laboratory exercise #1 time response & frequency...

Mechatronics Second-Order Dynamic System Response

K. Craig1

Mechatronics Time Response & Frequency Response

2nd-Order Dynamic System2-Pole, Low-Pass, Active Filter

R1

C2ein

eout

R3

R4

C5

R7

R6

+

-

+

-

Assignment: Perform a Complete Dynamic SystemInvestigation of the Two-Pole, Low-Pass, Active 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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R1

C2ein

eout

R3

R4

C5

R7

R6

+

-

+

-

Physical Model Ideal Transfer Function

( )7

6 1 3 2 5out

2in

3 2 1 2 4 2 3 4 2 5

R 1R R R C Ce s

e 1 1 1 1s sR C R C R C R R C C

=

+ + + +


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Zero-Order Dynamic 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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0n

2

1

2 0

0

0

a undamped natural frequencya

a damping ratio2 a abK steady-state gaina

ω =

ζ =

=

2nd-Order DynamicSystem Model

20 0

2 1 0 0 0 i2

20 0

0 i2 2n n

d q dqa a a q b qdt dtd q dq1 2 q Kqdt dt

+ + =

ζ+ + =

ω ω

Step Responseof a

2nd-Order System


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Step Responseof a

2nd-Order System

20 0

0 i2 2n n

d q dq1 2 q Kqdt dt

ς+ ζ + =

ω ω

Underdamped

( )n t 2 1 2o is n2

1q Kq 1 e sin 1 t sin 1 11

−ζω −

= − ω −ζ + −ζ ζ < − ζ

( ) n to is nq Kq 1 1 t e 1−ω = − +ω ζ = Critically Damped

( )

( )

2n

2n

2 1 t

2

o is2 1 t

2

11 e

2 1q Kq 1

1 e

2 1

−ζ+ ζ − ω

−ζ− ζ − ω

ζ + ζ −−

ζ − = ζ >

ζ − ζ − + ζ −

Over-damped


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Frequency Responseof a

2nd-Order System

( )o2

i2n n

Q Kss 2 sQ 1

=ζ

+ +ω ω

( ) 1o22i 2 2 n

2 nn n

Q K 2i tanQ

41

− ζω = ∠

ω ω − ω ζ ω − + ω ω ω ω

Laplace Transfer Function

Sinusoidal Transfer Function


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2nd-Order System


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-40 dB per decade slope


2nd-Order System


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Some Observations• When a physical system exhibits a natural

oscillatory behavior, a 1st-order model (or even a cascade of several 1st-order models) cannot provide the desired response. The simplest model that does possess that possibility is the 2nd-order dynamic system model.

• This system is very important in control design.– System specifications are often given assuming that the

system is 2nd order.– For higher-order systems, we can often use dominant

pole techniques to approximate the system with a 2nd-order transfer function.


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• Damping ratio ζ clearly controls oscillation; ζ < 1 is required for oscillatory behavior.

• The undamped case (ζ = 0) is not physically realizable (total absence of energy loss effects) but gives us, mathematically, a sustained oscillation at frequency ωn.

• Natural oscillations of damped systems are at the damped natural frequency ωd, and not at ωn.

• In hardware design, an optimum value of ζ = 0.64 is often used to give maximum response speed without excessive oscillation.

2d n 1ω = ω −ζ


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• Undamped natural frequency ωn is the major factor in response speed. For a given ζ response speed is directly proportional to ωn.

• Thus, when 2nd-order components are used in feedback system design, large values of ωn (small lags) are desirable since they allow the use of larger loop gain before stability limits are encountered.

• For frequency response, a resonant peak occurs for ζ < 0.707. The peak frequency is ωp and the peak amplitude ratio depends only on ζ.

2p n 2

K1 2 peak amplitude ratio2 1

ω = ω − ζ =ζ −ζ


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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.

– For a 2nd-order system:2 2 4

nBW 1 2 2 4 4= ω − ζ + − ζ + ζ


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– As ζ varies from 0 to 1, BW varies from 1.55ωn to 0.64ωn. For a value of ζ = 0.707, BW = ωn. For most design considerations, we assume that the bandwidth of a 2nd-order all pole system can be approximated by ωn.


