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1. s to Z-Domain Transfer Function

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1. s to Z-Domain Transfer Function

DiscreteZOH

Signals

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1. s to Z-Domain Transfer Function

DiscreteZOH

Signals

1. Get step responseof continuous trans-fer functionys(t).

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1. s to Z-Domain Transfer Function

DiscreteZOH

Signals

1. Get step responseof continuous trans-fer functionys(t).

2. Discretize step re-sponse: ys(nTs).

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1. s to Z-Domain Transfer Function

DiscreteZOH

Signals

1. Get step responseof continuous trans-fer functionys(t).

2. Discretize step re-sponse: ys(nTs).

1. Z-transform the step re-

sponse to obtain Ys(z).

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1. s to Z-Domain Transfer Function

DiscreteZOH

Signals

1. Get step responseof continuous trans-fer functionys(t).

2. Discretize step re-sponse: ys(nTs).

1. Z-transform the step re-

sponse to obtain Ys(z).2. Divide the result from

above by Z-transform of astep, namely,z/(z 1).

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1. s to Z-Domain Transfer Function

DiscreteZOH

Signals

1. Get step responseof continuous trans-fer functionys(t).

2. Discretize step re-sponse: ys(nTs).

1. Z-transform the step re-

sponse to obtain Ys(z).2. Divide the result from

above by Z-transform of astep, namely,z/(z 1).

Ga(s): Laplace transferfunction

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1. s to Z-Domain Transfer Function

DiscreteZOHSignals

1. Get step responseof continuous trans-fer functionys(t).

2. Discretize step re-sponse: ys(nTs).

1. Z-transform the step re-

sponse to obtain Ys(z).2. Divide the result from

above by Z-transform of astep, namely,z/(z 1).

Ga(s): Laplace transferfunction

G(z): Z-transfer function

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1. s to Z-Domain Transfer Function

DiscreteZOHSignals

1. Get step responseof continuous trans-fer functionys(t).

2. Discretize step re-sponse: ys(nTs).

1. Z-transform the step re-

sponse to obtain Ys(z).2. Divide the result from

above by Z-transform of astep, namely,z/(z 1).

Ga(s): Laplace transferfunction

G(z): Z-transfer function

G(z) =z 1

zZ

L1

Ga(s)

s

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1. s to Z-Domain Transfer Function

DiscreteZOHSignals

1. Get step responseof continuous trans-fer functionys(t).

2. Discretize step re-sponse: ys(nTs).

1. Z-transform the step re-

sponse to obtain Ys(z).2. Divide the result from

above by Z-transform of astep, namely,z/(z 1).

Ga(s): Laplace transferfunction

G(z): Z-transfer function

G(z) =z 1

zZ

L1

Ga(s)

s

Step Response Equivalence

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1. s to Z-Domain Transfer Function

DiscreteZOHSignals

1. Get step responseof continuous trans-fer functionys(t).

2. Discretize step re-sponse: ys(nTs).

1. Z-transform the step re-

sponse to obtain Ys(z).2. Divide the result from

above by Z-transform of astep, namely,z/(z 1).

Ga(s): Laplace transferfunction

G(z): Z-transfer function

G(z) =z 1

zZ

L1

Ga(s)

s

Step Response Equivalence = ZOH EquivalenceDigital Control 1 Kannan M. Moudgalya, Autumn 2007

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2. Important Result from Differentiation

Recall

1(n)an z

z a=

n=0

anzn,

Differentiating w.r.t. a,

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2. Important Result from Differentiation

Recall

1(n)an z

z a=

n=0

anzn,

Differentiating w.r.t. a,z

(z a)2=

n=0nan1zn

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2. Important Result from Differentiation

Recall

1(n)an z

z a=

n=0

anzn,

Differentiating w.r.t. a,z

(z a)2=

n=0nan1zn

nan11(n) z(z a)2

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2. Important Result from Differentiation

Recall

1(n)an z

z a=

n=0

anzn,

Differentiating w.r.t. a,z

(z a)2=

n=0nan1zn

nan11(n) z(z a)2

n(n 1)an21(n) 2z

(z a)3Digital Control 2 Kannan M. Moudgalya, Autumn 2007

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3. ZOH Equivalence of 1/s

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3. ZOH Equivalence of 1/s

The step response of

1/s is

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2.

