15.4.1 Simple Poles

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15.4.1 Simple Poles

Recall from Chapter 14 that a simple pole is a first-order pole. If F(s) has only simple poles, then D(s) becomes a product of factors, so that

then

$$D(s)$$

becomes a product of factors, so that
\n

$$F(s) = \frac{N(s)}{(s+p_1)(s+p_2)\cdots(s+p_n)}$$

\n(15.48)

where s = −p1, −p2, … , −pn are the simple poles, and pi ≠ pj for all i ≠ j (i.e., the poles are distinct). Assuming that the degree of N(s) is

Software packages such as MATLAB, Mathcad, and Maple are capable of finding partial fraction expansions quite easily.

Otherwise, we must first apply long division so that F(s) = N(s)∕D(s) = Q(s) + R(s)∕D(s), where the degree of ^R(s), the remainder of the long division, is less than the degree of D(s).

less than the degree of D(s), we use partial fraction expansion to decompose F(s) in Eq. (15.48) as

$$F(s) = \frac{k_1}{s + p_1} + \frac{k_2}{s + p_2} + \dots + \frac{k_n}{s + p_n}$$

(15.49)

The expansion coefficients k1, k2, … , kn are kno wn as the residues of F(s). There are man y ways of finding the expansion coefficients. One way is using the residue method. If we multiply both sides of Eq. (15.49) by (s + p1), we obtain

$$(s+p_1)F(s) = k_1 + \frac{(s+p_1)k_2}{s+p_2} + \dots + \frac{(s+p_1)k_n}{s+p_n}$$

(15.50)

Because pi ≠ pj, setting s = −p1 in Eq. (15.50) leaves only k1 on the righthand side of Eq. (15.50). Hence,

$$(s+p_1)F(s)\big|_{s=-p_1} = k_1\tag{15.51}$$

Thus, in general,

$$k_i = (s + p_i)F(s) \Big|_{s = -p_i}$$

(15.52)

This is known as Heaviside’s theorem. Once the values of ki are known, we proceed to find the inverse of F(s) using Eq. (15.49). Since the inverse transform of each term in Eq. (15.49) is ⁻¹ [k∕(s + a)] = ke⁻at u(t), then, from Table 15.2,

$$f(t) = (k_1 e^{-p_1 t} + k_2 e^{-p_2 t} + \dots + k_n e^{-p_n t}) u(t)$$

(15.53)

15.4.2 Repeated Poles

Suppose F(s) has n repeated poles at s = −p. Then we may represent F(s) as

$$F(s) = \frac{k_n}{(s+p)^n} + \frac{k_{n-1}}{(s+p)^{n-1}} + \dots + \frac{k_2}{(s+p)^2} + \frac{k_1}{s+p} + F_1(s)$$

\n(15.54)

where F1(s) is the remaining part of F(s) that does not ha ve a pole at s = −p. We determine the expansion coefficient kn as

$$k_n = (s+p)^n F(s) \Big|_{s=-p} \tag{15.55}$$

as we did above. To determine kn−1, we multiply each term in Eq. (15.54) by (s + p) n and differentiate to get rid of kn, then evaluate the result at s = −p to get rid of the other coefficients except kn−1. Thus, we obtain

$$k_{n-1} = \frac{d}{ds} [(s+p)^n F(s)] \Big|_{s=-p}$$

(15.56)

Repeating this gives

$$k_{n-2} = \frac{1}{2!} \frac{d^2}{ds^2} [(s+p)^n F(s)] \Big|_{s=-p}$$

(15.57)

Historical note: Named after Oliver Heaviside (1850–1925), an English engineer, the pioneer of operational calculus.

The mth term becomes

$$k_{n-m} = \frac{1}{m!} \frac{d^m}{ds^m} [(s+p)^n F(s)] \Big|_{s=-p}$$

(15.58)

where m = 1, 2, … , n − 1. One can expect the differentiation to be difficult to handle as m increases. Once we obtain the values of k1, k2, … , kn by partial fraction expansion, we apply the inverse transform

$$\mathcal{L}^{-1}\left[\frac{1}{(s+a)^n}\right] = \frac{t^{n-1}e^{-at}}{(n-1)!}u(t)$$

\n(15.59)

to each term on the right-hand side of Eq. (15.54) and obtain

$$f(t) = \left(k_1 e^{-pt} + k_2 t e^{-pt} + \frac{k_3}{2!} t^2 e^{-pt} + \dots + \frac{k_n}{(n-1)!} t^{n-1} e^{-pt} \right) u(t) + f_1(t)$$

\n(15.60)

15.4.3 Complex Poles

A pair of comple x poles is simple if it is not repeated; it is a double or multiple pole if repeated. Simple comple x poles may be handled the same way as simple real poles, but because complex algebra is involved the result is always cumbersome. An easier approach is a method known as completing the square. The idea is to express each complex pole pair (or quadratic term) in D(s) as a complete square such as (s + α) 2 + β² and then use Table 15.2 to find the inverse of the term.

