7.4 Singularity Functions

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7.4 Singularity Functions

Before going on with the second half of this chapter, we need to digress and consider some mathematical concepts that will aid our understand ing of transient analysis. A basic understanding of singularity functions will help us make sense of the response of first-order circuits to a sudden application of an independent dc voltage or current source.

Singularity functions (also called switching functions) are very useful in circuit analysis. They serve as good approximations to the switching signals that arise in circuits with switching operations. They are helpful in the neat, compact description of some circuit phenomena, especially the step response of RC or RL circuits to be discussed in the next sections. By definition,

Singularity functions are functions that either are discontinuous or have discontinuous derivatives.

Figure 7.21 A plot of i and io.

Figure 7.22 For Practice Prob. 7.5.

The three most widely used singularity functions in circuit analysis are the unit step, the unit impulse, and the unit ramp functions.

The unit step function ^u(t) is 0 for negative values of t and 1 for positive values of t.

In mathematical terms,

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

(7.24)

The unit step function is undefined at t = 0, where it changes abrupt ly from 0 to 1. It is dimensionless, lik e other mathematical functions such as sine and cosine. Figure 7.23 depicts the unit step function. If the abrupt change occurs at t = t0 (where t0 > 0) instead of t = 0, the unit step function becomes

$$u(t - t_0) = \begin{cases} 0, & t < t_0 \\ 1, & t > t_0 \end{cases}$$

(7.25)

which is the same as saying that u(t) is delayed by t0 seconds, as shown in Fig. 7.24(a). To get Eq. (7.25) from Eq. (7.24), we simply replace every t by t − t0. If the change is at t = −t0, the unit step function becomes

$$u(t+t_0) = \begin{cases} 0, & t < -t_0 \\ 1, & t > -t_0 \end{cases} \tag{7.26}$$

meaning that u(t) is advanced by t0 seconds, as shown in Fig. 7.24(b).

We use the step function to represent an abrupt change in voltage or current, like the changes that occur in the circuits of control systems and digital computers. For example, the voltage

$$v(t) = \begin{cases} 0, & t < t_0 \\ V_0, & t > t_0 \end{cases} \tag{7.27}$$

may be expressed in terms of the unit step function as

$$v(t) = V_0 u(t - t_0)$$

\n(7.28)

(b)

t = 0

b

a

If we let t0 = 0, then v(t) is simply the step v oltage V0 u(t). A voltage source of V0 u(t) is shown in Fig. 7.25(a); its equivalent circuit is shown in Fig. 7.25(b). It is e vident in Fig. 7.25(b) that terminals a-b are shortcircuited (v = 0) for t < 0 and that v = V0 appears at the terminals for

V0

= ⁺

Figure 7.25 (a) Voltage source of V0 u(t), (b) its equivalent circuit.

‒ ⁺

b

a

(a)

‒

V0u(t)

Alternatively, we may derive Eqs. (7.25) and (7.26) from Eq. (7.24) by writing u [f (t)] = 1, f (t) > 0, where ^f(t) may be t − ^t0 or t + t0.

Figure 7.24

The unit step function.

(a) The unit step function delayed by t0, (b) the unit step advanced by t0.

t > 0. Similarly, a current source of I0 u(t) is shown in Fig. 7.26(a), while its equivalent circuit is in Fig. 7.26(b). Notice that for t < 0, there is an open circuit (i = 0), and that i = I0 flows for t > 0.

Figure 7.26

(a) Current source of I0u(t), (b) its equivalent circuit.

The derivative of the unit step function u(t) is the unit impulse function δ(t), which we write as

$$\delta(t) = \frac{d}{dt}u(t) = \begin{cases} 0, & t < 0 \\ \text{Undefined}, & t = 0 \\ 0, & t > 0 \end{cases} \tag{7.29}$$

The unit impulse function—also known as the delta function—is shown in Fig. 7.27.

Figure 7.27 0 t δ(t) (∞)

The unit impulse function.

The unit impulse function δ(t) is zero everywhere except at t = 0, where it is undefined.

Impulsive currents and v oltages occur in electric circuits as a result of switching operations or impulsi ve sources. Although the unit im pulse function is not ph ysically realizable ( just like ideal sources, ideal resistors, etc.), it is a very useful mathematical tool.

The unit impulse may be regarded as an applied or resulting shock. It may be visualized as a very short duration pulse of unit area. This may be expressed mathematically as

$$\int_{0^{-}}^{0^{+}} \delta(t) dt = 1 \tag{7.30}$$

where t = 0⁻ denotes the time just before t = 0 and t = 0 ⁺ is the time just after t = 0. For this reason, it is customary to write 1 (denoting unit area) beside the arrow that is used to symbolize the unit impulse function, as in Fig. 7.27. The unit area is known as the strength of the impulse function. When an impulse function has a strength other than unity , the area of the impulse is equal to its strength. F or example, an impulse func tion 10δ(t) has an area of 10. Figure 7.28 sho ws the impulse functions 5δ(t + 2),10δ(t), and −4δ(t − 3).

