1. s to Z-Domain Transfer Function

(1)

(2)

Discrete ZOH Signals

(3)

1. Get

step

response

of continuous

trans-fer function

y

_s

(

t

)

.

(4)

1. Get

step

response

of continuous

trans-fer function

y

_s

(

t

)

.

2. Discretize step

(5)

1. Get

step

response

of continuous

trans-fer function

y

_s

(

t

)

.

2. Discretize step

re-sponse:

y

_s

(

nT

_s

)

.

1. Z-transform the step

re-sponse to obtain

Y

_s

(

z

)

.

(6)

1. Get

step

response

of continuous

trans-fer function

y

_s

(

t

)

.

2. Discretize step

re-sponse:

y

_s

(

nT

_s

)

.

re-sponse to obtain

Y

_s

(

z

)

.

2. Divide the result from

above by Z-transform of a

step, namely,

z/

(

z

−

1)

.

(7)

1. Get

step

response

of continuous

trans-fer function

y

_s

(

t

)

.

2. Discretize step

re-sponse:

y

_s

(

nT

_s

)

.

re-sponse to obtain

Y

_s

(

z

)

.

above by Z-transform of a

step, namely,

z/

(

z

−

1)

.

•

G

_a

(

s

)

: Laplace transfer

function

(8)

1. Get

step

response

of continuous

trans-fer function

y

_s

(

t

)

.

2. Discretize step

re-sponse:

y

_s

(

nT

_s

)

.

re-sponse to obtain

Y

_s

(

z

)

.

step, namely,

z/

(

z

−

1)

.

•

G

_a

(

s

)

: Laplace transfer

function

(9)

1. Get

step

response

of continuous

trans-fer function

y

_s

(

t

)

.

2. Discretize step

re-sponse:

y

_s

(

nT

_s

)

.

re-sponse to obtain

Y

_s

(

z

)

.

step, namely,

z/

(

z

−

1)

.

•

G

_a

(

s

)

: Laplace transfer

function

•

G

(

z

)

: Z-transfer function

G(z) = z − 1 z Z L−1Ga(s) s

(10)

1. Get

step

response

of continuous

trans-fer function

y

_s

(

t

)

.

2. Discretize step

re-sponse:

y

_s

(

nT

_s

)

.

re-sponse to obtain

Y

_s

(

z

)

.

step, namely,

z/

(

z

−

1)

.

•

G

_a

(

s

)

: Laplace transfer

function

•

G

(

z

)

G(z) = z − 1 z Z L−1Ga(s) s

(11)

1. Get

step

response

of continuous

trans-fer function

y

_s

(

t

)

.

2. Discretize step

re-sponse:

y

_s

(

nT

_s

)

.

re-sponse to obtain

Y

_s

(

z

)

.

step, namely,

z/

(

z

−

1)

.

•

G

_a

(

s

)

: Laplace transfer

function

•

G

(

z

)

G(z) = z − 1 z Z L−1Ga(s) s

Step Response Equivalence = ZOH Equivalence

(12)

(13)

2. Important Result from Differentiation

Recall

1(

n

)

a

n

↔

z

−

a

=

∞

X

n=0

a

n

z

−n

,

Differentiating w.r.t.

a

,

(14)

Recall

1(

n

)

a

n

↔

z

−

a

=

∞

X

n=0

a

n

z

−n

,

a

,

z

(

z

−

a

)

2

=

∞

X

n=0

na

n−1

z

−n

(15)

Recall

1(

n

)

a

n

↔

z

−

a

=

∞

X

n=0

a

n

z

−n

,

a

,

z

(

z

−

a

)

2

=

∞

X

n=0

na

n−1

z

−n

na

n−1

1(

n

)

↔

z

(

z

−

a

)

2

(16)

Recall

1(

n

)

a

n

↔

z

−

a

=

∞

X

n=0

a

n

z

−n

,

a

,

z

(

z

−

a

)

2

=

∞

X

n=0

na

n−1

z

−n

na

n−1

1(

n

)

↔

z

(

z

−

a

)

2

n

(

n

−

1)

a

n−2

1(

n

)

↔

2

z

(

z

−

a

)

3

(17)

(18)

3. ZOH Equivalence of 1/s

The step response of

(19)

The step response of

(20)

1

/s

is

1

/s

2

. In time

domain, it is,

(21)

1

/s

is

1

/s

2

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

(22)

1

/s

is

1

/s

2

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

(23)