Because N(s) and D(s) always have real coefficients and we know that the complex roots of polynomials with real coef ficients must occur in conjugate pairs, F(s) may have the general form

$$F(s) = \frac{A_1 s + A_2}{s^2 + as + b} + F_1(s)$$

\n(15.61)

where F1(s) is the remaining part of F(s) that does not ha ve this pair of complex poles. If we complete the square by letting

$$s^{2} + as + b = s^{2} + 2as + a^{2} + \beta^{2} = (s + a)^{2} + \beta^{2}$$

(15.62)

and we also let

$$A_1s + A_2 = A_1(s + \alpha) + B_1\beta \tag{15.63}$$

then Eq. (15.61) becomes

$$F(s) = \frac{A_1(s+\alpha)}{(s+\alpha)^2 + \beta^2} + \frac{B_1\beta}{(s+\alpha)^2 + \beta^2} + F_1(s)$$

(15.64)

From Table 15.2, the inverse transform is

$$f(t) = (A_1 e^{-\alpha t} \cos \beta t + B_1 e^{-\alpha t} \sin \beta t) u(t) + f_1(t)$$

(15.65)

The sine and cosine terms can be combined using Eq. (9.11).

Whether the pole is simple, repeated, or comple x, a general approach that can al ways be used in finding the expansion coefficients is the method of algebra, illustrated in Examples 15.9 to 15.11. To apply the method, we first set F(s) = N(s)∕D(s) equal to an expansion containing unknown constants. We multiply the result through by a common denominator. Then we determine the unkno wn constants by equating coefficients (i.e., by algebraically solving a set of simultaneous equations for these coefficients at like powers of s).

Another general approach is to substitute specific, convenient values of s to obtain as man y simultaneous equations as the number of unknown coefficients, and then solve for the unkno wn coefficients. We must make sure that each selected v alue of s is not one of the poles of F(s). Example 15.11 illustrates this idea.

Find the inverse Laplace transform of Example 15.8

$$F(s) = \frac{3}{s} - \frac{5}{s+1} + \frac{6}{s^2+4}$$

Solution:

The inverse transform is given by

$$f(t) = \mathcal{L}^{-1}[F(s)] = \mathcal{L}^{-1}\left(\frac{3}{s}\right) - \mathcal{L}^{-1}\left(\frac{5}{s+1}\right) + \mathcal{L}^{-1}\left(\frac{6}{s^2+4}\right)$$ $$= (3 - 5e^{-t} + 3\sin 2t)u(t), \qquad t \ge 0$$

where Table 15.2 has been consulted for the inverse of each term.

$$F(s) = 5 + \frac{6}{s+4} - \frac{7s}{s^2 + 25}$$

Answer: 5δ(t) + (6e⁻4*^t* − 7 cos(5t))u(t).

$$F(s) = \frac{s^2 + 12}{s(s+2)(s+3)}$$

Solution:

Unlike in the pre vious example where the partial fractions ha ve been provided, we first need to determine the partial fractions. Given that there are three poles, we let

les, we let
\n

$$\frac{s^2 + 12}{s(s+2)(s+3)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+3}$$

\n(15.9.1)

where A, B, and C are the constants to be determined. We can find the constants using two approaches.

