Unformatted text preview: ME375 Handouts Dynamic Response of Linear Systems
• Linear System Response
– Superposition Principle
– Responses to Specific Inputs • Dynamic Response of 1st Order Systems
– Characteristic Equation  Free Response
– Stable 1st Order System Response • Dynamic Response of 2nd Order Systems
– Characteristic Equation  Free Response
– Stable 2nd Order System Response • Transient and SteadyState Response
School of Mechanical Engineering
Purdue University ME375 Dynamic Response  1 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
u1 (t) Output Linear System
Linear System y1 (t)
y2 (t) u2 (t)
k1 u1 (t) + k2 u2 (t) The response of a linear system to a complicated input can be obtained by
obtained
studying how the system responds to simple inputs, such as zero input, unit
input,
impulse , unit step, and sinusoidal inputs.
inputs.
School of Mechanical Engineering
Purdue University ME375 Dynamic Response  2 1 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) – Sinusoidal Response – Unit Step Response
Response to unit step
input (u (t) = 1). Response to sinusoidal
inputs at different
frequencies.
The steady state sinusoidal
response is call the
Frequency Response.
Response. u(t) Time t Time t
School of Mechanical Engineering
Purdue University ME375 Dynamic Response  3 Dynamic Response of 1st Order Systems
y + ay = b u
• Characteristic Equation: s+a =0 ⇒ • Free Response [ yH(t)]: (u = 0)
)]: a>0
e. g. ⇒ yH (t ) = A e − a t a=0 y H ( t ) = A e − 4 ⋅t yH (t) e. g. a<0 y H ( t ) = A e 0 ⋅t y H ( 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  4 2 ME375 Handouts Response of Stable 1st Order System
• Stable 1st Order System
y + ay = b u ⇒ τy + y = Ku where τ : Time Constant
K : Static (Steady State, DC) Gain
– Unit Step Response ( u = 1 and zero ICs )
y(t) y (t ) = y H (t ) + y P (t )
= Ae −t τ yP(t) = K K +K IC : y (0) = A + K
Time t A= yH(t) = − K y (t ) = e t/τ School of Mechanical Engineering
Purdue University ME375 Dynamic Response  5 Response of Stable 1st Order System
Normalized Unit Step Response (u = 1 & zero ICs)
1 τy + y = Ku
⇒ y ( t ) = K (1 − e −tτ Normalized Response 0.9 ) Norm alized
(such that as t → ∞ , y n → 1) : ⇒ yn ( t ) = y(t )
−t
= (1 − e τ )
K 0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 τ Time t
( 1− e −t/ τ 2τ 1τ 3τ 2τ 3τ
Time [ t ] 4τ 4τ 5τ 6τ 5τ ) School of Mechanical Engineering
Purdue University ME375 Dynamic Response  6 3 ME375 Handouts Response of Stable 1st Order System
Effect of Time Constant τ :
τy + y = Ku Normalized: ⇒ yn ( t ) = 0.9 ) 0.8
Normalized Response ⇒ y ( t ) = K (1 − e 1
−tτ y(t )
−t
= (1 − e τ )
K Slope at t = 0:
d
yn ( t ) =
dt
d
yn ( 0 ) =
⇒
dt
⇒ 0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 Q: What is your conclusion ? 2 4
6
Time [sec] School of Mechanical Engineering
Purdue University 8 10 ME375 Dynamic Response  7 Example – Vehicle Acceleration
v F m
b StandingStart Acceleration; Lincoln Aviator SUV StandingStart Acceleration; Dodge Viper SRT10
160 140 140 120 120 100 100
Speed (MPH) 160 Speed (MPH) m
1
v + v = F = vmax
b
b 80 80 60 60 40 40 20 20 0
0 5 10 15 20
Time (sec) 25 30 35 40 0
0 5 School of Mechanical Engineering
Purdue University 10 15 20
Time (sec) 25 30 35 40 ME375 Dynamic Response  8 4 ME375 Handouts Response of Stable 1st Order System
Normalized Unit Step Response
τy + y = Ku ⇒ y ( t ) = K (1 − e −tτ 1
0.9 ) Normalized
(such that as t → ∞ , yn → 1): ⇒ yn ( t ) = y(t )
−t
= (1 − e τ )
K Normalized Response 0.8 Initial Slope 0.7
0.6
0.5
0.4
0.3
0.2 1
yn ( 0 ) =
τ
K
y(0 ) =
τ 0.1
0
0 1τ School of Mechanical Engineering
Purdue University 2τ 3τ
Time [ t ] 4τ 5τ 6τ ME375 Dynamic Response  9 Response of Stable 1st Order System
Q: How would you calculate the response
of a 1st order system to a unit pulse
(not unit impulse)?
impulse)? Q: How would you calculate the unit
impulse response of a 1st order system?
system? u(t) t1 Time t (Hint: superposition principle ?!) Q: How would you calculate the
sinusoidal response of a 1st order
system?