1

/s

is

1

/s

2

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

2

=

t

Sampling it with a

pe-riod of

T

_s

,

(24)

1

/s

is

1

/s

2

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

2

=

t

Sampling it with a

pe-riod of

T

_s

,

(25)

1

/s

is

1

/s

2

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

2

=

t

Sampling it with a

pe-riod of

T

_s

,

(26)

1

/s

is

1

/s

2

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

2

=

t

Sampling it with a

pe-riod of

T

_s

,

y

_s

(

nT

_s

) =

nT

_s

Taking Z-transforms

(27)

1

/s

is

1

/s

2

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

2

=

t

Sampling it with a

pe-riod of

T

_s

,

y

_s

(

nT

_s

) =

nT

_s

Taking Z-transforms

Y

_s

(

z

) =

T

s

z

(28)

1

/s

is

1

/s

2

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

2

=

t

Sampling it with a

pe-riod of

T

_s

,

y

_s

(

nT

_s

) =

nT

_s

Taking Z-transforms

Y

_s

(

z

) =

T

s

z

(

z

−

1)

2

Divide by

z/

(

z

−

1)

,

(29)

1

/s

is

1

/s

2

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

2

=

t

Sampling it with a

pe-riod of

T

_s

,

y

_s

(

nT

_s

) =

nT

_s

Taking Z-transforms

Y

_s

(

z

) =

T

s

z

(

z

−

1)

2

Divide by

z/

(

z

−

1)

, to

get the ZOH equivalent

discrete domain transfer

function

(30)

1

/s

is

1

/s

2

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

2

=

t

Sampling it with a

pe-riod of

T

_s

,

y

_s

(

nT

_s

) =

nT

_s

Taking Z-transforms

Y

_s

(

z

) =

T

s

z

(

z

−

1)

2

Divide by

z/

(

z

−

1)

, to

get the ZOH equivalent

discrete domain transfer

function

G

(

z

) =

T

s

z

−

1

(31)

(32)

4. ZOH Equivalence of 1/s2

(33)

(34)

1

/s

2

is

1

/s

3

. In time

domain, it is,

(35)

1

/s

2

is

1

/s

3

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

(36)

1

/s

2

is

1

/s

3

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

3

=

1

2

t

2

.

(37)

1

/s

2

is

1

/s

3

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

3

=

1

2

t

2

.

Sampling it with a

pe-riod of

T

_s

,

y

_s

(

nT

_s

) =

1

(38)

1

/s

2

is

1

/s

3

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

3

=

1

2

t

2

.

Sampling it with a

pe-riod of

T

_s

,

y

_s

(

nT

_s

) =

1

2

n

2

T

s2

(39)

1

/s

2

is

1

/s

3

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

3

=

1

2

t

2

.

Sampling it with a

pe-riod of

T

_s

,

y

_s

(

nT

_s

) =

1

2

n

2

T

s2

Take Z-transform

Y

_s

(

z

) =

T

s2

z

(

z

+ 1)

2(

z

−

1)

3

(40)

The step response of

1

/s

2

is

1

/s

3

. In time

domain, it is,

y

_s

(

t

) =

L

−1

1

s

3

=

1

2

t

2

.

Sampling it with a

pe-riod of

T

_s

,

y

_s

(

nT

_s

) =

1

2

n

2

T

s2

Take Z-transform

Y

_s

(

z

) =

T

s2

z

(

z

+ 1)

2(

z

−

1)

3

Dividing by

z/

(

z

−

1)

,

we get

G

(

z

) =

T

s2

(

z

+ 1)

2(

z

−

1)

2

(41)

(42)

5. ZOH Equivalent First Order Transfer Function Find the ZOH equivalent of K/(τ s + 1).

(43)

Ys(s) = 1 s K τ s + 1 = K " 1 s − 1 s + _τ1 #

(44)

Ys(s) = 1 s K τ s + 1 = K " 1 s − 1 s + _τ1 # ys(t) = K h 1 − e−t/τi , t ≥ 0

(45)

Ys(s) = 1 s K τ s + 1 = K " 1 s − 1 s + _τ1 # ys(t) = K h 1 − e−t/τi , t ≥ 0 ys(nTs) = K h 1 − e−nTs/τi , n ≥ 0

(46)

