Lecture 4
Step response and pole locations
4-1
Review
Definition of z-transform:
Discrete transfer function:
U (z) = Z{uk } =
1
X
uk z
k
k=0
Y (z)
= G(z) = Z{gk },
U (z)
gk = pulse response
Construct a discrete model of a continuous sampled-data system G(s) . . .
. . . by computing the pulse response gk and transforming to get G(z):
G(z) = (1
4-2
z
1
)Z
⇢
G(s)
s
Output response: Y (z) = G(z)U (z) () yk = gk ⇤ uk
Review
Analyse/design a discrete controller D(z):
by considering the purely discrete time system:
Closed loop system tranfer function:
Y (z)
G(z)D(z)
=
R(z)
1 + G(z)D(z)
How do the closed loop poles relate to
! stability?
! performance?
4-3
Response of 2nd order system
Consider the z-transform of a sinusoid multiplied by a an exponential signal:
y(t) = e
? sample:
at
cos(bt) U(t)
y(kT ) = rk cos(k✓) U(kT )
z
1
z
+
j✓
re )
2 (z re
z(z r cos ✓)
=
(z rej✓ )(z re j✓ )
? transform: Y (z) =
1
2 (z
? e.g. yk is the pulse response of G(z):
G(z) =
z(z r cos ✓)
(z rej✓ )(z re
n z = rej✓
z = re j✓
nz=0
zeros:
z = r cos ✓
poles:
4-4
j✓ )
(U(t) = unit step)
with r = e
j✓ )
aT
& ✓ = bT
Response of 2nd order system
Responses for varying r:
r<1
+
exponentially decaying
envelope
r = 0.7
✓ = ⇡/4
0.5
yk
.
1
0
−0.5
0
2
4
6
8
10
sample k
r=1
+
sinusoidal response
with 2⇡/✓ samples
per period
1
0.5
yk
.
0
r = 1.0
✓ = ⇡/4
−0.5
−1
0
2
4
6
8
10
sample k
r>1
+
exponentially increasing
envelope
10
5
yk
.
r = 1.3
✓ = ⇡/4
0
−5
0
2
4
6
8
10
sample k
4-5
Response of 2nd order system
1
.
✓=0
+
decaying exponential
yk
Responses for varying ✓:
r = 0.7
✓=0
0.5
0
0
2
4
6
8
10
sample k
1
✓ = ⇡/2
+
2⇡/✓ = 4 samples
per period
r = 0.7
✓ = ⇡/2
0.5
yk
.
0
−0.5
0
2
4
6
8
10
sample k
✓=⇡
+
2 samples per period
1
0.5
yk
.
0
r = 0.7
✓=⇡
−0.5
−1
0
4-6
2
4
6
sample k
8
10
Response of 2nd order system
Some special cases:
.
for ✓ = 0, Y (z) simplifies to:
Y (z) =
z
z
r
=) exponentially decaying response
.
when ✓ = 0 and r = 1:
Y (z) =
z
z
1
=) unit step
.
when r = 0:
Y (z) = 1
=) unit pulse
.
when ✓ = 0 and
1 < r < 0:
samples of alternating signs
4-7
Pole positions in the z-plane
Poles inside the unit circle
are stable
Im(z)
Poles outside the unit circle
are unstable
Poles on the unit circle
are oscillatory
Real poles at 0 < z < 1
give exponential response
Higher frequency of
oscillation for larger ✓
Lower apparent damping
for larger ✓ and r
4-8
Re(z)
Relationship with s-plane poles
If F (s) has a pole at s = a
then F (z) has a pole at z = e
aT
F (s)
f (kT )
1
s
1(kT )
1
s2
kT
1
s+a
e
1
(s + a)2
kT e
a
s(s + a)
1
"
consistent with z = esT
What about transfer functions?