system? School of Mechanical Engineering
Purdue University ME375 Dynamic Response  10 5 ME375 Handouts Dynamic 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 [ yH(t)]: (u = 0)
Determined by the roots of the characteristic equation:
– Real and Distinct [ s1 & s2 ]: y H ( t ) = A1 e s1 ⋅t + A 2 e s 2 ⋅t – Real and Identical [ s1 = s2 ]:
– Complex [ s1,2 = α ± jβ ] y H ( t ) = A 1 e s1 ⋅ t + A 2 t e s1 ⋅ t y H ( t ) = e α ⋅t ( A1 c o s ( β t ) + A 2 s in ( β t ) ) = A e α ⋅t c o s( β t + φ )
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Purdue University ME375 Dynamic Response  11 Dynamic Response of 2nd Order Systems
Free Response (Two distinct real roots)
roots)
y H (t ) = A1 e s1⋅t + A2 e s2 ⋅t
s1 < 0 & s2 < 0 y H (t ) = A1 e s1⋅t + A2 e s2 ⋅t
s1 < 0 & s2 = 0 Img. y H (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  12 6 ME375 Handouts Dynamic Response of 2nd Order Systems
Free Response (Two identical real roots )
y H (t ) = A1 e s1⋅t + A2 te s1⋅t
s1 = s2 < 0 y H (t ) = A1 e s1⋅t + A2 te s1⋅t
s1 = s2 = 0 y H (t ) = A1 e s1⋅t + A2 te s1⋅t
s1 = s2 > 0
Img. Img. Img. Real yH (t) yH (t) yH (t) Real Real Time (t) Time (t) Time (t) School of Mechanical Engineering
Purdue University ME375 Dynamic Response  13 Dynamic Response of 2nd Order Systems
Free Response (Two complex roots)
roots)
y H (t ) = A eα ⋅t cos( β t + φ )
s1,2 = α ± j β & α < 0 y H (t ) = A eα ⋅t cos( β t + φ )
s1,2 = ± j β & α = 0 y H (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  14 7 ME375 Handouts Example – Automotive Suspension
y
m my + by + ky = br + kr g
k b r Response to Initial Conditions
0.02 0 0.02
Amplitude for free response:
my + by + ky = 0
b
k
y+ y+ y =0
m
m
y + 28 y + 400 y = 0 0.04 0.06 0.08 0.1 0 0.05 0.1 School of Mechanical Engineering
Purdue University 0.15 0.2
Time (sec) 0.25 0.3 0.35 0.4 ME375 Dynamic Response  15 Dynamic 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” ? Q: For a 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
rate? School of Mechanical Engineering
Purdue University ME375 Dynamic Response  16 8 ME375 Handouts Response of Stable 2nd Order System
• Stable 2nd Order System
y + a1 y + a 0 y = b u
where ⇒ y + 2ζ ω 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
Characteristic 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  17 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 y (t) = School of Mechanical Engineering
Purdue University ME375 Dynamic Response  18 9 ME375 Handouts Response of Stable 2nd Order System
Unit Step Response of 2nd Order Systems
1.6K 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 10 11 tS School of Mechanical Engineering
Purdue University 12 13 14 15 ME375 Dynamic Response  19 Response of Stable 2nd Order System
• Peak Time (tP)
Time when output y(t) reaches its
maximum value yMAX. FG
H y MAX = y ( t P ) = K ⋅ 1 + e y (t ) =
⎛
⎡
⎤⎞
ζω n
K ⋅ ⎜ 1 − e −ζω n ⋅t ⎢ cos(ω d t ) +
sin(ω d t ) ⎥ ⎟
ωd
⎣
⎦⎠
⎝
d
y (t ) =
dt
Find t P such that y ( t P ) = 0
⇒ • Percent Overshoot (%OS)
At peak time tP the maximum
output
πζ
− 1− ζ 2 IJ
K The overshoot (OS) is:
(OS)
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  20 10 ME375 Handouts 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: % 1% 2% 5% tS 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? School of Mechanical Engineering
Purdue University ME375 Dynamic Response  21 In Class Exercise
• MassSpringDamper System
Mass SpringQ: What is the static (steadystate) gain of
(steadythe system ? x K
M f(t) B I/O Model: M x + Bx + K x = f (t ) Q: How would the physical parameters
(M, B, K) affect the response of the
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  22 11 ME375 Handouts Transient and Steady State Response
Ex: Let’s find the total response of a stable first order system:
Let’
y + 5 y = 10u
to a ramp input:
with IC:
– Total Response u(t) = 5t
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  23 Transient and Steady State Responses
In general, the total response of a stable LTI system
an y ( n) + an−1 y ( n−1) + + a1 y + a0 y = bm u ( m) + bm−1 u ( m−1) + + b1 u + b0 u m−1 bm s + bm−1s + + 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 an input u(t) can be decomposed into two parts:
G( s ) ≡ m 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
characteristic
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.
magnitude
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Purdue University ME375 Dynamic Response  24 12 ME375 Handouts Transient and Steady State Response
Ex: Let’s find the total response of a stable second order system:
Let’
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  25 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),
signal’
sF(
region),
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:
Ex: 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
converge to a constant value?
(2) If the output converges, what will be its steady state value?
value? School of Mechanical Engineering
Purdue University ME375 Dynamic Response  26 13 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
G (s) ≡ b m s m + bm − 1 s m − 1 +
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 the free response of the system is: YFree (s) = F(s)
=
an sn + an−1sn−1 + + a1s + a0 Apply FVT: School of Mechanical Engineering
Purdue University ME375 Dynamic Response  27 Steady State Response
• Steady State Value of the Unit Impulse Response
Y (s) = G (s) ⋅U (s) =
Apply FVT:
• Steady State Value of the Unit Step Response
Y (s) = G(s) ⋅U (s) =
Apply FVT: 1st Order Systems:
G(s) =
⇒ G(0) = 2nd Order Systems: b0
a1s + a0 G( s ) = b1s + b0
a2 s 2 + a1s + a0 ⇒ G(0) =
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Purdue University ME375 Dynamic Response  28 14 ...
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This note was uploaded on 08/28/2010 for the course ME 375 taught by Professor Meckle during the Spring '10 term at Purdue.
 Spring '10
 Meckle

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