Download First-Order and Second-Order Systems: Analysis and Response and more Lecture notes Law in PDF only on Docsity!
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
DEPARTMENT OF MECHANICAL ENGINEERING
2.151 Advanced System Dynamics and Control
Review of First- and Second-Order System Response^1
1 First-Order Linear System Transient Response
The dynamics of many systems of interest to engineers may be represented by a simple model containing one independent energy storage element. For example, the braking of an automobile, the discharge of an electronic camera flash, the flow of fluid from a tank, and the cooling of a cup of coffee may all be approximated by a first-order differential equation, which may be written in a standard form as
τ
dy dt
where the system is defined by the single parameter τ , the system time constant, and f (t) is a forcing function. For example, if the system is described by a linear first-order state equation and an associated output equation:
x˙ = ax + bu (2) y = cx + du. (3)
and the selected output variable is the state-variable, that is y(t) = x(t), Eq. (3) may be rearranged
dy dt
− ay = bu, (4)
and rewritten in the standard form (in terms of a time constant τ = − 1 /a), by dividing through by −a:
−
a
dy dt
u(t) (5)
where the forcing function is f (t) = (−b/a)u(t). If the chosen output variable y(t) is not the state variable, Eqs. (2) and (3) may be combined to form an input/output differential equation in the variable y(t):
dy dt − ay = d du dt
To obtain the standard form we again divide through by −a:
−
a
dy dt
du dt
ad − bc a
u(t). (7)
Comparison with Eq. (1) shows the time constant is again τ = − 1 /a, but in this case the forcing function is a combination of the input and its derivative
f (t) = −
d a
du dt
ad − bc a u(t). (8)
In both Eqs. (5) and (7) the left-hand side is a function of the time constant τ = − 1 /a only, and is independent of the particular output variable chosen. (^1) D. Rowell 10/22/
Example 1
A sample of fluid, modeled as a thermal capacitance Ct, is contained within an insulating vacuum flask. Find a pair of differential equations that describe 1) the temperature of the fluid, and 2) the heat flow through the walls of the flask as a function of the external ambient temperature. Identify the system time constant.
TC
Tambq walls Rt
fluid Ct
CtRt
TCT
heatflow ambTref
Figure 1: A first-order thermal model representing the heat exchange between a laboratory vacuum flask and the environment.
Solution: The walls of the flask may be modeled as a single lumped thermal resistance Rt and a linear graph for the system drawn as in Fig. 1. The environment is assumed to act as a temperature source Tamb(t). The state equation for the system, in terms of the temperature TC of the fluid, is dTC dt
RtCt
TC +
RtCt Tamb(t). (i)
The output equation for the flow qR through the walls of the flask is
qR =
Rt
TR
Rt
TC +
Rt
Tamb(t). (ii)
The differential equation describing the dynamics of the fluid temperature TC is found directly by rearranging Eq. (i):
RtCt dTC dt
from which the system time constant τ may be seen to be τ = RtCt. The differential equation relating the heat flow through the flask is dqR dt
RtCt qR =
Rt
dTamb dt
. (iv)
This equation may be written in the standard form by dividing both sides by 1/RtCt,
RtCt
dqR dt
dTamb dt , (v)
and by comparison with Eq. (7) it can be seen that the system time constant τ = RtCt and the forcing function is f (t) = CtdTamb/dt. Notice that the time constant is independent of the output variable chosen.
012345
00.20.40.60.
Normalized time
t/t
y(t)/y(0)
0.0490.3680.1350.
Figure 3: Normalized unforced response of a stable first-order system.
Time −t/τ y(t)/y(0) = e−t/τ^ ys(t) = 1 − e−t/τ 0 0.0 1.0000 0. τ 1.0 0.3679 0. 2 τ 2.0 0.1353 0. 3 τ 3.0 0.0498 0. 4 τ 4.0 0.0183 0.
Table 1: Exponential components of first-order system responses in terms of normalized time t/τ.
