1.
s to Z-Domain Transfer Function
1.
s to Z-Domain Transfer Function
Discrete
ZOH
Signals
1.
s to Z-Domain Transfer Function
Discrete
ZOH
Signals
1. Get step response
of continuous transfer function ys(t).
1.
s to Z-Domain Transfer Function
Discrete
ZOH
Signals
1. Get step response
of continuous transfer function ys(t).
2. Discretize step response: ys(nTs).
1.
s to Z-Domain Transfer Function
Discrete
ZOH
Signals
1. Get step response
of continuous transfer function ys(t).
2. Discretize step response: ys(nTs).
1. Z-transform the step response to obtain Ys(z).
1.
s to Z-Domain Transfer Function
Discrete
ZOH
Signals
1. Get step response
of continuous transfer function ys(t).
2. Discretize step response: ys(nTs).
1. Z-transform the step response to obtain Ys(z).
2. Divide the result from
above by Z-transform of a
step, namely, z/(z − 1).
1.
s to Z-Domain Transfer Function
Discrete
ZOH
Signals
1. Z-transform the step response to obtain Ys(z).
2. Divide the result from
above by Z-transform of a
step, namely, z/(z − 1).
1. Get step response
of continuous trans- • Ga(s): Laplace transfer
function
fer function ys(t).
2. Discretize step response: ys(nTs).
1.
s to Z-Domain Transfer Function
Discrete
ZOH
Signals
1. Z-transform the step response to obtain Ys(z).
2. Divide the result from
above by Z-transform of a
step, namely, z/(z − 1).
1. Get step response
of continuous trans- • Ga(s): Laplace transfer
function
fer function ys(t).
2. Discretize step re- • G(z): Z-transfer function
sponse: ys(nTs).
1.
s to Z-Domain Transfer Function
1. Z-transform the step response to obtain Ys(z).
2. Divide the result from
above by Z-transform of a
step, namely, z/(z − 1).
Discrete
ZOH
Signals
1. Get step response
of continuous trans- • Ga(s): Laplace transfer
function
fer function ys(t).
2. Discretize step re- • G(z): Z-transfer function
sponse: ys(nTs).
G(z) =
z−1
z
−1 Ga (s)
Z L
s
1.
s to Z-Domain Transfer Function
1. Z-transform the step response to obtain Ys(z).
2. Divide the result from
above by Z-transform of a
step, namely, z/(z − 1).
Discrete
ZOH
Signals
1. Get step response
of continuous trans- • Ga(s): Laplace transfer
function
fer function ys(t).
2. Discretize step re- • G(z): Z-transfer function
sponse: ys(nTs).
G(z) =
z−1
−1 Ga (s)
Z L
z
Step Response Equivalence
s
1.
s to Z-Domain Transfer Function
1. Z-transform the step response to obtain Ys(z).
2. Divide the result from
above by Z-transform of a
step, namely, z/(z − 1).
Discrete
ZOH
Signals
1. Get step response
of continuous trans- • Ga(s): Laplace transfer
function
fer function ys(t).
2. Discretize step re- • G(z): Z-transfer function
sponse: ys(nTs).
G(z) =
z−1
−1 Ga (s)
Z L
z
s
Step Response Equivalence = ZOH Equivalence
Digital Control
1
Kannan M. Moudgalya, Autumn 2007
2.
Important Result from Differentiation
2.
Important Result from Differentiation
Recall
1(n)an ↔
z
z−a
Differentiating w.r.t. a,
=
∞
X
n=0
anz −n,
2.
Important Result from Differentiation
Recall
1(n)an ↔
z
z−a
Differentiating w.r.t. a,
z
=
∞
X
anz −n,
n=0
∞
X
n−1z −n
=
na
(z − a)2
n=0
2.
Important Result from Differentiation
Recall
1(n)an ↔
z
z−a
Differentiating w.r.t. a,
z
=
∞
X
anz −n,
n=0
∞
X
n−1z −n
=
na
(z − a)2
n=0
z
n−1
na
1(n) ↔
(z − a)2
2.
Important Result from Differentiation
Recall
1(n)an ↔
z
z−a
Differentiating w.r.t. a,
=
∞
X
anz −n,
n=0
∞
X
z
n−1z −n
=
na
(z − a)2
n=0
z
n−1
na
1(n) ↔
(z − a)2
2z
n−2
n(n − 1)a
1(n) ↔
(z − a)3
Digital Control
2
Kannan M. Moudgalya, Autumn 2007
3.
ZOH Equivalence of 1/s
3.
ZOH Equivalence of 1/s
The step response of
1/s is
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2.
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2. In time
domain, it is,
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2. In time
domain, it is,
1
−1
ys(t) = L
s2
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2. In time
domain, it is,
1
−1
ys(t) = L
=t
2
s
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2. In time
domain, it is,
1
−1
ys(t) = L
=t
2
s
Sampling it with a period of Ts,
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2. In time
domain, it is,
1
−1
ys(t) = L
=t
2
s
Sampling it with a period of Ts,
ys(nTs) =
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2. In time
domain, it is,
1
−1
ys(t) = L
=t
2
s
Sampling it with a period of Ts,
ys(nTs) = nTs
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2. In time
domain, it is,
1
−1
ys(t) = L
=t
2
s
Sampling it with a period of Ts,
ys(nTs) = nTs
Taking Z-transforms
Ys(z) =
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2. In time
domain, it is,
1
−1
ys(t) = L
=t
2
s
Sampling it with a period of Ts,
ys(nTs) = nTs
Taking Z-transforms
Ts z
Ys(z) =
(z − 1)2
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2. In time
domain, it is,
1
−1
ys(t) = L
=t
2
s
Sampling it with a period of Ts,
ys(nTs) = nTs
Taking Z-transforms
Ts z
Ys(z) =
(z − 1)2
Divide by z/(z −1),
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2. In time
domain, it is,
1
−1
ys(t) = L
=t
2
s
Sampling it with a period of Ts,
ys(nTs) = nTs
Taking Z-transforms
Ts z
Ys(z) =
(z − 1)2
Divide by z/(z −1), to
get the ZOH equivalent
discrete domain transfer
function
3.