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2. In timedomain, it is,

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2. In timedomain, it is,

ys(t) = L11

s2

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2. In timedomain, it is,

ys(t) = L11

s2=t

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2. In timedomain, it is,

ys(t) = L11

s2=t

Sampling it with a pe-riod ofTs,

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2. In timedomain, it is,

ys(t) = L11

s2=t

Sampling it with a pe-riod ofTs,

ys(nTs) =

/

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2. In timedomain, it is,

ys(t) = L11

s2=t

Sampling it with a pe-riod ofTs,

ys(nTs) = nTs

ZO i l f /

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2. In timedomain, it is,

ys(t) = L11

s2=t

Sampling it with a pe-riod ofTs,

ys(nTs) = nTs

Taking Z-transforms

Ys(z) =

ZOH E i l f 1/

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2. In timedomain, it is,

ys(t) = L11

s2=t

Sampling it with a pe-riod ofTs,

ys(nTs) = nTs

Taking Z-transforms

Ys(z) = Tsz

(z 1)2

ZOH E i l f 1/

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2. In timedomain, it is,

ys(t) = L11

s2=t

Sampling it with a pe-riod ofTs,

ys(nTs) = nTs

Taking Z-transforms

Ys(z) = Tsz

(z 1)2

Divide by z/(z1),

3 ZOH E i l f 1/

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2. In timedomain, it is,

ys(t) = L11

s2=t

Sampling it with a pe-riod ofTs,

ys(nTs) = nTs

Taking Z-transforms

Ys(z) = Tsz

(z 1)2

Divide by z/(z1), toget the ZOH equivalentdiscrete domain transferfunction

3 ZOH E i l f 1/

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3. ZOH Equivalence of 1/s

The step response of

1/s is 1/s2. In timedomain, it is,

ys(t) = L11

s2=t

Sampling it with a pe-riod ofTs,

ys(nTs) = nTs

Taking Z-transforms

Ys(z) = Tsz

(z 1)2

Divide by z/(z1), toget the ZOH equivalentdiscrete domain transferfunction

G(z) = Ts

z 1

Digital Control 3 Kannan M. Moudgalya, Autumn 2007

4 ZOH E i l f 1/ 2

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4. ZOH Equivalence of 1/s2

4 ZOH Equivalence of 1/s2

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4. ZOH Equivalence of 1/s2

The step response of

1/s2 is

4 ZOH Equivalence of 1/s2

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4. ZOH Equivalence of 1/s2

The step response of

1/s2 is 1/s3.

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4 ZOH Equivalence of 1/s2

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4. ZOH Equivalence of 1/s

The step response of

1/s2 is 1/s3. In timedomain, it is,

ys(t) = L11

s3=

1

2

t2.

4 ZOH Equivalence of 1/s2

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4. ZOH Equivalence of 1/s

The step response of

1/s2 is 1/s3. In timedomain, it is,

ys(t) = L11

s3=

1

2

t2.

Sampling it with a pe-riod ofTs,

ys(nTs) =12

n2T2s

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4. ZOH Equivalence of 1/s2

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4. ZOH Equivalence of 1/s

The step response of

1/s2 is 1/s3. In timedomain, it is,

ys(t) = L11

s3=

1

2

t2.

Sampling it with a pe-riod ofTs,

ys(nTs) =12

n2T2s

Take Z-transform

Ys(z) =T2s z(z+ 1)

2(z 1)3

4. ZOH Equivalence of 1/s2

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q /

The step response of

1/s2 is 1/s3. In timedomain, it is,

ys(t) = L11

s3=

1

2

t2.

Sampling it with a pe-riod ofTs,

ys(nTs) =12

n2T2s

Take Z-transform

Ys(z) =T2s z(z+ 1)

2(z 1)3

Dividing byz/(z 1),we get

G(z) =T2s (z+ 1)

2(z 1)2

Digital Control 4 Kannan M. Moudgalya, Autumn 2007

5. ZOH Equivalent First Order Transfer Function

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q

5. ZOH Equivalent First Order Transfer Function

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Find the ZOH equivalent ofK/( s+ 1).

5. ZOH Equivalent First Order Transfer Function

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Find the ZOH equivalent ofK/( s+ 1).

Ys(s) =1

s

K

s+ 1= K

1

s

1

s+ 1

5. ZOH Equivalent First Order Transfer Function

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Find the ZOH equivalent ofK/( s+ 1).

Ys(s) =1

s

K

s+ 1= K

1

s

1

s+ 1

ys(t) =K1 et/ , t 0

5. ZOH Equivalent First Order Transfer Function

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Find the ZOH equivalent ofK/( s+ 1).

Ys(s) =1

s

K

s+ 1= K

1

s

1

s+ 1

ys(t) =K1 et/ , t 0ys(nTs) =K

1 enTs/

, n 0

5. ZOH Equivalent First Order Transfer Function

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Find the ZOH equivalent ofK/( s+ 1).

Ys(s) =1

s

K

s+ 1= K

1

s

1

s+ 1

ys(t) =K1 et/ , t 0ys(nTs) =K

1 enTs/

, n 0

Ys(z) =K zz 1

z

z eT

s/

5. ZOH Equivalent First Order Transfer Function

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Find the ZOH equivalent ofK/( s+ 1).

