(a)

← Back to LINEAR SYSTEMS AND SIGNALS Overview (a)

$$\mathcal{L}[\delta(t)] = \int_{0^-}^{\infty} \delta(t) e^{-st} dt$$

Using the sampling property [Eq. (1.11) with T = 0], we obtain

$$\mathcal{L}[\delta(t)] = 1 \qquad \text{for all } s$$

that is,

$$\delta(t) \Longleftrightarrow 1 \qquad \text{for all } s$$

^† The condition of Eq. (4.9) is sufficient but not necessary for the existence of the Laplace transform. For example, x(t) = 1/ ^√t is infinite at ^{t =} 0, and Eq. (4.9) cannot be satisfied; but the transform of 1/ ^√t exists and is given by ^√π/s.

^‡ However, if we consider a truncated (finite-duration) signal et 2 , the Laplace transform exists.

(b) To find the Laplace transform of u(t), recall that u(t) = 1 for t ≥ 0. Therefore,

$$\mathcal{L}[u(t)] = \int_{0^{-}}^{\infty} u(t)e^{-st} dt = \int_{0^{-}}^{\infty} e^{-st} dt = -\frac{1}{s}e^{-st} \Big|_{0^{-}}^{\infty}$$

= $\frac{1}{s}$ Re $s > 0$

We also could have obtained this result from Eq. (4.6) by letting a = 0. (c) Because cos ^ω0t u(t) ^{= 1 2} [ej^ω0*^t* +e−jω0*^t* ]u(t), we know that

$$\mathcal{L}[\cos \omega_0 t u(t)] = \frac{1}{2} \mathcal{L}[e^{j\omega_0 t} u(t) + e^{-j\omega_0 t} u(t)]$$

From Eq. (4.6), it follows that

$$\mathcal{L}[\cos \omega_0 t u(t)] = \frac{1}{2} \left[ \frac{1}{s - j\omega_0} + \frac{1}{s + j\omega_0} \right] \qquad \text{Re}\,(s \pm j\omega) = \text{Re}\,s > 0$$

\n

$$= \frac{s}{s^2 + \omega_0^2} \qquad \text{Re}\,s > 0 \tag{4.10}$$

For the unilateral Laplace transform, there is a unique inverse transform of X(s); consequently, there is no need to specify the ROC explicitly. For this reason, we shall generally ignore any mention of the ROC for unilateral transforms. Recall, also, that in the unilateral Laplace transform it is understood that every signal x(t) is zero for t < 0, and it is appropriate to indicate this fact by multiplying the signal by u(t).

DR ILL 4.1 Bilateral Laplace Transform of Gate Functions

By direct integration, find the Laplace transform X(s) and the region of convergence of X(s) for the gate functions shown in Fig. 4.2.

Figure 4.2 Gate functions for Drill 4.1.

ANSWERS

(a)

$$\frac{1}{s}(1-e^{-2s})$$

for all s
\n(b) $\frac{1}{s}(1-e^{-2s})e^{-2s}$ for all s

4.1-1 Finding the Inverse Transform

Finding the inverse Laplace transform by using Eq. (4.2) requires integration in the complex plane, a subject beyond the scope of this book (but see, e.g., [3]). For our purpose, we can find the inverse transforms from Table 4.1. All we need is to express X(s) as a sum of simpler functions of the forms listed in the table. Most of the transforms X(s) of practical interest are rational functions, that is, ratios of polynomials in s. Such functions can be expressed as a sum of simpler functions by using partial fraction expansion (see Sec. B.5).

Values of s for which X(s) = 0 are called the zeros of X(s); the values of s for which X(s)→∞ are called the poles of X(s). If X(s) is a rational function of the form P(s)/Q(s), the roots of P(s) are the zeros and the roots of Q(s) are the poles of X(s).