ZOH Equivalence of 1/s
The step response of
1/s is 1/s2. In time
domain, it is,
1
−1
ys(t) = L
=t
2
s
Sampling it with a period of Ts,
Taking Z-transforms
Ts z
Ys(z) =
(z − 1)2
Divide by z/(z −1), to
get the ZOH equivalent
discrete domain transfer
function
Ts
G(z) =
z−1
ys(nTs) = nTs
Digital Control
3
Kannan M. Moudgalya, Autumn 2007
4.
ZOH Equivalence of 1/s2
4.
ZOH Equivalence of 1/s2
The step response of
1/s2 is
4.
ZOH Equivalence of 1/s2
The step response of
1/s2 is 1/s3.
4.
ZOH Equivalence of 1/s2
The step response of
1/s2 is 1/s3. In time
domain, it is,
4.
ZOH Equivalence of 1/s2
The step response of
1/s2 is 1/s3. In time
domain, it is,
1
−1
ys(t) = L
=
3
s
4.
ZOH Equivalence of 1/s2
The step response of
1/s2 is 1/s3. In time
domain, it is,
1
1 2
−1
ys(t) = L
= t .
3
s
2
4.
ZOH Equivalence of 1/s2
The step response of
1/s2 is 1/s3. In time
domain, it is,
1
1 2
−1
ys(t) = L
= t .
3
s
2
Sampling it with a period of Ts,
1 2 2
ys(nTs) = n Ts
2
4.
ZOH Equivalence of 1/s2
The step response of
Take Z-transform
1/s2 is 1/s3. In time
domain, it is,
1
1 2
−1
ys(t) = L
= t .
3
s
2
Sampling it with a period of Ts,
1 2 2
ys(nTs) = n Ts
2
4.
ZOH Equivalence of 1/s2
The step response of
Take Z-transform
1/s2 is 1/s3. In time
Ts2z(z + 1)
Ys(z) =
domain, it is,
2(z − 1)3
1
1 2
−1
ys(t) = L
= t .
3
s
2
Sampling it with a period of Ts,
1 2 2
ys(nTs) = n Ts
2
4.
ZOH Equivalence of 1/s2
The step response of
1/s2 is 1/s3. In time
domain, it is,
1
1 2
−1
ys(t) = L
= t .
3
s
2
Sampling it with a period of Ts,
1 2 2
ys(nTs) = n Ts
2
Digital Control
4
Take Z-transform
Ts2z(z + 1)
Ys(z) =
2(z − 1)3
Dividing by z/(z − 1),
we get
G(z) =
Ts2(z + 1)
2(z − 1)2
Kannan M. Moudgalya, Autumn 2007
5.
ZOH Equivalent First Order Transfer Function
5.
ZOH Equivalent First Order Transfer Function
Find the ZOH equivalent of K/(τ s + 1).
5.
ZOH Equivalent First Order Transfer Function
Find the ZOH equivalent of K/(τ s + 1).
"
#
1
1
1 K
=K
−
Ys(s) =
sτs + 1
s
s + τ1
5.
ZOH Equivalent First Order Transfer Function
Find the ZOH equivalent of K/(τ s + 1).
"
#
1
1
1 K
=K
−
Ys(s) =
sτs + 1
s
s + τ1
h
i
ys(t) = K 1 − e−t/τ , t ≥ 0
5.
ZOH Equivalent First Order Transfer Function
Find the ZOH equivalent of K/(τ s + 1).
"
#
1
1
1 K
=K
−
Ys(s) =
sτs + 1
s
s + τ1
h
i
ys(t) = K 1 − e−t/τ , t ≥ 0
h
i
−nTs /τ
ys(nTs) = K 1 − e
, n≥0
5.
ZOH Equivalent First Order Transfer Function
Find the ZOH equivalent of K/(τ s + 1).
"
#
1
1
1 K
=K
−
Ys(s) =
sτs + 1
s
s + τ1
h
i
ys(t) = K 1 − e−t/τ , t ≥ 0
h
i
−nTs /τ
ys(nTs) = K 1 − e
, n≥0
Ys(z) = K
z
z−1
−
z
z − e−Ts/τ
5.
ZOH Equivalent First Order Transfer Function
Find the ZOH equivalent of K/(τ s + 1).
"
#
1
1
1 K
=K
−
Ys(s) =
sτs + 1
s
s + τ1
h
i
ys(t) = K 1 − e−t/τ , t ≥ 0
h
i
−nTs /τ
ys(nTs) = K 1 − e
, n≥0
Ys(z) = K
z
z−1
−
z
z − e−Ts/τ
=
Kz(1 − e−Ts/τ )
(z − 1)(z − e−Ts/τ )
5.
