Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Your Ultimate AI Essay Writer & Assistant.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Your Ultimate AI Essay Writer & Assistant.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 2

Q. 2.7.7

Two Complex Roots

Use two methods to obtain the inverse Laplace transform of

X(s)=\frac{3s+7}{4s^2+24s+136}= \frac{3s+7}{4(s^2+6s+34)}

Step-by-Step

Verified Solution

a. The denominator roots are s = −3 ± 5j. To avoid complex-valued coefficients, we note that the denominator of X(s) can be written as (s + 3)² + 5², and we can express X(s) as follows:

X(s)=\frac{1}{4}[\frac{3s+7}{(s+3)^2+5^2}] \quad (1)

which can be expressed as the sum of two terms that are proportional to entries 10 and 11 in Table 2.3.1.

X(s)=\frac{1}{4}[C_1\frac{5}{(s+3)^2+5^2}+C_2 \frac{s+3}{(s+3)^2+5^2}]

We can obtain the coefficients by noting that

C_1 \frac{5}{(s+3)^2+5^2}+C_2 \frac{s+3}{(s+3)^2+5^2}= \frac{5C_1+C_2(s+3)}{(s+3)^2+5^2} \quad (2)

Comparing the numerators of equations (1) and (2), we see that

5C_1+C_2(s+3)=C_2s+5C_1+3C_2=3s+7

which gives C_2 = 3 \text{ and } 5C_1 + 3C_2 = 7. \text{ Thus, } C_1 = −2/5. The inverse transform is

x(t)=\frac{1}{4}C_1e^{-3t} \sin 5t+\frac{1}{4}C_2e^{-3t} \cos5t = -\frac{1}{10} e^{-3t} \sin5t + \frac{3}{4}e^{-3t} \cos5t

b. The denominator roots are distinct and the expansion (2.8.3) gives

x(t)=(-0.0147-0.3912j)e^{(-3+5j)t}+(-0.0147-0.3912j)e^{(-3-5j)t}+0.0294 \quad (2.8.3)

X(s)=\frac{3s+7}{4s^2+24s+136}=\frac{3s+7}{4(s+3-5j)(s+3+5j)} \\ = \frac{C_1}{s+3-5j}+\frac{C_2}{s+3+5j}

where, from (2.8.4),

\frac{C+jD}{s+a-jb}+\frac{C-jD}{s+a+jb} \quad (2.8.4)

C_1= \underset{s \rightarrow -3+5j}{\text{lim}}(s+3-5j)X(s)= \underset{s \rightarrow -3+5j}{\text{lim}} \frac{3s+7}{4(s+3+5j)} \\ =\frac{-2+15j}{40j}=\frac{15+2j}{40}

This can be expressed in complex exponential form as follows (see Table 2.1.3):

C_1=|C_1|e^{j \phi}= |\frac{15+2j}{40}|e^{j \phi}=\frac{\sqrt{229}}{40}e^{j \phi}

where 𝜙 = tan^{−1}(2/15) = 0.1326 rad.

The second coefficient is

C_2=\underset{s \rightarrow -3-5j}{\text{lim}}(s+3+5j)X(s)= \underset{s \rightarrow -3-5j}{\text{lim}} \frac{3s+7}{4(s+3-5j)} \\ = \frac{2+15j}{40j}=\frac{15-2j}{40}

Note that C_1 and C_2 are complex conjugates. This will always be the case for coefficients of complex-conjugate roots in a partial-fraction expansion. Thus, C_2=|C_1|e^{-j \phi}= \sqrt{229}e^{-0.1326j}/40.

The inverse transform gives

x(t)=C_1e^{(-3+5j)t}+C_2e^{(-3-5j)t}=C_1e^{-3t}e^{5jt}+C_2e^{-3t}e^{-5jt} \\ =|C_1|e^{-3t}[e^{(5t+ \phi)j}+e^{-(5t+ \phi)j}]=2|C_1|e^{-3t} \cos(5t+\phi)

where we have used the relation e^{j \theta}+e^{-j \theta}=2 \cos \theta, which can be derived from the Euler identity (Table 2.1.4). Thus,

x(t)=\frac{\sqrt{229}}{20}e^{-3t} \cos (5t+0.1326)

This answer is equivalent to that found in part (a), as can be seen by applying the trigonometric identity cos(5t + 𝜙) = cos 5t cos 𝜙 − sin 5tsin 𝜙.

