30 + Interview Questions in Intoduction To The Laplace Transform in Network Theory Page 1 InterviewSolution

1.	The inverse transform of the function k/(s+a) is?(a) ke^-at u(t)(b) ke^at u(t)(c) ke^-at u(t-a)(d) ke^at u(t-a)I got this question in class test.This interesting question is from Inverse Transforms topic in division Intoduction to the Laplace Transform of Network Theory
Answer» The correct choice is (b) ke^at u(t) Best EXPLANATION: The inverse transform of the FUNCTION k/(s) is k. The inverse transform of the function k/(s+a) is ke^at u(t). k/(s+a)<—–> ke^at u(t).

Discussion

2.	For the function F (s) = (s+5)/s(s^2+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)tThe question was asked in an international level competition.Enquiry is from Inverse Transforms topic in section Intoduction to the Laplace Transform of Network Theory
Answer» RIGHT answer is (c) 1- 1/2 E^(-1+j2)t-1/2 e^(-1-j2)t The best explanation: 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.

Discussion

3.	For the function F (s) = (s+5)/s(s^2+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)The question was asked during an interview for a job.The question is from Inverse Transforms topic in chapter Intoduction to the Laplace Transform of Network Theory
Answer» The CORRECT answer is (a) 1/s-1/2(s+1-j2) -1/2(s+1+j2) To explain: The expression of F (s) after splitting into PARTIAL FRACTIONS is (s+5)/s(s^2+2s+5) = 1/s-1/2(s+1-j2) -1/2(s+1+j2).

Discussion

4.	For the function F (s) = (s+5)/s(s^2+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) 1/2The question was posed to me in an internship interview.My query is from Inverse Transforms in section Intoduction to the Laplace Transform of Network Theory
Answer» The correct answer is (c) -1/2 Easiest explanation: 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.

Discussion

5.	For the function F (s) = (s+5)/s(s^2+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) -1/4I had been asked this question in a job interview.This key question is from Inverse Transforms in portion Intoduction to the Laplace Transform of Network Theory
Answer» The correct answer is (b) -1/2 For EXPLANATION I would say: 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.

Discussion

6.	For the function F (s) = (s+5)/s(s^2+2s+5), after splitting this function into the partial fractions, 1/s co-efficient is?(a) 1(b) 2(c) 3(d) 4This question was addressed to me by my college director while I was bunking the class.I would like to ask this question from Inverse Transforms in portion Intoduction to the Laplace Transform of Network Theory
Answer» Right ANSWER is (a) 1 To explain: To obtain the CONSTANT A, multiply the GIVEN EQUATION with (s) and putting s = 0. A= sF(s)\|s=0 = (s+5)/((s^2+2s+5))=1.

Discussion

7.	For the function F (s) = (s^2+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)The question was posed to me in class test.I would like to ask this question from Inverse Transforms topic in section Intoduction to the Laplace Transform of Network Theory
Answer» CORRECT answer is (d) 1/15s+2.1/(s+5)-1.17/(s+3) To explain I WOULD say: The VALUES of A, B,C are A = 1/15, B = 2.1, C = -1.17. Partial fraction EXPANSION of the FUNCTION is (s^2+s+1)/s(s+5)(s+3) =1/15s+2.1/(s+5)-1.17/(s+3).

Discussion

8.	For the function F (s) = (s^2+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) 2.27The question was posed to me at a job interview.This intriguing question comes from Inverse Transforms in portion Intoduction to the Laplace Transform of Network Theory
Answer» The correct choice is (a) -1.17 Best EXPLANATION: To obtain the CONSTANT C, multiply the GIVEN equation with (s+3) and putting s = -3. C = (s + 3)F (s)\|s = -3 = (s^2+s+1)/s(s+5)\|s=-3 = -1.17.

Discussion

9.	For the function F (s) = (s^2+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) 4.1The question was posed to me in unit test.My doubt is from Inverse Transforms in section Intoduction to the Laplace Transform of Network Theory
Answer» The CORRECT OPTION is (b) 2.1 For explanation: To obtain the constant B, multiply the given equation with (s+5) and putting s = -5. B = (s + 5)F (s) \|s=-5 = (s^2+s+1)/s(s+3)\|s=-5 = 2.1.