ZOH Equivalent First Order Transfer Function
Find the ZOH equivalent of K/(τ s + 1).
"
#
1
1
1 K
=K
−
Ys(s) =
sτs + 1
s
s + τ1
h
i
ys(t) = K 1 − e−t/τ , t ≥ 0
h
i
−nTs /τ
ys(nTs) = K 1 − e
, n≥0
Ys(z) = K
z
z−1
−
z
z − e−Ts/τ
=
Kz(1 − e−Ts/τ )
(z − 1)(z − e−Ts/τ )
Dividing by z/(z − 1), we get
G(z) =
Digital Control
K(1 − e−Ts/τ )
z − e−Ts/τ
5
Kannan M. Moudgalya, Autumn 2007
6.
ZOH Equivalent First Order Transfer Function
- Example
6.
ZOH Equivalent First Order Transfer Function
- Example
Sample at Ts = 0.5
and find ZOH equivalent
trans. function of
10
Ga(s) =
5s + 1
6.
ZOH Equivalent First Order Transfer Function
- Example
Sample at Ts = 0.5
and find ZOH equivalent
trans. function of
10
Ga(s) =
5s + 1
Scilab Code:
6.
ZOH Equivalent First Order Transfer Function
- Example
Sample at Ts = 0.5
and find ZOH equivalent
trans. function of
10
Ga(s) =
5s + 1
Scilab Code:
Ga = tf(10,[5 1]);
6.
ZOH Equivalent First Order Transfer Function
- Example
Sample at Ts = 0.5
and find ZOH equivalent
trans. function of
10
Ga(s) =
5s + 1
Scilab Code:
Ga = tf(10,[5 1]);
G = ss2tf(dscr(Ga,0.5));
6.
ZOH Equivalent First Order Transfer Function
- Example
Sample at Ts = 0.5 Scilab output is,
and find ZOH equivalent
trans. function of
10
Ga(s) =
5s + 1
Scilab Code:
Ga = tf(10,[5 1]);
G = ss2tf(dscr(Ga,0.5));
6.
ZOH Equivalent First Order Transfer Function
- Example
Sample at Ts = 0.5 Scilab output is,
and find ZOH equivalent
0.9546
G(z) =
trans. function of
z − 0.9048
10
Ga(s) =
5s + 1
Scilab Code:
Ga = tf(10,[5 1]);
G = ss2tf(dscr(Ga,0.5));
6.
ZOH Equivalent First Order Transfer Function
- Example
Sample at Ts = 0.5 Scilab output is,
and find ZOH equivalent
0.9546
G(z) =
trans. function of
z − 0.9048
10
−0.1)
10(1
−
e
Ga(s) =
=
5s + 1
z − e−0.1
Scilab Code:
Ga = tf(10,[5 1]);
G = ss2tf(dscr(Ga,0.5));
6.
ZOH Equivalent First Order Transfer Function
- Example
Sample at Ts = 0.5
and find ZOH equivalent
trans. function of
10
Ga(s) =
5s + 1
Scilab Code:
Ga = tf(10,[5 1]);
G = ss2tf(dscr(Ga,0.5));
Digital Control
Scilab output is,
0.9546
G(z) =
z − 0.9048
10(1 − e−0.1)
=
z − e−0.1
In agreement with the
formula in the previous
slide
6
Kannan M. Moudgalya, Autumn 2007
7.
Discrete Integration
7.
Discrete Integration
u(n)
u(k − 1)
u(k)
n
7.
Discrete Integration
y(k) = blue shaded area
u(n)
u(k − 1)
u(k)
n
7.
Discrete Integration
y(k) = blue shaded area
+ red shaded area
u(n)
u(k − 1)
u(k)
n
7.
Discrete Integration
y(k) = blue shaded area
+ red shaded area
y(k) = y(k − 1)
u(n)
u(k − 1)
u(k)
n
7.
Discrete Integration
y(k) = blue shaded area
+ red shaded area
y(k) = y(k − 1) + red shaded area
u(n)
u(k − 1)
u(k)
n
7.
Discrete Integration
u(n)
u(k − 1)
u(k)
n
y(k) = blue shaded area
+ red shaded area
y(k) = y(k − 1) + red shaded area
Ts
[u(k) + u(k − 1)]
y(k) = y(k − 1) +
2
7.
Discrete Integration
u(n)
u(k − 1)
u(k)
n
y(k) = blue shaded area
+ red shaded area
y(k) = y(k − 1) + red shaded area
Ts
[u(k) + u(k − 1)]
y(k) = y(k − 1) +
2
Take Z-transform:
7.
Discrete Integration
u(n)
u(k − 1)
u(k)
n
y(k) = blue shaded area
+ red shaded area
y(k) = y(k − 1) + red shaded area
Ts
[u(k) + u(k − 1)]
y(k) = y(k − 1) +
2
Take Z-transform:
Ts −1
−1
Y (z) = z Y (z) +
U (z) + z U (z)
2
7.
Discrete Integration
u(n)
u(k − 1)
u(k)
n
y(k) = blue shaded area
+ red shaded area
y(k) = y(k − 1) + red shaded area
Ts
[u(k) + u(k − 1)]
y(k) = y(k − 1) +
2
Take Z-transform:
Ts −1
−1
Y (z) = z Y (z) +
U (z) + z U (z)
2
Bring all Y to left side:
7.
