Engineering Mathematics Questions and Answers for Campus interviews focuses on “Laplace Transform By Properties – 2”.
1. Transfer function may be defined as ____________
a) Ratio of out to input
b) Ratio of laplace transform of output to input
c) Ratio of laplace transform of output to input with zero initial conditions
d) None of the mentioned
Answer: c
Explanation: Transfer function may be defined as the ratio of laplace transform of output to input with zero initial conditions.
2. Poles of any transfer function is define as the roots of equation of denominator of transfer function.
a) True
b) False
Answer: a
Explanation: Let transfer function be defined as G(s)/H(s), then poles of transfer function may be defined as H(s)=0.
3. Zeros of any transfer function is define as the roots of equation of numerator of transfer function.
a) True
b) False
Answer: a
Explanation: Let transfer function be defined as G(s)/H(s), then zeros of transfer function may be defined as G(s)=0.
4. Find the poles of transfer function which is defined by input x(t)=5Sin(t)-u(t) and output y(t)=Cos(t)-u(t).
a) 4.79, 0.208
b) 5.73, 0.31
c) 5.89, 0.208
d) 5.49, 0.308
Answer: a
Explanation: Given ,y(t)=Cos(t) – u(t) and x(t) = 5Sin(t) – u(t),
Hence, transfer function H(s)=(frac{[frac{s}{s^2+1}-frac{1}{s}]}{[frac{5}{s^2+1}-frac{1}{s}]} =-frac{1}{frac{s(s^2+1)}{frac{(5s-s^2-1)}{s(s^2+1)}}}=frac{1}{S^2-5S+1})
Roots of equation s2 – 5s + 1 = 0 is s = 4.79, 0.208.
5. Find the equation of transfer function which is defined by y(t)-∫0t y(t)dt + d⁄dt x(t) – 5Sin(t) = 0.
a) (frac{s(e^{-as}-1)}{s-1})
b) (frac{(e^{-as}-s)}{s-1})
c) (frac{s(e^{-as}-s)}{s-1})
d) (frac{s(e^{-as}-s^2)}{s-1})
Answer: c
Explanation:
Given, (y(t)-∫_0^t y(t)dt+frac{d}{dt} x(t)-x(t-a)=0)
Taking Laplace, (Y(s)-frac{Y(s)}{s}+sX(s)-e^{-as} X(s)=0)
H(s)=Y(s)/X(s) =(frac{(e^{-as}-s)}{1-frac{1}{s}}=frac{s(e^{-as}-s)}{s-1})
6. Find the poles of transfer function given by system d2⁄dt2 y(t) – d⁄dt y(t) + y(t) – ∫0t x(t)dt = x(t).
a) 0, 0.7 ± 0.466
b) 0, 2.5 ± 0.866
c) 0, 0 .5 ± 0.866
d) 0, 1.5 ± 0.876
Answer: c
Explanation: We know that,
Given, (frac{d^2}{dt^2} y(t)-frac{d}{dt} y(t)+y(t)-int_0^t x(t)dt=x(t))
Now, Taking laplace, we get, ((s^2-s+1)Y(s)=(1+frac{1}{s})X(s))
H(s)=(frac{s+1}{s(s^2-s+1)})
Roots of s3-s2+s=0, are 0,.5±0.866
7. Find the transfer function of a system given by equation d2⁄dt2 y(t-a) + x(t) + 5 d⁄dt y(t) = x(t-a).
a) (e-as-s)/(1+e-as s2)
b) (e-as-5s)/(e-as s2)
c) (e-as-s)/(2+e-as s2)
d) (e-as-5s)/(1+e-as s2)
Answer: d
Explanation: Given, d2⁄dt2 y(t-a) + x(t) + 5 d⁄dt y(t) = x(t-a).
Taking laplace transform, s2 Y(s) e-sa + X(s) + 5sY(s) = e-as X(s)
Hence, H(s) = Y(s)⁄X(s) =(e-as-5s)/(1+e-as s2).