Answer :
Answer:
y(s) = [tex]\frac{5s-53}{s^{2} - 10s + 26}[/tex]
we will compare the denominator to the form [tex](s-a)^{2} +\beta ^{2}[/tex]
[tex]s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}[/tex]
comparing coefficients of terms in s
[tex]s^{2} :[/tex] 1
s: -2a = -10
a = -2/-10
a = 1/5
constant: [tex]a^{2}+\beta ^{2} = 26[/tex]
[tex](\frac{1}{5} )^{2} + \beta ^{2} = 26\\\\\beta^{2} = 26 - \frac{1}{10} \\\\\beta =\sqrt{26 - \frac{1}{10}} =5.09[/tex]
hence the first answers are:
a = 1/5 = 0.2
β = 5.09
Given that y(s) = [tex]A(s-a)+B((s-a)^{2} +\beta ^{2} )[/tex]
we insert the values of a and β
[tex]\\5s-53[/tex] = [tex]A(s-0.2)+B((s-0.2)^{2} + 5.09^{2} )[/tex]
to obtain the constants A and B we equate the numerators and we substituting s = 0.2 on both side to eliminate A
5(0.2)-53 = A(0.2-0.2) + B((0.2-0.2)²+5.09²)
-52 = B(26)
B = -52/26 = -2
to get A lets substitute s=0.4
5(0.4)-53 = A(0.4-0.2) + (-2)((0.4 - 0.2)²+5.09²)
-51 = 0.2A - 52.08
0.2A = -51 + 52.08
A = -1.08/0.2 = 5.4
the constants are
a = 0.2
β = 5.09
A = 5.4
B = -2
Step-by-step explanation:
- since the denominator has a complex root we compare with the standard form [tex]s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}[/tex]
- Expand and compare coefficients to obtain the values of a and β as shown above
- substitute the values gotten into the function
- Now assume any value for 's' but the assumption should be guided to eliminate an unknown, just as we've use s=0.2 above to eliminate A
- after obtaining the first constant, substitute the value back into the function and obtain the second just as we've shown clearly above
Thanks...