Discussion

10.	For the function F (s) = (s^2+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) 1/20This question was posed to me in an interview for internship.This intriguing question comes from Inverse Transforms topic in division Intoduction to the Laplace Transform of Network Theory
Answer» CORRECT answer is (c) 1/15 For explanation: 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 = (s^2+s+1)/(s+5)(s+3)\|s=0 =1/15.

Discussion

11.	If u (t) = 1 for t >= 0 and u (t) = 0 for t < 0, determine the Laplace transform of [u (t) – u (t – a)].(a) 1/s(1+e^(-as))(b) 1/s(1-e^(-as))(c) 1/s(1+e^as)(d) 1/s(1-e^as)I have been asked this question by my college director while I was bunking the class.My doubt stems from Operational Transforms in portion Intoduction to the Laplace Transform of Network Theory
Answer» The correct ANSWER is (b) 1/s(1-e^(-as)) To EXPLAIN I WOULD say: As u (t) = 1 for t >= 0 and u (t) = 0 for t < 0, the Laplace transform of [u (t) – u (t – a)] is L[u (t)– u (t – a)] = 1/s-e^(-as)1/s = 1/s (1-e^(-as)).

Discussion

12.	Find the Laplace transform of the function f (t) = tsin2t.(a) 4s/(s^2+4)^2(b) -4s/(s^2+4)^2(c) -4s/(s^2-4)^2(d) 4s/(s^2-4)^2The question was asked during an online interview.Question is from Operational Transforms in division Intoduction to the Laplace Transform of Network Theory
Answer» Right option is (a) 4s/(s^2+4)^2 Explanation: The LAPLACE transform of the FUNCTION of sin2t is L(sin2t)=2/(s^2+4). So the Laplace transform of the function F (t) = tsin2t is L(tsin2t) = -d/ds [2/(s^2+4)] = 4s/(s^2+4)^2.

Discussion

13.	Find the Laplace transform of ramp function r (t) = t.(a) 1/s(b) 1/s^2(c) 1/s^3(d) 1/s^4The question was asked in an online quiz.I'm obligated to ask this question of Operational Transforms in portion Intoduction to the Laplace Transform of Network Theory
Answer» CORRECT ANSWER is (B) 1/s^2 Easy EXPLANATION: We KNOW .

Discussion

14.	Find the Laplace transform of (t + 2)^2 e^t.(a) 2/(s-1)^3 – 2/(s-1)^2 + 4/(s-1)(b) 2/(s-1)^3 – 2/(s-1)^2 – 4/(s-1)(c) 2/(s-1)^3 + 2/(s-1)^2 + 4/(s-1)(d) 2/(s-1)^3 + 2/(s-1)^2 – 4/(s-1)I have been asked this question during an online exam.This interesting question is from Operational Transforms in portion Intoduction to the Laplace Transform of Network Theory
Answer» RIGHT ANSWER is (C) 2/(s-1)^3 + 2/(s-1)^2 + 4/(s-1) Best explanation: The LAPLACE transform of t^2+2t+4 is L(t^2+2t+4)=2/(s)^3 + 2/(s)^2+4/s. So the Laplace transform of (t + 2)^2 e^t is L((t + 2)^2 e^t) = 2/(s-1)^3 + 2/(s-1)^2 + 4/(s-1).

Discussion

15.	Find the Laplace transform of e^atsinbt.(a) b/((s-a)^2+b^2)(b) b/((s+a)^2+b^2)(c) b/((s+a)^2-b^2)(d) b/((s-a)^2-b^2)The question was posed to me in an online quiz.I'm obligated to ask this question of Operational Transforms in chapter Intoduction to the Laplace Transform of Network Theory
Answer» Correct ANSWER is (a) B/((s-a)^2+b^2) Easy explanation: The Laplace TRANSFORM of sinbt is L(sinbt)=b/(s^2+b^2). So the Laplace transform of e^atsinbt is L(EXP(at) sinbt)=b/((s-a)^2+b^2).

