Network Theory Multiple Choice Questions on “Inverse Transforms”.
1. For the function F (s) = (s2+s+1)/s(s+5)(s+3), after splitting this function into partial fractions, the co-efficient of the term 1/s is?
A. 1/5
B. 1/10
C. 1/15
D. 1/20
Answer: C
Clarification: To obtain the constant A, multiply the given equation with (s) and putting s = 0. The co-efficient of the term 1/s is A = sF (s)|s=0 = (s2+s+1)/(s+5)(s+3)|s=0 =1/15.
2. For the function F (s) = (s2+s+1)/s(s+5)(s+3), after splitting this function into partial fractions, the co-efficient of 1/(s+5) is?
A. 1.1
B. 2.1
C. 3.1
D. 4.1
Answer: B
Clarification: To obtain the constant B, multiply the given equation with (s+5) and putting s = -5. B = (s + 5)F (s) |s=-5 = (s2+s+1)/s(s+3)|s=-5 = 2.1.
3. For the function F (s) = (s2+s+1)/s(s+5)(s+3), after splitting this function into partial fractions, co-efficient of 1/(s+3) is?
A. -1.17
B. 1.17
C. -2.27
D. 2.27
Answer: A
Clarification: To obtain the constant C, multiply the given equation with (s+3) and putting s = -3. C = (s + 3)F (s)|s = -3 = (s2+s+1)/s(s+5)|s=-3 = -1.17.
4. For the function F (s) = (s2+s+1)/s(s+5)(s+3), after splitting this function into partial fractions. Find the partial fraction expansion of the function.
A. 1/15s-2.1/(s+5)+1.17/(s+3)
B. 1/15s-2.1/(s+5)-1.17/(s+3)
C. 1/15s+2.1/(s+5)+1.17/(s+3)
D. 1/15s+2.1/(s+5)-1.17/(s+3)
Answer: D
Clarification: The values of A, B,C are A = 1/15, B = 2.1, C = -1.17. Partial fraction expansion of the function is (s2+s+1)/s(s+5)(s+3) =1/15s+2.1/(s+5)-1.17/(s+3).
5. For the function F (s) = (s+5)/s(s2+2s+5), after splitting this function into the partial fractions, 1/s co-efficient is?
A. 1
B. 2
C. 3
D. 4
Answer: A
Clarification: To obtain the constant A, multiply the given equation with (s) and putting s = 0. A= sF(s)|s=0 = (s+5)/((s2+2s+5))=1.
6. For the function F (s) = (s+5)/s(s2+2s+5), after splitting this function into the partial fractions, the co-efficient of 1/(s+1-j2) is?
A. 1/2
B. -1/2
C. 1/4
D. -1/4
Answer: B
Clarification: To obtain the constant B, multiply the given equation with (s+1-j2) and putting s = -1+j2. B = (s + 1 – j2)F (s)|s = (-1+j2) = (s+5)/s(s+1+j2)|s=-1+j2 = -1/2.
7. For the function F (s) = (s+5)/s(s2+2s+5), after splitting this function into the partial fractions, determine the co-efficient of 1/(s+1-j2)?
A. -1/4
B. 1/4
C. -1/2
D. 1/2
Answer: C
Clarification: To obtain the constant B*, multiply the given equation with (s+1+j2) and putting s = -1-j2. B* = (s + 1 + j2)F (s)|s = -1 – j2 = (s+5)/s(s+1-j2)|s=-1-j2 = -1/2.
8. For the function F (s) = (s+5)/s(s2+2s+5), after splitting this function into the partial fractions. What is the expression of F (s) after splitting into partial fractions is?
A. 1/s-1/2(s+1-j2) -1/2(s+1+j2)
B. 1/s+1/2(s+1-j2) -1/2(s+1+j2)
C. 1/s+1/2(s+1-j2) +1/2(s+1+j2)
D. 1/s-1/2(s+1-j2) +1/2(s+1+j2)
Answer: A
Clarification: The expression of F (s) after splitting into partial fractions is (s+5)/s(s2+2s+5) = 1/s-1/2(s+1-j2) -1/2(s+1+j2).
9. For the function F (s) = (s+5)/s(s2+2s+5), after splitting this function into the partial fractions. What is the inverse transform of F (s) is?
A. 1+ 1/2 e(-1+j2)t-1/2 e(-1-j2)t
B. 1+ 1/2 e(-1+j2)t+1/2 e(-1-j2)t
C. 1- 1/2 e(-1+j2)t-1/2 e(-1-j2)t
D. 1- 1/2 e(-1+j2)t+1/2 e(-1-j2)t
Answer: C
Clarification: The inverse transform F(s) is f(t), f (t) = L-1(F (s)) = L-1(1/s-1/2(s+1-j2) – 1/2(s+1+j2)) = 1-1/2 e(-1+j2)t-1/2 e(-1-j2)t.
10. The inverse transform of the function k/(s+A. is?
A. ke-at u(t)
B. keat u(t)
C. ke-at u(t-A.
D. keat u(t-A.
Answer: B
Clarification: The inverse transform of the function k/(s) is k. The inverse transform of the function k/(s+A. is keat u(t). k/(s+A. <—–> keat u(t).