l}(t)=x(t)\delta_\Delta(t)\]

In Figure 1.39(a) we have depicted the two time functions \(x(t)\) and \(\delta_\Delta(t)\), and in Figure 1.39(b) we see an enlarged view of the nonzero portion of their product.

By construction, \(x_\text{l}(t)\) is zero outside of the interval \(0\le{t}\le\Delta\). For \(\Delta\) sufficiently small so that \(x(t)\) is approximately constant over this interval,

\[x(t)\delta_\Delta(t)\approx{x(0)}\delta_\Delta(t)\]

Since \(\delta(t)\) is the limit as \(\Delta\rightarrow0\) of \(\delta_\Delta(t)\), it follows that

\[\tag{1.76}x(t)\delta(t)={x(0)}\delta(t)\]

By the same argument, we have an analogous expression for an impulse concentrated at an arbitrary point, say \(t_0\). That is

\[x(t)\delta(t-t_0)={x(t_0)}\delta(t-t_0)\]

Although our discussion of the unit impulse in this section has been somewhat informal, it does provide us with some important intuition about this signal that will be useful throughout our tutorials.

As we have stated, the unit impulse should be viewed as an idealization. As we illustrate and discuss in more detail in a later tutorial, any real physical system has some inertia associated with it and thus does not respond instantaneously to inputs.

Consequently, if a pulse of sufficiently short duration is applied to such a system, the system response will not be noticeably influenced by the pulse's duration or by the details of the shape of the pulse, for that matter. Instead, the primary characteristic of the pulse that will matter is the net, integrated effect of the pulse - i.e., its area.

For systems that respond much more quickly than others, the pulse will have to be of much shorter duration before the details of the pulse shape or its duration no longer matter.

Nevertheless, for any physical system, we can always find a pulse that is "short enough." The unit impulse then is an idealization of this concept - the pulse that is short enough for any system.

As we will see in later tutorials, the response of a system to this idealized pulse plays a crucial role in signal and system analysis, and in the process of developing and understanding this role, we will develop additional insight into the idealized signal.

Example 1.7

Consider the discontinuous signal \(x(t)\) depicted in Figure 1.40(a).

Because of the relationship between the continuous-time unit impulse and unit step, we can readily calculate and graph the derivative of this signal. Specifically, the derivative of \(x(t)\) is clearly 0, except at the discontinuities.

In the case of the unit step, we have seen [eq. (1.72)] that differentiation gives rise to a unit impulse located at the point of discontinuity. Furthermore, by multiplying both sides of eq. (1.72) by any number \(k\), we see that the derivative of a unit step with a discontinuity of size \(k\) gives rise to an impulse of area \(k\) at the point of discontinuity.

This rule also holds for any other signal with a jump discontinuity, such as \(x(t)\) in Figure 1.40(a). Consequently, we can sketch its derivative \(\dot{x}(t)\), as in Figure 1.40(b), where an impulse is placed at each discontinuity of \(x(t)\), with area equal to the size of the discontinuity.

Note, for example, that the discontinuity in \(x(t)\) at \(t=2\) has a value of \(-3\), so that an impulse scaled by \(-3\) is located at \(t=2\) in the signal \(\dot{x}(t)\).

As a check of our result, we can verify that we can recover \(x(t)\) from \(\dot{x}(t)\). Specifically, since \(x(t)\) and \(\dot{x}(t)\) are both zero for \(t\le0\), we need only check that for \(t\gt0\),

\[\tag{1.77}x(t)=\displaystyle\int\limits_0^t\dot{x}(\tau)\text{d}\tau\]

As illustrated in Figure 1.40(c), for \(t\lt1\), the integral on the right-hand side of eq. (1.77) is zero, since none of the impulses that constitute \(\dot{x}(t)\) are within the interval of integration.

For \(1\lt{t}\lt2\), the first impulse (located at \(t=1\)) is the only one within the integration interval, and thus the integral in eq. (1.77) equals 2, the area of this impulse.

For \(2\lt{t}\lt4\), the first two impulses are within the interval of integration, and the integral accumulates the sum of both of their areas, namely, \(2-3=-1\).

Finally, for \(t\gt4\), all three impulses are within the integration interval, so that the integral equals the sum of all three areas - that is, \(2-3+2=+1\). The result is exactly the signal \(x(t)\) depicted in Figure 1.40(a).

The next tutorial introduces continuous-time and discrete-time systems and their properties.