laboratory exercise #1 time response & frequency response...

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


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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?


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Zero-Order Dynamic System Model


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Validation of a Zero-OrderDynamic System Model


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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


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( )

( )

( )

( )

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?


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– 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.


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– 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.


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• 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.


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• 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.


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System Frequency Response


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Bode Plotting of 1st-Order

Frequency Response

dB = 20 log10 (amplitude ratio)decade = 10 to 1 frequency changeoctave = 2 to 1 frequency change


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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 +


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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


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• 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


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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 +


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• 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.


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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


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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

laboratory exercise #1 time response & frequency response...

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