Discrete Integration
u(n)
u(k − 1)
u(k)
n
y(k) = blue shaded area
+ red shaded area
y(k) = y(k − 1) + red shaded area
Ts
[u(k) + u(k − 1)]
y(k) = y(k − 1) +
2
Take Z-transform:
Ts −1
−1
Y (z) = z Y (z) +
U (z) + z U (z)
2
Bring all Y to left side:
Ts −1
−1
Y (z) − z Y (z) =
U (z) + z U (z)
2
7.
Discrete Integration
u(n)
u(k − 1)
u(k)
n
Digital Control
y(k) = blue shaded area
+ red shaded area
y(k) = y(k − 1) + red shaded area
Ts
[u(k) + u(k − 1)]
y(k) = y(k − 1) +
2
Take Z-transform:
Ts −1
−1
Y (z) = z Y (z) +
U (z) + z U (z)
2
Bring all Y to left side:
Ts −1
−1
Y (z) − z Y (z) =
U (z) + z U (z)
2
Ts
−1
(1 − z )Y (z) = (1 + z −1)U (z)
2
7
Kannan M. Moudgalya, Autumn 2007
8.
Transfer Function for Discrete Integration
8.
Transfer Function for Discrete Integration
Recall from previous slide
(1 − z
−1
)Y (z) =
Ts
2
(1 + z −1)U (z)
8.
Transfer Function for Discrete Integration
Recall from previous slide
(1 − z
−1
)Y (z) =
Y (z) =
Ts
(1 + z −1)U (z)
2
Ts 1 + z −1
2 1 − z −1
U (z)
8.
Transfer Function for Discrete Integration
Recall from previous slide
(1 − z
−1
)Y (z) =
Ts
(1 + z −1)U (z)
2
Ts 1 + z −1
U (z)
2 1 − z −1
Ts z + 1
U (z)
=
2 z−1
Y (z) =
8.
Transfer Function for Discrete Integration
Recall from previous slide
(1 − z
−1
)Y (z) =
Ts
(1 + z −1)U (z)
2
Ts 1 + z −1
U (z)
2 1 − z −1
Ts z + 1
U (z)
=
2 z−1
Y (z) =
Integrator has a transfer function,
8.
Transfer Function for Discrete Integration
Recall from previous slide
(1 − z
−1
)Y (z) =
Ts
(1 + z −1)U (z)
2
Ts 1 + z −1
U (z)
2 1 − z −1
Ts z + 1
U (z)
=
2 z−1
Y (z) =
Integrator has a transfer function,
GI (z) =
Ts z + 1
2 z−1
8.
Transfer Function for Discrete Integration
Recall from previous slide
(1 − z
−1
)Y (z) =
Ts
(1 + z −1)U (z)
2
Ts 1 + z −1
U (z)
2 1 − z −1
Ts z + 1
U (z)
=
2 z−1
Y (z) =
Integrator has a transfer function,
GI (z) =
A low pass filter!
Ts z + 1
2 z−1
8.
Transfer Function for Discrete Integration
Recall from previous slide
(1 − z
−1
)Y (z) =
Ts
Im(z)
(1 + z −1)U (z)
2
Ts 1 + z −1
U (z)
2 1 − z −1
Ts z + 1
U (z)
=
2 z−1
Y (z) =
Integrator has a transfer function,
GI (z) =
A low pass filter!
Ts z + 1
2 z−1
× Re(z)
8.
Transfer Function for Discrete Integration
Recall from previous slide
(1 − z
−1
)Y (z) =
Ts
Im(z)
(1 + z −1)U (z)
2
Ts 1 + z −1
U (z)
2 1 − z −1
Ts z + 1
U (z)
=
2 z−1
Y (z) =
× Re(z)
1
s
↔
Ts z + 1
2 z−1
Integrator has a transfer function,
GI (z) =
Ts z + 1
2 z−1
A low pass filter!
Digital Control
8
Kannan M. Moudgalya, Autumn 2007
9.
Derivative Mode
9.
Derivative Mode
• Integral Mode:
1
s
↔
Ts z + 1
2 z−1
9.
Derivative Mode
• Integral Mode:
1
s
↔
Ts z + 1
2 z−1
2 z−1
• Derivative Mode: s ↔
Ts z + 1
9.
Derivative Mode
• Integral Mode:
1
s
↔
Ts z + 1
2 z−1
2 z−1
• Derivative Mode: s ↔
Ts z + 1
• High pass filter
9.
Derivative Mode
• Integral Mode:
1
s
↔
Ts z + 1
2 z−1
2 z−1
• Derivative Mode: s ↔
Ts z + 1
• High pass filter
• Has a pole at z = −1.
9.
Derivative Mode
• Integral Mode:
1
s
↔
Ts z + 1
2 z−1
2 z−1
• Derivative Mode: s ↔
Ts z + 1
• High pass filter
• Has a pole at z = −1. Hence produces in partial fraction
expansion, a term of the form
9.
Derivative Mode
• Integral Mode:
1
s
↔
Ts z + 1
2 z−1
2 z−1
• Derivative Mode: s ↔
Ts z + 1
• High pass filter
• Has a pole at z = −1. Hence produces in partial fraction
expansion, a term of the form
z
↔ (−1)n
z+1
9.
Derivative Mode
• Integral Mode:
1
s
↔
Ts z + 1
2 z−1
2 z−1
• Derivative Mode: s ↔
Ts z + 1
• High pass filter
• Has a pole at z = −1. Hence produces in partial fraction
expansion, a term of the form
z
↔ (−1)n
z+1
• Results in wildly oscillating control effort.
