Question

Find the value of A with R=4 for the following partial fraction expansion: Y(s)=25/s²+2Rs+25​=1/s​+A/s+R+j√25−R²​​+​A∗/s+R−j√25−R²

Answer 1

To find the value of A with R=4 in the given partial fraction expansion, we need to compare the coefficients of the like terms in the numerator of both sides of the equation. Equating the coefficients of (1/s) gives: 25 = A*(s+R+j(sqrt(25-R^2))) + A*(s+R-j(sqrt(25-R^2))). Therefore, A = 1/2.

Step-by-step explanation:

To find the value of A with R=4 in the given partial fraction expansion, we need to compare the coefficients of the like terms in the numerator of both sides of the equation.

Equating the coefficients of (1/s) gives: 25 = A*(s+R+j(sqrt(25-R^2))) + A*(s+R-j(sqrt(25-R^2)))

Since the term (s+R+j(sqrt(25-R^2))) and (s+R-j(sqrt(25-R^2))) have the same denominators, their numerators must be equal. Therefore: 1 = A + A

So, A = 1/2.