Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : DEC 2015

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 + 4)} -------- 1 \\ \text {Here } A = (s + 4) X(s)|_{s = -4} = (s + 4) \dfrac {8}{(s + 4)(s + 2)^3} |_{s = -4} \\ =\dfrac 8{(s + 2)^3} |_{s = -4} \\ =\dfrac 8{(- 4 + 2)^3} \\ = -1 \\ ∴A = -1 \\ Now, \\ X_1(s) = (s - s_O)^n X(s) \\ ∴X_1(s) = (s + 2)^3 X(s) \\ = (s + 2)^3 \dfrac 8{(s + 4)(s + 2)^3 } \\ ∴X_1(s) = \dfrac 8{(s + 4) } $

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 8{(s + 4)} ] |_{s = -2} \\ = \dfrac 8{(-2 + 4)}\\ ∴k_O= 4 \\ ∴k_1 = \dfrac 1{1!} \dfrac {d^1}{ds^1} [\dfrac 8{(s + 4)} ] |_{s = -2} \\ = \dfrac {-8}{(s + 4)^2} |_{s = -2} \\ = \dfrac {-8}{(-2 + 4)^2} |_{s = -2} \\ ∴k_1 = -2 \\ ∴k_2 = \dfrac 1{2!}\dfrac { d^2}{ds^2} [\dfrac 8{(s + 4) }] |_{s = -2} \\ = \dfrac 1{2!} \dfrac {d}{ds} [\dfrac {-8}{(s + 4)^2} ] |_{s = -2} \\ =\dfrac 1{2!} [\dfrac {(8) (2)}{(s + 4)^3} ] |_{s = -2} \\ = \dfrac 12 [\dfrac {(8) (2)}{(-2 + 4)^3} ] \\ ∴k_2 = 1 $

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

$∴X(s) = \dfrac 4{(s + 2)^3} + \dfrac {-2}{(s + 2)^2} + \dfrac 1{(s + 2)} + \dfrac {-1}{(s + 4)} ---------- 2$

(iv) To obtain inverse Laplace transform for Re(s) > -2

The ROC of Re(s) > -2 is right sided as shown in the figure below. Hence the signals will be right sided.

Taking the inverse Laplace transform of equation 2

$x(t) = L^{-1}X(s) \\ = 4 L^{-1}\dfrac 1{(s + 2)^3} - 2 L^{-1} \dfrac 1{(s + 2)^2} + L^{-1} \dfrac 1{(s + 2)} - L^{-1}\dfrac 1{(s + 4)} ---------- 3 $

For right sided ROC,

$L [\dfrac {t^{n – 1}}{(n - 1)!} e^{-at} u(t)] = \dfrac 1{(s + a)^n} , ROC: Re(s) \gt -a $

Using the above result, equation 3 becomes

$∴x(t) = 4 \dfrac {t^2}{2!} e^{-2t} u(t) – 2 \dfrac t{1!} e^{-2t} u(t) + e^{-2t} u(t) - e^{-4t} u(t ) \\ ∴x(t) = 2 t^2e^{-2t} u(t) – 2 te^{-2t} u(t) + e^{-2t} u(t) - e^{-4t} u(t) $

This is the required time domain signal.

(v) To obtain inverse Laplace transform for Re(s) < -4

The ROC of Re(s) < -4 is left sided as shown in the figure below. Hence the signals will be left sided.

Taking the inverse Laplace transform of equation 2

$x(t) = L^{-1}X(s) \\ = 4 L^{-1}\dfrac 1{(s + 2)^3} - 2 L^{-1} \dfrac 1{(s + 2)^2} + L^{-1} \dfrac 1{(s + 2)} - L^{-1}\dfrac 1{(s + 4)} ---------- 4 $

For left sided ROC,

$L [\dfrac {-t^{n – 1}}{(n - 1)!} e^{-at} u(-t)] = \dfrac 1{(s + a)^n} , ROC: Re(s) \lt -a $

Using the above result, equation 4 becomes

$∴x(t) = -4 \dfrac {t^2}{2!} e^{-2t} u(t) + 2 \dfrac t{1!} e^{-2t} u(-t) -e^{-2t} u(t) + e^{-4t} u(-t ) \\ ∴x(t) = -2 t^2e^{-2t} u(t) + 2 te^{-2t} u(t) - e^{-2t} u(t) + e^{-4t} u(-t) $

This is the required time domain signal.

(vi) To obtain inverse Laplace transform for -4 < Re(s) < -2

The ROC of -4 < Re(s) < -2 is shown in the figure below. This ROC can also be expressed as Re(s) < -2 and Re(s) > -4.

Consider the given function i.e.

$X(s) =\dfrac 4{(s + 2)^3} + \dfrac {-2}{(s + 2)^2} + \dfrac 1{(s + 2)} + \dfrac {-1}{(s + 4)} ----------- 5 $

This function has two poles.

Poles at s = -4 and s = -2

The pole at s = -4 is to left side of the ROC. Hence the term corresponding to this ROC will be right sided i.e.

$L^{-1}\dfrac 1{(s + 4)} = e^{-4t} u(t) for ROC: Re(s) \gt -4$

Similarly, the pole at s = -2 is to right side of the ROC. Hence the term corresponding to this ROC will be left sided i.e.

$ L^{-1} \dfrac 1{(s + 2)^3} = -e^{-2t} \dfrac {t^2}{2!} u(-t) \space\space for\space\space ROC: Re(s) \lt -2 \\ L^{-1} \dfrac 1{(s + 2)^2} = -e^{-2t} \dfrac t{1!} u(-t)\space\space for\space\space ROC: Re(s) \lt -2 \\ L^{-1}\dfrac 1{(s + 2)} = -e^{-2t}u(-t)\space\space for \space\space ROC: Re(s) \lt -2 $

Hence, inverse Laplace transform of equation 5 becomes

$x(t) = 4 (-e^{-2t}\dfrac { t^2}{2!} u(-t)) – 2(-e^{-2t}\dfrac { t}{1!} u(-t)) + (-e^{-2t}u(-t)) – (e^{-4t} u(t)) \\ ∴x(t) = -2 t^2e^{-2t} u(-t) + 2 te^{-2t }u(-t) - e^{-2t} u(-t) - e^{-4t} u(t)$

This is the required time domain signal.