EL 625 Lecture 2 1

EL 625 Lecture 2

State equations of finite dimensional linear systems

Continuous-time:

x(t) = A(t)x(t) + B(t)u(t)

y(t) = C(t)x(t) + D(t)u(t)

Discrete-time:

x(tk+1) = A(tk)x(tk) + B(tk)u(tk)

y(tk) = C(tk)x(tk) + D(tk)u(tk)

state x(t) - vector of length n 1input u(t) - r 1output y(t) - m 1

A,B,C and D are matrices of sizes n n, n r,m n and m rrespectively.

If A,B,C,D are functions of time timevaryingIf these matrices are constant with time time-invariant

EL 625 Lecture 2 2

State Differential equations of circuits

basic elements resistor, capacitor, inductor

Resistor:

vR(t) = R(t)iR(t)

vR(t) : the voltage across the resistor

iR(t) : the current through the resistor

R(t) : the resistance

A resistor is a zero-memory element.

BBB

BBBB

BBBB

BBBB

R

h h

+ vR

-iR

Any circuit with only resistors (a purely resistive network) has

zero-memory and is of zero order (needs no states to describe it).

EL 625 Lecture 2 3

Capacitor: iC(t) = qC(t) , qC(t) = C(t)vC(t) where:

iC(t) : the current through the capacitor

qC(t) : the electrical charge in the capacitor

vC(t) : the voltage across the capacitor

C(t) : the capacitance

C(t)dvC(t)dt = iC(t) dC(t)dt vC(t)If the capacitance does not change with time,

C dvC(t)dt = iC(t)

C

h h

+ vC

-iC

EL 625 Lecture 2 4

Inductor: vL(t) = (t) , (t) = L(t)iL(t) where:

iL(t) : the current through the inductor

vL(t) : the voltage across the inductor

(t) : the flux stored in the inductor

L(t) : the inductance

L(t)diL(t)dt = vL(t) dL(t)dt iL(t)If the inductance does not change with time,

LdiL(t)dt = vL(t)

L

h h

+ vL

-iL

EL 625 Lecture 2 5

Example:

LC1

PPPPPPPPP

PP

R1

C2

PPPPPPPPP

PP

R2

&%

'$

&%

'$

+

e1(t)

e2(t)

+

+

+ +

-

iL(t)

vL(t)

? ?i1(t) i2(t)

v2(t)

v1(t)

Applying Kirchoffs current and voltage laws,

iL = i1 + i2

e1 = vL + i1R1 + v1

e1 = vL + v2 + i2R2 + e2

From the terminal relationships of the capacitors and the inductor,

v1 =1C1i1

v2 =1C2i2

iL =1LvL

EL 625 Lecture 2 6

The inputs are e1 and e2. The output is the voltage across the

inductor, vL. Choosing as our states, iL, v1 and v2,

x =

iL

v1

v2

, u =

e1

e2

, y = [vL]

Need to express x in terms of x and the inputs, e1 and e2,. . .

iL =1LvL

But, vL is not a state variable. . . we need to express vL in terms of

the state variables and the inputs. . .

vL = iL R1R2R1 +R2

+v1 R2R1 + R2

+v2 R1R1 + R2

+e1+e2 R1R1 +R2

iL = iL

R1R2L(R1 + R2)

+ v1 R2L(R1 + R2)

+ v2 R1L(R1 +R2)

+e1

1L

+ e2 R1L(R1 + R2)

EL 625 Lecture 2 7

Similarly,

i1 = iL R2R1 +R2

+ v1 1R1 +R2

+ v2 1R1 +R2

+ e2 1R1 + R2

v1 =1C1i1

= iL

R2C1(R1 + R2)

+ v1 1C1(R1 + R2)

+ v2 1C1(R1 + R2)

+e2

1C1(R1 +R2)

i2 = iL R1R1 +R2

+ v1 1R1 +R2

+ v2 1R1 +R2

+ e2 1R1 + R2

v2 =1C2i2

= iL

R1C2(R1 + R2)

+ v1 1C2(R1 + R2)

+ v2 1C2(R1 + R2)