To illustrate how the impulse function affects other functions, let us evaluate the integral

$$\int_{a}^{b} f(t)\delta(t - t_0)dt$$

\n(7.31)

Three impulse functions.

where a < t0 < b. Since δ(t − t0) = 0 except at t = t0, the integrand is zero except at t0. Thus,

$$\int_{a}^{b} f(t)\delta(t - t_0) dt = \int_{a}^{b} f(t_0)\delta(t - t_0) dt$$ $$= f(t_0) \int_{a}^{b} \delta(t - t_0) dt = f(t_0)$$

or

$$\int_{a}^{b} f(t)\delta(t - t_0) dt = f(t_0)$$

\n(7.32)

This shows that when a function is integrated with the impulse function, we obtain the value of the function at the point where the impulse occurs. This is a highly useful property of the impulse function kno wn as the sampling or sifting property. The special case of Eq. (7.31) is for t0 = 0. Then Eq. (7.32) becomes

$$\int_{0^{-}}^{0^{+}} f(t) \delta(t) dt = f(0)$$

\n(7.33)

Integrating the unit step function u(t) results in the unit ramp function r(t); we write

$$r(t) = \int_{-\infty}^{t} u(\lambda)d\lambda = tu(t)$$

(7.34)

$$\overline{\text{or}}$$ $$r(t) = \begin{cases} 0, & t \le 0 \\ t, & t \ge 0 \end{cases} \tag{7.35}$$

The unit ramp function is zero for negative values of t and has a unit slope for positive values of t.

Figure 7.29 shows the unit ramp function. In general, a ramp is a func tion that changes at a constant rate.

The unit ramp function may be delayed or adv anced as sho wn in Fig. 7.30. For the delayed unit ramp function,

$$r(t - t_0) = \begin{cases} 0, & t \le t_0 \\ t - t_0, & t \ge t_0 \end{cases}$$

(7.36)

and for the advanced unit ramp function,

$$r(t + t_0) = \begin{cases} 0, & t \le -t_0 \\ t + t_0, & t \ge -t_0 \end{cases}$$

(7.37)

We should keep in mind that the three singularity functions (impulse, step, and ramp) are related by differentiation as

$$\delta(t) = \frac{du(t)}{dt}, \qquad u(t) = \frac{dr(t)}{dt}$$

(7.38)

The unit ramp function: (a) delayed by t0, (b) advanced by t0.

1

r(t ‒ t0)

or by integration as

$$u(t) = \int_{-\infty}^{t} \delta(\lambda) d\lambda, \quad r(t) = \int_{-\infty}^{t} u(\lambda) d\lambda$$

(7.39)

Although there are man y more singularity functions, we are only inter ested in these three (the impulse function, the unit step function, and the ramp function) at this point.

Express the voltage pulse in Fig. 7.31 in terms of the unit step. Calculate Example 7.6 its derivative and sketch it.

Solution:

The type of pulse in Fig. 7.31 is called the gate function. It may be regarded as a step function that switches on at one value of t and switches off at another value of t. The gate function shown in Fig. 7.31 switches on at t = 2 s and switches off at t = 5 s. It consists of the sum of two unit step functions as shown in Fig. 7.32(a). From the figure, it is evident that

$$v(t) = 10u(t - 2) - 10u(t - 5) = 10[u(t - 2) - u(t - 5)]$$

Taking the derivative of this gives

$$\frac{dv}{dt} = 10[\delta(t-2) - \delta(t-5)]$$

which is shown in Fig. 7.32(b). We can obtain Fig. 7.32(b) directly from Fig. 7.31 by simply observing that there is a sudden increase by 10 V at t = 2 s leading to 10 δ(t − 2). At t = 5 s, there is a sudden decrease by 10 V leading to −10V δ(t − 5).

Gate functions are used along with switches to pass or block another signal.

For Example 7.6.

Practice Problem 7.6

Express the current pulse in Fig. 7.33 in terms of the unit step. Find its integral and sketch it.

Answer: 10[u (t) − 2u (t − 2) + u (t − 4)] A, 10[r (t) − 2r (t − 2) + r(t − 4)] amp-sec. See Fig. 7.34.

Example 7.7 Express the sawtooth function shown in Fig. 7.35 in terms of singularity functions.

Solution:

There are three ways of solving this problem. The first method is by mere observation of the given function, while the other methods involve some graphical manipulations of the function.

■ METHOD 1 By looking at the sketch of v(t) in Fig. 7.35, it is not hard to notice that the given function v(t) is a combination of singularity functions. So we let

$$v(t) = v_1(t) + v_2(t) + \cdots$$

(7.7.1)

The function v1(t) is the ramp function of slope 5, shown in Fig. 7.36(a); that is,

$$v_1(t) = 5r(t) \tag{7.7.2}$$

Partial decomposition of v(t) in Fig. 7.35.

