Lecture – 2
First and Second Order
Linear Differential Equations
Dr. Radhakant Padhi
Asst. Professor
Dept. of Aerospace Engineering
Indian Institute of Science - Bangalore
Linear Differential Equations
Principle of superposition:
If x1 (t ) and x2 (t ) are solutions then α1 x1 (t ) + α 2 x2 (t )
is also a solution for any scalars
α1 and α 2
Example:
x + x = 1 (Non homogeneous)
x1 = 1 + cos t and x2 = 1 + sin t both are solutions.
But
2(1 + cos t ) or [(1 + cos t ) + (1 + sin t )] are not solutions.
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Linear Differential Equations
However, if
x + x = 0
(Homogeneous)
then both of the above are solutions.
Lesson: Principle of superposition holds good only for
homogeneous linear differential equations.
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Homogeneous First-order Equations
System dynamics: x + kx = 0, x (t0 ) = x0 , k = 1/ τ
Solution:
( dx / x ) = −k dt
ln x = −k t + ln c
ln ( x / c ) = −k t
Initial condition:
Hence,
x = e− k t c
x0 = e−kt0 c, c = ekt0 x0
x(t ) = e − k (t −t0 ) x0
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Homogeneous First-order Equations
Note:
1)
2 2
3 3
a
t
a
t
at
+
+"
e = 1 + at +
2!
3!
2) If t0 = 0, then the solution is
x(t ) = e − k t x0 = e
− (t / τ )
x0
x0
0.63 x0
(0, 0)
τ
t
(no overshoot)
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Non-homogeneous First-order
Equations: Response to Step Inputs
Consider:
x +
1
τ
x=A
(A: constant)
Homogeneous solution:
1
xh + xh = 0
τ
−
t
xh = e τ c
Particular solution: x p = B
Substitute:
1
τ
B=A
∴ xp = B = τ A
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Non-homogeneous First-order
Equations: Response to Step Inputs
x(t ) = xh + x p
−
t
x(t ) = e τ c + τ A
x0 = c + τ A
x(t ) = e
−
t
τ
⇒ c = x0 − τ A
( x0 − τ A) + τ A
xss = lim x(t ) = τ A
t →∞
(Note: xss ≠ A)
Complete solution: x(t ) = e
−
t
τ
( x0 − τ A) + τ A
Steady state error (wrt. input A):
ess = A − [ x(t ) ]t →∞ = A − τ A = (1 − τ ) A
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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First-order Equations in Special Form:
Unit Step Response
Consider:
⎛1⎞
⎛1⎞
x + ⎜ ⎟ x = ⎜ ⎟ A
⎝τ ⎠
⎝τ ⎠
Homogeneous:
xh +
(Special case in text books: A = 1)
1
τ
xh = 0
−
t
xh = e τ c
Particular solution: x p = B
Substitute:
1 ⎞
⎛1
=
B
A ⎟ ⇒ ( B = A)
⎜
τ ⎠
⎝τ
∴ xp = B = A
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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First-order Equations in Special Form:
Unit Step Response
x(t ) = xh + x p
−
t
−
t
x(t ) = e τ c + A
x0 = c + A ⇒ c = x0 − A
x(t ) = e
τ
( x0 − A) + A
xss = lim x(t ) = A
t →∞
Complete solution: x(t ) = e
−
t
τ
( x0 − A) + A
Steady state error (wrt. input A): ess = A − xss = A − A = 0
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Unit Step Response of a
First-Order System
x (t )
T =τ
A =1
Settling time: Ts = 4T = 4τ
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Second Order System
Homogeneous:
x + ax + bx = 0,
Guess:
(a = 2ξωn , b = ωn 2 )
x(t ) = eλt
( λ 2 + aλ + b ) e λ t = 0
eλt ≠ 0
⇒ λ 2 + aλ + b = 0
: characteristic equation
− a ± a 2 − 4b
λ1,2 =
2
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Second Order System
Note: a, b ∈ R
⇒ Arises three cases
Case 1: Distinct and real roots
Case 2: Complex conjugate roots
Case 3: Real and repeated roots
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Case-1: Distinct and real roots
x1 = e
λ1 t
x2 = e
,
λ2t
x ( t ) = c1 x1 ( t ) + c 2 x 2 ( t )
x ( t ) = c1e
λ1 t
+ c2e
λ2t
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Case-2: Complex conjugate roots
λ 1 = σ + jω ,
λ 2 = σ − jω
x 1 = e ( σ + j ω ) t = e σ t (c o s ω t + j s in ω t )
x 2 = e ( σ − j ω ) t = e σ t (c o s ω t − j s in ω t )
x ( t ) = c 1 e ( σ + jω ) t + c 2 e ( σ − jω ) t
However notice that:
Linearly
independent
terms
x1 + x 2
= eσt cosω t
2
x1 − x 2
= e σ t s in ω t
2 j
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Case-2: Complex conjugate roots
So, we can also write the solution as :
⎛ x1 + x2
x(t ) = A ⎜
⎝ 2
⎛ x1 − x2 ⎞
⎞
⎟
⎟ + B⎜
⎠
⎝ 2j ⎠
x(t ) = eσ t ( A cos ωt + B sin ωt )
