Unformatted text preview: ME375 Handouts Dynamic Response of Linear Systems
Dynamic response: total response of a dynamic system.
response total response of dynamic system.
It can be decomposed by two methods:
• Forced and Free Response
–
–
–
– • Forced response of 1st and 2nd order systems due to a unit
step input (i.e. unit step response)
Performance specifications of unit step response
Free response of 1st and 2nd order systems
Pole location and free response Transient and SteadyState Response
–
– Transient response decay to zero
SteadySteadystate response takes the form of the forcing input
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Purdue University ME375 Dynamic Response  1 1 ME375 Handouts Linear System Response
y ( n ) + a n −1 y ( n −1) + + a 2 y + a1 y + a 0 y = bm u ( m ) + + b1u + b0 u • Superposition Principle
Input Output Linear System
u1 (t) y1 (t) u2 (t) y2 (t) k1 u1 (t) + k2 u2 (t)
The response of a linear system to a complicated input can be obtained by
studying how the system responds to simple inputs, such as zero input, unit
impulse , unit step, and sinusoidal inputs. School of Mechanical Engineering
Purdue University ME375 Dynamic Response  2 2 ME375 Handouts Typical Responses
• Free (Natural) Response
– Response due to nonzero initial conditions (ICs) and zero input.
non • Forced Response
– Response to nonzero input with zero ICs.
non– Unit Impulse Response
Response to unit impulse
input. u(t) – Unit Step Response
Response to unit step
input (u (t) = 1).
u(t) Time t – Sinusoidal Response
Response to sinusoidal
inputs at different
frequencies.
The steady state sinusoidal
response is call the
Frequency
Frequency Response. Time t
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Purdue University ME375 Dynamic Response  3 3 ME375 Handouts Forced Response of Stable 1st Order System
• Stable 1st Order System
y + ay = b u
where ⇒ τy + y = Ku τ : Time Constant
Time
K : Static (Steady State, DC) Gain
Static (force
and
– Unit Step Response (force response: u = 1 and zero ICs )
Y (s) = K
(τ s + 1) s y(t)
yP(t) = K K τK
K
−
s τ s +1
K
K
=
−
1
s
s+
= τ y (t ) = K − K e − Time t
yH(t) = − K e t/τ t τ
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Purdue University ME375 Dynamic Response  4 4 ME375 Handouts Forced Response of Stable 1st Order System
Normalized Unit Step Response (u = 1 & zero ICs)
zero
1 τy + y = Ku
⇒ y ( t ) = K (1 − e −tτ Normalized Response 0.9 ) Normalized
(such t hat as t → ∞ , y n → 1) :
that a s
1) y(t )
−t
⇒ yn ( t ) =
= (1 − e τ )
K 0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 1τ 0 τ Time t
( 1− e − t/ τ 2τ 3τ 2τ 3τ
Time [ t ] 4τ 4τ 5τ 6τ 5τ ) School of Mechanical Engineering
Purdue University ME375 Dynamic Response  5 5 ME375 Handouts Forced Response of Stable 1st Order System
Normalized Unit Step Response
τy + y = Ku
−tτ 0.9 ) 0.8 Normalized
(such that as t → ∞ , yn → 1 ): y(t )
−t
⇒ yn ( t ) =
= (1 − e τ )
K
Initial Slope 1
yn ( 0 ) =
τ
K
y(0 ) =
τ Normalized Response ⇒ y ( t ) = K (1 − e 1 0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Time [sec] School of Mechanical Engineering
Purdue University ME375 Dynamic Response  6 6 ME375 Handouts Forced Response of Stable 1st Order System
Q: How would you calculate the response
of
of a 1st order system to a unit pulse
(not unit impulse)?
u(t) t1 Time t (Hint: superposition principle) School of Mechanical Engineering
Purdue University ME375 Dynamic Response  7 7 ME375 Handouts Forced Response of Stable 2nd Order System
• Stable 2nd Order System
y + a1 y + a 0 y = b u
where ⇒ y + 2ζω n y + ω n 2 y = K ω n 2 u ωn > 0 : Natural Frequency [rad/s]
ζ > 0 : Damping Ratio
K : Static (Steady State, DC) Gain
Static Characteristic roots
roots
s 2 + 2ζω n s + ω n2 = 0
⇒ s = −ζω n ± ω n Img. ωn ( ζ 2 − 1) ⎧ζ > 1 :
⎪
⎨ζ = 1 :
⎪ζ < 1 :
⎩ Real −ωn
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Purdue University ME375 Dynamic Response  8 8 ME375 Handouts Forced Response of Stable 2nd Order System
Unit Step Response of Underdamped 2nd Order Systems ( u = 1 and zero ICs )
Undery + 2ζ ω n y + ω n 2 y = K ω n 2 u
Characteristic equation:
s 2 + 2ζ ω n s + ω n 2 = 0
⇒ s = −ζ ω n ± j ω n (1 − ζ 2 ) ωd ⎡
⎛
⎞⎤
⎢1 − e−ζωnt ⎜ cos ωd t + ζ
y (t ) = K
sin ωd t ⎟ ⎥
2
⎜
⎟⎥
⎢
1− ζ
⎝
⎠⎦
⎣ Response:
Response: = K ⎡1 − e−ζωnt cos (ωd t + φ ) ⎤
⎣
⎦
Question: how does it compare to the response of first order system?
