Review: step response of 1st order systems we’ve seen
•
Inertia with bearings (viscous friction)
Step input Ts (t) = T0 u(t) ⇒ Step response∗
•
•
´
T0 ³
−t/τ
,
ω(t) =
1−e
b
where τ =
J
.
b
RC circuit (charging of a capacitor)
R
+
Step input vi (t) = V0 u(t) ⇒ Step response
³
´
−t/τ
vC (t) = V0 1 − e
, where τ = RC.
vi
−
+
C
vC
−
DC motor with inertia load, bearings and negligible inductance
Step input vs (t) = V0 u(t) ⇒ Step response
´
Km ³
−t/τ
ω(t) =
,
V0 1 − e
R
where
τ= Ã
2.004 Fall ’07
J
b+
Km Kv
R
!.
∗
When writing the step response,
we may omit the step function u(t)
implying the result holds for t > 0 only.
Lecture 06 – Monday, Sept. 17
Goals for today
•
•
•
•
First-order systems response
– pole, zero definitions
– the significance of poles and zeros:
from s-domain representation to transient characteristics
DC motor dynamics:
– angular velocity (1st order: 1 pole)
– current (1st order: 1 pole, 1 zero)
Two new properties of the Laplace transform:
– final value theorem
– initial value theorem
Next two lectures:
– 2nd order systems
2.004 Fall ’07
Lecture 06 – Monday, Sept. 17
Current dynamics in DC motor system
Recall combined equations of motion
)
LsI(s) + RI(s) + Kv Ω(s) = Vs (s)
⇒
JsΩ(s) + bΩ(s) = Km I(s)
⎧ ·
µ
¶
µ
¶¸
LJ
Lb
K
Km
K
⎪
m
v
⎨
s2 +
+J s+ b+
Ω(s) =
Vs (s)
R
R
R
R
⎪
⎩ (Js + b) Ω(s) = K I(s)
m
Neglecting the DC motor’s inductance (i.e., assuming L/R ≈ 0), we find
⎧
Km
Transfer function for the angular velocity
⎪
⎪
⎪
⎪
Ω(s)
is of the form
RJ
⎪
⎪
=
⎪
µ
¶
⎪
µ
¶
Vs (s)
⎪
K
K
1
⎪
m
v
K
1
K
A
m
v
⎪
⎪
b+
s+
⎪
b+
,
p=
pole
⎪
⎨
J
R
s+p
J
R
µ
¶
⎪
⎪
b
1
Transfer function for the current
⎪
⎪
zero
s
+
⎪
⎪
I(s)
is
of
the
form
⎪
R
J
⎪
⎪
=
⎪
µ
¶
µ
¶
⎪
V
(s)
⎪
s
K
K
1
K
K
b
1
⎪
m
v
B
(s
+
z)
m
v
⎪
⎩
b+
s+
b+
, z=
,
p=
J
R
s+p
J
R
J
2.004 Fall ’07
Lecture 06 – Monday, Sept. 17
Poles and zeros in the s-plane
jω
right half—plane
imaginary axis
left half—plane
s—plane
real axis
2.004 Fall ’07
−p
−z
pole
zero
Lecture 06 – Monday, Sept. 17
σ
DC motor step response: numerical example
Km
L ≈ 0, R = 6Ω, Kv = 6V · sec,
= 6N · m/A, J = 2kg · m2 , b = 4kg · m2 · Hz.
We will compute the system’s response
(both angular velocity and current)
to the step input vs (t) = 30u(t) V.
Substituting the numerical values into the system TF,
⎧
1
1
Ω(s)
⎪
⎪
=
⎪
⎪
⎨ Vs (s)
2 s + 5 [Hz]
whereas the Laplace transform of the input is
we find
⎪
⎪
1 (s + 2 [Hz])
I(s)
h
i
h
i 30
⎪
⎪
=
⎩
Vs (s) ≡ L vs (t) = L 30u(t) =
.