+e2

1C2(R1 +R2)

EL 625 Lecture 2 8

x = Ax + Bu

y = Cx + Du

where

A =

R1R2L(R1+R2)

R2L(R1+R2)

R1L(R1+R2)

R2C1(R1+R2)

1C1(R1+R2)

1C1(R1+R2)

R1C2(R1+R2)

1C2(R1+R2)

1C2(R1+R2)

B =

1L

R1L(R1+R2)

0 1C1(R1+R2)

0 1C2(R1+R2)

C =[R1R2R1+R2

R2R1+R2

R1R1+R2

]

D =[

1 R1R1+R2]

EL 625 Lecture 2 9

State Differential Equations of Mechanical Systems

Spring:

fK(t) = K(t)[z2(t) z1(t)]Hookes Law where:

z1(t) and z2(t) : the displacements of the two ends of the spring

fK(t) : the force applied

K(t) : the spring constant K-

z1 z2 fK

Damping Element:

fD(t) = D(t)[v2(t) v1(t)]where:

v1(t) and v2(t) : the velocities of the two ends of the damping element

fD(t) : the force applied

D(t) : the damping coefficient.

D

-v1 v2fD

EL 625 Lecture 2 10

Mass:

M(t)dv(t)dt = fM(t) dM(t)dt v(t)where:

v(t) : the velocity of the mass

fM(t) : the force applied

M(t) : the mass.If mass does not change with time,

M dv(t)dt = fM(t)

.

M -v

fM

EL 625 Lecture 2 11

Example:

D

M1 KM2

- -x1 x2

- f(t)

f (t) is the input and x2(t) is the output. u = [f (t)],y = [x2]

Choose as the state variables

x1, x2, v1 and v2 where

v1 = x1 and v2 = x2.

= x =

x1

x2

v1

v2

M2v2 = f(t) + K(x1 x2)M1v1 = K(x2 x1) + D(v1)

EL 625 Lecture 2 12

x = Ax +Bu

y = Cx + Du

where

A =

0 0 1 0

0 0 0 1

KM1

KM1

DM1

0

KM2

KM2

0 0

B =

0

0

0

1M2

C =

[0 1 0 0

]D = [0]

EL 625 Lecture 2 13

Choice of state variables is not unique

Let xa(t) be a valid set of state variables with

xa = Aaxa +Bau

y = Caxa +Dau. (1)

Consider xb(t) = Txa(t) where T is an nn nonsingular matrix.xb = T xa = TAaT1xb + TBau

y = CaT1xb + Dau. (2)

xb(t) is also a valid set of state variables!

xb = Abxb + Bbu

y = Cbxb +Dbu (3)

where:

Ab = TAaT1

Bb = TBb

Cb = CaT1

Db = Da (4)

Similarity Transformation: xb = Txa

EL 625 Lecture 2 14

Convenient choice of state variables :

inductor currents and capacitor voltages for fixed net-

works

inductor fluxes and capacitor charges for time-varying net-

works

differences in displacements of the ends of springs

from their equilibrium positions and velocities of masses

for mechanical systems

Simulation Diagrams:

Basic Simulator Elements:

1. Dynamic element

(a) integrator : for analog systems

-

-u(t) y(t) = y(t0) + tt0u() d

y = D1u

(b) delay : for discrete-time systems

- -u(tk) y(tk) = u(tk1)

y = E1u

EL 625 Lecture 2 15

2. Summing element - adder

&%

'$

--+u1(t)1+u2(t) ... 6+

ur(t)

y(t) =ri=1

ui(t)

3. Scaling Element (Amplifier or Attenuator)

-

HHHHH -u(t) y(t) = K(t) u(t)K(t)

Minimal Realization: Fewest possible number of dynamic el-

ements.