Determine the inverse Laplace transform of Practice Problem 15.8

Find f(t) given that Example 15.9

■ METHOD 1 Residue method:

PROOF Residue method:
\n

$$A = sF(s) \Big|_{s=0} = \frac{s^2 + 12}{(s+2)(s+3)} \Big|_{s=0} = \frac{12}{(2)(3)} = 2$$

\n

$$B = (s+2)F(s) \Big|_{s=-2} = \frac{s^2 + 12}{s(s+3)} \Big|_{s=-2} = \frac{4+12}{(-2)(1)} = -8$$

\n

$$C = (s+3)F(s) \Big|_{s=-3} = \frac{s^2 + 12}{s(s+2)} \Big|_{s=-3} = \frac{9+12}{(-3)(-1)} = 7$$

■ METHOD 2 Algebraic method: Multiplying both sides of Eq. (15.9.1) by s(s + 2)(s + 3) gives

$$s^2 + 12 = A(s+2)(s+3) + Bs(s+3) + Cs(s+2)$$

or

$$s^2 + 12 = A(s^2 + 5s + 6) + B(s^2 + 3s) + C(s^2 + 2s)$$

Equating the coefficients of like powers of s gives

Constant:

$$12 = 6A \Rightarrow A = 2$$

\ns: $0 = 5A + 3B + 2C \Rightarrow 3B + 2C = -10$

\ns: $1 = A + B + C \Rightarrow B + C = -1$

Thus, A = 2, B = −8, C = 7, and Eq. (15.9.1) becomes

$$F(s) = \frac{2}{s} - \frac{8}{s+2} + \frac{7}{s+3}$$

By finding the inverse transform of each term, we obtain

$$f(t) = (2 - 8e^{-2t} + 7e^{-3t})u(t)$$

Practice Problem 15.9 Find f(t) if

$$F(s) = \frac{48(s+2)}{(s+1)(s+3)(s+4)}$$

Answer:

$$f(t) = (8e^{-t} + 24e^{-3t} - 32e^{-4t})u(t)$$

.

Example 15.10 Calculate v(t) given that

$$V(s) = \frac{10s^2 + 4}{s(s+1)(s+2)^2}$$

Solution:

While the pre vious e xample is on simple roots, this e xample is on repeated roots. Let

$$V(s) = \frac{10s^2 + 4}{s(s+1)(s+2)^2}$$

= $\frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+2)^2} + \frac{D}{s+2}$ (15.10.1)

■ METHOD 1 Residue method:

THEOREM 1 Residue method:
\n

$$A = sV(s) \Big|_{s=0} = \frac{10s^2 + 4}{(s+1)(s+2)^2} \Big|_{s=0} = \frac{4}{(1)(2)^2} = 1$$

\n

$$B = (s+1)V(s) \Big|_{s=-1} = \frac{10s^2 + 4}{s(s+2)^2} \Big|_{s=-1} = \frac{14}{(-1)(1)^2} = -14$$

\n

$$C = (s+2)^2 V(s) \Big|_{s=-2} = \frac{10s^2 + 4}{s(s+1)} \Big|_{s=-2} = \frac{44}{(-2)(-1)} = 22$$

\n

$$D = \frac{d}{ds} [(s+2)^2 V(s)] \Big|_{s=-2} = \frac{d}{ds} \left( \frac{10s^2 + 4}{s^2 + s} \right) \Big|_{s=-2} = \frac{(s^2 + s)(20s) - (10s^2 + 4)(2s + 1)}{(s^2 + s)^2} \Big|_{s=-2} = \frac{52}{4} = 13$$

■ METHOD 2 Algebraic method: Multiplying Eq. (15.10.1) by s(s + 1)(s + 2)² , we obtain

$$10s2 + 4 = A(s + 1)(s + 2)2 + Bs(s + 2)2 + Cs(s + 1) + Ds(s + 1)(s + 2)$$

or

$$10s2 + 4 = A(s3 + 5s2 + 8s + 4) + B(s3 + 4s2 + 4s)$$

• C(s² + s) + D(s³ + 3s² + 2s)

Equating coefficients,

Constant:

$$4 = 4A$$

$\Rightarrow$ $A = 1$

\ns: $0 = 8A + 4B + C + 2D$ $\Rightarrow$ $4B + C + 2D = -8$

\ns²: $10 = 5A + 4B + C + 3D$ $\Rightarrow$ $4B + C + 3D = 5$

\ns³: $0 = A + B + D$ $\Rightarrow$ $B + D = -1$

Solving these simultaneous equations gi ves A = 1, B = −14, C = 22, D = 13, so that

$$V(s) = \frac{1}{s} - \frac{14}{s+1} + \frac{13}{s+2} + \frac{22}{(s+2)^2}$$

Taking the inverse transform of each term, we get

$$v(t) = (1 - 14e^{-t} + 13e^{-2t} + 22te^{-2t})u(t)$$

Obtain g(t) if Practice Problem 15.10

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

Answer: (2 − 3.25e^−t − 1.5te^−t + 2.25e⁻3*^t* )u(t).