Vs (s)
6 s + 5 [Hz]
s
I. Angular velocity
15
Ω(s) =
= 15
s (s + 5)
Ω(s) = 3
µ
µ
1
1
−
s s+5
K1
K2
+
s
s+5
¶
¶
⇒
where
¯
¯
1 ¯¯
1
1 ¯¯
1
=
=
=
−
K1 =
,
K
⇒
2
s + 5 ¯s=0
5
s ¯s=−5
5
¢
¡
ω(t) = 3 1 − e−5t u(t)
Forced response
2.004 Fall ’07
Lecture 06 – Monday, Sept. 17
rad
.
sec
Homogeneous response
(Natural response)
DC motor step response: numerical example
Km
L ≈ 0, R = 6Ω, Kv = 6V · sec,
= 6N · m/A, J = 2kg · m2 , b = 4kg · m2 · Hz.
We will compute the system’s response
(both angular velocity and current)
to the step input vs (t) = 30u(t) V.
Substituting the numerical values into the system TF,
⎧
1
1
Ω(s)
⎪
⎪
=
⎪
⎪
⎨ Vs ( s )
2 s + 5 [Hz]
whereas the Laplace transform of the input is
we find
⎪
⎪
1 (s + 2 [Hz])
I (s )
h
i
h
i 30
⎪
⎪
=
⎩
Vs (s) ≡ L vs (t) = L 30u(t) =
.
Vs ( s )
6 s + 5 [Hz]
s
II. Current
5 (s + 2)
=5
I(s) =
s (s + 5)
I(s) =
µ
µ
K2 0
K1 0
+
s
s+5
3
2
+
s s+5
¶
¶
⇒
K1 0
where
¯
s + 2 ¯¯
2
,
=
=
s + 5 ¯s=0
5
¢
¡
i(t) = 2 + 3e−5t u(t) A.
Forced response
2.004 Fall ’07
Lecture 06 – Monday, Sept. 17
K2 0
¯
s + 2 ¯¯
3
⇒
=
=
s ¯s=−5
5
Homogeneous response
(Natural response)
DC motor system in the s-plane
Input: Voltage source
Output: Angular velocity
system
representation
Gω (s)
30
Vs (s) =
s
1 1
2s+5
Ω(s)
Input: Voltage source
Output: Current
30
Vs (s) =
s
2.004 Fall ’07
system
representation
Gi (s)
1s+2
6s+5
I(s)
Lecture 06 – Monday, Sept. 17
DC motor step response (angular velocity)
Image removed due to copyright restrictions.
Please see: Fig. 4.1 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
2.004 Fall ’07
Lecture 06 – Monday, Sept. 17
DC motor step response (current)
Image removed due to copyright restrictions.
Please see: Fig. 4.1 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
2.004 Fall ’07
Lecture 06 – Monday, Sept. 17
1st order system response from s-plane representation
•
•
•
•
Pole in the input function generates forced response
Pole in the transfer function generates natural response
Zero in the transfer function does not alter the speed of settling to
steady state (i.e. the time constant) but it does alter the relative
amplitudes of the forced and natural responses
Pole at –α generates response
e–αt (exponentially decreasing if
pole on the right half-plane;
increasing if on the left half-plane)
jω
−αt
e
•
Pole at zero generates step
function
jω
u(t)
σ
σ
0
−α
Time constant τ=1/α
2.004 Fall ’07
Lecture 06 – Monday, Sept. 17
u(t)
1st order system: transient response properties
Step response
in the s—domain
a
;
s(s + a)
in the time domain
¢
¡
1 − e−at u(t);
time constant
τ=
1
;
a
rise time (10%→90%)
2.2
;
Tr =
a
settling time (98%)
4
Ts = .
a
c(t)
Initial slope =
1
=a
time constant
1.0
0.9
0.8
0.7
63% of final value
at t = one time constant
0.6
0.5
0.4
0.3
0.2
0.1
0
1
a
2
a
Tr
3
a
4
a
5
a
Ts
Figure by MIT OpenCourseWare.