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Complex conjugate roots:
Typical response plot
x (t )
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Case-3: Real and repeated roots
λ1 = λ 2 = λ
x1 (t ) = e
λt
How to find
x 2 (t )
Method of variation of parameters
Assume: x2 (t ) = u (t ) x1 (t )
Then
x2 = u x1 + u x1
x2 = u x1 + u x1 + u x1 + u x1
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Case-3: Real and repeated roots
Substituting back and rearranging the terms, it leads to
u ( x1 + a x1 + b ) + u ( 2 x1 + a x1 ) + u x1 = 0
However,
( x1 + a x1 + b ) = 0
( since x1 is a solution )
⎛a⎞
Moreover, x1 = λ eλ t = λ x1 = − ⎜ ⎟ x1
⎝2⎠
( 2 x1 + a x1 ) = 0
a⎞
⎛
since
=
−
λ
⎜
⎟
2⎠
⎝
This leads to u x1 = 0. Moreover x1 = eλt ≠ 0. Hence u = 0
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Case-3: Real and repeated roots
u (t ) = t
One solution
∴ x (t ) = t x (t ) = te λ
2
t
1
x(t ) = c1 x1 (t ) + c 2 x 2 (t )
λt
x (t) = (c1 + c2 t) e
Corollary: When λ = 0
(double poles at origin)
x(t ) = c + c t → ∞ as t → ∞
1
2
Double pole (in general, multiple poles) at origin is de-stabilizing..!
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Summary
Case
1
Roots of characteristic
equation
Distinct & Real λ1 , λ 2
Complex Conjugate
2
3
λ = σ + jω
Real double Roots
Basics
General Solution
e λ1t , e λ2t
c1e λ1t + c 2 e λ2t
eσ t cos ωt
eσ t sin ωt
λt
e , te
eσ t ( A cos ωt + B sin ωt )
λt
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
(c1 + c 2 t )e λt
21
Example
x − 4 x + 4 x = 0
λ 2 − 4λ + 4 = 0
x(0) = 3, x (0) = 1
λ = 2, 2
x(t ) = (c1 + c2t ) e
(Real, repeated)
2t
x (t ) = 2 (c1 + c2 t ) e 2 t + c2 e 2 t
Solving for the initial conditions
c1 = 3
c 2 = −5
x(t ) = (3 − 5t ) e 2t
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Particular Solution: Method of
Undetermined Coefficients
x + ax + bx = f (t )
Terms in f(t)
ke
pt
Choice of xp(t)
ce
pt
k tn
k cos ωt
kn t n + kn −1 t n −1 + " + k1t + k0
k sin ωt
A cos ωt + B sin ωt
A cos ωt + B sin ωt
Substitute these expressions back in the differential
equation and determine the unknown coefficients
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Example
Find
x p (t ) for
Choose
x − 3 x + 2 x = 4t + e3t
x p (t ) = (k1t + k0 ) + ce3t
Substitute and equate the coefficients
1
k1 = 2, k 0 = 3, c =
2
1 3t
∴ x p (t ) = (2t + 3) + e
2
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Second order system in
Standard form
c + 2ςωn c + ω c = ω u
2
n
2
n
ωn : Natural frequency
ς : Damping ratio
Transfer function (will be studied later):
C (s)
ωn2
= 2
2
U ( s ) s + 2ςωn s + ωn
Roots: poles
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Unit Step Response of a
Second-order System
ξ=0
0<ξ<1
ξ=1
ξ>1
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Under-damped system response
specifications
Ref: N. S. Nise:
Control Systems Engineering,
4th Ed., Wiley, 2004
Tr : Rise time
Tp : Pick time
Ts : Settling time
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Transient Response
Specifications
π −β
R is e tim e : T r =
,
ωd
π
P e a k tim e : T p =
ωd
D a m p e d n a tu ra l fre q u e n c y: ω d = ω n 1 − ζ
M a x im u m o v e r s h o o t: M
S e ttlin g tim e : T s =
=
4
ξω n
3
ξω n
p
= e
⎛ − πξ
⎜
⎜
⎝
1− ζ
2
⎞
⎟
⎟
⎠
(2%
c r ite r io n )
(5 %
c r ite r io n )
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Percentage over-shoot as a
function of damping ratio
Ref: N. S. Nise:
Control Systems Engineering,
4th Ed., Wiley, 2004
Good to increase damping ratio.
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Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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Rise time as a function of
damping ratio
Ref: N. S. Nise:
Control Systems Engineering,
4th Ed., Wiley, 2004
Increasing damping ratio too much is Bad..!
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Second-order System Response
as a Function of Damping Ratio
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ADVANCED CONTROL SYSTEM DESIGN
Dr. Radhakant Padhi, AE Dept., IISc-Bangalore
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