EXAMPLE 4.3 Inverse Unilateral Laplace Transform

Find the inverse unilateral Laplace transforms of

(a)

$$\frac{7s-6}{s^2-s-6}$$

\n(b)

$$\frac{2s^2+5}{s^2+3s+2}$$

\n(c)

$$\frac{6(s+34)}{s(s^2+10s+34)}$$

\n(d)

$$\frac{8s+10}{(s+1)(s+2)^3}$$

In no case is the inverse transform of these functions directly available in Table 4.1. Rather, we need to expand these functions into partial fractions, as discussed in Sec. B.5-1. Today, it is very easy to find partial fractions via software such as MATLAB. However, just as the availability of a calculator does not obviate the need for learning the mechanics of arithmetical operations (addition, multiplication, etc.), the widespread availability of computers does not eliminate the need to learn the mechanics of partial fraction expansion.

$$\left( \mathbf{a}\right)$$ $$X(s) = \frac{7s - 6}{(s + 2)(s - 3)} = \frac{k_1}{s + 2} + \frac{k_2}{s - 3}$$

To determine k1, corresponding to the term (s + 2), we cover up (conceal) the term (s + 2) in X(s) and substitute s = −2 (the value of s that makes s + 2 = 0) in the remaining expression (see Sec. B.5-2):

$$k_1 = \frac{7s - 6}{(s + 2)(s - 3)}\bigg|_{s = -2} = \frac{-14 - 6}{-2 - 3} = 4$$

Similarly, to determine k² corresponding to the term (s − 3), we cover up the term (s − 3) in X(s) and substitute s = 3 in the remaining expression

$$k_2 = \frac{7s - 6}{(s + 2)(s - 3)}\bigg|_{s=3} = \frac{21 - 6}{3 + 2} = 3$$

Therefore,

$$X(s) = \frac{7s - 6}{(s + 2)(s - 3)} = \frac{4}{s + 2} + \frac{3}{s - 3}$$

(4.11)

CHECKING THE ANSWER

It is easy to make a mistake in partial fraction computations. Fortunately it is simple to check the answer by recognizing that X(s) and its partial fractions must be equal for every value of s if the partial fractions are correct. Let us verify this assertion in Eq. (4.11) for some convenient value, say, s = 0. Substitution of s = 0 in Eq. (4.11) yields†

$$1 = 2 - 1 = 1$$

We can now be sure of our answer with a high margin of confidence. Using pair 5 of Table 4.1 in Eq. (4.11), we obtain

$$x(t) = \mathcal{L}^{-1}\left(\frac{4}{s+2} + \frac{3}{s-3}\right) = (4e^{-2t} + 3e^{3t})u(t)$$

(b)

$$X(s) = \frac{2s^2 + 5}{s^2 + 3s + 2} = \frac{2s^2 + 5}{(s+1)(s+2)}$$

Observe that X(s) is an improper function with M = N. In such a case, we can express X(s) as a sum of the coefficient of the highest power in the numerator plus partial fractions corresponding to the poles of X(s) (see Sec. B.5-5). In the present case, the coefficient of the highest power in the numerator is 2. Therefore,

$$X(s) = 2 + \frac{k_1}{s+1} + \frac{k_2}{s+2}$$

where

$$k_1 = \frac{2s^2 + 5}{(s+1)(s+2)}\bigg|_{s=-1} = \frac{2+5}{-1+2} = 7$$

and

$$k_2 = \frac{2s^2 + 5}{(s+1)(s+2)}\bigg|_{s=-2} = \frac{8+5}{-2+1} = -13$$

Therefore,

$$X(s) = 2 + \frac{7}{s+1} - \frac{13}{s+2}$$

From Table 4.1, pairs 1 and 5, we obtain

$$x(t) = 2\delta(t) + (7e^{-t} - 13e^{-2t})u(t)$$

^† Because ^X(s) = ∞ at its poles, we should avoid the pole values (−2 and 3 in the present case) for checking. The answers may check even if partial fractions are wrong. This situation can occur when two or more errors cancel their effects. But the chances of this problem arising for randomly selected values of s are extremely small.