Ys(s) = 1 s K τ s + 1 = K " 1 s − 1 s + _τ1 # ys(t) = K h 1 − e−t/τi , t ≥ 0 ys(nTs) = K h 1 − e−nTs/τi , n ≥ 0 Ys(z) = K z z − 1 − z z − e−Ts/τ

(47)

Ys(s) = 1 s K τ s + 1 = K " 1 s − 1 s + _τ1 # ys(t) = K h 1 − e−t/τi , t ≥ 0 ys(nTs) = K h 1 − e−nTs/τi , n ≥ 0 Ys(z) = K z z − 1 − z z − e−Ts/τ = Kz(1 − e−Ts/τ) (z − 1)(z − e−Ts/τ)

(48)

Ys(s) = 1 s K τ s + 1 = K " 1 s − 1 s + _τ1 # ys(t) = K h 1 − e−t/τi , t ≥ 0 ys(nTs) = K h 1 − e−nTs/τi , 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) = K(1 − e−Ts/τ) z − e−Ts/τ

(49)

6. ZOH Equivalent First Order Transfer Function - Example

(50)

Sample at

T

_s

= 0

.

5

and find ZOH equivalent

trans. function of

G

_a

(

s

) =

10

(51)

Sample at

T

_s

= 0

.

5

trans. function of

G

_a

(

s

) =

10

5

s

+ 1

(52)

Sample at

T

_s

= 0

.

5

trans. function of

G

_a

(

s

) =

10

5

s

+ 1

Scilab Code:

(53)

Sample at

T

_s

= 0

.

5

trans. function of

G

_a

(

s

) =

10

5

s

+ 1

Scilab Code:

Ga = tf(10,[5 1]);

G = ss2tf(dscr(Ga,0.5));

(54)

Sample at

T

_s

= 0

.

5

trans. function of

G

_a

(

s

) =

10

5

s

+ 1

Scilab Code:

Ga = tf(10,[5 1]);

(55)

Sample at

T

_s

= 0

.

5

trans. function of

G

_a

(

s

) =

10

5

s

+ 1

Scilab Code:

Ga = tf(10,[5 1]);

Scilab output is,

G

(

z

) =

0

.

9546

(56)

Sample at

T

_s

= 0

.

5

trans. function of

G

_a

(

s

) =

10

5

s

+ 1

Scilab Code:

Ga = tf(10,[5 1]);

Scilab output is,

G

(

z

) =

0

.

9546

z

−

0

.

9048

=

10(1

−

e

−0.1

)

(57)

Sample at

T

_s

= 0

.

5

trans. function of

G

_a

(

s

) =

10

5

s

+ 1

Scilab Code:

Ga = tf(10,[5 1]);

Scilab output is,

G

(

z

) =

0

.

9546

z

−

0

.

9048

=

10(1

−

e

−0.1

)

z

−

e

−0.1

In agreement with the

formula in the previous

slide

(58)

(59)

7. Discrete Integration

u(k)

n u(n)

(60)

u(k)

n u(n)

u(k−1)

(61)

u(k)

n u(n)

u(k−1)

y(k) = blue shaded area

(62)

u(k)

n u(n)

u(k−1)

+ red shaded area

(63)

u(k)

n u(n)

u(k−1)

+ red shaded area

(64)

u(k)

n u(n)

u(k−1)

+ red shaded area

y(k) = y(k − 1) + red shaded area

y(k) = y(k − 1) + Ts

(65)

u(k)

n u(n)

u(k−1)

+ red shaded area

y(k) = y(k − 1) + Ts

2 [u(k) + u(k − 1)] Take Z-transform:

(66)

u(k)

n u(n)

u(k−1)

+ red shaded area

y(k) = y(k − 1) + Ts 2 [u(k) + u(k − 1)] Take Z-transform: Y (z) = z−1Y (z) + Ts 2 U(z) + z−1U(z)

(67)

u(k)

n u(n)

u(k−1)

+ red shaded area

(68)

u(k)

n u(n)

u(k−1)

+ red shaded area

Bring all Y to left side: Y (z) − z−1Y (z) = Ts

2

(69)

u(k)

n u(n)

u(k−1)

+ red shaded area

Bring all Y to left side: Y (z) − z−1Y (z) = Ts 2 U(z) + z−1U(z) (1 − z−1)Y (z) = Ts 2 (1 + z−1)U(z)

(70)

(71)

8. Transfer Function for Discrete Integration Recall from previous slide

(1 − z−1)Y (z) = Ts

(72)