Table 2.3.1 Table of Laplace transform pairs.
X(s) x(t), t ≥ 0
1.                1 \delta(t), \text{unit impulse}
2.                \frac{1}{s} u_s(t), \text{unit step}
3.                \frac{c}{s} \text{constant, c}
4.                \frac{e^{-sD}}{s} u_s(t-D), \text{ shifted unit step}
5.                \frac{n!}{s^{n+1}} t^n
6.                \frac{1}{s+a} e^{-at}
7.                \frac{1}{(s+a)^n} \frac{1}{(n-1)!}t^{n-1}e^{-at}
8.                \frac{b}{s^2+b^2} \sin \ bt
9.                \frac{s}{s^2+b^2} \cos \ bt
10.                \frac{b}{(s+a)^2+b^2} e^{-at} \sin \ bt
11.                \frac{s+a}{(s+a)^2+b^2} e^{-at} \cos \ bt
12.                \frac{a}{s(s+a)} 1-e^{-at}
13.                \frac{1}{(s+a)(s+b)} \frac{1}{b-a}(e^{-at}-e^{-bt})
14.                \frac{s+p}{(s+a)(s+b)} \frac{1}{b-a}[(p-a)e^{-at}-(p-b)e^{-bt}]
15.                \frac{1}{(s+a)(s+b)(s+c)} \frac{e^{-at}}{(b-a)(c-a)}+\frac{e^{-bt}}{(c-b)(a-b)}+\frac{e^{-ct}}{(a-c)(b-c)}
16.                \frac{s+p}{(s+a)(s+b)(s+c)} \frac{(p-a)e^{-at}}{(b-a)(c-a)} + \frac{(p-b)e^{-bt}}{(c-b)(a-b)}+\frac{(p-c)e^{-ct}}{(a-c)(b-c)}
17.                \frac{b}{s^2-b^2} \sinh \ bt
18.                \frac{s}{s^2-b^2} \cosh \ bt
19.                \frac{a^2}{s^2(s+a)} at-1+e^{-at}
20.                \frac{a^2}{s(s+a)^2} 1-(at+1)e^{-at}
21.                \frac{\omega_n^2}{s^2+2\zeta \omega_ns+\omega_n^2} \frac{\omega_n}{\sqrt{1- \zeta^2}}e^{-\zeta \omega_nt} \sin \omega_n \sqrt{1-\zeta^2}t
22.                \frac{s}{s^2+2\zeta \omega_ns+\omega_n^2} -\frac{1}{\sqrt{1-\zeta^2}} e^{-\zeta \omega_n t} \sin (\omega_n \sqrt{1-\zeta^2} t- \phi), \phi= \tan^{-1} \frac{\sqrt{1-\zeta^2}}{\zeta}
23.                \frac{\omega_n^2}{s(s^2+2\zeta \omega_ns + \omega_n^2)} 1-\frac{1}{\sqrt{1-\zeta^2}} e^{-\zeta \omega_n t} \sin (\omega_n \sqrt{1-\zeta^2}t+\phi), \phi= \tan^{-1} \frac{\sqrt{1-\zeta^2}}{\zeta}
24.                \frac{1}{s[(s+a)^2+b^2]} \frac{1}{a^2+b^2}[1-(\frac{a}{b} \sin \ bt + \cos \ bt) e^{-at}]
25.                \frac{b^2}{s(s^2+b^2)} 1- \cos \ bt
26.                \frac{b^3}{s^2(s^2+b^2)} bt – \sin \ bt
27.                \frac{2b^3}{(s^2+b^2)^2} \sin \ bt – bt \ \cos \ bt
28.                \frac{2bs}{(s^2+b^2)^2} t \ \sin \ bt
29.                \frac{s^2-b^2}{(s^2+b^2)^2} t \ \cos \ bt
30.                \frac{s}{(s^2+b_1^2)(s^2+b_2^2)} \frac{1}{b_2^2-b_1^2}(\cos b_1t- \cos \ b_2t), \quad (b_1^2 \neq b_2^2)
31.                \frac{s^2}{(s^2+b^2)^2} \frac{1}{2b}(\sin \ bt + bt \ \cos bt)

 

Table 2.1.4 The exponential function.
Taylor series
e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+…+ \frac{x^n}{n!}+…
Euler’s identities
e^{j \theta}= \cos \theta + j \sin \theta
e^{-j \theta}= \cos \theta – j \sin \theta
Limits
\underset{x \rightarrow \infty}{\text{lim}} xe^{-x}=0 \quad \text{if x is real.}
\underset{x \rightarrow \infty}{\text{lim}} e^{-st}=0 \quad \text{if the real part of s is positive.}
\text{If a is real and positive,}
e^{-at} < 0.02 \ if \ t > 4/a.
e^{-at} < 0.01 \ if \ t > 5/a.
\text{The time constant is} \ \tau=1/a.
Table 2.1.3 Roots and complex numbers.
The quadratic formula
The roots of as² + bs + c = 0 are given by
s=\frac{-b \pm \sqrt{b^2-4ac}}{2a}
For complex roots, s = −𝜎 ± j𝜔, the quadratic can be expressed as
as² + bs + c = a [ (s +𝜎)² + 𝜔²] = 0
Complex numbers
Rectangular representation:
z = x + jy, j = \sqrt{-1}
Complex conjugate:
\overset{-}{z} = x − jy
Magnitude and angle:
|z| = \sqrt{x^2+y^2} \quad \theta = \angle z = \tan ^{-1} \frac{y}{x}  (See Figure 2.1.11)
Polar and exponential representation:
z = |z| \angle \theta = |z| e^{j \theta}
Equality: If z_1 = x_1 + jy_1 and z_2 = x_2 + jy_2, then
z_1 = z_2 if x_1 = x_2 and y_1 = y_2
Addition:
z_1 + z_2 = (x_1 + x_2) + j(y_1 + y_2)
Multiplication:
z_1z_2 = |z_1||z_2|∠(\theta_1 + \theta_2)
z_1z_2 = (x_1x_2− y_1y_2) + j(x_1y_2 + x_2y_1)
Complex-conjugate multiplication:
(x + jy)(x − jy) = x² + y²
Division:
\frac{1}{z}= \frac{1}{x+yj} = \frac{x-jy}{x²+y²}
\frac{z_1}{z_2}=\frac{|z_1|}{|z_2|} \angle (\theta_1-\theta_2)
\frac{z_1}{z_2}=\frac{x_1+jy_1}{x_2+jy_2}=\frac{x_1+jy_1}{x_2+jy_2} \frac{x_2-jy_2}{x_2-jy_2}=\frac{(x_1+jy_1)(x_2-jy_2)}{x_2^2+y_2^2}
2.1.11