(c)

$$X(s) = \frac{6(s+34)}{s(s^2+10s+34)} = \frac{6(s+34)}{s(s+5-j3)(s+5+j3)}$$ $$= \frac{k_1}{s} + \frac{k_2}{s+5-j3} + \frac{k_2^*}{s+5+j3}$$

Note that the coefficients (k² and k^{∗ 2}) of the conjugate terms must also be conjugate (see Sec. B.5). Now

$$k_1 = \frac{6(s+34)}{s(s^2+10s+34)}\bigg|_{s=0} = \frac{6 \times 34}{34} = 6$$ $$k_2 = \frac{6(s+34)}{s(s+5-j3)(s+5+j3)}\bigg|_{s=-5+j3} = \frac{29+j3}{-3-j5} = -3+j4$$

Therefore,

$$k_2^* = -3 - j4$$

To use pair 10b of Table 4.1, we need to express k² and k^{∗ 2} in polar form.

$$-3 + j4 = (\sqrt{3^2 + 4^2}) e^{j \tan^{-1}(4/4)} = 5 e^{j \tan^{-1}(4/4)}$$

Observe that tan−¹(4/−3) = tan−¹(−4/3). This fact is evident in Fig. 4.3. For further discussion of this topic, see Ex. B.1.

Figure 4.3 Visualizing tan−1(−4/3) = tan−1(4/−3).

From Fig. 4.3, we observe that

$$k_2 = -3 + j4 = 5e^{j126.9^{\circ}}$$

so

$$k_2^* = 5e^{-j126.9^\circ}$$

Therefore,

$$X(s) = \frac{6}{s} + \frac{5e^{j126.9^{\circ}}}{s+5-j3} + \frac{5e^{-j126.9^{\circ}}}{s+5+j3}$$

From Table 4.1 (pairs 2 and 10b), we obtain

$$x(t) = [6 + 10e^{-5t}\cos(3t + 126.9^\circ)]u(t)$$

ALTERNATIVE METHOD USING QUADRATIC FACTORS

The foregoing procedure involves considerable manipulation of complex numbers. Pair 10c (Table 4.1) indicates that the inverse transform of quadratic terms (with complex conjugate poles) can be found directly without having to find first-order partial fractions. We discussed such a procedure in Sec. B.5-2. For this purpose, we shall express X(s) as

$$X(s) = \frac{6(s+34)}{s(s^2+10s+34)} = \frac{k_1}{s} + \frac{As+B}{s^2+10s+34}$$

We have already determined that k¹ = 6 by the (Heaviside) “cover-up” method. Therefore,

$$\frac{6(s+34)}{s(s^2+10s+34)} = \frac{6}{s} + \frac{As+B}{s^2+10s+34}$$

Clearing the fractions by multiplying both sides by s(s² +10s+34) yields

$$6(s+34) = (6+A)s2 + (60+B)s + 204$$

Now, equating the coefficients of s² and s on both sides yields

$$A = -6 \qquad \text{and} \qquad B = -54$$

and

$$X(s) = \frac{6}{s} + \frac{-6s - 54}{s^2 + 10s + 34}$$

We now use pairs 2 and 10c to find the inverse Laplace transform. The parameters for pair 10c are ^A = −6, ^B = −54, ^{a =} 5, ^{c =} 34, ^{b = √} c−a² = 3, and

$$r = \sqrt{\frac{A^2c + B^2 - 2ABA}{c - a^2}} = 10 \qquad \theta = \tan^{-1} \frac{Aa - B}{A\sqrt{c - a^2}} = 126.9^{\circ}$$

Therefore,

$$x(t) = [6 + 10e^{-5t}\cos(3t + 126.9^\circ)]u(t)$$

which agrees with the earlier result.