Figure 4.3
2.004 Fall ’07
steady state
(final value)
Lecture 06 – Monday, Sept. 17
t
DC motor step responses
¡
ω(t) = 3 1 − e
−5t
ω(∞) = 3
¢
u(t) rad/sec
rad/sec
dω
(0+) = 3 × 5 = 15
dt
2.004 Fall ’07
¡
¢
−5t
i(t) = 2 + 3e
u(t)
rad/sec2
Lecture 06 – Monday, Sept. 17
i(∞) = 2 A
di
(0+) = ∞
dt
A
The Final Value theorem: steady-state
We will now learn two additional properties of the Laplace transform, which we
will quote without proof. Let F (s) denote the Laplace transform of the function
f (t). The first property is the
Final Value theorem:
f (∞) = lims→0 sF (s);
Let us see how this applies to the step response of a general 1st —order system
with a pole at −a and without a zero (e.g., the angular velocity response of the
DC motor.) We select the system gain such that the steady—state will equal 1.
The step response in the s—domain then is
F1 (s) =
1
1
a
= −
;
s(s + a)
s s+a
also,
sF1 (s) =
a
.
s+a
Using the partial fraction expansion above, we find the time domain step response as
¡
¢
(as advertised.)
f1 (t) = 1 − e−at u(t) ⇒ f1 (∞) = 1
Applying the final value theorem, we find, consistently,
f1 (∞) = lims→0 sF1 (s) = lims→0
2.004 Fall ’07
a
= 1.
s+a
Lecture 06 – Monday, Sept. 17
The Initial Value theorem: initial slope
The second property of the Laplace transform is the
Initial Value theorem
f (0+) = lims→∞ sF (s).
Let us use this property to compute the initial slope of the step response, i.e.
the value of the derivative of the step response at t = 0+ for the same general
1st —order system with steady state equal to unity, a pole at −a and without a
zero. Since we are interested in the derivative of f (t), the Laplace transform of
interest is
·
¸
df1 (t)
as
H1 (s) = L
= sF1 (s) ⇒ sH1 (s) = s2 F1 (s) =
.
dt
s+a
Applying the final value theorem,
as
a
df1
(0+) = lims→∞
= lims→∞
= a.
dt
s+a
1 + a/s
Again, this is consistent with the result we get directly from the time domain:
¢
¡
df1
df1
(t) = − (−a) e−at ⇒
(0+) = a.
f1 (t) = 1 − e−at ⇒
dt
dt
2.004 Fall ’07
Lecture 06 – Monday, Sept. 17
Initial and final value of 1st-order system with a zero
We now consider the same 1st —order system with unity steady state, a pole at
−a, but we also add a zero at −z. In that case, the Laplace transforms of the
step response and its derivative are
F2 (s) = a
z a−z
s+z
= +
;
s(s + a)
s s+a
sF2 (s) = a
s+z
;
s+a
s2 F2 (s) = a
s(s + z)
.
s+a
We can readily see that
f2 (∞) = lims→0 sF2 (s) = z;
f2 (0+) = lims→∞ sF2 (s) = a;
df2
(0+) = lims→∞ s2 F2 (s) = ∞.
dt
You should verify that these results are consistent with the time—domain solution
for this system.
We can see that the effects of the zero −z on the 1st —order system are (in
comparison to a system with the same pole at −a but without the zero)
• amplify the steady—state response by z;
• raise the initial value from zero to A;
• raise the initial slope to infinity.
The infinite initial slope is non—physical; in the case of the DC motor, it occurs
because we neglected the inductance L.
2.004 Fall ’07
Lecture 06 – Monday, Sept. 17
How the zero acts
Comparing the system response F1 (s) (without a zero) and the system response
F2 (s) (with a zero at −z), we can see that
F2 (s) = (s + z)F1 (s) = sF1 (s) + zF1 (s) ⇒
df1 (t)
+ zf1 (t).
dt
That is, the zero results in derivative action and amplification.
Both of these results are qualitatively consistent with our observations from the
previous page.
f2 (t) =
DC motor
Note: In this case,
there is an additional
amplification factor of 1/3
from ω(t) to i(t).
tive
a
v
i
r
de
2.004 Fall ’07
n
actio
Lecture 06 – Monday, Sept. 17
amp
lifica
tion