Ys(s) =1

s

K

s+ 1= K

1

s

1

s+ 1

ys(t) =K1 et/ , t 0ys(nTs) =K

1 enTs/

, n 0

Ys(z) =K z

z 1

z

z eT

s/ =

Kz(1 eTs/)

(z 1)(z eT

s/

)

5. ZOH Equivalent First Order Transfer Function

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Find the ZOH equivalent ofK/( s+ 1).

Ys(s) =1

s

K

s+ 1= K

1

s

1

s+ 1

ys(t) =K1 et/ , t 0ys(nTs) =K

1 enTs/

, n 0

Ys(z) =K z

z 1

z

z eTs/ =

Kz(1 eTs/)

(z 1)(z eTs/

)Dividing byz/(z 1), we get

G(z) =K(1 eTs/)

z eTs/

Digital Control 5 Kannan M. Moudgalya, Autumn 2007

6. ZOH Equivalent First Order Transfer FunctionExample

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- Example

6. ZOH Equivalent First Order Transfer FunctionExample

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- Example

Sample at Ts = 0.5and find ZOH equivalenttrans. function of

Ga(s) =

10

5s+ 1

6. ZOH Equivalent First Order Transfer FunctionExample

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- Example

Sample at Ts

= 0.5and find ZOH equivalenttrans. function of

Ga(s) =

10

5s+ 1

Scilab Code:

6. ZOH Equivalent First Order Transfer Function- Example

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- Example

Sample at Ts

= 0.5and find ZOH equivalenttrans. function of

Ga(s) =

10

5s+ 1

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

6. ZOH Equivalent First Order Transfer Function- Example

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- Example

Sample at Ts

= 0.5and find ZOH equivalenttrans. function of

Ga(s) =

10

5s+ 1

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

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

6. ZOH Equivalent First Order Transfer Function- Example

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Example

Sample at Ts

= 0.5and find ZOH equivalenttrans. function of

Ga(s) =

10

5s+ 1

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

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

Scilab output is,

6. ZOH Equivalent First Order Transfer Function- Example

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Example

Sample at Ts

= 0.5and find ZOH equivalenttrans. function of

Ga(s) =

10

5s+ 1

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

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

Scilab output is,

G(z) = 0.9546

z 0.9048

6. ZOH Equivalent First Order Transfer Function- Example

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Example

Sample at Ts

= 0.5and find ZOH equivalenttrans. function of

Ga(s) =

10

5s+ 1

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

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

Scilab output is,

G(z) = 0.9546

z 0.9048

=10(1 e0.1)

z e0.1

6. ZOH Equivalent First Order Transfer Function- Example

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p

Sample at Ts

= 0.5and find ZOH equivalenttrans. function of

Ga(s) =

10

5s+ 1Scilab Code:Ga = tf(10,[5 1]);

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

Scilab output is,

G(z) = 0.9546

z 0.9048

=10(1 e0.1)

z e0.1

In agreement with theformula in the previousslide

Digital Control 6 Kannan M. Moudgalya, Autumn 2007

7. Discrete Integration

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7. Discrete Integration

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u(k)

n

u(n)

u

(k

1)

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7. Discrete Integration

y(k) = blue shaded area

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u(k)

n

u(n)

u

(k

1)

y(k) =blue shaded area

+red shaded area

7. Discrete Integration

y(k) = blue shaded area

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u(k)

n

u(n)

u

(k

1)

y(k) =blue shaded area

+red shaded area

y(k) =y(k 1)

7. Discrete Integration

( ) y(k) = blue shaded area

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u(k)

n

u(n)

u

(k

1)

y(k) =blue shaded area

+red shaded area

y(k) =y(k 1) +red shaded area

7. Discrete Integration

( ) y(k) = blue shaded area

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u(k)

n

u(n)

u(k 1)

y(k) =blue shaded area

+red shaded area

y(k) =y(k 1) +red shaded area

y(k) =y(k 1) +Ts

2[u(k) +u(k 1)]

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7. Discrete Integration

u(n) y(k) = blue shaded area

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u(k)

n

u(n)

u(k 1)

y(k) blue shaded area

+red shaded area

y(k) =y(k 1) +red shaded area

y(k) =y(k 1) +Ts

2[u(k) +u(k 1)]

Take Z-transform:

Y(z) =z1Y(z) +Ts

2

U(z) +z1U(z)

7. Discrete Integration

u(n) y(k) =blue shaded area

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u(k)

n

u(n)

u(k 1)

y( )

+red shaded area

y(k) =y(k 1) +red shaded area

y(k) =y(k 1) +Ts

2[u(k) +u(k 1)]