(1 − z−1)Y (z) = Ts 2 (1 + z−1)U(z) Y (z) = Ts 2 1 + z−1 1 − z−1U(z)

(73)

(1 − z−1)Y (z) = Ts 2 (1 + z−1)U(z) Y (z) = Ts 2 1 + z−1 1 − z−1U(z) = Ts 2 z + 1 z − 1U(z)

(74)

(1 − z−1)Y (z) = Ts 2 (1 + z−1)U(z) Y (z) = Ts 2 1 + z−1 1 − z−1U(z) = Ts 2 z + 1 z − 1U(z) Integrator has a transfer function,

(75)

GI(z) = Ts 2

z + 1 z − 1

(76)

GI(z) = Ts 2

z + 1 z − 1 A low pass filter!

(77)

GI(z) = Ts 2

× Im(z)

(78)

GI(z) = Ts 2

× Im(z) Re(z) 1 s ↔ Ts 2 z + 1 z − 1

(79)

(80)

9. Derivative Mode • Integral Mode: 1 s ↔ Ts 2 z + 1 z − 1

(81)

9. Derivative Mode • Integral Mode: 1 s ↔ Ts 2 z + 1 z − 1 • Derivative Mode: s ↔ 2 Ts z − 1 z + 1

(82)

(83)

• High pass filter

(84)

• Has a pole at z = −1. Hence produces in partial fraction expansion, a term of the form

(85)

z

(86)

z

z + 1 ↔ (−1)n

• Results in wildly oscillating control effort.

(87)

(88)

10. Derivative Mode - Other Approximations

(89)

Backward difference: y(k) = y(k − 1) + T_su(k)

(90)

(1 − z−1)Y (z) = TsU(z) Y (z) = Ts 1

(91)

(1 − z−1)Y (z) = TsU(z) Y (z) = Ts 1

1 − z−1 = Ts

z

(92)

(1 − z−1)Y (z) = TsU(z) Y (z) = Ts 1 1 − z−1 = Ts z z − 1U(z) 1 s ↔

(93)

(1 − z−1)Y (z) = TsU(z) Y (z) = Ts 1 1 − z−1 = Ts z z − 1U(z) 1 s ↔ Ts z z − 1

(94)

(95)

Forward difference: y(k) = y(k − 1) + T_su(k − 1)

(96)

(1 − z−1)Y (z) = Tsz−1U(z) Y (z) = Ts z

−1

(97)

(1 − z−1)Y (z) = Tsz−1U(z) Y (z) = Ts z −1 1 − z−1U(z) = Ts z − 1U(z)

(98)

(1 − z−1)Y (z) = Tsz−1U(z) Y (z) = Ts z −1 1 − z−1U(z) = Ts z − 1U(z) 1 s ↔

(99)

(1 − z−1)Y (z) = Tsz−1U(z) Y (z) = Ts z −1 1 − z−1U(z) = Ts z − 1U(z) 1 s ↔ Ts z − 1

(100)

(1 − z−1)Y (z) = Tsz−1U(z) Y (z) = Ts z −1 1 − z−1U(z) = Ts z − 1U(z) 1 s ↔ Ts z − 1 Both derivative modes are high pass,

(101)

(102)

Both derivative modes are high pass, no oscillations, same gains

(103)

(104)

11. PID Controller

(105)

11. PID Controller

Proportional Mode: Most popular control mode. Increase in proportional mode generally results in

(106)

11. PID Controller

(107)

11. PID Controller

(108)

11. PID Controller

• Decreased steady state offset and increased oscillations Integral Mode: Used to remove steady state offset.

(109)

11. PID Controller

• Decreased steady state offset and increased oscillations Integral Mode: Used to remove steady state offset. Increase

(110)

11. PID Controller

in integral mode generally results in

(111)

11. PID Controller

• Zero steady state offset

(112)

11. PID Controller

• Increased oscillations

(113)

11. PID Controller

Derivative Mode: Mainly used for prediction purposes. Increase in derivative mode generally results in

(114)

11. PID Controller

(115)

11. PID Controller

• Decreased oscillations and improved stability

(116)

11. PID Controller

• Decreased oscillations and improved stability

• Sensitive to noise

The most popular controller in industry.