SHORTCUTS

The partial fractions with quadratic terms also can be obtained by using shortcuts. We have

$$X(s) = \frac{6(s+34)}{s(s^2+10s+34)} = \frac{6}{s} + \frac{As+B}{s^2+10s+34}$$

342 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

We can determine A by eliminating B on the right-hand side. This step can be accomplished by multiplying both sides of the equation for X(s) by s and then letting s→ ∞. This procedure yields

$$0 = 6 + A \quad \Longrightarrow \quad A = -6$$

Therefore,

$$\frac{6(s+34)}{s(s^2+10s+34)} = \frac{6}{s} + \frac{-6s+16}{s^2+10s+34}$$

To find B, we let s take on any convenient value, say, s = 1, in this equation to obtain

$$\frac{210}{45} = 6 + \frac{B - 6}{45} \implies B = -54$$

a result that agrees with the answer found earlier.

(d)

$$X(s) = \frac{8s+10}{(s+1)(s+2)^3} = \frac{k_1}{s+1} + \frac{a_0}{(s+2)^3} + \frac{a_1}{(s+2)^2} + \frac{a_2}{s+2}$$

where

$$k_1 = \frac{8s + 10}{(s+1)(s+2)^3}\Big|_{s=-1} = 2$$

\n

$$a_0 = \frac{8s + 10}{(s+1)(s+2)^3}\Big|_{s=-2} = 6$$

\n

$$a_1 = \left\{\frac{d}{ds}\left[\frac{8s + 10}{(s+1)(s+2)^3}\right]\right\}_{s=-2} = -2$$

\n

$$a_2 = \frac{1}{2}\left\{\frac{d^2}{ds^2}\left[\frac{8s + 10}{(s+1)(s+2)^3}\right]\right\}_{s=-2} = -2$$

Therefore,

$$X(s) = \frac{2}{s+1} + \frac{6}{(s+2)^3} - \frac{2}{(s+2)^2} - \frac{2}{s+2}$$

and

$$x(t) = [2e^{-t} + (3t^2 - 2t - 2)e^{-2t}]u(t)$$

ALTERNATIVE METHOD:AHYBRID OF HEAVISIDE AND CLEARING FRACTIONS

In this method, the simpler coefficients k¹ and a⁰ are determined by the Heaviside “cover-up” procedure, as discussed earlier. To determine the remaining coefficients, we use the clearing-fraction method. Using the values k¹ = 2 and a⁰ = 6 obtained earlier by the Heaviside “cover-up” method, we have

$$\frac{8s+10}{(s+1)(s+2)^3} = \frac{2}{s+1} + \frac{6}{(s+2)^3} + \frac{a_1}{(s+2)^2} + \frac{a_2}{s+2}$$

We now clear fractions by multiplying both sides of the equation by (s + 1)(s + 2)3. This procedure yields†

$$8s + 10 = 2(s + 2)3 + 6(s + 1) + a1(s + 1)(s + 2) + a2(s + 1)(s + 2)2$$

= (2 + a₂)s³ + (12 + a₁ + 5a₂)s² + (30 + 3a₁ + 8a₂)s + (22 + 2a₁ + 4a₂)

Equating coefficients of s³ and s² on both sides, we obtain

$$0 = (2 + a_2) \implies a_2 = -2 0 = 12 + a_1 + 5a_2 = 2 + a_1 \implies a_1 = -2$$

We can stop here if we wish, since the two desired coefficients a¹ and a² have already been found. However, equating the coefficients of s¹ and s⁰ serves as a check on our answers. This step yields

$$8 = 30 + 3a_1 + 8a_2$$ $$10 = 22 + 2a_1 + 4a_2$$

Substitution of a¹ = a² = −2, obtained earlier, satisfies these equations. This step confirms the correctness of our answers.