Take Z-transform:

Y(z) =z1Y(z) +Ts

2

U(z) +z1U(z)

Bring allY to left side:

7. Discrete Integration

u(n) y(k) =blue shaded area

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u(k)

n

u(n)

u(k 1)

y( )

+red shaded area

y(k) =y(k 1) +red shaded area

y(k) =y(k 1) +Ts

2[u(k) +u(k 1)]

Take Z-transform:

Y(z) =z1Y(z) +Ts

2

U(z) +z1U(z)

Bring allY to left side:

Y(z) z1Y(z) =Ts

2

U(z) +z1U(z)

7. Discrete Integration

u(n) y(k) =blue shaded area

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u(k)

n

( )

u(k 1)

y( )

+red shaded area

y(k) =y(k 1) +red shaded area

y(k) =y(k 1) +Ts

2[u(k) +u(k 1)]

Take Z-transform:

Y(z) =z1Y(z) +Ts

2

U(z) +z1U(z)

Bring allY to left side:

Y(z) z1Y(z) =Ts

2

U(z) +z1U(z)

(1 z1)Y(z) =

Ts

2(1 +z1)U(z)

Digital Control 7 Kannan M. Moudgalya, Autumn 2007

8. Transfer Function for Discrete Integration

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8. Transfer Function for Discrete Integration

Recall from previous slide

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Recall from previous slide

(1 z1)Y(z) =T

s2 (1 +z

1)U(z)

8. Transfer Function for Discrete Integration

Recall from previous slide

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Recall from previous slide

(1 z1)Y(z) =T

s2 (1 +z

1)U(z)

Y(z) =Ts

2

1 +z1

1 z1U(z)

8. Transfer Function for Discrete Integration

Recall from previous slide

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Recall from previous slide

(1 z1)Y(z) =T

s2 (1 +z

1)U(z)

Y(z) =Ts

2

1 +z1

1 z1U(z)

= Ts2

z+ 1z 1

U(z)

8. Transfer Function for Discrete Integration

Recall from previous slide

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p

(1 z1)Y(z) =Ts

2 (1 +z1)U(z)

Y(z) =Ts

2

1 +z1

1 z1U(z)

= Ts2

z+ 1z 1

U(z)

Integrator has a transfer function,

8. Transfer Function for Discrete Integration

Recall from previous slide

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p

(1 z1)Y(z) =Ts

2 (1 +z1)U(z)

Y(z) =Ts

2

1 +z1

1 z1U(z)

= Ts2

z+ 1z 1

U(z)

Integrator has a transfer function,

GI(z) =Ts

2

z+ 1

z 1

8. Transfer Function for Discrete Integration

Recall from previous slide

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p

(1 z1)Y(z) =Ts

2 (1 +z1)U(z)

Y(z) =Ts

2

1 +z1

1 z1U(z)

= Ts2

z+ 1z 1

U(z)

Integrator has a transfer function,

GI(z) =Ts

2

z+ 1

z 1

A low pass filter!

8. Transfer Function for Discrete Integration

Recall from previous slide

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p

(1 z1)Y(z) =Ts

2 (1 +z1)U(z)

Y(z) =Ts

2

1 +z1

1 z1U(z)

= Ts2

z+ 1z 1

U(z)

Integrator has a transfer function,

GI(z) =Ts

2

z+ 1

z 1

A low pass filter!

Im(z)

Re(z)

8. Transfer Function for Discrete Integration

Recall from previous slide

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(1 z1)Y(z) =Ts

2 (1 +z1)U(z)

Y(z) =Ts

2

1 +z1

1 z1U(z)

=T

s2

z+ 1z 1

U(z)

Integrator has a transfer function,

GI(z) =Ts

2

z+ 1

z 1

A low pass filter!

Im(z)

Re(z)

1

s

Ts

2

z+ 1

z 1

Digital Control 8 Kannan M. Moudgalya, Autumn 2007

9. Derivative Mode

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9. Derivative Mode

1 T z + 1

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Integral Mode: 1

s

Ts

2

z+ 1

z 1

9. Derivative Mode

1 T z + 1

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Integral Mode: 1

s

Ts

2

z+ 1

z 1

Derivative Mode: s 2

Ts

z 1

z+ 1

9. Derivative Mode

1 T z + 1

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Integral Mode: 1

s

Ts

2

z+ 1

z 1

Derivative Mode: s 2

Ts

z 1

z+ 1

High pass filter

9. Derivative Mode

1 Ts z + 1

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Integral Mode: 1

s

Ts

2

z+ 1

z 1

Derivative Mode: s 2

Ts

z 1

z+ 1

High pass filter

Has a pole at z = 1.

