Digital Controls & Digital Filters Lectures 15 & 16 15-16...Closed-Loop Systems-Signal Flow Graph Closed-Loop Systems-State-Space Representation Digital Controls & Digital Filters

Closed-Loop Systems-Signal Flow Graph Closed-Loop Systems-State-Space Representation

Digital Controls & Digital FiltersLectures 15 & 16

M.R. Azimi, Professor

Department of Electrical and Computer EngineeringColorado State University

Spring 2017

M.R. Azimi Digital Control & Digital Filters


Sampled Signal Flow Graph

Sampled Signal Flow Graph (SSFG) together with the Mason’s Gain Formulaprovide:

a Systematic approach for finding closed loop transfer function,

b Reliable for multi-loop and multi-sampler systems.

Steps

Draw original SFG (OSFG) from block diagram.

Write equations for all non-input nodes in OSFG as a function of inputvariables and take starred transform.Note: Actual inputs and outputs of samplers are considered as InputNodes.

Form sampled SFG (SSFG) and apply Mason’s gain to obtain transferfunction.

If desired to find C(s) or C(z,m), combine OSFG and SSFG by connectingoutput of samplers from SSFG to OSFG using branches with unity gain(Composite SFG or CSFG). Then apply Mason’s gain to CSFG for C(s) asoutput.



Sampled Signal Flow Graph-Cont.

Example 1Consider

Draw OSFG from block diagram:

Input nodes: R,E∗

Output nodes: E,CC = GE∗ star =⇒ C∗ = G∗E∗

E = R−HGE∗ star =⇒ E∗ = R∗ −HG∗E∗

Draw SSFG from these Eqs. and apply Mason’sgain:

SSFG

C∗(s) = G∗R∗

1+GH∗ =⇒ C(z) = G(z)R(z)

1+GH(z)




To find C(s) and C(z,m) form CSFG (C(s) as output node):

C(s) = R∗(s)G(s)

1+GH∗(s)

Thus,C(z,m) = G(z,m)R(z)

1+GH(z)




Example 2: Use SSFG method to find expressions for C(z), C(s) and C(z,m).

Find C(z), C(s) and C(z,m) using SSFG method.

Input: R,E∗

Non input: E,CC = R+G2E

∗ =⇒ C∗ = R∗ +G∗2E

∗

E = RG1 −HG2E∗ −HR =⇒ E∗ = RG1

∗ −HG2∗E∗ −HR∗




From SSFG:

C∗ =(RG1

∗−HR∗)G∗

2+R∗(1+G2H∗)

1+G2H∗ =

R∗ +(RG1

∗−HR∗)G∗

2

1+G2H∗

C(z) = R(z) + (RG1(z)−HR(z))G2(z)

1+G2H(z)

From CSFG:

C(s) = R(1+G2H∗)+(RG1

∗−HR∗)G2

1+G2H∗

=⇒ C(s) = R(s) + (RG1∗−HR

∗)G2(s)

1+G2H∗

C(z,m) = R(z,m) +(

RG1(z)−HR(z)

1+G2H(z)

)G2(z,m)




Example 3: Use SSFG method to find expressions for C(z), C(s) and C(z,m).

Note: This block diagram can be slightly simplified by removing one of thesamplers since

as A = DE∗ =⇒ A∗ = D∗E∗

Using this property, OSFG can be put together very easily.




Inputs: R, E∗

Outputs: C, Y, EE = R

1+G2− D∗G1G2E

∗

1+G2

Y = R1+G2

+ D∗G1E∗

1+G2

C = RG2

1+G2+ D∗G1G2E

∗

1+G2

E∗ =[

R1+G2

]∗−D∗E∗

[G1G2

1+G2

]∗Y ∗ =

[R

1+G2

]∗+D∗E∗

[G1

1+G2

]∗C∗ =

[RG2

1+G2

]∗+D∗E∗

[G1G2

1+G2

]∗




From SSFG

C∗ =D∗[

G1G21+G2

]∗[R

1+G2

]∗1+D∗

[G1G21+G2

]∗ +

[RG21+G2

]∗(1+D∗

[G1G21+G2

]∗)

1+D∗[

G1G21+G2

]∗or

C(z) =D(z)Z

[G1G21+G2

]Z[

R1+G2

]1+D(z)Z

[G1G21+G2

] + Z[

RG2

1+G2

]




From CSFG and noting that we have two non-touching loops with gains

L11 = −G2 and L12 = −D∗[G1G2

1+G2

]∗, the ∆ of the CSFG will contain their

products L11L12 = G2D∗[G1G2

1+G2

]∗i.e.