The third column of Table 1 summarizes the homogeneous response after periods t = τ, 2 τ,... After a period of one time constant (t/τ = 1) the output has decayed to y(τ ) = e−^1 y(0) or 36.8% of its initial value, after two time constants the response is y(2τ ) = 0. 135 y(0). Several first-order mechanical and electrical systems and their time constants are shown in Fig.
- For the mechanical mass-damper system shown in Fig. 4a, the velocity of the mass decays from any initial value in a time determined by the time constant τ = m/B, while the unforced deflection of the spring shown in Fig. 4b decays with a time constant τ = B/K. In a similar manner the voltage on the capacitor in Fig. 4c will decay with a time constant τ = RC, and the current in the inductor in Fig. 4d decays with a time constant equal to the ratio of the inductance to the resistance τ = L/R. In all cases, if SI units are used for the element values, the units of the time constant will be seconds.
vmF(t)mB
mBF(t) K B
V(t) BKV(t)v= 0refv= 0ref
RC+-V(t) V(t)V= 0refRC LRI(t) I(t)V= 0ref LR
Figure 4: Time constants of some typical first-order systems.
Example 2
A water tank with vertical sides and a cross-sectional area of 2 m^2 , shown in Fig. 5, is fed from a constant displacement pump, which may be modeled as a flow source Qin(t). A valve, represented by a linear fluid resistance Rf , at the base of the tank is always open and allows water to flow out. In normal operation the tank is filled to a depth of
valve Rf
tank Cf
cp(t) Q(t)out
Q(t)p= prefR C ffin
Q(t)in
atm
Figure 5: Fluid tank example
1.0 m. At time t = 0 the power to the pump is removed and the flow into the tank is disrupted.
If the flow through the valve is 10−^6 m^3 /s when the pressure across it is 1 N/m^2 , determine the pressure at the bottom of the tank as it empties. Estimate how long it takes for the tank to empty.
The first-order homogeneous solution is of the form of an exponential function yh(t) = e−λt where λ = 1/τ. The total response y(t) is the sum of two components
y(t) = yh(t) + yp(t) = Ce−t/τ^ + yp(t) (14)
where C is a constant to be found from the initial condition y(0) = 0, and yp(t) is a particular solution for the given forcing function f (t). In the following sections we examine the form of y(t) for the ramp, step, and impulse singularity forcing functions.
1.2.1 The Characteristic Unit Step Response
The unit step us(t) is commonly used to characterize a system’s response to sudden changes in its input. It is discontinuous at time t = 0:
f (t) = us(t) =
{ 0 t < 0 , 1 t ≥ 0.
The characteristic step response ys(t) is found by determining a particular solution for the step input using the method of undetermined coefficients. From Table 8.2, with a constant input for t > 0, the form of the particular solution is yp(t) = K, and substitution into Eq. (13) gives K = 1. The complete solution ys(t) is ys(t) = Ce−t/τ^ + 1. (15)
The characteristic response is defined when the system is initially at rest, requiring that at t = 0, ys(0) = 0. Substitution into Eq. (14) gives 0 = C + 1, so that the resulting constant C = −1. The unit step response of a system defined by Eq. (13) is:
ys(t) = 1 − e−t/τ^. (16)
Equation (16) shows that, like the homogeneous response, the time dependence of the step response depends only on τ and may expressed in terms of a normalized time scale t/τ. The unit step char- acteristic response is shown in Fig. 6, and the values at normalized time increments are summarized in the fourth column of Table 1. The response asymptotically approaches a steady-state value
yss = lim t→∞ ys(t) = 1. (17)
It is common to divide the step response into two regions,
(a) a transient region in which the system is still responding dynamically, and
(b) a steady-state region, in which the system is assumed to have reached its final value yss.
