Lecture 10: From Analog to Digital Controllers A Digital Control ...

Lecture 10:
From Analog to Digital Controllers
u( t )
A Digital Control System
y ( t )
• The sampling theorem
• Aliasing and presampling filters
• Unsampling signals
• Discrete time approximation of continuous time controller
– Frequency domain
– State space
u k
u( t )
t
Process
Hold
Sampler
u k
y k
D-A Computer A-D y k
t
t
y(
t
)
t
The Sampling Theorem (Shannon 1949)
Proof idea
A continuous-time signal with Fourier transform F(ω ) = 0 for
|ω | > ω 0 , is uniquely defined by its sampled values, provided
the sampling frequency ω s = 2π /h satisfies ω s /2 > ω 0 .
The signal can be reconstructed by the interpolation formula
f (t) =
∞∑
k=−∞
f (kh) sinω s(t − kh)/2
ω s (t − kh)/2
−2ω N
−ωN
0
F(ω)
ω N
2ω N
F s (ω)
−2ω N −ω N 0 ω N 2ω N
Let F(ω ) be the Fourier transform of f and expand the periodic
function F s (ω ) as a Fourier series
F s (ω ) = 1 h
∞∑
k=−∞
F(ω + kω s ) =
∞∑
k=−∞
C k e −ikhω
It is straightforward to verify that
The frequency ω s /2 is denoted ω N and called the Nyquist
frequency.
C k = f (kh) k = 0, ±1, ±2, . . .
Hence {f (kh)} → F s (ω ) → F(ω ) → f (t)
c Dept. of Automatic Control, Lund 1

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Aliasing and frequency folding
1
Aliasing and presampling filters
0
0 5 10
Time
F (ω ) F s (ω )
−2ω N
−ωN
0
ω N
2ω N
−2ω N −ω N 0 ω N 2ω N
ω N = ω s /2 Nyquist frequency
New frequencies since the system is NOT time-invariant
ω sampled = ω + nω s n = 0, ±1, ±2, . . .
Mini-problem
Example — Feedwater heating in a ship boiler
A wheel, rotating 90 turns per second, is filmed using a
camera with sampling time 10ms. What will the wheel rotation
look like on the film?
Feed
water
Pressure
Pump
Steam
Valve
To boiler
Temperature
38 min
2 min
Condensed
water
Temperature
Pressure
2.11 min
Time
c Dept. of Automatic Control, Lund 2

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Pre- and postsampling filters
Example – Prefiltering
Typical control problem:
(a)
1
(b)
1
"Decrease the influence of a low-frequency process disturbance
despite high-frequency measurement noise."
0
0
• Frequency separation
• Prefilter ω N
– Bessel, Butterworth, ITAE filters
(c)
0 10 20 30
1
(d)
0 10 20 30
1
• Postsampling filters
0
0
– Avoid exciting mechanical resonances
– Higher order hold
0 10 20 30
Time
0 10 20 30
Time
ω d = 0.9, ω N = 0.5, ω alias = 0.1 (6th order Bessel filter)
Consequence of using prefilter
Pre- and postfilter should be included in the process model
Exception: Fast sampling
A Besselfilter can be approximated with a delay:
• Shannon
Unsampling signals
f (t) =
∞∑
k=−∞
f (kh) sin(ω s(t − kh)/2)
ω s (t − kh)/2
Gain
1
0.01
0.1 1 10
0
• Zero order hold
• First order hold
Phase
• Predictive first order hold
0.1 1 10
Frequency, rad/s
6th order Bessel (solid line), time delay (dashed line)
c Dept. of Automatic Control, Lund 3

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Shannon reconstruction
f (t) =
∞∑
k=−∞
The exact formula is not causal !
f (kh) sin(ω s(t − kh)/2)
ω s (t − kh)/2
Approximation with information from d time steps ahead:
ˆ f (nh + τ ) =
∑n+d
k=−∞
f (kh)h(nh + τ − kh) h(τ ) = sinω sτ /2
ω s τ /2
Zero and first order hold
t 0 t 1 t 2 t 3 t 4 t 5
t 6 Time
1
0
sinπ t
π t
0 10
Time
t 0 t 1 t 2 t 3 t 4 t 5
t 6 Time
Predictive first order hold
Sinusoidal signal with h = 1 and h = 0.5
• Use forward difference instead of backward difference
u(kh + τ ) = u(kh) + τ (u(kh + h) − u(kh))
h
• Use model of the controller.
Zero order hold
h=1
1
0
0 5 10
Zero order hold
h=0.5
1
0
0 5 10
First order hold
1
0
First order hold
1
0
0 5 10
0 5 10
t 0 t 1
t 2 t 3
t 4
t 5
t 6 Time
Predictive hold
1
0
Predictive hold
1
0
0 5 10
Time
0 5 10
Time
c Dept. of Automatic Control, Lund 4

Implementing a controller using a computer
H(z) ≈ G (s)
Discrete time approximation of
continuous time controller
— Frequency domain
u(t)
Want to get
Methods:
A-D
{ u( kh )} { y( kh )}
Algorithm
Clock
D-A
A/D + Algorithm + D/A ≈ G(s)
• Approximate s i.e G(s) → H(z)
• Pole-zero matching
• Use sample/hold to get H(z) from G(s)
y(t)
Matlab
SYSD = C2D(SYSC,TS,METHOD) converts the continuous
system SYSC to a discrete-time system SYSD with
sample time TS. The string METHOD selects the
discretization method among the following:
’zoh’ Zero-order hold on the inputs.
’foh’ Linear interpolation of inputs
(triangle appx.)
’tustin’ Bilinear (Tustin) approximation.
’prewarp’ Tustin approximation with frequency
prewarping.
The critical frequency Wc is specified
last as in C2D(SysC,Ts,’prewarp’,Wc)
’matched’ Matched pole-zero method
(for SISO systems only).
Approximation methods
Forward difference (Euler’s method)
dx(t)
dt
Backward difference
dx(t)
dt
≈
≈
x(t + h) − x(t)
h
x(t) − x(t − h)
h
Trapezoidal method (Tustin, bilinear)
ẋ(t + h) + ẋ(t)
2
≈
= q − 1
h
= 1 − q−1
h
x(t + h) − x(t)
h
x(t)
x(t)
c Dept. of Automatic Control, Lund 5