Digital Control
9
Kannan M. Moudgalya, Autumn 2007
10.
Derivative Mode - Other Approximations
10.
Derivative Mode - Other Approximations
Backward difference:
y(k) = y(k − 1) + Tsu(k)
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
Y (z) = Ts
1 − z −1
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
1
↔
s
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
z
1
↔ Ts
s
z−1
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
z
1
↔ Ts
s
z−1
Forward difference: y(k) = y(k − 1) + Tsu(k − 1)
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
z
1
↔ Ts
s
z−1
Forward difference: y(k) = y(k − 1) + Tsu(k − 1)
(1 − z −1)Y (z) = Tsz −1U (z)
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
z
1
↔ Ts
s
z−1
Forward difference: y(k) = y(k − 1) + Tsu(k − 1)
(1 − z −1)Y (z) = Tsz −1U (z)
z −1
Y (z) = Ts
U (z)
−1
1−z
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
z
1
↔ Ts
s
z−1
Forward difference: y(k) = y(k − 1) + Tsu(k − 1)
(1 − z −1)Y (z) = Tsz −1U (z)
z −1
Ts
Y (z) = Ts
U (z) =
U (z)
−1
1−z
z−1
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
z
1
↔ Ts
s
z−1
Forward difference: y(k) = y(k − 1) + Tsu(k − 1)
(1 − z −1)Y (z) = Tsz −1U (z)
z −1
Ts
Y (z) = Ts
U (z) =
U (z)
−1
1−z
z−1
1
↔
s
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
z
1
↔ Ts
s
z−1
Forward difference: y(k) = y(k − 1) + Tsu(k − 1)
(1 − z −1)Y (z) = Tsz −1U (z)
z −1
Ts
Y (z) = Ts
U (z) =
U (z)
−1
1−z
z−1
1
Ts
↔
s
z−1
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
z
1
↔ Ts
s
z−1
Forward difference: y(k) = y(k − 1) + Tsu(k − 1)
(1 − z −1)Y (z) = Tsz −1U (z)
z −1
Ts
Y (z) = Ts
U (z) =
U (z)
−1
1−z
z−1
1
Ts
↔
s
z−1
Both derivative modes are high pass,
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
z
1
↔ Ts
s
z−1
Forward difference: y(k) = y(k − 1) + Tsu(k − 1)
(1 − z −1)Y (z) = Tsz −1U (z)
z −1
Ts
Y (z) = Ts
U (z) =
U (z)
−1
1−z
z−1
1
Ts
↔
s
z−1
Both derivative modes are high pass, no oscillations,
10.
Derivative Mode - Other Approximations
Backward difference: y(k) = y(k − 1) + Tsu(k)
(1 − z −1)Y (z) = TsU (z)
1
z
U (z)
Y (z) = Ts
= Ts
−1
1−z
z−1
z
1
↔ Ts
s
z−1
Forward difference: y(k) = y(k − 1) + Tsu(k − 1)
(1 − z −1)Y (z) = Tsz −1U (z)
z −1
Ts
Y (z) = Ts
U (z) =
U (z)
−1
1−z
z−1
1
Ts
↔
s
z−1
Both derivative modes are high pass, no oscillations, same gains
Digital Control
10
Kannan M. Moudgalya, Autumn 2007
11.
PID Controller
11.
PID Controller
Proportional Mode: Most popular control mode.
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
• Decreased steady state offset
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
• Decreased steady state offset and increased oscillations
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
• Decreased steady state offset and increased oscillations
Integral Mode: Used to remove steady state offset.
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
• Decreased steady state offset and increased oscillations
Integral Mode: Used to remove steady state offset. Increase
in integral mode generally results in
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
• Decreased steady state offset and increased oscillations
Integral Mode: Used to remove steady state offset. Increase
in integral mode generally results in
• Zero steady state offset
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
• Decreased steady state offset and increased oscillations
Integral Mode: Used to remove steady state offset. Increase
in integral mode generally results in
• Zero steady state offset
• Increased oscillations
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
• Decreased steady state offset and increased oscillations
Integral Mode: Used to remove steady state offset. Increase
in integral mode generally results in
• Zero steady state offset
• Increased oscillations
Derivative Mode: Mainly used for prediction purposes.
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
• Decreased steady state offset and increased oscillations
Integral Mode: Used to remove steady state offset. Increase
in integral mode generally results in
• Zero steady state offset
• Increased oscillations
Derivative Mode: Mainly used for prediction purposes. Increase
in derivative mode generally results in
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
• Decreased steady state offset and increased oscillations
Integral Mode: Used to remove steady state offset. Increase
in integral mode generally results in
• Zero steady state offset
• Increased oscillations
Derivative Mode: Mainly used for prediction purposes. Increase
in derivative mode generally results in
• Decreased oscillations and improved stability
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
• Decreased steady state offset and increased oscillations
Integral Mode: Used to remove steady state offset. Increase
in integral mode generally results in
• Zero steady state offset
• Increased oscillations
Derivative Mode: Mainly used for prediction purposes. Increase
in derivative mode generally results in
• Decreased oscillations and improved stability
• Sensitive to noise
11.