9. Derivative Mode

1 Ts z + 1

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Integral Mode: 1

s

Ts

2

z+ 1

z 1

Derivative Mode: s 2

Ts

z 1

z+ 1

High pass filter

Has a pole at z = 1. Hence produces in partial fractionexpansion, a term of the form

9. Derivative Mode

1 Ts z + 1

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Integral Mode: 1

s

Ts

2

z+ 1

z 1

Derivative Mode: s 2

Ts

z 1

z+ 1

High pass filter

Has a pole at z = 1. Hence produces in partial fractionexpansion, a term of the form

z

z+ 1 (1)n

9. Derivative Mode

1 Ts z+ 1

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Integral Mode:

s

s

2

+

z 1

Derivative Mode: s 2

Ts

z 1

z+ 1

High pass filter

Has a pole at z = 1. Hence produces in partial fractionexpansion, a term of the form

z

z+ 1

(1)n

Results in wildly oscillating control effort.

Digital Control 9 Kannan M. Moudgalya, Autumn 2007

10. Derivative Mode - Other Approximations

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10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)

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y( ) y( ) + s ( )

10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)

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y( ) y( ) s ( )

(1 z1

)Y(z) =TsU(z)

10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)

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y( ) y( ) ( )

(1 z1

)Y(z) =TsU(z)

Y(z) =Ts1

1 z1

10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)

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( ) ( ) ( )

(1 z1

)Y(z) =TsU(z)

Y(z) =Ts1

1 z1=Ts

z

z 1U(z)

10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)

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(1 z1

)Y(z) =TsU(z)

Y(z) =Ts1

1 z1=Ts

z

z 1U(z)

1

s

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10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)

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(1 z1

)Y(z) =TsU(z)Y(z) =Ts

1

1 z1=Ts

z

z 1U(z)

1

s Ts

z

z 1Forward difference: y(k) =y(k 1) +Tsu(k 1)

10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)

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(1 z1

)Y(z) =TsU(z)Y(z) =Ts

1

1 z1=Ts

z

z 1U(z)

1

s Ts

z

z 1Forward difference: y(k) =y(k 1) +Tsu(k 1)

(1 z1)Y(z) =Tsz1U(z)

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10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)1

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(1 z1

)Y(z) =TsU(z)Y(z) =Ts

1

1 z1=Ts

z

z 1U(z)

1

s Ts

z

z 1Forward difference: y(k) =y(k 1) +Tsu(k 1)

(1 z1)Y(z) =Tsz1U(z)

Y(z) =Tsz1

1 z1U(z) =

Ts

z 1U(z)

10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)1

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(1 z1

)Y(z) =TsU(z)Y(z) =Ts

1

1 z1=Ts

z

z 1U(z)

1

s Ts

z

z 1Forward difference: y(k) =y(k 1) +Tsu(k 1)

(1 z1)Y(z) =Tsz1U(z)

Y(z) =Tsz1

1 z1U(z) =

Ts

z 1U(z)1

s

10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)1

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(1 z1

)Y(z) =TsU(z)Y(z) =Ts

1

1 z1=Ts

z

z 1U(z)

1

s Ts

z

z 1Forward difference: y(k) =y(k 1) +Tsu(k 1)

(1 z1)Y(z) =Tsz1U(z)

Y(z) =Tsz1

1 z1U(z) =

Ts

z 1U(z)1

s

Ts

z 1

10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)1

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(1 z1

)Y(z) =TsU(z)Y(z) =Ts

1

1 z1=Ts

z

z 1U(z)

1

s Ts

z

z 1Forward difference: y(k) =y(k 1) +Tsu(k 1)

(1 z1)Y(z) =Tsz1U(z)

Y(z) =Tsz1

1 z1U(z) =

Ts

z 1U(z)1

s

Ts

z 1Both derivative modes are high pass,

10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)1

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(1 z1

)Y(z) =TsU(z)Y(z) =Ts

1

1 z1=Ts

z

z 1U(z)

1

s Ts

z

z 1Forward difference: y(k) =y(k 1) +Tsu(k 1)

(1 z1)Y(z) =Tsz1U(z)

Y(z) =Tsz1

1 z1U(z) =

Ts

z 1U(z)1

s

Ts

z 1Both derivative modes are high pass, no oscillations,

10. Derivative Mode - Other Approximations

Backward difference: y(k) =y(k 1) +Tsu(k)1

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(1 z1

)Y(z) =TsU(z)Y(z) =Ts

1

1 z1=Ts

z

z 1U(z)

1

s

Tsz

z 1Forward difference: y(k) =y(k 1) +Tsu(k 1)

(1 z1)Y(z) =Tsz1U(z)

Y(z) =Tsz1

1 z1U(z) =

Ts

z 1U(z)1

s

Ts

z 1Both derivative modes are high pass, no oscillations, same gainsDigital Control 10 Kannan M. Moudgalya, Autumn 2007

11. PID Controller

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11. PID Controller

Proportional Mode:Most popular control mode.