There is no clear division between these regions but the time t = 4τ , when the response is within 2% of its final value, is often chosen as the boundary between the transient and steady-state responses. The initial slope of the response may be found by differentiating Eq. (16) to yield: dy dt
∣∣ ∣∣ t=
τ
The step response of a first-order system may be easily sketched with knowledge of (1) the system time constant τ , (2) the steady-state value yss, (3) the initial slope ˙y(0), and (4) the fraction of the final response achieved at times equal to multiples of τ.
012345
00.20.40.60.811.2 t/ t
y(t)/ysss
yss
Step responseNormalized time
Figure 6: The step response of a first-order system described by τ y˙ + y = us(t).
1.2.2 The Characteristic Impulse Response
The impulse function δ(t) is defined as the limit of a pulse of duration T and amplitude 1/T as T approaches zero, and is used to characterize the response of systems to brief transient inputs. The impulse may be considered as the derivative of the unit step function. The derivative property of linear systems allows us to find the characteristic impulse response yδ(t) by simply differentiating the characteristic step response ys(t). When the forcing function f (t) = δ(t) the characteristic response is
yδ(t) = dys dt
d dt
( 1 − e−t/τ^
)
τ e−t/τ^ for t ≥ 0. (19)
The characteristic impulse response is an exponential decay, similar in form to the homogeneous response. It is discontinuous at time t = 0 and has an initial value y(0+) = 1/τ , where the super- script 0+^ indicates a time incrementally greater than zero. The response is plotted in normalized form in Fig. 7.
1.2.3 The Characteristic Ramp Response
The unit ramp ur(t) = t for t ≥ 0 is the integral of the unit step function us(t):
ur(t) =
∫ (^) t
0
us(t)dt. (20)
The integration property of linear systems (Section 8.4.4) allows the characteristic response yr(t) to a ramp forcing function f (t) = ur(t) to be found by integrating the step response ys(t):
yr(t) =
∫ (^) t
0
ys(t)dt =
∫ (^) t
0
( 1 − e−t/τ^
) dt
= t − τ
( 1 − e−t/τ^
) (21)
012345
012345
t
Time
ty(t)r
Ramp response
Figure 8: The ramp response of a first-order system described by τ y˙ + y = ur(t).
The characteristic responses yu(t) are by definition zero for time t < 0. If there is a discontinuity in yu(t) at t = 0, as in the case for the characteristic impulse response yδ(t) (Eq. (19)), the derivative dyu/dt contains an impulse component, for example
d dt yδ(t) =
τ δ(t) −
τ 2 e−t/τ^ (26)
and if q 1 6 = 0 the response y(t) will contain an impulse function.
1.3.1 The Input/Output Step Response
The characteristic response for a unit step forcing function, f (t) = us(t), is (Eq. (16)):
ys(t) =
( 1 − e−t/τ^
) for t > 0.
The system input/output step response is found directly from Eq. (25):
y(t) = q 1 d dt
( 1 − e−t/τ^
)
( 1 − e−t/τ^
)
= q 0
[ 1 −
( 1 − q 1 q 0 τ
) e−t/τ
]
. (27)
If q 1 6 = 0 the output is discontinuous at t = 0, and y(0+) = q 1 /τ. The steady-state response yss is
yss = lim t→∞ y(t) = q 0. (28)
The output moves from the initial value to the final value with a time constant τ.
1.3.2 The Input/Output Impulse Response
The characteristic impulse response yδ(t) found in Eq. (19) is
yδ(t) =
τ
e−t/τ^ fort ≥ 0
Input u(t) Characteristic Response Input/Output Response y(t) for t ≥ 0
u(t) = 0 y(t) = y(0)e−t/τ
u(t) = ur(t) yr(t) = t − τ
( 1 − e−t/τ^
) y(t) =
[ q 0 t + (q 1 − q 0 τ )
( 1 − e−t/τ^
)]
u(t) = us(t) ys(t) = ys(t) = 1 − e−t/τ^ y(t) =
[ q 0 −
( q 0 −
q 1 τ
) e−t/τ
]
u(t) = δ(t) yδ(t) =
τ e−t/τ^ y(t) =
q 1 τ δ(t) +
( q 0 τ
q 1 τ 2
) e−t/τ
Table 2: The response of the first-order linear system τ y˙ + y = q 1 u˙ + q 0 u for the singularity inputs.