Discussion

16.	Find the Laplace transform of the function f (t) = 3t^4 – 2t^3 + 4e^-3t – 2sin5t + 3cos2t.(a) 72/s^5 – 12/s^4 + 4/(s+3)+10/(s^2+25)+3s/(s^2+4)(b) 72/s^5 – 12/s^4 + 4/(s+3)-10/(s^2+25)+3s/(s^2+4)(c) 72/s^5 – 12/s^4 – 4/(s+3)+10/(s^2+25)+3s/(s^2+4)(d) 72/s^5 – 12/s^4 – 4/(s+3)-10/(s^2+25)+3s/(s^2+4)I had been asked this question during an online exam.My doubt stems from Operational Transforms topic in portion Intoduction to the Laplace Transform of Network Theory
Answer» Right answer is (b) 72/s^5 – 12/s^4 + 4/(s+3)-10/(s^2+25)+3s/(s^2+4) Best explanation: L (3t^4 -2t^3+4e^-3t – 2sin5t +3cos2t) = 3 L (t^4)-2L (t^3)+4L (e^-3t)-2L (sin5t) + 3L (cos2t) = 72/s^5) -12/s^4 +4/(s+3)-10/(s^2+25)+3s/(s^2+4).

Discussion

17.	Find the Laplace transform of the function f(t) = cos2t.(a) (2s^2+4)/2s(s^2-4)(b) (2s^2-4)/2s(s^2-4)(c) (2s^2-4)/2s(s^2+4)(d) (2s^2+4)/2s(s^2+4)This question was posed to me in an international level competition.This intriguing question originated from Operational Transforms topic in chapter Intoduction to the Laplace Transform of Network Theory
Answer» Right CHOICE is (d) (2S^2+4)/2s(s^2+4) The explanation: The Laplace transform of the FUNCTION f(t) = cos2t is L (cos2t) = L((1+cos2t)/2) = L(1/2)+L(cos2t/2)= 1/2[L(1)+L(cos2t)] = (2s^2+4)/2s(s^2+4).

Discussion

18.	The Laplace transform of f1 (t) + f2 (t) is?(a) F1(s) + F2(s)(b) F1(s) – F2(s)(c) F1(s) – 2F2(s)(d) F1(s) + 2F2(s)The question was posed to me by my college director while I was bunking the class.Question is taken from Operational Transforms in division Intoduction to the Laplace Transform of Network Theory
Answer» Correct ANSWER is (a) F1(s) + F2(s) For explanation: ADDITION or subtraction in time domain translates into addition or subtraction in FREQUENCY domain. L (f1 (t) + f2 (t)) = F1(s) + F2(s).

Discussion

19.	Find the Laplace transform of the function f (t) = 4t^3 + t^2 – 6t + 7.(a) 24/s^4 + 2/s^3 + 6/s^2 + 7/s(b) 24/s^4 – 2/s^3 – 6/s^2 + 7/s(c) 24/s^4 + 2/s^3 – 6/s^2 + 7/s(d) 24/s^4 – 2/s^3 + 6/s^2 + 7/sThe question was asked during a job interview.The query is from Operational Transforms topic in chapter Intoduction to the Laplace Transform of Network Theory
Answer» Correct choice is (c) 24/s^4 + 2/s^3 – 6/s^2 + 7/s For EXPLANATION: L (4t^3 + T^2 -6T +7) = 4L (t^3) + L(t^2)-6L (t) + 7L(1) = 4×3!/s^4 + 2!/s^3 – 6 (1!)/(s^2)+71/s = 24/s^4 + 2/s^3 -6/s^2 + 7/s.

Discussion

20.	Find the function f (t) in terms of unit step function in the graph shown below.(a) 4t [u (t) – u (t + 5)](b) 4t [u (t) + u (t + 5)](c) 4t [u (t) – u (t – 5)](d) 4t [u (t) + u (t – 5)]I had been asked this question in quiz.This key question is from Definition of the Laplace Transform in section Intoduction to the Laplace Transform of Network Theory
Answer» The CORRECT answer is (c) 4t [U (t) – u (t – 5)] Easy explanation: The WAVEFORM shown in the figure starts at t = 0 and ends at t = 5 sec. In terms of unit step function the waveform can be expressed as f (t) = 4t [u (t) – u (t – 5)].

