This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ME375 Handouts Dynamic Response of Linear Systems
• Linear System Response
–
– • Dynamic Response of 1st Order Systems
–
– • Characteristic Equation  Free Response
Stable 1st Order System Response Dynamic Response of 2nd Order Systems
–
– • Superposition Principle
Responses to Specific Inputs Characteristic Equation  Free Response
Stable 2nd Order System Response Transient and SteadyState Response
SteadySchool of Mechanical Engineering
Purdue University ME375 Dynamic Response  1 Linear System Response
&
&
y ( n ) + a n −1 y ( n −1) + L + a2 && + a1 y + a 0 y = bm u ( m ) + L + b1u + b0 u
y • 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 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) – 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
response is call the
Frequency
Frequency Response. Time t
School of Mechanical Engineering
Purdue University ME375 Dynamic Response  3 Dynamic Response of 1st Order Systems
• Characteristic Equation: &
y + ay = b u 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 y H (t ) e.g. a<0 y Free ( t ) = A e 0 ⋅ t y Free ( t ) = A e − ( − 4 ) ⋅ t y H (t ) y H (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 ?
School of Mechanical Engineering
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
Ti
K : Static (Steady State, DC) Gain
Static
– Unit Step Response (force response: u = 1 and zero ICs )
(force
and
Y (s) = K
(τ s + 1) s y(t)
y P (t ) = K K K
τK
−
s τ s +1
K
K
=
−
1
s
s+
= Time t
yH(t) = − K τ y (t ) = K − K e − e t/τ 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)
zero
1 &
τy + y = Ku
⇒ y ( t ) = K (1 − e −tτ Normalized Response 0.9 ) Normalized
(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 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  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 y(t )
−t
= (1 − e τ )
K Slope at t = 0:
0:
d
yn ( t ) =
dt
d
yn ( 0 ) =
⇒
dt
⇒ Normalized Response ⇒ y ( t ) = K (1 − e 1
−tτ 0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 Q: Can y ( 0 ) = 0 ?
Can & 0 2 4
6
Time [sec] School of Mechanical Engineering
Purdue University 8 10 ME375 Dynamic Response  7 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): y(t )
−t
⇒ yn ( t ) =
= (1 − e τ )
K
Initial Slope 1
&
yn ( 0 ) =
τ
K
&
y(0 ) =
τ Normalized Response 0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 1τ School of Mechanical Engineering
Purdue University 2τ 3τ
Time [ t ] 4τ 5τ 6τ
ME375 Dynamic Response  8 4 ME375 Handouts 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)?
u(t) t1 Time t (Hint: superposition principle ?!) School of Mechanical Engineering
Purdue University ME375 Dynamic Response  9 Dynamic Response of 2nd Order Systems
&
&& + a1 y + a0 y = b1 u + b0 u
&
y • Characteristic Equation: s 2 + a1 s + a 0 = 0
⇒ • 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 ]:
– Complex [ s1,2 = α ± jβ ]
y Free ( t ) = e α ⋅t y F r e e ( t ) = A 1 e s1 ⋅ t + A 2 t e s1 ⋅ 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  10 5 ME375 Handouts Dynamic 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 y H (t ) Real Real
y H (t ) y H (t ) Time (t) Time (t) Time (t) School of Mechanical Engineering
Purdue University ME375 Dynamic Response  11 Dynamic 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 y H (t ) y H (t ) y H (t ) Time (t) Real Real 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 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 y H (t ) Real y H (t ) y H (t ) Time (t) Time (t) Time (t) School of Mechanical Engineering
Purdue University ME375 Dynamic Response  13 Example – Automotive Suspension
y
m && &
&
my + by + ky = br + kr g
k b r Response to Initial Conditions
0.02 b
k
&& + y + y = 0
&
y
m
m
&& + 28 y + 400 y = 0
&
y 0 0.02
Amplitude for free response:
&& &
my + by + ky = 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  14 7 ME375 Handouts 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  15 Response of Stable 2nd Order System