Mini-problem
Compare the differential equation
ẏ(t) + 10y(t) = 0
to the Euler approximation
y(k + 1) − y(k) + 10y(k) = 0
Is there any qualitative difference?
What is the source of the problem?
Properties of the approximation H(z) ≈ G(s)
s = z − 1
h
s = z − 1
s = 2 h
zh
z − 1
z + 1
(Forward difference or Euler’s method)
(Backward difference)
(Tustin’s or bilinear approximation)
Where do stable poles of G get mapped?
Forward differences Backward differences Tustin
Prewarping to reduce frequency distortion
s
iω ′
−iω ′
Choose one fix-point ω 1
s =
z
e iω ′
ω 1
tan(ω 1 h/2) ⋅ z − 1
z + 1
e −iω ′
Approximation
Pole-zero matching
1. All poles of G(s) are mapped according to z = e sh
2. All finite zeros are also mapped as z = e sh
3. The zeros of G(s) at s = ∞ are mapped into z = −1. One
of the zeros of G(s) at s = ∞ is mapped into z = ∞, i.e.
deg A(z) − deg B(z) = 1
4. The gain of H(z) is matched the gain of G(s) at one
frequency, center frequency or at steady-state, s = 0
This implies that H ( e ) iω 1h
= G(iω 1 ) Tustin is obtained for
ω 1 = 0 since tan ( )
ω 1 h
2 ≈
ω 1 h
for small ω .
2
c Dept. of Automatic Control, Lund 6

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Comparison of approximations
(s + 1) 2 (s 2 + 2s + 400)
G(s) =
(s + 5) 2 (s 2 + 2s + 100)(s 2 + 3s + 2500)
h = 0.03 gives ω N = 105 rad/s
Magnitude
10
10
10
10
10 0 10 1 10 2
Selection of sampling interval
The Zero-Order-Hold has the transfer function
For small h
1 − e −sh
sh
G ZoH (s) = 1 − e−sh
sh
≈ 1 − 1 + sh − (sh)2 /2 + ⋅ ⋅ ⋅
sh
exp(−sh/2) ≈ 1 − sh 2 + ⋅ ⋅ ⋅
= 1 − sh 2 + ⋅ ⋅ ⋅
Phase
0
10 0 10 1 10 2
Frequency, rad/s
G ZoH (s) can be approximated as a delay h/2
Assume the phase margin can be decreased by 5 ○ to 15 ○ then
hω c ≈ 0.15 − 0.5
G(iω ) (full), zoh (dashed), foh (dot-dashed), Tustin (dotted)
This leads to ω N 5–20 larger than ω c
Digital redesign of lead compensator
A lead compensator G(s) = 4(s + 1)/(s + 2) for the process
P(s) = 1/(s 2 + s) gives ω c = 1.6. Hence choose h = 0.1 − −0.3.
Euler’s method: H E (z) = 4
Output
1
z−1
h + 1 z − (1 − h)
= 4
+ 2 z − (1 − 2h)
z−1
h
Discrete time approximation of
continuous time controller
— State space
0
0 10
4
Input
2
0 10
Time
h = 0.1 (dash-dotted), 0.25 (full), 0.5 (dotted), G(iω ) (dashed)
c Dept. of Automatic Control, Lund 7

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State-feedback redesign
The system ẋ = Ax + Bu with continuous-time state feedback
u(t) = Mu c (t) − Lx(t)
gives ẋ = (A − B L) x + BMu
} {{ }
c = A c x + BMu c Sampling gives
A c
x(kh + h) = Φ c x(kh) + Γ c Mu c (kh)
Alternatively, sampling first, then applying discrete-time state
feedback
u(kh) = ˜Mu c (kh) − ˜Lx(kh)
gives
x(kh + h) = (Φ − Γ ˜L)x(kh) + Γ ˜Mu c (kh)
State feedback redesign cont’d
With ˜L = L 0 + L 1 h/2 the two system matrices
Φ c = I + (A − B L)h + ( A 2 − B LA − AB L + (B L) 2) h 2 /2 + ⋅ ⋅ ⋅
Φ − Γ ˜L = I + (A − B L 0 )h + ( A 2 − AB L 0 − B L 1
) h 2 /2 + ⋅ ⋅ ⋅
are identical of order h 2 provided that L 0 = L, L 1 = A − B L.
To make the steady-state gain correct, let ˜M = M 0 + M 1 h/2
and compare
Γ c M = B M h + (A − B L)B M h 2 /2 + ⋅ ⋅ ⋅
Γ ˜M = B M 0 h + (B M 1 + AB M 0 )h 2 /2 + ⋅ ⋅ ⋅
The expressions are identical of order h 2 provided that
M 0 = M, M 1 = −LBM.
State-feedback redesign – Example
Double integrator with h = 0.5, L = [1 1], and M = 1
Position
1
Position
1
0
0 10
0
0 10
Velocity
0.5
0
Velocity
0.5
0
0 10
0 10
1
1
Input
0
Input
0
0 10
Time
0 10
Time
˜L =
⎧
⎩ 1 − 0.5h 1
⎫
⎭ ˜M = 1 − 0.5h
c Dept. of Automatic Control, Lund 8