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11. PID Controller

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

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11. PID Controller

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

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Decreased steady state offset

11. PID Controller

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

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Decreased steady state offset and increased oscillations

11. PID Controller

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

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Decreased steady state offset and increased oscillations

Integral Mode:Used to remove steady state offset.

11. PID Controller

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

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Decreased steady state offset and increased oscillations

Integral Mode:Used to remove steady state offset. Increasein integral mode generally results in

11. PID Controller

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

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Decreased steady state offset and increased oscillations

Integral Mode:Used to remove steady state offset. Increasein integral mode generally results in

Zero steady state offset

11. PID Controller

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

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Decreased steady state offset and increased oscillations

Integral Mode:Used to remove steady state offset. Increasein integral mode generally results in

Zero steady state offset Increased oscillations

11. PID Controller

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

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Decreased steady state offset and increased oscillations

Integral Mode:Used to remove steady state offset. Increasein 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. Increasein proportional mode generally results in

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Decreased steady state offset and increased oscillations

Integral Mode:Used to remove steady state offset. Increasein integral mode generally results in

Zero steady state offset Increased oscillations

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

11. PID Controller

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

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Decreased steady state offset and increased oscillations

Integral Mode:Used to remove steady state offset. Increasein integral mode generally results in

Zero steady state offset Increased oscillations

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

Decreased oscillations and improved stability

11. PID Controller

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

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Decreased steady state offset and increased oscillations

Integral Mode:Used to remove steady state offset. Increasein integral mode generally results in

Zero steady state offset Increased oscillations

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

Decreased oscillations and improved stability

Sensitive to noise

11. PID Controller

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

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Decreased steady state offset and increased oscillations

Integral Mode:Used to remove steady state offset. Increasein integral mode generally results in

Zero steady state offset Increased oscillations

Derivative Mode:Mainly used for prediction purposes. Increasein 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

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12. PID Controller - Basic Design

Let input to controller byE(z)

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12. PID Controller - Basic Design

Let input to controller byE(z)and output from it beU(z).

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12. PID Controller - Basic Design

Let input to controller byE(z)and output from it beU(z). Ifgain isK,

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12. PID Controller - Basic Design

Let input to controller byE(z)and output from it beU(z). Ifgain isK, i is integral time

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12. PID Controller - Basic Design

Let input to controller byE(z)and output from it beU(z). Ifgain isK, i is integral time and d is derivative time,

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12. PID Controller - Basic Design

Let input to controller byE(z)and output from it beU(z). Ifgain isK, i is integral time and d is derivative time,

1

t d (t)

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u(t) =K

e(t) + 1

i

t

0

e(t)dt+dde(t)

dt

12. PID Controller - Basic Design

Let input to controller byE(z)and output from it beU(z). Ifgain isK, i is integral time and d is derivative time,

1

t de(t)

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u(t) =K

e(t) + 1

i

t

0

e(t)dt+dde(t)

dt

U(s) =K(1 + 1

is

+ds)E(s)

12. PID Controller - Basic Design

Let input to controller byE(z)and output from it beU(z). Ifgain isK, i is integral time and d is derivative time,

1

t de(t)

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u(t) =K

e(t) + 1

i

t

0

e(t)dt+dde(t)

dt

U(s) =K(1 + 1

is

+ds)E(s)

U(s) =

Sc(s)

Rc(s)E(s)

12. PID Controller - Basic Design

Let input to controller byE(z)and output from it beU(z). Ifgain isK, i is integral time and d is derivative time,

1

t de(t)

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u(t) =K

e(t) + 1

i

t

0

e(t)dt+dde(t)

dt

U(s) =K(1 +

1

is+ds)E(s)

U(s) =

Sc(s)

Rc(s)E(s)

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

12. PID Controller - Basic Design

Let input to controller byE(z)and output from it beU(z). Ifgain isK, i is integral time and d is derivative time,

1

t de(t)

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u(t) =K

e(t) + 1i

t

0

e(t)dt+dde(t)

dt

U(s) =K(1 +

1

is+ds)E(s)

U(s) =

Sc(s)

Rc(s)E(s)

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

Filteredderivative mode:

12. PID Controller - Basic Design

Let input to controller byE(z)and output from it beU(z). Ifgain isK, i is integral time and d is derivative time,

( )

( ) 1

t( ) de(t)

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u(t) =K

e(t) + 1i

t

0

e(t)dt+dde(t)

dt

U(s) =K(1 +

1

is+ds)E(s)