ANOTHER ALTERNATIVE:AHYBRID OF HEAVISIDE AND SHORTCUTS

In this method, the simpler coefficients k¹ and a⁰ are determined by the Heaviside “cover-up” procedure, as discussed earlier. The usual shortcuts are then used to determine the remaining coefficients. Using the values k¹ = 2 and a⁰ = 6, determined earlier by the Heaviside method, we have

$$\frac{8s+10}{(s+1)(s+2)^3} = \frac{2}{s+1} + \frac{6}{(s+2)^3} + \frac{a_1}{(s+2)^2} + \frac{a_2}{s+2}$$

There are two unknowns, a¹ and a2. If we multiply both sides by s and then let s → ∞, we eliminate a1. This procedure yields

0 = 2+a² ⇒ a² = −2

Therefore,

$$\frac{8s+10}{(s+1)(s+2)^3} = \frac{2}{s+1} + \frac{6}{(s+2)^3} + \frac{a_1}{(s+2)^2} - \frac{2}{s+2}$$

There is now only one unknown, a1. This value can be determined readily by setting s equal to any convenient value, say, s = 0. This step yields

$$\frac{10}{8} = 2 + \frac{3}{4} + \frac{a_1}{4} - 1 \implies a_1 = -2$$

^† We could have cleared fractions without finding k¹ and a0. This alternative, however, proves more laborious because it increases the number of unknowns to 4. By predetermining k¹ and a0, we reduce the unknowns to 2. Moreover, this method provides a convenient check on the solution. This hybrid procedure achieves the best of both methods.

EXAMPLE 4.4 Inverse Laplace Transform with MATLAB

Using the MATLAB residue command, determine the inverse Laplace transform of each of the following functions:

(a)

$$X_a(s) = \frac{2s^2 + 5}{s^2 + 3s + 2}$$

\n(b) $X_b(s) = \frac{2s^2 + 7s + 4}{(s+1)(s+2)^2}$
\n(c) $X_c(s) = \frac{8s^2 + 21s + 19}{(s+2)(s^2 + s + 7)}$

In each case, we use the MATLAB residue command to perform the necessary partial fraction expansions. The inverse Laplace transform follows using Table 4.1.

(a)
>> num = [2 0 5]; den = [1 3 2];
>> [r, p, k] = residue(num,den)
    r = -13
          7
    p = -2

Therefore, Xa(s) = −13/(s+2) +7/(s+1) +2 and xa(t) = (−13e−2*^t* +7e−^t )u(t)+2δ(t).

(b)

-1 k= 2

>> num = [2 7 4]; den = [conv([1 1],conv([1 2],[1 2]))];
>> [r, p, k] = residue(num,den)
    r= 3
         2
        -1
    p = -2
        -2
        -1
    k = []

Therefore, Xb(s) = 3/(s+2) +2/(s+2)² −1/(s+1) and xb(t) = (3e−2*^t* +2te−2*^t* −e−^t )u(t).

(c) In this case, a few calculations are needed beyond the results of the residue command so that pair 10b of Table 4.1 can be utilized.

num = [8 21 19]; den = [conv([1 2],[1 1 7])]; >> [r, p, k]= residue(num,den)

r = 3.5000-0.48113i
         3.5000+0.48113i
         1.0000
    p = -0.5000+2.5981i
        -0.5000-2.5981i
        -2.0000
    k = []
>> ang = angle(r), mag = abs(r)
    ang = -0.13661
           0.13661
           0
    mag = 3.5329
           3.5329
           1.0000

Thus,

$$X_c(s) = \frac{1}{s+2} + \frac{3.5329e^{-j0.13661}}{s+0.5-j2.5981} + \frac{3.5329e^{j0.13661}}{s+0.5+j2.5981}$$

and

$$x_c(t) = [e^{-2t} + 1.7665e^{-0.5t}\cos(2.5981t - 0.1366)]u(t).$$

EXAMPLE 4.5 Symbolic Laplace and Inverse Laplace Transforms with MATLAB

Using MATLAB’s symbolic math toolbox, determine the following:

• (a) the direct unilateral Laplace transform of xa(t) = sin(at)+cos(bt)
• (b) the inverse unilateral Laplace transform of Xb(s) = as²/(s² +b²)

(a) Here, we use the sym command to symbolically define our variables and expression for xa(t), and then we use the laplace command to compute the (unilateral) Laplace transform.

syms a b t; x_a = sin(a*t)+cos(b*t); >> X_a = laplace(x_a); X_a = a/(a^2 + s^2) + s/(b^2 + s^2)

Therefore, Xa(s) ⁼ *^{a s}*2+a^{2 +} *^{s s}*2+b² . It is also easy to use MATLAB to determine Xa(s) in standard rational form.