(117)

(118)

12. PID Controller - Basic Design Let input to controller by E(z)

(119)

12. PID Controller - Basic Design

(120)

Let input to controller by E(z) and output from it be U(z). If gain is K,

(121)

Let input to controller by E(z) and output from it be U(z). If gain is K, τ_i is integral time

(122)

Let input to controller by E(z) and output from it be U(z). If gain is K, τ_i is integral time and τ_d is derivative time,

(123)

u(t) = K e(t) + 1 τi Z t 0 e(t)dt + τd de(t) dt

(124)

u(t) = K e(t) + 1 τi Z t 0 e(t)dt + τd de(t) dt U(s) = K(1 + 1 τis + τds)E(s)

(125)

u(t) = K e(t) + 1 τi Z t 0 e(t)dt + τd de(t) dt U(s) = K(1 + 1 τis + τds)E(s) U(s) =4 Sc(s) Rc(s)E(s)

(126)

(127)

If integral mode is present, R_c(0) = 0.

(128)

Filtered derivative mode:

u(t) = K 1 + 1 τis + τds 1 + τds N ! e(t)

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Filtered derivative mode:

u(t) = K 1 + 1 τis + τds 1 + τds N ! e(t) N is a large number, of the order of 100.

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13. Reaction Curve Method - Ziegler Nichols Tun-ing

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• Applicable only to stable systems

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• Give a unit step input to a stable system and get

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1. the time lag after which the system starts responding (L), 2. the steady state gain (K) and

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3. the time the output takes to reach the steady state, after it starts responding (τ)

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3. the time the output takes to reach the steady state, after it starts responding (τ)

R=K/τ

L τ K

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R=K/τ

L τ K

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R=K/τ

L τ K

• Let the slope of the response be calculated as R = K τ .

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R=K/τ

L τ K

• Let the slope of the response be calculated as R = K τ . Then the PID settings are given below:

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R=K/τ

L τ K

Kp τi τd P 1/RL

PI 0.9/RL 3L

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R=K/τ

L τ K

Kp τi τd P 1/RL

PI 0.9/RL 3L

PID 1.2/RL 2L 0.5L

Consistent units should be used

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15. Stability Method - Ziegler Nichols Tuning

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Another way of finding the PID tuning parameters is as follows.

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• Close the loop with a proportional controller

• Gain of controller is increased until the closed loop system becomes unstable

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• At the verge of instability, note down the gain of the controller (K_u) and the period of oscillation (P_u)

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• PID settings are given below:

Kp τi τd

P 0.5K_u

PI 0.45K_u P_u/1.2

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• PID settings are given below:

Kp τi τd

P 0.5K_u

PI 0.45K_u P_u/1.2

PID 0.6Ku Pu/2 Pu/8

Consistent units should be used

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16. Design Procedure

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A common procedure to design discrete PID controller:

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• Tune continuous PID controller by any popular technique

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• Get continuous PID settings

• Discretize using the method discussed now or the ZOH equiv-alent method discussed earlier

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• Get continuous PID settings

• Discretize using the method discussed now or the ZOH equiv-alent method discussed earlier

• Direct digital design techniques

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17. 2-DOF Controller y Tc Rc G = B A Sc Rc r u −

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17. 2-DOF Controller y Tc Rc G = B A Sc Rc r u − u = Tc Rc r − Sc Rc y

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It is easy to arrive at the following relation between r and y. y = Tc

Rc

B/A

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It is easy to arrive at the following relation between r and y. y = Tc Rc B/A 1 + BSc/ARcr = BTc ARc + BScr

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It is easy to arrive at the following relation between r and y. y = Tc Rc B/A 1 + BSc/ARcr = BTc ARc + BScr Error e, given by r − y is given by

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e = 1 − BTc ARc + BSc r

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e = 1 − BTc ARc + BSc r = ARc + BSc − BTc ARc + BSc r

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18. Offset-Free Tracking of Steps with Integral

E(z) = A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z)

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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)

lim

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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) lim n→∞e(n) = limz→1 z − 1 z A(z)Rc(z) + B(z)Sc(z) − B(z)Tc(z) A(z)Rc(z) + B(z)Sc(z) z z − 1

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Because the controller has an integral action, R_c(1) = 0: e(∞) =

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Because the controller has an integral action, R_c(1) = 0: e(∞) = Sc(z) − Tc(z) Sc(z) z=1

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Because the controller has an integral action, R_c(1) = 0: e(∞) = Sc(z) − Tc(z) Sc(z) z=1 = Sc(1) − Tc(1) Sc(1)

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This condition can be satisfied if one of the following is met: Tc = Sc

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Tc = Sc(1) Tc(1) = Sc(1)