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2n

2 2n n

21,2 n n

1,2 d

KG(s)s 2 s

s i 1s i

ω=

+ ςω +ω

= −ςω ± ω −ς

= −σ± ω

( )

( )

2

rn

sn

1p

1.8t rise time

4.6t settling time

M e 0 1 overshoot

1 0 0.60.6

−πς

−ς

≈ω

≈ςω

= ≤ ζ <

ζ ≈ − ≤ ζ ≤

Location of PolesOf

Transfer Function

General All-Pole2nd-Order

Step Response

( ) td d

d

y t 1 e cos t sin t−σ σ= − ω + ω ω


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Time-Response Specifications vs. Pole-Location Specifications

nr

1.8t

ω ≥

( )p0.6 1 M 0 0.6ζ ≥ − ≤ ζ ≤

s

4.6t

σ ≥


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Experimental Determination of ζ and ωn

• ζ and ωn can be obtained in a number of ways from step or frequency-response tests.

• For an underdamped second-order system, the values of ζ and ωn may be found from the relations:

( )

21p 2

e p

2 dd n n 2 2

1M e

1log M

211 T 1

πζ−

−ζ= ζ = π

+

ω πω = ω −ζ ω = =

⇒

−ζ −ζ⇒

d

2T π=ω


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( )( ) ( )

( )

nTn

n n

2 2d n

2 2

1

n 1

x tln ln e T

x t T

2 2 2

1 1

2

B1 lnn B

ζω

+

δ = = = ζω +

ζω π ζω π πζ= = =

ω ω −ζ − ζ

δζ =

π + δ

δ =

Free Response of a 2nd-Order System

( ) ( )n tdx t Be sin t−ζω= ω + φ

d

2T π=ω

• Logarithmic Decrement δ is the natural logarithm of the ratio of two successive amplitudes.


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– If several cycles of oscillation appear in the record, it is more accurate to determine the period T as the average of as many distinct cycles as are available rather than from a single cycle.

– If a system is strictly linear and second-order, the value of n is immaterial; the same value of ζ will be found for any number of cycles. Thus if ζ is calculated for, say, n= 1, 2, 4, and 6 and different numerical values of ζ are obtained, we know that the system is not following the postulated mathematical model.

• For overdamped systems (ζ > 1.0), no oscillations exist, and the determination of ζ and ωn becomes more difficult. Usually it is easier to express the system response in terms of two time constants.


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– For the overdamped step response:

– where

( )

( )

2n

2n

1 2

2 1 t

2

o is2 1 t

2

t to 1 2

is 2 1 2 1

11 e

2 1q Kq 1

1 e

2 1

q e e 1Kq

−ζ+ ζ − ω

−ζ− ζ − ω

− −τ τ

ζ + ζ −−

ζ − = ζ >

ζ − ζ − + ζ −

τ τ= − +τ − τ τ − τ

( ) ( )1 22 2

n n

1 1 1 1

τ τζ − ζ − ω ζ + ζ − ω


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– To find τ1 and τ2 from a step-function response curve, we may proceed as follows:

• Define the percent incomplete response Rpi as:

• Plot Rpi on a logarithmic scale versus time t on a linear scale. This curve will approach a straight line for large t if the system is second-order. Extend this line back to t = 0, and note the value P1where this line intersects the Rpi scale. Now, τ1 is the time at which the straight-line asymptote has the value 0.368P1.

• Now plot on the same graph a new curve which is the difference between the straight-line asymptote and Rpi. If this new curve is not a straight line, the system is not second-order. If it is a straight line, the time at which this line has the value 0.368(P1-100) is numerically equal to τ2.

• Frequency-response methods may also be used to find τ1 and τ2.

opi

is

qR 1 100

Kq −


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Step-Response Test for

Overdamped Second-Order

Systems


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Frequency-Response Test of Second-Order

Systems


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Hardware Parameters

1

2

3

4

5

6

7

R 2.4 kC 0.047 F, 0.002 F R 9.1 kR 2.4 kC 0.00047 F, 0.001 F R 10.0 kR 10.0 k

== µ µ=== µ µ==

Ω

ΩΩ

ΩΩ

laboratory exercise #1 time response & frequency...

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