PID Controller
Proportional Mode: Most popular control mode. Increase
in proportional mode generally results in
• Decreased steady state offset and increased oscillations
Integral Mode: Used to remove steady state offset. Increase
in integral mode generally results in
• Zero steady state offset
• Increased oscillations
Derivative Mode: Mainly used for prediction purposes. Increase
in derivative mode generally results in
• Decreased oscillations and improved stability
• Sensitive to noise
The most popular controller in industry.
Digital Control
11
Kannan M. Moudgalya, Autumn 2007
12.
PID Controller - Basic Design
12.
PID Controller - Basic Design
Let input to controller by E(z)
12.
PID Controller - Basic Design
Let input to controller by E(z) and output from it be U (z).
12.
PID Controller - Basic Design
Let input to controller by E(z) and output from it be U (z). If
gain is K,
12.
PID Controller - Basic Design
Let input to controller by E(z) and output from it be U (z). If
gain is K, τi is integral time
12.
PID Controller - Basic Design
Let input to controller by E(z) and output from it be U (z). If
gain is K, τi is integral time and τd is derivative time,
12.
PID Controller - Basic Design
Let input to controller by E(z) and output from it be U (z). If
gain is K, τi is integral time and τd is derivative time,
Z
de(t)
1 t
u(t) = K e(t) +
e(t)dt + τd
τi 0
dt
12.
PID Controller - Basic Design
Let input to controller by E(z) and output from it be U (z). If
gain is K, τi is integral time and τd is derivative time,
Z
de(t)
1 t
u(t) = K e(t) +
e(t)dt + τd
τi 0
dt
1
U (s) = K(1 +
+ τds)E(s)
τi s
12.
PID Controller - Basic Design
Let input to controller by E(z) and output from it be U (z). If
gain is K, τi is integral time and τd is derivative time,
Z
de(t)
1 t
u(t) = K e(t) +
e(t)dt + τd
τi 0
dt
1
U (s) = K(1 +
+ τds)E(s)
τi s
4 Sc (s)
U (s) =
E(s)
Rc(s)
12.
PID Controller - Basic Design
Let input to controller by E(z) and output from it be U (z). If
gain is K, τi is integral time and τd is derivative time,
Z
de(t)
1 t
u(t) = K e(t) +
e(t)dt + τd
τi 0
dt
1
U (s) = K(1 +
+ τds)E(s)
τi s
4 Sc (s)
U (s) =
E(s)
Rc(s)
If integral mode is present, Rc(0) = 0.
12.
PID Controller - Basic Design
Let input to controller by E(z) and output from it be U (z). If
gain is K, τi is integral time and τd is derivative time,
Z
de(t)
1 t
u(t) = K e(t) +
e(t)dt + τd
τi 0
dt
1
U (s) = K(1 +
+ τds)E(s)
τi s
4 Sc (s)
U (s) =
E(s)
Rc(s)
If integral mode is present, Rc(0) = 0.
Filtered derivative mode:
12.
PID Controller - Basic Design
Let input to controller by E(z) and output from it be U (z). If
gain is K, τi is integral time and τd is derivative time,
Z
de(t)
1 t
u(t) = K e(t) +
e(t)dt + τd
τi 0
dt
1
U (s) = K(1 +
+ τds)E(s)
τi s
4 Sc (s)
U (s) =
E(s)
Rc(s)
If integral mode is present, Rc(0) = 0.
Filtered derivative mode:
!
1
τds
u(t) = K 1 +
+
e(t)
τd s
τis
1+ N
12.
PID Controller - Basic Design
Let input to controller by E(z) and output from it be U (z). If
gain is K, τi is integral time and τd is derivative time,
Z
de(t)
1 t
u(t) = K e(t) +
e(t)dt + τd
τi 0
dt
1
U (s) = K(1 +
+ τds)E(s)
τi s
4 Sc (s)
U (s) =
E(s)
Rc(s)
If integral mode is present, Rc(0) = 0.
Filtered derivative mode:
!
1
τds
u(t) = K 1 +
+
e(t)
τd s
τis
1+ N
N is a large number, of the order of 100.
Digital Control
12
Kannan M. Moudgalya, Autumn 2007
13.
Reaction Curve Method - Ziegler Nichols Tuning
13.
Reaction Curve Method - Ziegler Nichols Tuning
• Applicable only to stable systems
13.
Reaction Curve Method - Ziegler Nichols Tuning
• Applicable only to stable systems
• Give a unit step input to a stable system and get
13.
Reaction Curve Method - Ziegler Nichols Tuning
• Applicable only to stable systems
• Give a unit step input to a stable system and get
1. the time lag after which the system starts responding (L),
13.
Reaction Curve Method - Ziegler Nichols Tuning
• Applicable only to stable systems
• Give a unit step input to a stable system and get
1. the time lag after which the system starts responding (L),
2. the steady state gain (K) and
13.
Reaction Curve Method - Ziegler Nichols Tuning
• Applicable only to stable systems
• Give a unit step input to a stable system and get
1. the time lag after which the system starts responding (L),
2. the steady state gain (K) and
3. the time the output takes to reach the steady state, after
it starts responding (τ )
13.
Reaction Curve Method - Ziegler Nichols Tuning
• Applicable only to stable systems
• Give a unit step input to a stable system and get
1. the time lag after which the system starts responding (L),
2. the steady state gain (K) and
3. the time the output takes to reach the steady state, after
it starts responding (τ )
R = K/τ
K
L
Digital Control
τ
13
Kannan M. Moudgalya, Autumn 2007
14.