with a discontinuity at time t = 0. Substituting into Eq. (25)
y(t) = q 1
dyδ dt
=
q 1 τ δ(t) +
( (^) q 0 τ
q 1 τ 2
) e−t/τ^ , (29)
where the impulse is generated by the discontinuity in yδ(t) at t = 0 as shown in Eq. (26).
1.3.3 The Input/Output Ramp Response
The characteristic response to a unit ramp r(t) = t is
yr(t) = t − τ
( 1 − e−t/τ^
)
and using Eq. (21) the response is:
y(t) = q 1 d dt
[( t − τ
( 1 − e−t/τ^
)) us(t)
]
( t − τ
( 1 − e−t/τ^
)) us(t)
=
[ q 0 t + (q 1 − q 0 τ )
( 1 − e−t/τ^
)] us(t). (30)
1.4 Summary of Singularity Function Responses
Table 2 summarizes the homogeneous and forced responses of the first-order linear system de- scribed by the classical differential equation
τ dy dx
for the three commonly used singularity inputs. The response of a system with a non-zero initial condition, y(0), to an input u(t) is the sum of the homogeneous component due to the initial condition, and a forced component computed with zero initial condition, that is ytotal(t) = y(0)e−t/τ^ + yu(t), (32)
where yu(t) is the response of the system to the given input u(t) if the system was originally at rest.
produces a constant torque of Tin = 10 N-m. The shaft bearings may be modeled as viscous rotary dampers with a damping coefficient of BR = 0.1 N-m-s/rad. If the flywheel is at rest at t = 0 and the power is suddenly applied to the motor, compute and plot the variation in speed of the flywheel, and find the maximum angular velocity of the flywheel.
T(t)
W refJB = 0
in R
T(t)in W
bearingflywheel
JB
R
Figure 10: Rotary flywheel system and its linear graph
Solution: The state equation for the system may be found directly from the linear graph in Fig. 10: dΩJ dt
BR
J
ΩJ +
J
Tin(t), (i)
which in the standard form is
J BR
dΩJ dt
+ ΩJ =
BR
Tin(t). (ii)
For the flywheel J = mr^2 /2 = 1 kg-m^2 , and the time constant is
τ =
J
BR
= 10 s. (iii)
The characteristic response to a unit step in the forcing function is
ys(t) = 1 − e−t/^10 (iv)
and by the principle of superposition, when the forcing function is scaled so that f (t) = (Tin/BR)us(t), the output is similarly scaled:
ΩJ (t) = Tin BR
( 1 − e−(BR/J)t
) = 100
( 1 − e−t/^10
)
. (v)
The steady-state angular velocity is
Ωss = lim t→∞ ΩJ (t) = Tin/BR = 100 rad/s (vi)
and the angular velocity reaches 98% of this value in t = 4τ = 40 seconds. The step response is shown in Fig. 11.
t
W J (t)
01020304050
020406080100120
Time `(sec)
W SS
Figure 11: Response of the rotary flywheel system to a constant torque input, with initial condition ΩJ (0) = 0, in Example 4
Example 5
During normal operation the flywheel drive system described in Example 4 is driven by a programmed torque source that produces a torque profile as shown in Fig. 12. The torque is ramped up to a maximum of 20 N-m over a period of 100 seconds, held at a constant value for 25 seconds and then reduced to zero. Find the resulting angular velocity of the shaft.
0
5
10
15
200
50 100 150
T(t)int
time (sec)
T(t)in W
bearingflywheel
JB
R
Input (N.m)
Figure 12: Rotary flywheel system and the input torque function specified in Example 5.
