Find the poles and residues of the following transfer function. (s+24)(s+23) (s+

Question

Find the poles and residues of the following transfer function.     (s+24)(s+23)     
(s+5)(s+10)(s+14)     (s+24)(s+23)     
(s+5)(s+10)(s+14) =     a1    
(s+p1) +     a2    
(s+p2) +     a3    
(s+p3) a1= 1 p1= 2 a2= 3 p2= 4 a3= 5 p3= 6     (s+24)(s+23)     
(s+5)(s+10)(s+14)     (s+24)(s+23)     
(s+5)(s+10)(s+14) =     a1    
(s+p1) +     a2    
(s+p2) +     a3    
(s+p3)

Explanation / Answer

Given H = (s+24)(s+23)/(s+5)(s+10)(s+14) Since numerator is of order s2 and denominator of orders3 , the above is a proper fraction. Hence we have thefollowing Expanding H in partial fractions H = A/(s+5) + B/(s+10) + C/(s+14) Comparing the numerators of two values of H (s+24)(s+23) = A(s+10)(s+14) + B(s+5)(s+14) + C (s+5)(s+10) put s = -5 => (24-5)(23-5) = A (10-5)(14-5) =>   A = 38/5 = 7.6 put s = -10 => (24-10)(23-10) = B(-10+5)(-10+14) => B =-91/10 = -9.1 put s = -14 => (24-14)(23-14) = C (-14+5)(-14+10) => C = 5/2= 2.5 There fore H = 7.6 / (s+5) -9.1/(s+10) + 2.5/(s+14) Comparing it with a1/(s+p1) + a2/(s+p2) + a3(s+p3) we get   a1 =7.6        p1 = 5            a2 = -9.1        p2 = 10             a3 = 2.5         p3 =14 I hope this helps you

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