U(s) =

Sc(s)

Rc(s)E(s)

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

Filteredderivative mode:u(t) =K

1 +

1

is+

ds

1 + dsN

e(t)

12. PID Controller - Basic Design

Let input to controller byE(z)and output from it beU(z). Ifgain isK, i is integral time and d is derivative time,

( )

( ) 1

t( )d de(t)

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u(t) =K

e(t) + 1i

0

e(t)dt+dde(t)

dt

U(s) =K(1 +

1

is+ds)E(s)

U(s) =

Sc(s)

Rc(s)E(s)

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

Filteredderivative mode:u(t) =K

1 +

1

is+

ds

1 + dsN

e(t)

N is a large number, of the order of 100.Digital Control

12 Kannan M. Moudgalya, Autumn 2007

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13. Reaction Curve Method - Ziegler Nichols Tun-ing

Applicable only to stable systems

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13. Reaction Curve Method - Ziegler Nichols Tun-ing

Applicable only to stable systems

Gi i i bl d

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Give a unit step input to a stable system and get

13. Reaction Curve Method - Ziegler Nichols Tun-ing

Applicable only to stable systems

Gi it t i t t t bl t d t

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Give a unit step input to a stable system and get

1. the time lag after which the system starts responding (L),

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13. Reaction Curve Method - Ziegler Nichols Tun-ing

Applicable only to stable systems

Give a unit step input to a stable system and get

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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, afterit starts responding ()

13. Reaction Curve Method - Ziegler Nichols Tun-ing

Applicable only to stable systems

Give a unit step input to a stable system and get

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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, afterit starts responding ()

R= K/

L

K

Digital Control 13 Kannan M. Moudgalya, Autumn 2007

14. Reaction Curve Method - Ziegler Nichols Tun-ing

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14. Reaction Curve Method - Ziegler Nichols Tun-ing

R= K/

K

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L

14. Reaction Curve Method - Ziegler Nichols Tun-ing

R= K/

K

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L

Let the slope of the response be calculated as R = K

.

14. Reaction Curve Method - Ziegler Nichols Tun-ing

R= K/

K

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L

Let the slope of the response be calculated as R = K

.

Then the PID settings are given below:

14. Reaction Curve Method - Ziegler Nichols Tun-ing

R= K/

K

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L

Let the slope of the response be calculated as R = K

.

Then the PID settings are given below:

Kp i d

P 1/RLPI 0.9/RL 3LPID 1.2/RL 2L 0.5L

14. Reaction Curve Method - Ziegler Nichols Tun-ing

R= K/

K

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L

Let the slope of the response be calculated as R = K

.

Then the PID settings are given below:

Kp i d

P 1/RLPI 0.9/RL 3LPID 1.2/RL 2L 0.5L

Consistent units should be usedDigital Control 14 Kannan M. Moudgalya, Autumn 2007

15. Stability Method - Ziegler Nichols Tuning

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15. Stability Method - Ziegler Nichols Tuning

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

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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

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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

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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

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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

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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 i dP 0.5KuPI 0.45Ku Pu/1.2PID 0.6Ku Pu/2 Pu/8

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16. Design Procedure

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16. Design Procedure

A common procedure to design discrete PID controller:

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16. Design Procedure

A common procedure to design discrete PID controller:

Tune continuous PID controller by any popular technique

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16. Design Procedure

A common procedure to design discrete PID controller:

Tune continuous PID controller by any popular technique

Get continuous PID settings

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16. Design Procedure

A common procedure to design discrete PID controller:

Tune continuous PID controller by any popular technique

Get continuous PID settings

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

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

Direct digital design techniques

Digital Control 16 Kannan M. Moudgalya, Autumn 2007

17. 2-DOF Controller

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17. 2-DOF Controller

yTc

RcG =

B

A

Sc

Rc

r u

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17. 2-DOF Controller

yTc

RcG =

B

A

Sc

Rc

r u

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u= Tc

Rc

r Sc

Rc

y

17. 2-DOF Controller

yTc

RcG =

B

A

Sc

Rc

r u

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u= Tc

Rc

r Sc

Rc

y

It is easy to arrive at the following relation between r andy.