Solution: From Example 4 the differential equation describing the system is
J BR
dΩJ dt
+ ΩJ =
BR
Tin(t), (i)
W J (t)
050100150200
050100150200250 Time (sec)
t
Figure 13: Response of the rotary flywheel system to the torque input profile Tin(t) = 0. 2 ur(t) −
- 2 ur(t − 100) − 20 us(t) N-m, with initial condition ΩJ (0) = 0 rad/s.
RC+-V(t) 1R2V (t)o V(t)V= 0RC in
R12ref
Figure 14: Electrical lead network and its linear graph.
and the output equation is
vo(t) = vR 2 = −vc + Vin(t), (ii)
The input/output differential equation is
R 1 R 2 C R 1 + R 2
dvo dt
R 1 R 2 C
R 1 + R 2
dVin dt
R 1
R 1 + R 2
Vin. (iii)
with the system time constant τ = R 1 R 2 C/(R 1 + R 2 ) = 5 × 10 −^3 seconds. The input pulse duration (10 ms) is comparable to the system time constant, and therefore it is not valid to approximate the input as an impulse. The pulse input can, however, be written as the sum of two unit step functions
Vin(t) = us(t) − us(t − 0 .01) (iv)
and the response determined in two separate intervals (1) 0 ≤ t < 0 .01 s where the input is us(t), and (2) t ≥ 0 .01 s, where both components contribute.
The input/output unit step response is given by Eq. (27),
vo(t) =
R 2
R 1 + R 2
( R 2 R 1 + R 2
) e−t/τ
=
R 2
R 1 + R 2
R 1
R 1 + R 2
e−t/τ
=
( 0 .5 + 0. 5 e−t/^0.^005
) for t ≥ 0. (v)
At time t = 0+^ the initial response is vo(0+) = 1 volt, and the steady-state response (v 0 )ss = 0.5 volt. The settling time is approximately 4τ , or about 20 ms. The response to the 10 ms duration pulse may be found from Eqs. (iv) and (v) by using the principle of superposition:
vpulse(t) = vo(t) − v 0 (t − .01). (vi)
In the interval 0 ≤ t < 0 .01, the initial condition is zero and the response is:
vpulse(t) =
( 0 .5 + 0. 5 e−t/^0.^005
) , (vii)
in the second interval t ≥ .01 , when the input is Vin = us(t) − us(t − .01), the response is the sum of two step responses:
vpulse(t) =
( 0 .5 + 0. 5 e−t/^0.^005
) −
( 0 .5 + 0. 5 e−(t−^0 .01)/^0.^005
)
( et/.^005 − e−(t−.01)/.^005
)
= 0. 5 et/^0.^005
( 1 − e^2
) = − 3. 195 e−t/.^005 V. (viii)
The step response (Eq. (v)) and the pulse response described by Eqs. (vii) and (viii) are plotted in Fig. 15.
2 Second-Order System Transient Response
Second-order state determined systems are described in terms of two state variables. Physical second-order system models contain two independent energy storage elements which exchange stored energy, and may contain additional dissipative elements; such models are often used to represent the exchange of energy between mass and stiffness elements in mechanical systems; be- tween capacitors and inductors in electrical systems, and between fluid inertance and capacitance elements in hydraulic systems. In addition second-order system models are frequently used to rep- resent the exchange of energy between two independent energy storage elements in different energy domains coupled through a two-port element, for example energy may be exchanged between a mechanical mass and a fluid capacitance (tank) through a piston, or between an electrical induc- tance and mechanical inertia as might occur in an electric motor. Engineers often use second-order system models in the preliminary stages of design in order to establish the parameters of the energy storage and dissipation elements required to achieve a satisfactory response. Second-order systems have responses that depend on the dissipative elements in the system. Some systems are oscillatory and are characterized by decaying, growing, or continuous oscillations.
