MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig1
Studio Exercise Time Response & Frequency Response
1st-Order Dynamic SystemRC Low-Pass Filter
R
Cein eout
iin ioutAssignment:
Perform a Complete Dynamic System Investigation
of theRC Low-Pass Filter
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig2
P h y s ic a lS y s t e m
P h y s ic a lM o d e l
M a t hM o d e l
M o d e lP a r a m e t e r
I D
A c t u a lD y n a m icB e h a v io r
C om p a r eP r e d ic t e dD y n a m icB e h a v io r
M a k eD e s ig n
D e c is io n s
D e s ig nC om p le t e
Measurements,Calculations,
Manufacturer's Specifications
Assumptionsand
Engineering Judgement
Physical Laws
ExperimentalAnalysis
Equation Solution:Analytical
and NumericalSolution
Model Adequate,Performance Adequate
Model Adequate,Performance Inadequate
Modify or
Augment
Model Inadequate:Modify
D y n a m ic S y s t e m I n v e s t ig a t io n
Which Parameters to Identify?What Tests to Perform?
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig3
Zero-Order Dynamic System Model
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig4
Validation of a Zero-OrderDynamic System Model
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig5
1st-Order Dynamic System Model
tis
oKqq e
−τ=
τis
o t 0
Kqq==
τ
Slope at t = 0
t = τ
τ = time constantK = steady-state gain
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig6
( )
( )
( )
( )
t
o is
to is
is
to
is
o10 10
is
q t Kq 1 e
q t Kqe
Kqq t
1 eKq
q t t tlog 1 log e 0.4343Kq
−τ
−τ
−τ
= −
−
= −
− =
− = − = − τ τ
Straight-Line Plot:
( )o10
is
q tlog 1 vs. t
Kq −
Slope = -0.4343/τ
• How would you determine if an experimentally-determined step response of a system could be represented by a first-order system step response?
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig7
– This approach gives a more accurate value of τ since the best line through all the data points is used rather than just two points, as in the 63.2% method. Furthermore, if the data points fall nearly on a straight line, we are assured that the instrument is behaving as a first-order type. If the data deviate considerably from a straight line, we know the system is not truly first order and a τ value obtained by the 63.2% method would be quite misleading.
– An even stronger verification (or refutation) of first-order dynamic characteristics is available from frequency-response testing. If the system is truly first-order, the amplitude ratio follows the typical low- and high-frequency asymptotes (slope 0 and –20 dB/decade) and the phase angle approaches -90° asymptotically.
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig8
– If these characteristics are present, the numerical value of τ is found by determining ω (rad/sec) at the breakpoint and using τ = 1/ωbreak. Deviations from the above amplitude and/or phase characteristics indicate non-first-order behavior.
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig9
• What is the relationship between the unit-step response and the unit-ramp response and between the unit-impulse response and the unit-step response?– For a linear time-invariant system, the response to the
derivative of an input signal can be obtained by differentiating the response of the system to the original signal.
– For a linear time-invariant system, the response to the integral of an input signal can be obtained by integrating the response of the system to the original signal and by determining the integration constants from the zero-output initial condition.
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig10
• Unit-Step Input is the derivative of the Unit-Ramp Input.
• Unit-Impulse Input is the derivative of the Unit-Step Input.
• Once you know the unit-step response, take the derivative to get the unit-impulse response and integrate to get the unit-ramp response.
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig11
System Frequency Response
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig12
Bode Plotting of 1st-Order
Frequency Response
dB = 20 log10 (amplitude ratio)decade = 10 to 1 frequency changeoctave = 2 to 1 frequency change
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig13
R
Cein eout
iin ioutAnalog Electronics:RC Low-Pass FilterTime Response &
Frequency Response
outin
outin
outout
in
ee RCs 1 Rii Cs 1
e 1 1 when i 0e RCs 1 s 1
+ − = −
= = =+ τ +
MechatronicsStudio Exercise – 1st-Order Dynamic System
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Time Response to Unit Step Input
0 1 2 3 4 5 6 7 8x 10
-4
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (sec)
Ampl
itude
R = 15 KΩC = 0.01 µF
Time Constant τ = RC
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig15
• Time Constant τ– Time it takes the step response to reach 63% of
the steady-state value• Rise Time Tr = 2.2 τ
– Time it takes the step response to go from 10% to 90% of the steady-state value
• Delay Time Td = 0.69 τ– Time it takes the step response to reach 50% of
the steady-state value
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig16
R = 15 KΩ
C = 0.01 µFFrequency Response
102
103
104
105
-25
-20
-15
-10
-5
0
Frequency (rad/sec)
Gai
n dB
102
103
-100
-80
-60
-40
-20
0
Frequency (ra
Phas
e (d
egre
es)
Bandwidth = 1/τ
( )( ) ( )
1out2 22 1 2in
e K K 0 Ki tane i 1 1 tan 1
−
−
∠ω = = = ∠− ωτ
ωτ+ ωτ + ∠ ωτ ωτ +
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig17
• Bandwidth– The bandwidth is the frequency where the
amplitude ratio drops by a factor of 0.707 = -3dB of its gain at zero or low-frequency.
– For a 1st -order system, the bandwidth is equal to 1/ τ.
– The larger (smaller) the bandwidth, the faster (slower) the step response.
– Bandwidth is a direct measure of system susceptibility to noise, as well as an indicator of the system speed of response.
MechatronicsStudio Exercise – 1st-Order Dynamic System
K. Craig18
MatLab / Simulink DiagramFrequency Response for 1061 Hz Sine Input
t
time
output
output
input
input
Sine Wave
1
tau.s+1
First-OrderPlant
Clock
τ = 1.5E-4 sec
MechatronicsStudio Exercise – 1st-Order Dynamic System
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0 0.5 1 1.5 2 2.5 3 3.5 4
x 10-3
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time (sec)
ampl
itude
Response to Input 1061 Hz Sine Wave
Amplitude Ratio = 0.707 = -3 dB Phase Angle = -45°
Input
Output