⇢
G(s)
1
G(z) = (1 z )Z
s
#
b 1
e
(s + a)(s + b)
If G(s) has poles s = ai
a
+ a2
F (z)
z
z
Tz
(z 1)2
then G(z) has poles z = eai T
but the zeros are unrelated
b
(s + a)2 + b2
e
z
e
akT
z
e
e
akT
aT
aT )2
z(1 e aT )
(z 1)(z e aT )
akT
akT
aT
T ze
(z e
akT
(e
sin bkT
aT
bT
(z
e
e aT )(z
z2
z sin aT
(2 cos aT )z + 1
bkT
sin akT
s2
1
ze
z2
2e
aT
)z
e bT )
sin bT
aT(cos bT )z
+e
4-9
The mapping from s-plane to z-plane
Locus of s =
+ j! under the mapping z = esT :
? imaginary axis (s = j!,
= 0) ! unit circle (|z| = 1)
? left-half plane ( < 0) ! inside of unit circle (|z| < 1)
? right-half plane ( > 0) ! outside of unit circle (|z| > 1)
? region of s-plane within the Nyquist rate (|!| < ⇡/T ) ! entire z-plane
s-plane
z-plane
Im(s)
Im(z)
✓
! = ⇡/T
z = esT
-
!=
⇡/T
@
R
@
4 - 10
Re(s)
6
! = ±⇡/T
Re(z)
2aT
The mapping from s-plane to z-plane
s-plane
z-plane
Im(s)
Im(z)
z = esT
Re(s)
@
I
@
s = + j!
= constant
Re(z)
z=e
|z| = e
Im(s)
T j!T
T
e
= constant
Im(z)
z = esT
Re(s)
Re(z)
s = + j!
! = constant
z = e T ej!T
arg(z) = !T constant
4 - 11
The mapping from s-plane to z-plane
Pole locations for constant damping ratio ⇣ < 1
Im(s)
p
1 ⇣ 2 !0
s2 + ⇣!0 s + !02 = 0
+
s=
⇣!0 ± j
⇣ = 0.5
⇣ = 0.7
p
✓
1
⇣2
!0
Re(s)
⇣!0
cos ✓ = ⇣
Im(s)
Im(z)
⇣ = 0.5
z = esT
Re(s)
4 - 12
⇣ = 0.7
⇣ = 0.5
p
s = ⇣!0 + j 1
⇣ 2 !0 : ⇣ = constant
Re(z)
@
I
@ ⇣ = 0.7
z=e
⇣!0 T
e
j
p
1 ⇣ 2 !0 T
The mapping from s-plane to z-plane
4 - 13
The mapping from s-plane to z-plane
!0 = 0.5⇡/T
⇣ = 0.2
@
R
@
↵
⇡
!0 = 0.3⇡/T
⇣ = 0.5
4 - 14
System specifications
Second order step responses (e.g. see HLT)
Design criteria based on step response:
?
Damping ratio ⇣ in range 0.5 – 0.9
[application-dependent]
?
Natural frequency !0 as large as possible
[for fastest response]
4 - 15
System specifications
Typical specifications for the step response:
Rise time (10% ! 90%):
Peak overshoot:
tr ⇡ 1.8/!0 p
?
Settling time (to 1%):
ts = 4.6/(⇣!0 )
?
Steady state error to unit step:
?
Phase margin:
?
?
4 - 16
Mp ⇡ e
⇡⇣/
ess
PM
⇡ 100⇣
1 ⇣2
System specifications
Typical specifications for the step response:
tr , M p
!
⇣, !0
!
ts
!
radius of poles: |z| < 0.01T /ts
ess
!
