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Characteristic Equation, Characteristic Roots and Characteristic Modes

This assignment is worth 10% of the unit mark. This assignment has 3 questions. Total 60 marks.

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Added on  2022-10-19

Characteristic Equation, Characteristic Roots and Characteristic Modes

This assignment is worth 10% of the unit mark. This assignment has 3 questions. Total 60 marks.

   Added on 2022-10-19

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a. Find the characteristic equation, characteristic roots and characteristic modes of
this system:
Characteristic equation:
d2y(t)/dt2 + 4.dy(t)/d(t) +4y(t) = dx(t)/dt + 0.5 x(t)
Taking Laplace transfome:
L{d2y(t)/dt2} + 4L {dy(t)/d(t)} + 4L {y(t)} = L{dx(t)/dt} + 0.5L {x(t)}
Transfer Function = Laplace transform of output / Laplace Transform of input
Initial conditions sets to be zero:
y(0) = 0
x(0) = 0
H(s) = Y(s)/X(s) = (s+0.5)/ (s2 + 4s + 4)
Characteristics Roots:
(s +0.5) = 0
S= -0.5
Poles:
s+0.5 = 0,
s= -0.5,
Zeroes:
(s2 + 4s + 4) = 0
Taking the partial fraction:
s2 + (2 +2) s+ 4=0
s2 + 2s+2s+ 4=0
(s+2) (s+2) = 0
s = -2,-2.
Characteristic modes:
Characteristic Equation, Characteristic Roots and Characteristic Modes_1
If the number of poles are less than the number of zeros i.e. Z > P then the value of transfer
function becomes infinity. Hence it shows that there are poles at infinity and the number of such
poles is (Z-P)
Poles < Zeroes
b. Comment on the stability of the system.
H(s) = Y(s)/X(s) = (s+0.5)/ (s2 + 4s + 4)
Poles:
s+0.5 = 0,
s= -0.5
Zeroes:
(s+2) (s+2) = 0
s = -2,-2.
When the poles of the system are located in the left-half plane (LHP) and the system is not
improper, the system is shown to be stable.
Characteristic Equation, Characteristic Roots and Characteristic Modes_2
c. Find y0(t), the zero-input component of the response For t ≥ 0, if the initial
conditions are y(0) = 3 and y (0) = −4.
Characteristic equation:
d2y(t)/dt2 + 4.dy(t)/d(t) +4y(t) = dx(t)/dt + 0.5 x(t)
d2y(t)/dt2 + 4.dy(t)/d(t) +4y(t) = 0
λ2 + 4 λ + 4 = 0
(λ +2) (λ +2) = 0
λ = -2,-2
yh (t) = C1 e-2t + C2 e-2t
Characteristic Equation, Characteristic Roots and Characteristic Modes_3
y0(t) = 3, 3 = C1 +C2
y’0(t) = dy0(t)/dt
y’0(t) = -2 e-2t - 2 e-2t
y’0(t) = -4, -4= -2C1 - 2C2
4 = 2 C1 + 2C2
C1 = 1
C2 = 4
Y0 (t) = e-2t + 4 e-2t
d. Mathematically derive an expression for (t) , the impulse response of the system.
H(s) = (s+0.5)/ (s2 + 4s + 4)
h(t) = L-1H(s)
h(t) = L-1 {(s+0.5)/ (s2 + 4s + 4)}
h(t) = L-1 {(s+0.5)/ (s+2)2}
(s+0.5)/ (s+2)2 = a 0
( s+2) + a 1
( s+2)2
For the denominator root:
a0 = 1
a1 = -1.5
h(t) = L-1{ 1
(s+2) - 1.5
( s+2) 2}
Use the linearity property of inverse Laplace transform:
Characteristic Equation, Characteristic Roots and Characteristic Modes_4

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