Discussion

21.	The Laplace transform of kf(t) is?(a) F(s)(b) kF(s)(c) F(s)/k(d) k^2 F(s)The question was posed to me in a job interview.The query is from Operational Transforms topic in chapter Intoduction to the Laplace Transform of Network Theory
Answer» Right CHOICE is (b) kF(s) To elaborate: Operational transforms indicate how mathematical operations performed in EITHER f(t) or F(s) are converted into the OPPOSITE DOMAIN. Linearity property STATES that L (kf (t)) = kF (s).

Discussion

22.	In the graph shown below, find the expression f (t).(a) 2t(b) 3t(c) 4t(d) 5tI have been asked this question in an interview for internship.I want to ask this question from Definition of the Laplace Transform topic in division Intoduction to the Laplace Transform of Network Theory
Answer» The correct answer is (c) 4T The best explanation: The waveform shown in the figure STARTS at t = 0 and ends at t = 5 SEC. The equation for the above waveform is f (t) = 4t.

Discussion

23.	Find the function f3 (t) from the time t = 3 to 4 sec.(a) (20t – 40) [u (t-3) – u (t-4)](b) (20t + 40) [u (t-3) – u (t-4)](c) (20t + 40) [u (t-3) + u (t-4)](d) (20t – 40) [u (t-3) + u (t-4)]I got this question during an interview.This question is from Definition of the Laplace Transform topic in section Intoduction to the Laplace Transform of Network Theory
Answer» Correct answer is (a) (20T – 40) [u (t-3) – u (t-4)] For EXPLANATION: From the graph, f3 (t) = 20t – 40 for 3 < t < 4. In terms of unit step FUNCTION, f3 (t) = (20t – 40) [u (t-3) – u (t-4)]. This function TURN off at t = 3, turn off at t = 4.

Discussion

24.	Find the expression of f (t) in the graph shown below.(a) 10t [u (t) – u (t – 1)] – (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)](b) 10t [u (t) – u (t – 1)] – (-10t+20) [u (t-1) – u (t-3)] – (20t – 40) [u (t-3) – u (t-4)](c) 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)](d) 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] – (20t – 40) [u (t-3) – u (t-4)]The question was posed to me during an online exam.My query is from Definition of the Laplace Transform topic in section Intoduction to the Laplace Transform of Network Theory
Answer» The CORRECT ANSWER is (c) 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)] The explanation: We use the step FUNCTION to initiate and terminate these linear SEGMENTS at the proper TIMES. The expression of f (t) is f (t) = 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)].

Discussion

25.	The total period of the function shown in the figure is 4 sec and the amplitude is 10. Find the function f1 (t) from t = 0 to 1 in terms of unit step function.(a) 10t [u (t) – u (t + 1)](b) 10t [u (t) + u (t – 1)](c) 10t [u (t) + u (t + 1)](d) 10t [u (t) – u (t – 1)]I got this question in final exam.I'd like to ask this question from Definition of the Laplace Transform topic in portion Intoduction to the Laplace Transform of Network Theory
Answer» CORRECT choice is (d) 10t [u (t) – u (t – 1)] The BEST explanation: The function shown in the figure is made up of LINEAR segments with break points at 0, 1, 3 and 4 seconds. From the graph, f1 (t) = 10t FOR0 < t < 1. In terms of unit step function, f1 (t) = 10t [u (t) – u (t – 1)].

Discussion

26.	Find the function f2 (t) from the time t = 1 to 3 sec.(a) (-10t+20) [u (t-1) +u (t-3)](b) (-10t+20) [u (t-1) – u (t-3)](c) (-10t-20) [u (t-1) + u (t-3)](d) (-10t-20) [u (t-1) – u (t-3)]I had been asked this question in unit test.My doubt stems from Definition of the Laplace Transform in division Intoduction to the Laplace Transform of Network Theory
Answer» Right option is (b) (-10T+20) [u (t-1) – u (t-3)] The EXPLANATION: From the graph, f2 (t) = -10t + 20 for 1 < t < 3. In terms of unit step FUNCTION, f2 (t) = (-10t+20) [u (t-1) – u (t-3)]. This function turn off at t = 1, turn off at t = 3.

Discussion

27.	The unit step is not defined at t =?(a) 0(b) 1(c) 2(d) 3The question was asked during an online exam.The query is from Definition of the Laplace Transform topic in portion Intoduction to the Laplace Transform of Network Theory
Answer» Right choice is (a) 0 Best explanation: If K is 1, the FUNCTION is DEFINED as unit step function. And the unit step is not defined at t = 0.