τ y + y = Ku ⇒ y ( t ) = K (1 − e −t τ) School of Mechanical Engineering
Purdue University ME375 Dynamic Response  9 9 ME375 Handouts Forced Response of Stable 2nd Order System
How the solution was derived:
Y (s) = 2
K ωn
2
s ( s 2 + 2ζω n s + ω n ) = Ks + 2 K ζω n
K
−2
2
s s + 2ζω n s + ω n = K ( s + ζ ωn ) + K ζ ωn
K
−
s ( s + ζω )2 + (ω 1 − ζ 2 ) 2
n
n y (t ) = K − Ke−ζωn cos ωn 1 − ζ 2 t − K ζ e−ζωn sin ωn 1 − ζ 2 t 1− ζ 2
⎡
⎛
⎞⎤
ζ
−ζωnt
⎜ cos ωd t +
= K ⎢1 − e
sin ωd t ⎟ ⎥
2
⎜
⎟⎥
⎢
1− ζ
⎝
⎠⎦
⎣
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Purdue University ME375 Dynamic Response  10 10 ME375 Handouts Forced Response of Stable 2nd Order System
Unit Step Response of stable underdamped 2nd Order Systems
under1.6K yMAX
1.4K OS
Unit Step Response 1.2K
K
0.8K Td
0.6K
0.4K
0.2K
0 0 1 tP 2 3 4 5 6 7
8
Time [sec] 9 School of Mechanical Engineering
Purdue University 10 11 tS 12 13 14 15 ME375 Dynamic Response  11 11 ME375 Handouts Forced Response of Stable 2nd Order System
• Peak Time (tP)
Time when output y(t) reaches its
maximum value yMAX. Percent Overshoot (%OS)
At peak time tP the maximum
output
πζ FG
H y MAX = y ( t P ) = K ⋅ 1 + e y (t ) =
⎛
⎡
⎤⎞
ζω n
sin(ω d t ) ⎥ ⎟
K ⋅ ⎜ 1 − e −ζω n ⋅t ⎢ cos(ω d t ) +
ωd
⎣
⎦⎠
⎝
d
y (t ) =
dt
Find t P such that y ( t P ) = 0
⇒ • − 1− ζ 2 IJ
K The overshoot (OS) is:
O S = y M AX − y SS = The percent overshoot is:
% OS = tP = FG
Hy IJ
K OS
100%
SS − y ( 0 ) = School of Mechanical Engineering
Purdue University ME375 Dynamic Response  12 12 ME375 Handouts Forced Response of Stable 2nd Order System
• Settling Time (tS)
Time required for the response to
be within a specific percent of the
final (steadystate) value.
(steadySome typical specifications for
settling time are: 5%, 2% and 1%.
Look at the envelope of the response: %
tS 1% 2% 5% Q: What parameters in a 2nd order system
affect the peak time?
Q: What parameters in a 2nd order system
affect the % OS?
Q: What parameters in a 2nd order system
affect the settling time?
Q: Can you obtain the formula for a 3%
settling time?
settling time? School of Mechanical Engineering
Purdue University ME375 Dynamic Response  13 13 ME375 Handouts Pole location and Performance specifications
Img. ωn Real −ωn School of Mechanical Engineering
Purdue University ME375 Dynamic Response  14 14 ME375 Handouts System parameters and performance specifications
• MassSpringMassSpringDamper System
Q: What is the static (steadystate) gain of
(steadythe system ? x K
M f(t) B I/O Model: M x + B x + K x = f (t ) Q: How would the physical parameters
(M
(M, B, K) affect the response of the
K) affect the response of the
system
system ?
