ME 413 Systems Dynamics & Control Chapter 10: Time‐Domain Analysis and Design of Control Systems
Chapter 10: Time‐Domain Analysis of and Design of Control Systems A. Bazoune 10.5 Transient Response Specifications of a Second Order System Because systems that stores energy cannot respond instantaneously, they exhibit a transient response when they are subjected to inputs or disturbances. Consequently, the transient response characteristics constitute one of the most important factors in system design. In many practical cases, the desired performance characteristics of control systems can be given in terms of transient‐response specifications. Frequently, such performance characteristics are specified in terms of the transient response to unit‐step input, since such an input is easy to generate and is sufficiently drastic. (If the response of a linear system to a step input is known, it is mathematically possible to compute the system’s response to any input). The transient response of a system to a unit step‐input depends on initial conditions. For convenience in comparing the transient responses of various systems, it is common practice to use standard initial conditions: The system is at rest initially, with its output and all time derivatives thereof zero. Then the response characteristics can be easily compared. Transient‐Response Specifications. The transient response of a practical control system often exhibits damped oscillations before reaching a steady state. In specifying the transient‐response characteristics of a control system to a unit‐step input, it is common to name the following: 1. Delay time, Td 2. Rise time, Tr 3. Peak time, Tp 4. Maximum overshoot, M p 5. Settling time, Ts These specifications are defined next and are shown in graphically in Figure 10‐21. 1/9
ME 413 Systems Dynamics & Control Chapter 10: Time‐Domain Analysis and Design of Control Systems
Delay Time. The delay time Td is the time needed for the response to reach half of its final value the very first time. Rise Time. The rise time Tr is the time required for the response to rise from 10% to 90%, 5% to 95%, or 0% to 100% of its final value. For underdamped second order systems, the 0% to 100% rise time is normally used. For overdamped systems, the 10% to 90% rise time is common. Peak Time. The peak time Tp is the time required for the response to reach the first peak of the overshoot. The maximum percent overshoot M p is the Maximum (percent Overshoot). maximum peak value of the response curve [the curve of c ( t ) versus t ], measured from c ( ∞ ) . If c ( ∞ ) = 1 , the maximum percent overshoot is M p × 100% . If the final steady state value c ( ∞ ) of the response differs from unity, then it is common practice to use the following definition of the maximum percent overshoot: Maximum percent overshoot =
( )
C tp − C ( ∞ )
C (∞)