Reaction Curve Method - Ziegler Nichols Tuning
14.
Reaction Curve Method - Ziegler Nichols Tuning
R = K/τ
K
L
τ
14.
Reaction Curve Method - Ziegler Nichols Tuning
R = K/τ
K
L
τ
• Let the slope of the response be calculated as R =
K
τ
.
14.
Reaction Curve Method - Ziegler Nichols Tuning
R = K/τ
K
L
τ
K
.
• Let the slope of the response be calculated as R =
τ
Then the PID settings are given below:
14.
Reaction Curve Method - Ziegler Nichols Tuning
R = K/τ
K
L
τ
K
.
• Let the slope of the response be calculated as R =
τ
Then the PID settings are given below:
Kp
τi τd
P
1/RL
PI 0.9/RL 3L
PID 1.2/RL 2L 0.5L
14.
Reaction Curve Method - Ziegler Nichols Tuning
R = K/τ
K
L
τ
K
.
• Let the slope of the response be calculated as R =
τ
Then the PID settings are given below:
Kp
τi τd
P
1/RL
PI 0.9/RL 3L
PID 1.2/RL 2L 0.5L
Consistent units should be used
Digital Control
14
Kannan M. Moudgalya, Autumn 2007
15.
Stability Method - Ziegler Nichols Tuning
15.
Stability Method - Ziegler Nichols Tuning
Another way of finding the PID tuning parameters is as follows.
15.
Stability Method - Ziegler Nichols Tuning
Another way of finding the PID tuning parameters is as follows.
• Close the loop with a proportional controller
15.
Stability Method - Ziegler Nichols Tuning
Another way of finding the PID tuning parameters is as follows.
• Close the loop with a proportional controller
• Gain of controller is increased until the closed loop system
becomes unstable
15.
Stability Method - Ziegler Nichols Tuning
Another way of finding the PID tuning parameters is as follows.
• Close the loop with a proportional controller
• Gain of controller is increased until the closed loop system
becomes unstable
• At the verge of instability, note down the gain of the controller
(Ku) and the period of oscillation (Pu)
15.
Stability Method - Ziegler Nichols Tuning
Another way of finding the PID tuning parameters is as follows.
• Close the loop with a proportional controller
• Gain of controller is increased until the closed loop system
becomes unstable
• At the verge of instability, note down the gain of the controller
(Ku) and the period of oscillation (Pu)
• PID settings are given below:
15.
Stability Method - Ziegler Nichols Tuning
Another way of finding the PID tuning parameters is as follows.
• Close the loop with a proportional controller
• Gain of controller is increased until the closed loop system
becomes unstable
• At the verge of instability, note down the gain of the controller
(Ku) and the period of oscillation (Pu)
• PID settings are given below:
Kp
τi
τd
P
0.5Ku
PI 0.45Ku Pu/1.2
PID 0.6Ku Pu/2 Pu/8
15.
Stability Method - Ziegler Nichols Tuning
Another way of finding the PID tuning parameters is as follows.
• Close the loop with a proportional controller
• Gain of controller is increased until the closed loop system
becomes unstable
• At the verge of instability, note down the gain of the controller
(Ku) and the period of oscillation (Pu)
• PID settings are given below:
Kp
τi
τd
P
0.5Ku
PI 0.45Ku Pu/1.2
PID 0.6Ku Pu/2 Pu/8
Consistent units should be used
Digital Control
15
Kannan M. Moudgalya, Autumn 2007
16.
Design Procedure
16.
Design Procedure
A common procedure to design discrete PID controller:
16.
Design Procedure
A common procedure to design discrete PID controller:
• Tune continuous PID controller by any popular technique
16.
Design Procedure
A common procedure to design discrete PID controller:
• Tune continuous PID controller by any popular technique
• Get continuous PID settings
16.
Design Procedure
A common procedure to design discrete PID controller:
• Tune continuous PID controller by any popular technique
• Get continuous PID settings
• Discretize using the method discussed now or the ZOH equivalent method discussed earlier
16.
Design Procedure
A common procedure to design discrete PID controller:
• Tune continuous PID controller by any popular technique
• Get continuous PID settings
• Discretize using the method discussed now or the ZOH equivalent method discussed earlier
• Direct digital design techniques
Digital Control
16
Kannan M. Moudgalya, Autumn 2007
17.
2-DOF Controller
17.
2-DOF Controller
r
u
Tc
Rc
−
G=
Sc
Rc
B
A
y
17.
2-DOF Controller
r
u
Tc
Rc
−
G=
Sc
Rc
u=
Tc
Rc
r−
Sc
Rc
y
B
A
y
17.
2-DOF Controller
r
u
Tc
Rc
−
G=
B
y
A
Sc
Rc
u=
Tc
r−
Sc
y
Rc
Rc
It is easy to arrive at the following relation between r and y.
17.
2-DOF Controller
r
u
Tc
Rc
−
G=
B
y
A
Sc
Rc
u=
Tc
r−
Sc
y
Rc
Rc
It is easy to arrive at the following relation between r and y.
Tc
B/A
y=
r
Rc 1 + BSc/ARc
17.
2-DOF Controller
r
u
Tc
Rc
−
G=
B
y
A
Sc
Rc
u=
Tc
r−
Sc
y
Rc
Rc
It is easy to arrive at the following relation between r and y.
Tc
B/A
BTc
y=
r=
r
Rc 1 + BSc/ARc
ARc + BSc
17.