final value theorem: ess = lim (z
locations of dominant poles
z!1
1)E(z)
4 - 16
System specifications
Example – A continuous system with transfer function
1
G(s) =
s(10s + 1)
is controlled by a discrete control system with a ZOH
The closed loop system is required to have:
– step response overshoot: Mp < 16%
– step response settling time (1%): ts < 10 s
– steady state error to unit ramp: ess < 1
Check these specifications if T = 1 s and the controller is
uk =
4 - 17
0.5uk
1
+ 13(ek
0.88ek
1)
System specifications
1. (a) Find the pulse transfer function of G(s) plus the ZOH
G(z) = (1
z
1
)Z
n G(s) o
s
=
(z
z
1) n
Z
e.g. look up Z{a/s2 (s + a)} in tables:
⇣
0.1
)z + (1
(z 1) z (0.1 1 + e
G(z) =
z
0.1(z 1)2 (z
0.0484(z + 0.9672)
=
(z 1)(z 0.9048)
o
0.1
s2 (s + 0.1)
e
0.1
e
0.1 )
0.1e
0.1
)
⌘
(b) Find the controller transfer function (using z = shift operator):
U (z)
(1 0.88z 1 )
(z 0.88)
= D(z) = 13
=
13
E(z)
(1 + 0.5z 1 )
(z + 0.5)
4 - 18
System specifications
2. Check the steady state error ess when rk = unit ramp
ess = lim ek = lim (z
k!1
z!1
1)E(z)
E(z)
1
=
R(z)
1 + D(z)G(z)
R(z) =
ess
o
Tz
1
= lim (z 1)
= lim
z!1
z!1 (z
(z 1)2 1 + D(z)G(z)
T
10
= lim
z!1
0.0484(z + 0.9672)
(z 1)
D(1)
8
(z 1)(z 0.9048)
=
1 0.9048
= 0.96
0.0484(1 + 0.9672)D(1)
=) ess < 1
(as required)
Output y and reference r
so
n
T
1)D(z)G(z)
6
4
2
0
0
4 - 19
Tz
(z 1)2
5
Time (sec)
10
System specifications
3. Step response: overshoot Mp < 16% =) ⇣ > 0.5
settling time ts < 10 =) |z| < 0.011/10 = 0.63
The closed loop poles are the roots of 1 + D(z)G(z) = 0, i.e.
1 + 13
(z 0.88) 0.0484(z + 0.9672)
=0
(z + 0.5) (z 1)(z 0.9048)
=) z = 0.88,
0.050 ± j0.304
But the pole at z = 0.88 is cancelled by controller zero at z = 0.88, and
⇢
r = 0.31, ✓ = 1.73
±j✓
z = 0.050 ± j0.304 = re
=)
⇣ = 0.56
Output y and input u/10
1.5
1
0.5
0
−0.5
all specs satisfied!
−1
−1.5
0
1
2
3
4 - 20
4
5
6
Time (sec)
7
8
9
10
Fast sampling revisited
For small T :
z = esT = 1 + sT + (sT )2 /2 + · · · ⇡ 1 + sT
=)
s⇡
z
1
T
Hence the image of the unit circle under the map from z to s-plane becomes
⇥
⇤
Im(z 1)
Im (z 1)/T
z-plane loci of
constant ⇣ & !0
Re(z
1)
⇥
Re (z
1)/T
⇤
⇡ s-plane loci
near z = 1
but the dominant poles lie near z = 1. . .
. . . so the discrete response tends to the continuous response as T ! 0
4 - 21
Summary
Dependence of system pulse response on pole locations
For a sampled data system with a ZOH:
if s = ai is a pole of G(s), then z = eai T is a pole of G(z)
Locus of s =
+ j! under the mapping z = esT :
? the left half plane ( < 0) maps to the unit disk (|z| < 1)
? s-plane poles with damping ratio ⇣, natural frequency !0 map to
z-plane poles with:
|z| = e ⇣!0 T
p
arg(z) = 1 ⇣ 2 !0 T
Design specifications (rise time, settling time, overshoot)
imply constraints on locations of dominant poles
4 - 22
What you should know. . .
1. How control systems are a↵ected by the presence of a computer in the control loop.
[L1]
2. How to approximate fast sampling continuous systems (filters or controllers) using discrete
time approximations to continuous time derivatives.
[L1]
3. The e↵ects of sample rate on a computer-controlled system.
[L1]
4. How to describe di↵erence equations using transfer functions between signals represented by
z-transforms.
[L2]
5. The transfer function is the z-transform of the pulse response and the system output is the
convolution of the pulse response and the input.
[L2]
6. How to derive the discrete time model of a continuous time sampled data system using
transform techniques.
[L3]
7. How to compute the dynamic response of a sampled data system.
[L3]
8. Properties of z-transforms: linearity, initial and final value theorems, multiplication,
convolution and di↵erentiation.
[L3]
9. How the location of z-plane poles a↵ects the step response of a second order system.
[L4]
10. How the poles of sampled data systems map from the s-plane to the z-plane.
[L4]
11. How to relate specifications on damping and speed of response to specifications on z-plane
pole locations.
[L4]
4 - 23