X_a = collect(X_a) X_a = (a^2*s + a*b^2 + a*s^2 + s^3)/(s^4 + (a^2 + b^2)*s^2 + a^2*b^2)

Thus, we also see that Xa(s) ^{= s}3+as2+a2s+ab^{2 s}4+(a2+b2)s2+a2b²

(b) A similar approach is taken for the inverse Laplace transform, except that the ilaplace command is used rather than the laplace command.

>> syms a b s; X_b = (a*s^2)/(s^2+b^2);
>> x_b = ilaplace(X_b)
    x_b = a*dirac(t) - a*b*sin(b*t)

Therefore, xb(t) = aδ(t)−absin(bt)u(t).

DR ILL 4.2 Laplace Transform

Show that the Laplace transform of 10e−3*^t* cos (4t + 53.13◦) is (6s − 14)/(s² + 6s + 25). Use Table 4.1.

DR ILL 4.3 Inverse Laplace Transform

Find the inverse Laplace transform of the following:

(a)

$$\frac{s+17}{s^2+4s-5}$$

\n(b)

$$\frac{3s-5}{(s+1)(s^2+2s+5)}$$

\n(c)

$$\frac{16s+43}{(s-2)(s+3)^2}$$

ANSWERS

(a)

$$(3e^t - 2e^{-5t})u(t)$$

• (b) ⁻2e−^{t + 5 2} e−^t cos(2t −36.87◦) u(t)
• (c) [3e2*^t* +(t −3)e−3*^t* ]u(t)

A HISTORICAL NOTE: MARQUIS PIERRE-SIMON DE LAPLACE (1749–1827)

The Laplace transform is named after the great French mathematician and astronomer Laplace, who first presented the transform and its applications to differential equations in a paper published in 1779.

Laplace developed the foundations of potential theory and made important contributions to special functions, probability theory, astronomy, and celestial mechanics. In his Exposition du système du monde (1796), Laplace formulated a nebular hypothesis of cosmic origin and tried to explain the universe as a pure mechanism. In his Traité de mécanique céleste (celestial mechanics), which completed the work of Newton, Laplace used mathematics and physics to subject the solar system and all heavenly bodies to the laws of motion and the principle of gravitation. Newton had

Pierre-Simon de Laplace and Oliver Heaviside

been unable to explain the irregularities of some heavenly bodies; in desperation, he concluded that God himself must intervene now and then to prevent such catastrophes as Jupiter eventually falling into the sun (and the moon into the earth), as predicted by Newton’s calculations. Laplace proposed to show that these irregularities would correct themselves periodically and that a little patience—in Jupiter’s case, 929 years—would see everything returning automatically to order; thus there was no reason why the solar and the stellar systems could not continue to operate by the laws of Newton and Laplace to the end of time [4].

Laplace presented a copy of Mécanique céleste to Napoleon, who, after reading the book, took Laplace to task for not including God in his scheme: “You have written this huge book on the system of the world without once mentioning the author of the universe.” “Sire,” Laplace retorted, “I had no need of that hypothesis.” Napoleon was not amused, and when he reported this reply to another great mathematician-astronomer, Louis de Lagrange, the latter remarked, “Ah, but that is a fine hypothesis. It explains so many things” [5].

Napoleon, following his policy of honoring and promoting scientists, made Laplace the minister of the interior. To Napoleon’s dismay, however, the new appointee attempted to bring “the spirit of infinitesimals” into administration, and so Laplace was transferred hastily to the Senate.