(This is equivalent to asking you for the
relationship between the physical
parameters and the damping ratio, natural
frequency and the static gain.) School of Mechanical Engineering
Purdue University ME375 Dynamic Response  15 15 ME375 Handouts Example School of Mechanical Engineering
Purdue University ME375 Dynamic Response  16 16 ME375 Handouts Summary on unit step response School of Mechanical Engineering
Purdue University ME375 Dynamic Response  17 17 ME375 Handouts Summary on pole location and performance specifications
performance
Img. ωn Real −ωn School of Mechanical Engineering
Purdue University ME375 Dynamic Response  18 18 ME375 Handouts Free Response of 1st Order Systems
y + ay = b u
• Characteristic Equation: s+a =0 ⇒
• Free Response [ yFree(t)]: (u = 0)
0) a>0
e.g. ⇒ y Free ( t ) = A e − a t = y ( 0 ) e − a t a=0 y Free ( t ) = A e − 4 ⋅t yH (t) e.g. a<0 y Free ( t ) = A e 0 ⋅t y Free ( t ) = A e − ( − 4 ) ⋅ t yH (t) yH (t) Time (t) e.g. Time (t) Time (t) Q: What determines whether the free response will converge to zero ?
Q: How does the coefficient, a, affect the converging rate ?
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Purdue University ME375 Dynamic Response  19 19 ME375 Handouts Free Response of 2nd Order Systems
y + a1 y + a0 y = b1 u + b0 u
• Characteristic Equation:
s 2 + a1 s + a 0 = 0
⇒ • Free Response
Free Response [ yFree(t)]: (u = 0)
0)
Determined by the roots of the characteristic equation:
– Real and Distinct [ s1 & s2 ]: y F r e e ( t ) = A1 e s1 ⋅t + A 2 e s 2 ⋅t – Real and Identical [ s1 = s2 ]: y F r e e ( t ) = A 1 e s1 ⋅ t + A 2 t e s1 ⋅ t – Complex [ s1,2 = α ± jβ ]
y Free ( t ) = e α ⋅t ( A1 cos( β t ) + A 2 sin( β t ) ) = A e α School of Mechanical Engineering
Purdue University ⋅t cos( β t + φ ) ME375 Dynamic Response  20 20 ME375 Handouts Free Response of 2nd Order Systems
Free Response (Two distinct real roots)
y Free (t ) = A1 e s1⋅t + A2 e s2 ⋅t
s1 < 0 & s2 < 0 y Free (t ) = A1 e s1⋅t + A2 e s2 ⋅t
s1 < 0 & s2 = 0 Img. y Free (t ) = A1 e s1⋅t + A2 e s2 ⋅t
s1 < 0 & s2 > 0
Img. Img. Real yH (t) yH (t) Time (t) Real Real
yH (t) Time (t) School of Mechanical Engineering
Purdue University Time (t) ME375 Dynamic Response  21 21 ME375 Handouts Free Response of 2nd Order Systems
Free Response (Two identical real roots )
y Free (t ) = A1 e s1⋅t + A2 te s1⋅t
s1 = s2 < 0 y Free (t ) = A1 e s1⋅t + A2 te s1⋅t
s1 = s2 = 0 y Free (t ) = A1 e s1⋅t + A2 te s1⋅t
s1 = s2 > 0
Img. Img. Img. Real yH (t) yH (t) yH (t) Time (t) Real Real Time (t)
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Purdue University Time (t) ME375 Dynamic Response  22 22 ME375 Handouts Free Response of 2nd Order Systems
Free Response (Two complex roots)
y Free (t ) = A eα ⋅t cos( β t + φ )
s1,2 = α ± j β & α < 0 y Free (t ) = A eα ⋅t cos( β t + φ )
s1,2 = ± j β & α = 0 y Free (t ) = A eα ⋅t cos( β t + φ )
s1,2 = α ± j β & α > 0 Img. Img. Img. Real Real yH (t) yH (t) yH (t) Time (t) Real Time (t) School of Mechanical Engineering
Purdue University Time (t) ME375 Dynamic Response  23 23 ME375 Handouts Free Response of 2nd Order Systems
Q: What part of the characteristic roots determines whether the free response is bounded,
converging to zero or “blowing up” ?