2-DOF Controller
r
u
Tc
Rc
−
G=
B
y
A
Sc
Rc
u=
Tc
r−
Sc
y
Rc
Rc
It is easy to arrive at the following relation between r and y.
Tc
B/A
BTc
y=
r=
r
Rc 1 + BSc/ARc
ARc + BSc
Error e, given by r − y is given by
17.
2-DOF Controller
r
u
Tc
Rc
−
G=
B
y
A
Sc
Rc
u=
Tc
r−
Sc
y
Rc
Rc
It is easy to arrive at the following relation between r and y.
Tc
B/A
BTc
y=
r=
r
Rc 1 + BSc/ARc
ARc + BSc
Error e, given by r − y is given by
BTc
e= 1−
r
ARc + BSc
17.
2-DOF Controller
r
u
Tc
Rc
G=
−
B
y
A
Sc
Rc
u=
Tc
r−
Sc
y
Rc
Rc
It is easy to arrive at the following relation between r and y.
Tc
B/A
BTc
y=
r=
r
Rc 1 + BSc/ARc
ARc + BSc
Error e, given by r − y is given by
BTc
ARc + BSc − BTc
e= 1−
r=
r
ARc + BSc
ARc + BSc
Digital Control
17
Kannan M. Moudgalya, Autumn 2007
18.
Offset-Free Tracking of Steps with Integral
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
A(z)Rc(z) + B(z)Sc(z)
R(z)
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
lim e(n) =
n→∞
A(z)Rc(z) + B(z)Sc(z)
R(z)
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
A(z)Rc(z) + B(z)Sc(z)
lim e(n) = lim
n→∞
z→1
R(z)
z − 1 A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
z
A(z)Rc(z) + B(z)Sc(z)
z
z−1
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
A(z)Rc(z) + B(z)Sc(z)
lim e(n) = lim
n→∞
z→1
R(z)
z − 1 A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
z
A(z)Rc(z) + B(z)Sc(z)
Because the controller has an integral action,
z
z−1
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
A(z)Rc(z) + B(z)Sc(z)
lim e(n) = lim
n→∞
z→1
R(z)
z − 1 A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
z
A(z)Rc(z) + B(z)Sc(z)
Because the controller has an integral action, Rc(1) = 0:
z
z−1
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
A(z)Rc(z) + B(z)Sc(z)
lim e(n) = lim
n→∞
z→1
R(z)
z − 1 A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
z
A(z)Rc(z) + B(z)Sc(z)
Because the controller has an integral action, Rc(1) = 0:
e(∞) =
z
z−1
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
A(z)Rc(z) + B(z)Sc(z)
lim e(n) = lim
n→∞
z→1
R(z)
z − 1 A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
z
A(z)Rc(z) + B(z)Sc(z)
Because the controller has an integral action, Rc(1) = 0:
Sc(z) − Tc(z) e(∞) =
Sc(z)
z=1
z
z−1
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
A(z)Rc(z) + B(z)Sc(z)
lim e(n) = lim
n→∞
z→1
R(z)
z − 1 A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
z
A(z)Rc(z) + B(z)Sc(z)
Because the controller has an integral action, Rc(1) = 0:
Sc(z) − Tc(z) Sc(1) − Tc(1)
e(∞) =
=
Sc(z)
Sc(1)
z=1
z
z−1
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
A(z)Rc(z) + B(z)Sc(z)
lim e(n) = lim
n→∞
z→1
R(z)
z − 1 A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
z
A(z)Rc(z) + B(z)Sc(z)
z
z−1
Because the controller has an integral action, Rc(1) = 0:
Sc(z) − Tc(z) Sc(1) − Tc(1)
e(∞) =
=
Sc(z)
Sc(1)
z=1
This condition can be satisfied if one of the following is met:
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
A(z)Rc(z) + B(z)Sc(z)
lim e(n) = lim
n→∞
R(z)
z − 1 A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
z→1
z
A(z)Rc(z) + B(z)Sc(z)
z
z−1
Because the controller has an integral action, Rc(1) = 0:
Sc(z) − Tc(z) Sc(1) − Tc(1)
e(∞) =
=
Sc(z)
Sc(1)
z=1
This condition can be satisfied if one of the following is met:
T c = Sc
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
A(z)Rc(z) + B(z)Sc(z)
lim e(n) = lim
n→∞
z→1
R(z)
z − 1 A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
z
A(z)Rc(z) + B(z)Sc(z)
z
z−1
Because the controller has an integral action, Rc(1) = 0:
Sc(z) − Tc(z) Sc(1) − Tc(1)
e(∞) =
=
Sc(z)
Sc(1)
z=1
This condition can be satisfied if one of the following is met:
T c = Sc
Tc = Sc(1)
18.
Offset-Free Tracking of Steps with Integral
E(z) =
A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
A(z)Rc(z) + B(z)Sc(z)
lim e(n) = lim
n→∞
z→1
R(z)
z − 1 A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)
z
A(z)Rc(z) + B(z)Sc(z)
z
z−1
Because the controller has an integral action, Rc(1) = 0:
Sc(z) − Tc(z) Sc(1) − Tc(1)
e(∞) =
=
Sc(z)
Sc(1)
z=1
This condition can be satisfied if one of the following is met:
T c = Sc
Tc = Sc(1)
Tc(1) = Sc(1)
Digital Control
18
Kannan M. Moudgalya, Autumn 2007