det [sI − A] X(s) =
[ s − a 22 a 12 a 21 s − a 11
] [ b 1 b 2
] U (s)
from which
d^2 x 1 dt^2 − (a 11 + a 22 )
dx 1 dt
- (a 11 a 22 − a 12 a 21 ) x 1 = b 1
du dt
- (a 12 b 2 − a 22 b 1 )u. (35)
and d^2 x 2 dt^2 − (a 11 + a 22 )
dx 2 dt
- (a 11 a 22 − a 12 a 21 ) x 2 = b 2
du dt
- (a 21 b 1 − a 11 b 2 )u. (36)
which can be written in terms of the two parameters ωn and ζ
d^2 x 1 dt^2
dx 1 dt
du dt
- (a 12 b 2 − a 22 b 1 )u (37) d^2 x 2 dt^2
- 2ζωn
dx 2 dt
du dt
- (a 21 b 1 − a 11 b 2 )u. (38)
where ωn is defined to be the undamped natural frequency with units of radians/second, and ζ is defined to be the system (dimensionless) damping ratio. These definitions may be compared to Eqs. (35) and (36), to give the following relationships:
ωn =
a 11 a 22 − a 12 a 21 (39) ζ = −
2 ωn (a 11 + a 22 )
= − (a 11 + a 22 ) 2
a 11 a 22 − a 12 a 21
The undamped natural frequency and damping ratio play important roles in defining second-order system responses, similar to the role of the time constant in first-order systems, since they com- pletely define the system homogeneous equation.
Example 7
Determine the differential equations in the state variables x 1 (t) and x 2 (t) for the system [ x ˙ 1 x ˙ 2
]
[ − 1 − 2 2 − 3
] [ x 1 x 2
]
[ 1 0
] u. (i)
Find the undamped natural frequency ωn and damping ratio ζ for this system. Solution: For this system [sI − A] =
[ s + 1 2 − 2 s + 3
] (ii)
and det [sI − A] = s^2 + 4s + 7 and therefore for state variable x 1 (t):
d^2 x 1 dt^2
dx 1 dt
and for x 2 (t): d^2 x 2 dt^2
dx 2 dt
By inspection of either Eq. (iii) or Eq. (iv), ω^2 n = 7, and 2ζωn = 4, giving ωn =
rad/s, and ζ = 2/
2.0.2 Generation of a Differential Equation in an Output Variable
The output equation y = Cx + Du for any system variable is a single algebraic equation:
y(t) =
[ c 1 c 2
] [ x 1 x 2
]
= c 1 x 1 (t) + c 2 x 2 (t) + du(t). (41)
and in the Laplace domain
Y (s) =
( C(sI − A)−^1 B + D
) U (s)
=
det [sI − A] (Cadj(sI − A) + det [sI − A] D)
The determinants may be expanded and the resulting equation written as a differential equation:
d^2 y dt^2 − (a 11 + a 22 )
dy dt
- (a 11 a 22 − a 12 a 21 ) y = q 2
d^2 u dt^2
du dt
or in terms of the standard system parameters
d^2 y dt^2
2ζωn dy dt
ω n^2 y = q 2 d^2 u dt^2
q 1 du dt
q 0 u (43)
where the coefficients q 0 , q 1 , and q 2 are
q 0 = c 1 (−a 22 b 1 + a 12 b 2 ) + c 2 (−a 11 b 2 + a 21 b 1 ) + d (a 11 a 22 − a 12 a 21 ) q 1 = c 1 b 1 + c 2 b 2 − d (a 11 + a 22 ) q 2 = d. (44)
Notice that the left hand side of the differential equation is the same for all system variables, and that the only difference between any of the differential equations describing any system variable is in the constant coefficients q 2 , q 1 and q 0 on the right hand side.
Example 8
A rotational system consists of an inertial load J mounted in viscous bearings B, and driven by an angular velocity source Ωin(t) through a long light shaft with significant torsional stiffness K, as shown in the Fig. 16. Derive a pair of second-order differential equations for the variables ΩJ and ΩK.