Since v1(t) goes to infinity, we need another function at t = 2s in order to get v(t). We let this function be v2, which is a ramp function of slope −5, as shown in Fig. 7.36(b); that is,

$$v_2(t) = -5r(t-2)$$

\n(7.7.3)

Adding v1 and v2 gives us the signal in Fig. 7.36(c). Obviously, this is not the same as v(t) in Fig. 7.35. But the difference is simply a constant 10 units for t > 2 s. By adding a third signal v3, where

$$v_3 = -10u(t - 2) \tag{7.7.4}$$

we get v(t), as shown in Fig. 7.37. Substituting Eqs. (7.7.2) through (7.7.4) into Eq. (7.7.1) gives

Complete decomposition of v(t) in Fig. 7.35.

■ METHOD 2 A close observation of Fig. 7.35 reveals that v(t) is a multiplication of two functions: a ramp function and a gate function. Thus,

$$v(t) = 5t[u(t) - u(t-2)]$$

= 5tu(t) - 5tu(t - 2)
= 5r(t) - 5(t - 2 + 2)u(t - 2)
= 5r(t) - 5(t - 2)u(t - 2) - 10u(t - 2)
= 5r(t) - 5r(t - 2) - 10u(t - 2)

the same as before.

■ METHOD 3 This method is similar to Method 2. We observe from Fig. 7.35 that v(t) is a multiplication of a ramp function and a unit step function, as shown in Fig. 7.38. Thus,

$$v(t) = 5r(t)u(-t+2)$$

If we replace u(−t) by 1 −u(t), then we can replace u(−t + 2) by 1 −u(t − 2). Hence,

$$v(t) = 5r(t)[1 - u(t - 2)]$$

which can be simplified as in Method 2 to get the same result.

Decomposition of v(t) in Fig. 7.35.

For Practice Prob. 7.7.

Example 7.8 Given the signal

	3,	t < 0
g(t) =	−2,	0 < t < 1
	{ 2t − 4,	t > 1

express g(t) in terms of step and ramp functions.

Solution:

The signal g(t) may be regarded as the sum of three functions specified within the three intervals t < 0, 0 < t < 1, and t > 1.

For t < 0, g(t) may be regarded as 3 multiplied by u(−t), where u(−t) = 1 for t < 0 and 0 for t > 0. Within the time interval 0 < t < 1, the function may be considered as −2 multiplied by a gated function [u(t) − u(t − 1)]. For t > 1, the function may be regarded as 2t − 4 multiplied by the unit step function u(t − 1). Thus,

$$g(t) = 3u(-t) - 2[u(t) - u(t-1)] + (2t - 4)u(t-1)$$

= 3u(-t) - 2u(t) + (2t - 4 + 2)u(t-1)
= 3u(-t) - 2u(t) + 2(t-1)u(t-1)
= 3u(-t) - 2u(t) + 2r(t-1)

One may avoid the trouble of using u(−t) by replacing it with 1 − u(t). Then

$$g(t) = 3[1 - u(t)] - 2u(t) + 2r(t - 1) = 3 - 5u(t) + 2r(t - 1)$$

Alternatively, we may plot g(t) and apply Method 1 from Example 7.7.

If

$$h(t) = \begin{cases} 0, & t < 0 \\ -4, & 0 < t < 2 \\ 3t - 8, & 2 < t < 6 \\ 0, & t > 6 \end{cases}$$

Practice Problem 7.8

express h(t) in terms of the singularity functions.

Answer:

$$-4u(t) + 2u(t-2) + 3r(t-2) - 10u(t-6) - 3r(t-6)$$

.

Evaluate the following integrals involving the impulse function: Example 7.9

$$\int_0^{10} (t^2 + 4t - 2)\delta(t - 2) dt$$ $$\int_{-\infty}^{\infty} [\delta(t - 1)e^{-t} \cos t + \delta(t + 1)e^{-t} \sin t] dt$$

Solution:

For the first integral, we apply the sifting property in Eq. (7.32).

$$\int_0^{10} (t^2 + 4t - 2) \delta(t - 2) dt = (t^2 + 4t - 2)|_{t=2} = 4 + 8 - 2 = 10$$

Similarly, for the second integral,

$$\int_{-\infty}^{\infty} \left[ \delta(t-1)e^{-t} \cos t + \delta(t+1)e^{-t} \sin t \right] dt$$

= $e^{-t} \cos t|_{t=1} + e^{-t} \sin t|_{t=-1}$
= $e^{-t} \cos 1 + e^{t} \sin (-1) = 0.1988 - 2.2873 = -2.0885$

Evaluate the following integrals:

$$\int_{-\infty}^{\infty} (t^3 + 5t^2 + 10)\delta(t+3) \, dt, \qquad \int_{0}^{10} \delta(t-\pi) \cos 3t \, dt$$

Answer: 28, −1.