Question 2.7.7: Two Complex Roots Use two methods to obtain the inverse Lapl...

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.