OLIVER HEAVISIDE (1850–1925)

Although Laplace published his transform method to solve differential equations in 1779, the method did not catch on until a century later. It was rediscovered independently in a rather awkward form by an eccentric British engineer, Oliver Heaviside (1850–1925), one of the tragic figures in the history of science and engineering. Despite his prolific contributions to electrical engineering, he was severely criticized during his lifetime and was neglected later to the point that

348 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

hardly a textbook today mentions his name or credits him with contributions. Nevertheless, his studies had a major impact on many aspects of modern electrical engineering. It was Heaviside who made transatlantic communication possible by inventing cable loading, but few mention him as a pioneer or an innovator in telephony. It was Heaviside who suggested the use of inductive cable loading, but the credit is given to M. Pupin, who was not even responsible for building the first loading coil.† In addition, Heaviside was [6]:

• The first to find a solution to the distortionless transmission line.
• The innovator of lowpass filters.
• The first to write Maxwell’s equations in modern form.
• The codiscoverer of rate energy transfer by an electromagnetic field.
• An early champion of the now-common phasor analysis.
• An important contributor to the development of vector analysis. In fact, he essentially created the subject independently of Gibbs [7].
• An originator of the use of operational mathematics used to solve linear integro-differential equations, which eventually led to rediscovery of the ignored Laplace transform.
• The first to theorize (along with Kennelly of Harvard) that a conducting layer (the Kennelly–Heaviside layer) of atmosphere exists, which allows radio waves to follow earth’s curvature instead of traveling off into space in a straight line.
• The first to posit that an electrical charge would increase in mass as its velocity increases, an anticipation of an aspect of Einstein’s special theory of relativity [8]. He also forecast the possibility of superconductivity.

Heaviside was a self-made, self-educated man. Although his formal education ended with elementary school, he eventually became a pragmatically successful mathematical physicist. He began his career as a telegrapher, but increasing deafness forced him to retire at the age of 24. He then devoted himself to the study of electricity. His creative work was disdained by many professional mathematicians because of his lack of formal education and his unorthodox methods.

Heaviside had the misfortune to be criticized both by mathematicians, who faulted him for lack of rigor, and by men of practice, who faulted him for using too much mathematics and thereby confusing students. Many mathematicians, trying to find solutions to the distortionless transmission line, failed because no rigorous tools were available at the time. Heaviside succeeded because he used mathematics not with rigor, but with insight and intuition. Using his much maligned operational method, Heaviside successfully attacked problems that the rigid mathematicians could not solve, problems such as the flow-of-heat in a body of spatially varying conductivity. Heaviside brilliantly used this method in 1895 to demonstrate a fatal flaw in Lord Kelvin’s determination of the geological age of the earth by secular cooling; he used the same flow-of-heat theory as for his cable analysis. Yet the mathematicians of the Royal Society remained unmoved and were not the least impressed by the fact that Heaviside had found the answer to problems no one else could solve. Many mathematicians who examined his work dismissed it

^† Heaviside developed the theory for cable loading, George Campbell built the first loading coil, and the telephone circuits using Campbell’s coils were in operation before Pupin published his paper. In the legal fight over the patent, however, Pupin won the battle: he was a shrewd self-promoter, and Campbell had poor legal support.

with contempt, asserting that his methods were either complete nonsense or a rehash of known ideas [6].

Sir William Preece, the chief engineer of the British Post Office, a savage critic of Heaviside, ridiculed Heaviside’s work as too theoretical and, therefore, leading to faulty conclusions. Heaviside’s work on transmission lines and loading was dismissed by the British Post Office and might have remained hidden, had not Lord Kelvin himself publicly expressed admiration for it [6].

Heaviside’s operational calculus may be formally inaccurate, but in fact it anticipated the operational methods developed in more recent years [9]. Although his method was not fully understood, it provided correct results. When Heaviside was attacked for the vague meaning of his operational calculus, his pragmatic reply was, “Shall I refuse my dinner because I do not fully understand the process of digestion?”

Heaviside lived as a bachelor hermit, often in near-squalid conditions, and died largely unnoticed, in poverty. His life demonstrates the persistent arrogance and snobbishness of the intellectual establishment, which does not respect creativity unless it is presented in the strict language of the establishment.