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).
8. Any system is said to be stable if and only if ____________
a) It poles lies at the left of imaginary axis
b) It zeros lies at the left of imaginary axis
c) It poles lies at the right of imaginary axis
d) It zeros lies at the right of imaginary axis
Answer: a
Explanation: Any system is said to be stable if and only if it poles lies at the left of imaginary axis.
9. The system given by equation 5 d3⁄dt3 y(t) + 10 d⁄dt y(t) – 5y(t) = x(t) + ∫0t x(t)dt, is?
a) Stable
b) Unstable
c) Has poles 0, 0.455, -0.236±1.567
d) Has zeros 0, 0.455, -0.226±1.467
Answer: a
Explanation:
Given, 5 d3⁄dt3 y(t) + 10 d⁄dt y(t) – 5y(t) = x(t) + ∫0t x(t)dt,
By laplace transform, 5s3 Y(s)+10sY(s)-5Y(s)=X(s)+1/s X(s)
Hence,
(frac{Y(s)}{X(s)} =frac{[1+1/s]}{5s^3+10s-5}=frac{[s+1]}{s(5s^3+10s-5)})
Poles of (frac{Y(s)}{X(s)}) are, 0 ,0.455,-0.226±1.467
Hence system is Unstable.
10. Find the laplace transform of input x(t) if the system given by d3⁄dt3 y(t) – 2 d2⁄dt2 y(t) –d⁄dt y(t) + 2y(t) = x(t), is stable.
a) s + 1
b) s – 1
c) s + 2
d) s – 2
Answer: b
Explanation: d3⁄dt3 y(t) – 2 d2⁄dt2 y(t) – d⁄dt y(t) + 2y(t) = x(t),
Taking laplace transform,
(s3 – 2s2 – s + 2)Y(s) = X(s)
H(s) = Y(s)⁄X(s) = 1⁄(s-1)(s+1)(s+2)
For the system to be stable, X(s) = s – 1.
11. The system given by equation y(t – 2a) – 3y(t – a) + 2y(t) = x(t – a) is?
a) Stable
b) Unstable
c) Marginally stable
d) 0
Answer: a
Explanation: Given, y(t-2a)-3y(t-a)+2y(t)=x(t-a), then
Taking Laplace transform, e-2as Y(s)-3e-as Y(s)+2Y(s)=e-as X(s),
Hence, (H(s)=frac{e^{-as}}{(1-e^{-as})(2-e^{-as})})
To find the stability, we should have, (∫_{-∞}^∞ H(s)ds>0)
Hence, (int_{-infty}^infty frac{e^{-as}}{(1-e^{-as})(2-e^{-as})} ds)
Let, (e^{-as}=z=>-ae^{-as} ds=dz)
(frac{1}{a} int_0^∞ frac{1}{(1-z)(2-z)} dz=frac{1}{a} int_0^∞ left [frac{1}{(z-1) }-frac{1}{z-2} right ]dz=frac{1}{a} lnleft [frac{z-1}{z-2}right ])
putting, z=0, (frac{1}{a} lnleft [frac{z-1}{z-2}right ]=frac{1}{a} ln(frac{1}{2}))
putting, z=∞, (frac{1}{a} lnleft [frac{z-1}{z-2}right ]=frac{1}{a} ln[frac{1-1/z}{1-2/z}])=0
hence, (int_{-infty}^infty frac{e^{-as}}{(1-e^{-as})(2-e^{-as})} ds=frac{1}{a} ln(frac{1}{2}))
It is stable.