Find the in verse transform of the frequenc y-domain function in Example 15.11 Example 15.7: ^H(s) = _________________ 20 (^s + 3)(^s

$$H(s) = \frac{20}{(s+3)(s^2+8s+25)}$$

Solution:

In this example, H(s) has a pair of complex poles at s 2 + 8s + 25 = 0 or s = −4 ± j3. We let

is example,

$$H(s)$$

has a pair of complex poles at $s^2 + 8s + 25 = 0$ or
-4 ± j3. We let

$$H(s) = \frac{20}{(s+3)(s^2+8s+25)} = \frac{A}{s+3} + \frac{Bs+C}{(s^2+8s+25)}$$

(15.11.1)

We now determine the expansion coefficients in two ways.

■ METHOD 1 Combination of methods: We can obtain A using the method of residue,

$$A = (s + 3)H(s)\Big|_{s=-3} = \frac{20}{s^2 + 8s + 25}\Big|_{s=-3} = \frac{20}{10} = 2$$

Although B and C can be obtained using the method of residue, we will not do so, to a void comple x algebra. Rather , we can substitute two specific values of s [say s = 0, 1, which are not poles of F(s)] into Eq. (15.11.1). This will give us two simultaneous equations from which to find B and C. If we let s = 0 in Eq. (15.11.1), we obtain

$$\frac{20}{75} = \frac{A}{3} + \frac{C}{25}$$

or

$$20 = 25A + 3C \tag{15.11.2}$$

Because A = 2, Eq. (15.11.2) gi ves C = −10. Substituting s = 1 into Eq. (15.11.1) gives

$$\frac{20}{(4)(34)} = \frac{A}{4} + \frac{B+C}{34}$$

or

$$20 = 34A + 4B + 4C \tag{15.11.3}$$

But A = 2, C = −10, so that Eq. (15.11.3) gives B = −2.

■ METHOD 2 Algebraic method: Multiplying both sides of Eq. (15.11.1) by (s + 3)(s 2 + 8s + 25) yields

$$20 = A(s2 + 8s + 25) + (Bs + C)(s + 3)$$

= A(s² + 8s + 25) + B(s² + 3s) + C(s + 3) (15.11.4)

Equating coefficients gives

s 2 : 0 = A + B ⇒ A = −B s: 0 = 8A + 3B + C = 5A + C ⇒ C = −5A Constant: 20 = 25A + 3C = 25A − 15A ⇒ A = 2

That is, B = −2, C = −10. Thus,

$$B = -2, C = -10. \text{ Thus,}$$

\n

$$H(s) = \frac{2}{s+3} - \frac{2s+10}{(s^2+8s+25)} = \frac{2}{s+3} - \frac{2(s+4)+2}{(s+4)^2+9}$$

\n

$$= \frac{2}{s+3} - \frac{2(s+4)}{(s+4)^2+9} - \frac{2}{3} \frac{3}{(s+4)^2+9}$$

Taking the inverse of each term, we obtain

$$h(t) = \left(2e^{-3t} - 2e^{-4t}\cos 3t - \frac{2}{3}e^{-4t}\sin 3t\right)u(t) \tag{15.11.5}$$

It is alright to lea ve the result this w ay. However, we can combine the cosine and sine terms as

$$h(t) = (2e^{-3t} - Re^{-4t}\cos(3t - \theta))u(t)$$

(15.11.6)

To obtain Eq. (15.11.6) from Eq. (15.11.5), we apply Eq. (9.11). Next, we determine the coefficient R and the phase angle θ:

$$R = \sqrt{2^2 + \left(\frac{2}{3}\right)^2} = 2.108, \qquad \theta = \tan^{-1} \frac{\frac{2}{3}}{2} = 18.43^{\circ}$$

Thus,

$$h(t) = (2e^{-3t} - 2.108e^{-4t}\cos(3t - 18.43^\circ))u(t)$$ $$G(s) = \frac{20}{(s+1)(s^2+4s+13)}$$

Answer: 2e^−t − 2e⁻2*^t* cos 3t − 0.6667e⁻2*^t* sin 3t, t ≥ 0.