Discussion

28.	Laplace transform changes the ____ domain function to the _____ domain function.(a) time, time(b) time, frequency(c) frequency, time(d) frequency, frequencyI got this question in quiz.This intriguing question originated from Definition of the Laplace Transform topic in portion Intoduction to the Laplace Transform of Network Theory
Answer» Correct ANSWER is (b) time, frequency To explain I would say: Laplace transform changes the time domain FUNCTION F (t) to the frequency domain function F(s). Similarly Laplace TRANSFORMATION converts frequency domain function F(s) to the time domain function f(t).

Discussion

29.	The Laplace transform of a function f (t) is?(a) \(\int_0^{\infty}\) f(t) e^-st(b) \(\int_{-\infty}^0\) f(t) e^-st(c) \(\int_0^{\infty}\) f(t) e^st(d) \(\int_{-\infty}^0\) f(t) e^stI have been asked this question in an online interview.My question is from Definition of the Laplace Transform in portion Intoduction to the Laplace Transform of Network Theory
Answer» The correct choice is (a) \(\int_0^{\infty}\) f(t) e^-st Easiest EXPLANATION: The Laplace TRANSFORM is a powerful analytical technique that is WIDELY used to study the BEHAVIOR of linear, lumped parameter circuits. L(f(t)) = F (s)

Discussion

30.	In the bilateral Laplace transform, the lower limit is?(a) 0(b) 1(c) ∞(d) – ∞This question was posed to me in an international level competition.This interesting question is from Definition of the Laplace Transform in chapter Intoduction to the Laplace Transform of Network Theory
Answer» RIGHT OPTION is (d) – ∞ To explain: If the lower limit is 0, then the TRANSFORM is REFERRED to as one-sided or unilateral Laplace transform. In the two-sided or bilateral Laplace transform, the lower limit is – ∞.

Discussion

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The inverse transform of the function k/(s+a) is?(a) ke^-at u(t)(b) ke^at u(t)(c) ke^-at u(t-a)(d) ke^at u(t-a)I got this question in class test.This interesting question is from Inverse Transforms topic in division Intoduction to the Laplace Transform of Network Theory

Find the Laplace transform of the function f (t) = tsin2t.(a) 4s/(s^2+4)^2(b) -4s/(s^2+4)^2(c) -4s/(s^2-4)^2(d) 4s/(s^2-4)^2The question was asked during an online interview.Question is from Operational Transforms in division Intoduction to the Laplace Transform of Network Theory

Find the Laplace transform of ramp function r (t) = t.(a) 1/s(b) 1/s^2(c) 1/s^3(d) 1/s^4The question was asked in an online quiz.I'm obligated to ask this question of Operational Transforms in portion Intoduction to the Laplace Transform of Network Theory

Find the Laplace transform of e^atsinbt.(a) b/((s-a)^2+b^2)(b) b/((s+a)^2+b^2)(c) b/((s+a)^2-b^2)(d) b/((s-a)^2-b^2)The question was posed to me in an online quiz.I'm obligated to ask this question of Operational Transforms in chapter Intoduction to the Laplace Transform of Network Theory

The Laplace transform of kf(t) is?(a) F(s)(b) kF(s)(c) F(s)/k(d) k^2 F(s)The question was posed to me in a job interview.The query is from Operational Transforms topic in chapter Intoduction to the Laplace Transform of Network Theory

In the graph shown below, find the expression f (t).(a) 2t(b) 3t(c) 4t(d) 5tI have been asked this question in an interview for internship.I want to ask this question from Definition of the Laplace Transform topic in division Intoduction to the Laplace Transform of Network Theory

The unit step is not defined at t =?(a) 0(b) 1(c) 2(d) 3The question was asked during an online exam.The query is from Definition of the Laplace Transform topic in portion Intoduction to the Laplace Transform of Network Theory

In the bilateral Laplace transform, the lower limit is?(a) 0(b) 1(c) ∞(d) – ∞This question was posed to me in an international level competition.This interesting question is from Definition of the Laplace Transform in chapter Intoduction to the Laplace Transform of Network Theory