• Stable 2nd Order System
&& + a1 y + a 0 y = b u
&
y
where ⇒ && + 2 ζ ω n y + ω n 2 y = K ω n 2 u
&
y ωn > 0 : Natural Frequency [rad/s]
ζ > 0 : Damping Ratio
K : Static (Steady State, DC) Gain
Static 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
School of Mechanical Engineering
Purdue University ME375 Dynamic Response  16 8 ME375 Handouts Response of Stable 2nd Order System
Unit Step Response of Underdamped 2nd Order Systems ( u = 1 and zero ICs )
Under&& + 2 ζ ω n y + ω n 2 y = K ω n 2 u
&
y
Characteristic equation:
equation:
s 2 + 2ζ ω n s + ω n 2 = 0
⇒ s = −ζ ω n ± j ω n (1 − ζ 2 )
1 4243 ωd y (t) = School of Mechanical Engineering
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 )
UnderHow 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
s s 2 + 2ζω n s + ω n = K ( s + ζωn ) + K ζωn
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
1− ζ 2 School of Mechanical Engineering
Purdue University e −ζωn sin ωn 1 − ζ 2 t 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:
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: %
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? School of Mechanical Engineering
Purdue University ME375 Dynamic Response  21 Pole location and Transient Response
Img. ωn Real −ωn School of Mechanical Engineering
Purdue University ME375 Dynamic Response  22 11 ME375 Handouts In Class Exercise
• MassSpringDamper System
MassSpringQ: 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, 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  23 School of Mechanical Engineering
Purdue University ME375 Dynamic Response  24 Examples 12 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 a ramp input:
with IC:
– Total Response u(t) = 5t
y(0) = 2 Y ( s) = ⋅ U ( s) + ⋅ y(0), where 14 3
24 U (s) = L [5 t ] = Transfer Function G(s) ∴ Y ( s) = ⋅ + – PFE: Y (s) = A3
A1
A2
+
+
+
1444 24444
4
3 14 244
4
3 y (t ) =
School of Mechanical Engineering
Purdue University ME375 Dynamic Response  25 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
+ L+ a1 y + a0 y = bm u ( m) + bm−1 u ( m−1) + L+ b1 u + b0 u
bm s m + bm−1s m−1 + L + b1s + b0 N (s) bm (s − z1 )(s − z2 )L(s − zm )
=
=
an s n + an−1s n−1 + L + a1s + a0 D(s) an (s − p1 )(s − p2 )L(s − pn )
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.
School of Mechanical Engineering
Purdue University ME375 Dynamic Response  26 13 ME375 Handouts Transient and Steady State Response
Ex: Let’s find the total response of a stable second order system:
&& + 4 y + 3 y = 6 u
&
y 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 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
then f(t) converges to a constant value f(∞). f(∞) can be obtained without
converges to constant value
).
can be obtained without
knowing
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: && + 4 y + 1 2 y = 4 u + 3 u
&
&
y (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?
converge to constant value?
(2) If the output converges, what will be its steady state value? School of Mechanical Engineering
Purdue University ME375 Dynamic Response  28 14 ME375 Handouts Steady State Response
Given a stable LTI system &
&
an y(n) + an−1 y(n−1) +L+ a1 y + a0 y = bmu(m) + bm−1u(m−1) +L+ b1u + b0 u
The corresponding transfer function is
corresponding transfer function is
G (s) ≡ bm s m + bm −1 s m −1 + L + b1 s + b0
b ( s − z1 )( s − z 2 ) L ( s − z m )
=m
a n s n + a n −1 s n −1 + L + a1 s + a 0 a n ( s − p1 )( s − p 2 ) L ( s − p n ) • Steady State Value of the Free Response
Recall the free response of the system is: YFree (s) = F(s)
=
an sn + an−1sn−1 +L+ a1s + a0 Apply FVT: School of Mechanical Engineering
Purdue University ME375 Dynamic Response  29 Steady State Response
• Steady State Value of the Unit Impulse Response
Y (s) = G (s) ⋅U (s) =
Apply FVT:
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) =
School of Mechanical Engineering
Purdue University ME375 Dynamic Response  30 15 ...
View
Full
Document
 Fall '10
 Meckle

Click to edit the document details