× 100% Settling Time. The settling time Ts is the time required for the response curve to reach and stay within 2% of the final value. In some cases, 5% instead of 2% , is used as the percentage of the final value. The settling time is the largest time constant of the system. Comments. If we specify the values of Td , Tr , Tp , Ts and M p , the shape of the response curve is virtually fixed as shown in Figure 10.22. Figure 10‐22 Specifications of transient‐response curve. A Few Comments on Transient Response‐Specifications. In addition of requiring a dynamic system to be stable, i.e., its response does not increase unbounded with time (a condition that is satisfied for a second order system provided that ζ ≥ 0 , we also require the response: • to be fast • does not excessively overshoot the desired value (i.e., relatively stable) and • to reach and remain close to the desired reference value in the minimum time possible. 2/9
ME 413 Systems Dynamics & Control Chapter 10: Time‐Domain Analysis and Design of Control Systems
Second‐Order Systems and Transient‐Response‐Specifications. The response for a unit step input of an underdamped second order system ( 0 < ζ < 1) is given by ζ
c (t ) = 1 −
1−ζ
e
2
− ζω n t
sin ωd t − e
− ζω n t
cos ωd t
⎧ ζ
⎪⎫
− ζω t ⎪
=1−e
sin ωd t + cos ωd t ⎬
⎨
2
⎩⎪ 1 − ζ
⎭⎪
(10‐13) (10‐14) n
or 2
⎧⎪
1−ζ ⎫
⎪
−1
c (t ) = 1 −
sin ⎨ωd t + tan
⎬ 2
ζ
1−ζ
⎩⎪
⎭⎪
e
− ζω n t
A family of curves c ( t ) plotted against t with various values of ζ is shown in Figure 10‐24. Step Response
1.6
ζ = 0.2
1.4
0.5
1.2
0.7
u (t )
1
1
1
−
−−−
142
4
3−
In p u t
t
Amplitude
ωn2
s2 + 2 ζ ωns + ωn2
0.8
2
0.6
5
0.4
0.2
0
0
2
4
6
8
10
12
14
16
18
Time (sec)
_______________
144
42444
3 Output
Figure 10‐24 Unit step response curves for a second order system. Delay Time. We define the delay time by the following approximate formula: 1 + 0.7ζ
Td =
ωn
Rise Time. We find the rise time Tr by letting c ( Tr ) = 1 in Equation (10‐13), or c (T r ) = 1 = 1 − e
Since e
− ζω n t
− ζω nT r
⎧⎪ ζ
⎫⎪
sin ωdT r + cos ωdT r ⎬ ⎨
2
⎩⎪ 1 − ζ
⎭⎪
≠ 0 , Equation (10‐15) yields 3/9
(10‐15) 20
ME 413 Systems Dynamics & Control ζ
1−ζ
2
Chapter 10: Time‐Domain Analysis and Design of Control Systems
sin ωdT r + cos ωdT r = 0 or tan ωdT r = −
1−ζ
ζ
2
Thus, the rise Tr is ⎛ 1−ζ 2
T r = tan ⎜ −
⎜
ωd
ζ
⎝
1
−1
⎞ π −β
⎟=
⎟
ωd
⎠
(10‐16) where β is defined in Figure 10‐25. Clearly to obtain a large value of Tr we must have a large value of β . jω
jω d
ωn 1 − ζ
2
β = cos−1 (ζ )
or β = sin−1
ωn
(
1 −ζ 2
)
⎛ 1 −ζ 2
or β = tan−1 ⎜
⎜ ζ
⎝
β
σ
−σ
⎞
⎟
⎟
⎠
ζω n
Figure 10‐25 Definition of angle β Peak Time. We obtain the peak time Tp by differentiating c ( t ) in Equation (10‐13), with respect to time and letting this derivative equal zero. That is, dc (t )
ωn
− ζω t
=
e
sin ωd t = 0 2
dt
1−ζ
n
It follows that or sin ωd t = 0 ωd t = 0, π , 2π , 3π ,... = nπ , n = 0,1,2..... Since the peak time Tp corresponds to the first peak overshoot ( n = 1) , we have ωdT p
Tp =
π
π
=
ωd ωn 1 − ζ 2
4/9
=π
. Then (10‐17) ME 413 Systems Dynamics & Control Chapter 10: Time‐Domain Analysis and Design of Control Systems
The peak time Tp corresponds to one half‐cycle of the frequency damped oscillations. Maximum Overshoot M p The maximum overshoot M p occurs at the peak T p = π ωd
. Thus, from Equation (10‐13), ⎧
⎫
⎪
⎪
−ζωn (π ωd ) ⎪
ζ
⎪
M p = c Tp − 1 = − e
sin
π
cos
π
+
⎨
{⎬ 2
1− ζ
⎪ 14
4244
3 = −1 ⎪
⎪⎩
=0
⎭⎪
( )
or M p = e −πζ
1−ζ 2
(10‐18) Since c ( ∞ ) = 1 , the maximum percent overshoot is M p% = e−πζ
