$(i) \space X(s)=\dfrac {s-3}{s^2+4s+13} \ (ii)\space X(s) = \dfrac {5s^2-15s-11}{(s+1)(s-2)^3}$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2015

$$(i)\space X(s) = \dfrac {s - 3}{s^2 + 4s + 13}$$

Denominator can be written as

$X(s) = \dfrac {s - 3}{s^2 + 4s + 4 + 9 } \\ ∴X(s) = \dfrac {s - 3}{(s + 2)^2 + 9 } \\ X(s) =\dfrac { s - 3}{(s + 2)^2 + 3^2 } $

Numerator can be written as

$X(s) = \dfrac {(s + 2) - 5}{(s + 2)^2 + 3^2 } \\ ∴X(s) = \dfrac {s + 2}{(s + 2)^2 + 3^2} – \dfrac 5{(s + 2)^2 + 3^2} ---------- 1 \\ ∵L-1 \dfrac {(s + b)}{(s + b)^2 + a^2} = e^{-bt} \cos(at) \\ And \\ L-1 \dfrac {1}{(s + b)^2 + a^2} =\dfrac { e^-{bt}}a \sin(at)$

Therefore, the Inverse Laplace transform of equation 1 is

$L-1[X(s)] = e^{-2t} \cos(3t) – (5) \dfrac {e^{-2t}}3 \sin(3t) \\ ∴L-1[X(s)] = e^{-2t} \cos(3t) – \dfrac {5e^{-2t}}3 \sin(3t) $

$$ (ii)\space X(s) = \dfrac {5s^2 – 15s – 11}{(s + 1)(s - 2)^3} $$

The given function can be expressed as

$X(s) =\dfrac {k_O}{(s - 2)^3} + \dfrac {k_1}{(s - 2)^2} + \dfrac {k_2}{(s - 2)} + \dfrac {A}{(s + 1)} -------- 1 \\ \text {Here}\space A = (s + 1) X(s)|_{s = -1} = (s + 1) \dfrac {5s^2 – 15s – 11}{(s + 1)(s - 2)^3} |s = -1 \\ = \dfrac {5s^2 – 15s – 11}{(s - 2)^3} |s = -1 \\ = \dfrac {5(-1)^2 – 15(-1) – 11}{(-1 - 2)^3 } \\ =\dfrac 13 \\ ∴A =\dfrac 13 \\ Now, \\ X_1(s) = (s - s_O)^n X(s) \\ ∴X_1(s) = (s - 2)^3 X(s) \\ = (s - 2)^3 \dfrac {5s^2 – 15s – 11}{(s + 1)(s - 2)^3 } \\ ∴X_1(s) = \dfrac {5s^2 – 15s – 11}{(s + 1) } $

The value of k is obtained by the formula

$k_j = \dfrac 1{j!} \dfrac {dj}{dsj} X_1(s) |_{s = so}\space\space\space j = 0, 1, 2, 3…..n – 1 \\ ∴k_O = \dfrac 1{0!}\dfrac { d^0}{ds^0} [\dfrac {5s^2 – 15s – 11}{(s + 1)} ] |_{s = 2} \\ = \dfrac {5(2)^2 – 15(2) – 11}{(2 + 1) } \\ ∴k_O= -7 \\ ∴k_1 = \dfrac 1{1!} \dfrac {d^1}{ds^1} [\dfrac {5s^2 – 15s – 11}{(s + 1)} ] |_{s = 2} \\ = \dfrac {(s + 1) (10s - 15) – (5s^2 – 15s - 11) (1)}{(s + 1)^2 }|_{s = 2} \\ =\dfrac { 5s^2 + 10s – 4}{(s + 1)^2} |_{s = 2} \\ = \dfrac {5(2)^2 + 10(2) – 4}{(2 + 1)^2 } \\ ∴k_1 = 4 \\ ∴k_2 = \dfrac 1{2!} \dfrac {d^2}{ds^2} [\dfrac {5s^2 – 15s – 11}{(s + 1)} ] |_{s = 2} \\ = \dfrac 1{2!} \dfrac {d}{ds} [\dfrac {(s + 1) (10s - 15) – (5s^2 – 15s - 11) (1)}{(s + 1)^2} ] |_{s = 2} \\ = \dfrac 1{2!} \dfrac {d}{ds} [\dfrac {5s^2 + 10s – 4}{(s + 1)^2} ] |_{s = 2} \\ = \dfrac {1}{2!} [\dfrac {(s + 1)^2 (10s + 10) - (5s^2 + 10s - 4) (2(s + 1))}{(s + 1)^4 }] |_{s = 2} \\ =\dfrac 1{2!} \dfrac {18}{(s + 1)^3} |_{s = 2} \\ = \dfrac {1}{2!} \dfrac { 18}{(2 + 1)^3 } \\ ∴k_2 = \dfrac 13 $

Substituting the values of $k_O, k_1$ and $k_2$ in equation 1

$∴X(s) = \dfrac {-7}{(s - 2)^3} + \dfrac {4}{(s - 2)^2} + \dfrac {1/3}{(s - 2)} + \dfrac {1/3}{(s + 1)} --------- 2 \\ ∵L^{-1} \dfrac 1{(s - b)^n} = e^{bt} \dfrac {t^{n – 1}}{(n - 1)! } $

Therefore, Inverse Laplace transform of equation 2 is given by

$∴L-1[X(s)] = \dfrac {-7e^{2t} t^2}2 + 4e^{2t} t + \dfrac 13 e^{2t} + \dfrac 13 e^{-t}$