∆ = 1− L11 − L12 + L11L12 = G2 +D∗[G1G2

1+G2

]∗+G2D

∗[G1G2

1+G2

]∗Thus C(s) is,

C(s) =RG2

(1+D∗

[G1G21+G2

]∗)+[

R1+G2

]∗D∗G1G2

1+G2+D∗[

G1G21+G2

]∗+G2D∗

[G1G21+G2

]∗

=[

RG2

1+G2

]+

[G1G21+G2

]D∗[

R1+G2

]∗1+D∗

[G1G21+G2

]∗and

C(z,m) = Zm

[RG2

1+G2

]+

Zm

[G1G21+G2

]D(z)Z

[R

1+G2

]1+D(z)Z

[G1G21+G2

]



State-Space Representation-Closed-Loop Systems

So far, we utilized the SSFG method to find closed-loop transfer function. Nowsuppose that the discrete-time version of the plant state equations are given by:{

x1(n+ 1) = A1x1(n) +B1u(n)c(n) = C1x1(n)

where x1(n) is the plant state vector (discrete). We want to find the stateequations for the typical closed-loop system below.

From transfer function D(z) = M(z)E(z) , we obtain discrete state equation for

digital controller.{x2(n+ 1) = A2x2(n) +B2e(n)m(n) = u(n) = C2x2(n)

where x2(n) is the state vector of digital controller (discrete).




Note: Since both plant and controller are represented by discrete-time stateequations, one may ignore hold device and assume digital input/outputs.

Now augment these equations such that the input is the reference signal r(n)(discrete version) and output is c(n).

e(n) = r(n)− c(n)x1(n+ 1) = A1x1(n) +B1C2x2(n)x2(n+ 1) = A2x2(n) +B2(r(n)− c(n)) = A2x2(n)−B2C1x1(n) +B2r(n)

State-space equations for closed-loop system are:[x1(n+ 1)x2(n+ 1)

]︸︷︷︸

x(n+1)

=

[A1 B1C2

−B2C1 A2

]︸︷︷︸

A

[x1(n)x2(n)

]︸︷︷︸

x(n)

+

[0B2

]︸︷︷︸

B

r(n)

c(n) =[C1 0

]︸︷︷︸C

[x1(n)x2(n)

]




Example: Consider the system below where D(z) = 11−0.9z−1 and

Gp(s) = 6.4(s+1)(s+0.2) .

a Find state equations of the plant and its discrete version,b State equations of closed-loop system.

First convert Gp(s) to state-space. Since real and distinct poles, apply PFE

Gp(s) = C(z)U(z) = 6.4

(s+1)(s+0.2) = −8s+1 + 8

s+0.2

Thus,

v(t) =

[−1 00 −0.2

]︸︷︷︸

Ac

v(t) +

[11

]︸︷︷︸

Bc

u(t)

c(t) =[−8 8

]︸︷︷︸Cc

v(t)




We then convert these to discrete-time state equations using

A1 = eAcT = φc(T ) =

[e−T 0

0 e−0.2T

]=

[e−0.25 0

0 e−0.05

]B1 = A−1

c (A1 − I)Bc =

[0.210.25

]C1 = Cc x(n+ 1) =

[0.79 0

0 0.95

]x(n) +

[0.210.25

]u(n)

c(n) =[−8 8

]x(n)

For digital controller D(z) = U(z)E(z) = 1

1−0.9z−1 , difference equation is

u(n) = 0.9u(n− 1) + e(n)




Draw state diagram of discrete version of plant and combine with SFG of digitalcontroller to form overall SFG. Then assign states.

x1(n+ 1) = 0.79x1(n) + 0.21u(n)x2(n+ 1) = 0.95x2(n) + 0.25u(n)x3(n+ 1) = 0.9x3(n) + e(n) (digital controller state i.e. x3(n) = u(n− 1))u(n) = x3(n+ 1) = 0.9x3(n) + e(n)e(n) = r(n)− 0.03c(n)c(n) = −8x1(n) + 8x2(n)e(n) = r(n) + 0.24x1(n)− 0.24x2(n)

x1(n+ 1) = 0.84x1(n)− 0.05x2(n) + 0.19x3(n) + 0.21r(n)x2(n+ 1) = 0.06x1(n) + 0.89x2(n) + 0.225x3(n) + 0.25r(n)x3(n+ 1) = 0.24x1(n)− 0.24x2(n) + 0.9x3(n) + r(n)c(n) = −8x1(n) + 8x2(n)


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Digital Controls & Digital Filters Lectures 15 & 16 15-16...Closed-Loop Systems-Signal Flow Graph Closed-Loop Systems-State-Space Representation Digital Controls & Digital Filters