“bl Q: For a second order system, what conditions will guarantee the system to be stable ?
second order system what conditions will guarantee the system to be stable
(Hint: Check the characteristic roots ) Q: If the free response of the system converges to zero, what determines the convergence
the free response of the system converges to zero what determines the convergence
rate? School of Mechanical Engineering
Purdue University ME375 Dynamic Response  24 24 ME375 Handouts Transient and Steady State Response
Ex: Let’s find the total response of a stable first order system:
y + 5 y = 10u
to
to a ramp input:
with IC:
– Total Response u(t) = 5t
y(0) = 2 Y ( s) = ⋅ U ( s) + ⋅ y(0), where U ( s) = L [ 5 t ] = Transfer Function G( s) ∴ Y ( s) = ⋅ + – PFE: Y ( s) = + A1 + A2 + A3 y (t ) =
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Purdue University ME375 Dynamic Response  25 25 ME375 Handouts Transient and Steady State Responses
In general, the total response of a stable LTI system
total
stable
(n)
( n −1)
an y + an−1 y
+ + a1 y + a0 y = bm u ( m) + bm−1 u ( m−1) + + b1 u + b0 u bm s m + bm−1s m−1 + + b1s + b0 N (s) bm (s − z1 )(s − z2 ) (s − zm )
=
=
an s n + an−1s n−1 + + a1s + a0 D(s) an (s − p1 )(s − p2 ) (s − pn )
to
to an input u(t) can be decomposed into two parts:
G( s ) ≡ y (t ) = yT (t ) + ySS (t )
Transient
where
Response
• Transient Response (yT(t)) Steady State
Response – Contains the free response yFree(t) of the system plus a portion of the forced response.
– Will decay to zero at a rate that is determined by the characteristic roots (poles) of the
system. • Steady State Response (ySS(t))
– will take the same form as the forcing input.
– Specifically, for a sinusoidal input, the steady state response will be a sinusoidal signal
with the same frequency as the input but with different magnitude and phase.
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Purdue University ME375 Dynamic Response  26 26 ME375 Handouts Transient and Steady State Response
Ex: Let’s find the total response of a stable second order system:
y + 4y + 3y = 6u to a step input:
with IC:
– Total Response u(t) = 5
y ( 0 ) = 0 and y( 0) = 2 – PFE: School of Mechanical Engineering
Purdue University ME375 Dynamic Response  27 27 ME375 Handouts Steady State Response
• Final Value Theorem (FVT)
Given a signal’s LT F(s), if the poles of sF(s) all lie in the LHP (stable region),
the
all
then f(t) converges to a constant value f(∞). f(∞) can be obtained without
knowing f(t) by using the FVT: f ( ∞ ) = lim f (t ) = lim sF ( s )
t →∞ s→0 Ex: A model of a linear system is determined to be: y + 4 y + 1 2 y = 4u + 3u (1) if a constant input u = 5 is applied at t = 0, determine whether the output y(t) will
0,
converge to a constant value?
(2) If the output converges, what will be its steady state value? School of Mechanical Engineering
Purdue University ME375 Dynamic Response  28 28 ME375 Handouts Steady State Response
Given a stable LTI system an y(n) + an−1 y(n−1) + + a1 y + a0 y = bmu(m) + bm−1u(m−1) + + b1u + b0 u
The corresponding transfer function is
b m s m + bm −1 s m − 1 +
G (s) ≡
a n s n + a n −1 s n −1 + • + b1 s + b0
b ( s − z1 )( s − z 2 )
=m
+ a1 s + a 0 a n ( s − p1 )( s − p 2 ) (s − zm )
( s − pn ) Steady State Value of the Free Response
Recall
Recall the free response of the system is: YFree (s) = F(s)
=
an s + an−1s + + a1s + a0
n n−1 For stable system, apply FVT: School of Mechanical Engineering
Purdue University ME375 Dynamic Response  29 29 ME375 Handouts Steady State Response
• Steady State Value of the Unit Impulse Response
Y ( s) = G ( s) ⋅U ( s) =
For stable system, apply FVT: • Steady State Value of the Unit Step Response Y (s) = G(s) ⋅U (s) =
For stable system, apply FVT: 1st Order Systems:
G(s) =
⇒ G(0) = 2nd Order Systems: b0
a1s + a0 G( s ) = b1s + b0
a2 s + a1s + a0
2 ⇒ G(0) =
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Purdue University ME375 Dynamic Response  30 30 ME375 Handouts Steady State Response
• Steady State Value of the Harmonic Response
For forced response due to harmonic input, u(t) = sinωt ⎡ω⎤
Y(s) = G(s)U(s) = G(s) ⎢ 2 2 ⎥
⎣ s +ω ⎦
For stable systems, apply FVT:
FV
For ⎡ sω ⎤
yss = limsY (s) = limG(s) ⎢ 2 2 ⎥ = ??
s→0
s→0
⎣ s +ω ⎦ School of Mechanical Engineering
Purdue University ME375 Dynamic Response  31 31 ...
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 Fall '10
 Meckle
 Impulse response, Purdue University, School of Mechanical Engineering

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