1−ζ 2
× 100% The relationship between the damping ratio ζ and the maximum percent overshoot is shown in Figure 10‐26. Notice that no overshoot for ζ ≥ 1 and overshoot becomes negligible for ζ > 0.7 . Figure 10‐26 Relationship between the maximum percent overshoot M p % and damping ratio ζ . Settling Time Ts Based on 2% criterion the settling time Ts is defined as: e−ζωnTs = 0.02
−ζωnTs = ln ( 0.02 ) ⇒ Ts =
5/9
ln ( 0.02 )
−ζωn
≈
4 ζωn
ME 413 Systems Dynamics & Control Ts =
4
Chapter 10: Time‐Domain Analysis and Design of Control Systems
( 2% Criterion ) ζωn
(10‐19) (10‐20) Similarly for 5% we can get Ts =
3
( 5% Criterion )
ζωn
REVIEW AND SUMMARY
TRANSIENT RESPONSE SPECIFICATIONS OF A SECOND ORDER SYSTEM TABLE 1. Useful Formulas and Step Response Specifications for the Linear Second‐Order Model m &x& + c x& + k x = f (t ) where m, c, k constants 1. Roots 2. Damping ratio or − c ± c 2 − 4mk
s1, 2 =
2m
ζ = c / 2 mk 3. Undamped natural frequency ω n =
k
m
4. Damped natural frequency 5. Time constant ωd = ωn 1−ζ 2 τ = 2m / c = 1 / ζω n
if ζ ≤ 1 6. Stability Property Stable if, and only if, both roots have negative real parts, this occurs if and only if , m, c, and k have the same sign. 7. Maximum Percent Overshoot: The maximum % overshoot M p is the maximum peak value of the response curve. M p = 100e −πζ /
1−ζ 2
8. Peak time: Time needed for the response to reach the first peak of the overshoot Tp = π /ωn 1 − ζ 2
9. Delay time: Time needed for the response to reach 50% of its final value the first time Td ≈
1 + 0.7ζ
ωn
10. Settling time: Time needed for the response curve to reach and stay within 2% of the final value
Ts =
4
ζωn
11. Rise time: Time needed for the response to rise from (10% to 90%) or (0% to 100%) or (5% to 95%) of its final value Tr =
π −β
(See Figure 10‐25) ωd
6/9
ME 413 Systems Dynamics & Control Chapter 10: Time‐Domain Analysis and Design of Control Systems
SOLVED PROBLEMS █ Example 1 Figure 4‐20 (for Example 1) 7/9
ME 413 Systems Dynamics & Control █ Chapter 10: Time‐Domain Analysis and Design of Control Systems
Example 2 Figure 4‐21 (for Example 2) █ Solution First The transfer function of the system is 8/9
ME 413 Systems Dynamics & Control █ Chapter 10: Time‐Domain Analysis and Design of Control Systems
Example 3 (Example 10‐2in the Textbook Page 520‐521) Determine the values of Td , Tr , Tp , Ts when the control system shown in Figure 10‐28 is subject to a unit step input 1
s ( s + 1)
R ( s)
Figure 10‐28 C (s)
Control System █ Solution The closed‐loop transfer function of the system is 1
C (s)
s ( s + 1)
1
=
= 2
1
R (s) 1+
s + s +1
s ( s + 1)
Notice that ωn = 1 rad/s and ζ = 0.5 for this system. So ωd = ωn 1 − ζ 2 = 1 − 0.52 = 0.866 Tr =
Rise Time. π −β
ωd
jω
where β = sin −1 (ωd ωn ) = sin −1 ( 0.866 1) = 1.05 rad
jω d
or β = cos −1 (ζωn ωn ) = cos −1 (ζ ) = cos−1 ( 0.5 )
= 1.05 rad
ωn
ωn 1 − ζ 2
Therfore, Tr =
π − 1.05
0.866
Peak Time. Delay Time. Td =
β
= 2.41 s Tp =
ζω n
π
π
=
= 3.63 s ωd 0.866
1 + 0.7ζ
ωn
=
σ
−σ
1 + 0 .7 ( 0 .5 )
= 1.35 s 1
Mp = e
Maximum Overshoot : Settling time: Ts =
4
ζωn
=
−πζ
1−ζ 2
4
= 8 s 0.5 × 1
9/9
=e
−π ×0.5 1−0.52
=e
−1.81
= 0.163 = 16.3%