Signals & Systems Multiple Choice Questions on “Common Laplace Transforms – 1”.
1. The Laplace transform of f(t) = (e-2t – 1)2 is ________________
A. (frac{2}{s+2} + frac{1}{s})
B. (frac{1}{s+4})
C. (frac{1}{s+4} – frac{2}{s+2} + frac{1}{s})
D. –(frac{2}{s+2} + frac{1}{s})
Answer: C
Clarification: (e-2t – 1)2 = e-4t – 2e-2t + 1
We know that, L {e-at} = (frac{1}{s+a})
L {1} = (frac{1}{s})
∴L {(e-2t – 1)2} = L {e-4t – 2e-2t + 1} = (frac{1}{s+4} – frac{2}{s+2} + frac{1}{s}).
2. Given f (t) = t2e-2x cos (3t). The value of L {f(t)} is __________________
A. (frac{2(s+2)(s^2+4s-23)}{(s^2+4s+13)^3})
B. (frac{2(s-2)(s^2-4s-23)}{(s^2+4s+13)^3})
C. (frac{2(s+2)(s^2+4s+23)}{(s^2+4s+13)^3})
D. (frac{2(s-2)(s^2+4s-23)}{(s^2+4s-13)^3})
Answer: A
Clarification: Let g (t) = cos (3t); h (t) = e-2x cos (3t) = e-2x g (t)
Then, f (t) = t2h (t)
Let G(s) = L {g (t)}, H(s) = L {h (t)}, F(s) = L {f (t)}
So, G(s) = (frac{s}{s^2+9})
And H(s) = (frac{s+2}{(s+2)^2+9})
∴ F(s) = (-frac{d}{ds}[-frac{d}{ds} H(s)] = frac{2(s+2)(s^2+4s-23)}{(s^2+4s+13)^3}).
3. The inverse Laplace transform of F(s) = (frac{2}{s+c} e^{-bs}) is _________________
A. 2 e-k (t-B.
B. 2 e-k (t-B. u (t-B.
C. e-k (t-B. u (t-B.
D. 2 u (t-B.
Answer: B
Clarification: Let G(s) = (frac{2}{s+c})
Or, g (t) = L-1 {G(s)} = 2e-ct
Again, F(s) = L-1 {G(s) e-bs} = 2 e-k (t-B. u (t-B..
4. The inverse Laplace transform of F(s) = (frac{2}{s^2+3s+2}) is ______________
A. -2e-2t + 2e-t
B. 2e-2t + 2e-t
C. -2e-2t – 2e-t
D. 2e-t + e-2t
Answer: A
Clarification: s2 + 3s + 2 = (s+2) (s+1)
Now, F(s) = (frac{A}{(s+2)} + frac{B}{(s+1)})
Hence, A = (s+2) F(s) |s=-2
= (frac{2}{s+1})|s=-2 = -2
And, B = (s+1) F(s) |s=-1
= (frac{2}{s+2})| s=-1 = 2
∴ F(s) = (frac{-2}{(s+2)} + frac{2}{(s+1)})
∴ F (t) = L-1{F(s)}
= -2e-2t + 2e-t for t≥0
5. The Laplace transform of signal u(t-2) is ___________
A. (frac{-e^{-2s}}{s})
B. (frac{e^{-2s}}{s})
C. (frac{e^{-2s}}{1+s})
D. Zero
Answer: B
Clarification: X(s) = (int_0^∞ x(t) e^{-st} ,dt)
Here, the given signal is u (t-2).
Hence, u (t-2) = 1 for all t>2 and =0 for all t<2.
So, limit is from 2 to ∞
∴ Laplace of signal= (int_2^∞ e^{-st} ,dt = frac{e^{-2s}}{s}).
6. The Laplace transform of the signal u (t+2) is _________
A. (frac{1}{s})
B. (frac{-1}{s})
C. (frac{e^{-2s}}{s})
D. (frac{-e^{-2s}}{s})
Answer: A
Clarification: X(s) = (int_0^∞ x(t) e^{-st} ,dt)
Here, the given signal is u (t+2).
Hence, u (t+2) = 1 for all t>-2 and = 0 for all t<-2.
So, limit is from 0 to ∞
∴ Laplace of signal = (int_0^∞ e^{-st} ,dt = frac{1}{s}).
7. The Laplace transform of the signal e-2tu(t+1) is ___________
A. (frac{1}{s+2})
B. (frac{e^{-s}}{s+2})
C. (frac{e^{-(s+2)}}{s+2})
D. (frac{-e^{-s}}{s+2})
Answer: A
Clarification: X(s) = (int_0^∞ x(t) e^{-st} ,dt)
Here, the given signal is u (t+1).
Hence, u (t+1) = 1 for all t>-1 and = 0 for all t<-1.
So, limit is from 0 to ∞
∴ Laplace of signal = (int_0^∞ e^{-2t} e^{-st} ,dt = frac{1}{s+2}).
8. The Laplace transform of the signal e2tu(-t+2) is ___________
A. (frac{e^{2(s-2)-1}}{s-2})
B. (frac{e^{-2s}}{s+2})
C. (frac{1-e^{-2(s-2)}}{s-2})
D. (frac{e^{-2s}}{s-2})
Answer: C
Clarification: X(s) = (int_0^∞ x(t) e^{-st} ,dt)
= (int_0^∞ e^{2t} u(-t+2) e^{-st} ,dt)
= (int_0^2 e^{t(2-s)} ,dt)
= (frac{e^{2(2-s)-1}}{2-s})
= (frac{1-e^{-2(s-2)}}{s-2}).
9. The Laplace transform of the signal sin 5t is _____________
A. (frac{5}{s^2+5})
B. (frac{s}{s^2+5})
C. (frac{5}{s^2+25})
D. (frac{s}{s^2+25})
Answer: C
Clarification: We know that, sin 5t = (frac{(e^{j5t}-e^{-j5t})}{2j} )
So, X(s) = (int_0^∞ frac{(e^{j5t}-e^{-j5t})}{2j} e^{-st} ,dt )
= (frac{5}{s^2+25}).
10. The Laplace transform of the signal u(t) – u(t-2) is ______________
A. (frac{e^{-2s}-1}{s})
B. (frac{-e^{-2s}+1}{s})
C. (frac{2}{s})
D. (frac{-2}{s})