17. 2-DOF ControlleryTc

RcG =

B

A

Sc

Rc

r u

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u= Tc

Rc

r Sc

Rc

y

It is easy to arrive at the following relation between r andy.

y = Tc

Rc

B/A

1 +BSc/ARcr

17. 2-DOF ControlleryTc

RcG =

B

A

Sc

Rc

r u

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u= Tc

Rc

r Sc

Rc

y

It is easy to arrive at the following relation between r andy.

y = Tc

Rc

B/A

1 +BSc/ARcr =

BTc

ARc+BScr

17. 2-DOF ControlleryTc

RcG =

B

A

Sc

Rc

r u

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u= Tc

Rc

r Sc

Rc

y

It is easy to arrive at the following relation between r andy.

y = Tc

Rc

B/A

1 +BSc/ARcr =

BTc

ARc+BScr

Errore, given byr y is given by

17. 2-DOF ControlleryTc

RcG =

B

A

Sc

Rc

r u

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u= Tc

Rcr

Sc

Rcy

It is easy to arrive at the following relation between r andy.

y = Tc

Rc

B/A

1 +BSc/ARcr =

BTc

ARc+BScr

Errore, given byr y is given by

e =

1

BTc

ARc+BSc

r

17. 2-DOF ControlleryTc

RcG =

B

A

Sc

Rc

r u

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u= Tc

Rcr

Sc

Rcy

It is easy to arrive at the following relation between r andy.

y = Tc

Rc

B/A

1 +BSc/ARcr =

BTc

ARc+BScr

Errore, given byr y is given by

e =

1

BTc

ARc+BSc

r =

ARc+BSc BTc

ARc+BScr

Digital Control 17 Kannan M. Moudgalya, Autumn 2007

18. Offset-Free Tracking of Steps with Integral

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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)

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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 e(n) =

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limn

e(n) =

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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 e(n) = limz 1 A(z)Rc(z) +B(z)Sc(z)B(z)Tc(z)

( ) ( ) ( ) ( )

z

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n( )

z1 z A(z)Rc(z) +B(z)Sc(z) z 1

Because the controller has an integral action,

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 e(n) = limz 1 A(z)Rc(z) +B(z)Sc(z)B(z)Tc(z)

A( )R ( ) B( )S ( )

z

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n( )

z1 z A(z)Rc(z) +B(z)Sc(z) z 1

Because the controller has an integral action, Rc(1) = 0:

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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 e(n) = lim1

z 1 A(z)Rc(z) +B(z)Sc(z)B(z)Tc(z)

A( )R ( ) + B( )S ( )

z

1

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n( )

z1 z A(z)Rc(z) +B(z)Sc(z) z 1

Because the controller has an integral action, Rc(1) = 0:

e() = Sc(z) Tc(z)

Sc(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)

R(z)

limn

e(n) = lim1

z 1 A(z)Rc(z) +B(z)Sc(z)B(z)Tc(z)

A( )R ( ) + B( )S ( )

z

1

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n z1 z A(z)Rc(z) +B(z)Sc(z) z 1

Because the controller has an integral action, Rc(1) = 0:

e() = Sc(z) Tc(z)

Sc(z)

z=1

=Sc(1) Tc(1)

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)

R(z)

limn

e(n) = limz1

z 1

z

A(z)Rc(z) +B(z)Sc(z)B(z)Tc(z)

A(z)R (z) + B(z)S (z)

z

z 1

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n z1 z A(z)Rc(z) +B(z)Sc(z) z 1

Because the controller has an integral action, Rc(1) = 0:

e() = Sc(z) Tc(z)

Sc(z)

z=1

=Sc(1) Tc(1)

Sc(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)

R(z)

limn

e(n) = limz1

z 1

z

A(z)Rc(z) +B(z)Sc(z)B(z)Tc(z)

A(z)R (z) + B(z)S (z)

z

z 1

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n z1 z A(z)Rc(z) +B(z)Sc(z) z 1

Because the controller has an integral action, Rc(1) = 0:

e() = Sc(z) Tc(z)

Sc(z)

z=1

=Sc(1) Tc(1)

Sc(1)

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

Tc =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)

R(z)

limn

e(n) = limz1

z 1

z

A(z)Rc(z) +B(z)Sc(z)B(z)Tc(z)

A(z)R (z) + B(z)S (z)

z

z 1

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n z1 z A(z)Rc(z) +B(z)Sc(z) z 1

Because the controller has an integral action, Rc(1) = 0:

e() = Sc(z) Tc(z)

Sc(z)

z=1

=Sc(1) Tc(1)

Sc(1)

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

Tc =ScTc =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)

R(z)

limn

e(n) = limz1

z 1

z

A(z)Rc(z) +B(z)Sc(z)B(z)Tc(z)

A(z)R (z) + B(z)S (z)

z

z 1

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n z1 z A(z)Rc(z) +B(z)Sc(z) z 1

Because the controller has an integral action, Rc(1) = 0:

e() = Sc(z) Tc(z)

Sc(z)

z=1

=Sc(1) Tc(1)

Sc(1)

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

Tc =ScTc =Sc(1)

Tc(1) =Sc(1)Digital Control 18 Kannan M. Moudgalya, Autumn 2007