Convenient choice of state variables Outputs of integrators

and delay elements

Example:

y... + 3ty + 2y + (t)y = u + etu + u

D3y + 3tD2y + 2Dy + (t)y = D2u + etDu + u

(D is the derivative operator)

y = D3{D2u 3tD2y + etDu 2Dy + u (t)y}

y = D1{D2(D2u 3tD2y) + D1{D1(etDu 2Dy)

EL 625 Lecture 2 16

+D1{u (t)y}}}= D1

{uD2(3tD2y) + D1{D1(etDu) 2y

+D1{u (t)y}}}

Using integration by parts,

D2(3tD2y) = D1(3tDy D1(3Dy))= D1(3tDy 3y)= 3ty D1(3y)D1(3y)= 3ty D1(6y)

D1(etDu) = etu +D1(uet)

Thus,

y = D1{u 3ty +D16y + D1{etu + D1(uet) 2y

+D1{u (t)y}}}= D1

{u 3ty + D1{etu + +4y +D1{u + uet (t)y}}}

EL 625 Lecture 2 17

Simulation Diagram:

- -

HHHHHH

1 + et

-

- -

- -

-

?

?

AAAAAA

(t)?

?

AAAAAA

4?

?

AAAAAA

3t

6

6

6

AAAAAA

et

u y+ + + +

+ +x1x2x3

Choosing the outputs of the integrators as the states, we have

x1 = u 3tx1 + x2x2 = 4x1 + x3 + etu

x3 = u(1 + et) (t)x1y = x1

x =

x1

x2

x3

x = A(t)x + B(t)u

y = C(t)x + D(t)u

A(t) =

3t 1 04 0 1

(t) 0 0

; B(t) =

1

et

1 + et

; C(t) =

1

0

0

T

; D(t) = [0]

EL 625 Lecture 2 18

Example:

y(k + 3) + 3ky(k + 2) + 2y(k + 1) + (k)y(k) = u(k + 2) + eku(k + 1)

+u(k)

E3y(k) + 3kE2y(k) + 2Ey(k) + (k)y(k) = E2u(k) + ekEu(k)

+u(k)

where E is the delay operator.

y(k) = E3{E2u(k) 3kE2y(k) + ekEu(k) 2Ey(k) + u(k)

(k)y(k)}= E1

{E2(E2u(k) 3kE2y(k)) + E1{E1(ekEu(k) 2Ey(k))

+E1{u(k) (k)y(k)}}}

E2(3kE2y(k)) = 3(k 2)y(k)E1(ekEu(k)) = ek1u(k)

Thus,

y(k) = E1{u(k) 3(k 2)y(k) + E1{e(k1)u(k) 2y(k)

+E1{u(k) (k)y(k)}}}

EL 625 Lecture 2 19

Simulation Diagram:

-

-- -- --

?

?

AAAAAA

(k)?

?

AAAAAA

2?

?

AAAAAA

3(k 2)

6

6

6

AAAAAA

e(k1)

u(k) y(k)+ + +

+ +x1x2x3

Choosing the outputs of the delay elements as the states, we have

x1(k + 1) = u(k) 3(k 2)x1(k) + x2(k)x2(k + 1) = e(k1)u(k) 2x1(k) + x3(k)x3(k + 1) = u(k) (k)x1(k)

y(k) = x1(k)

x(k) =

x1(k)

x2(k)

x3(k)

x(k + 1) = A(k)x(k) + B(k)u(k)

y(k) = C(k)x(k) + D(k)u(k)

EL 625 Lecture 2 20

A(k) =

3(k 2) 1 02 0 1(k) 0 0

B(k) =

1

e(k1)

1

C(k) =[

1 0 0]

D(k) = [0]

EL 625 Lecture 2 21

Simpler method if no derivatives of the input are in the equation:

y(n)+n1(t)y(n1)+n2(t)y(n2)+. . .+1(t)y(1)+0(t)y = (t)u(t)

(y(i) ith derivative of y(t))Choose,

x1 = y

x2 = y(1)

x3 = y(2)

...

xn = y(n1)

x =

x1

x2

x3...

xn

x1 = x2

x2 = x3...

xn1 = xn

xn = n1(t)xn n2(t)xn1 . . . 0(t)x1+(t)u(t)

y = x1

EL 625 Lecture 2 22

x =

0 1 0 . . . 0... . . . . . . . . . ...... . . . . . . 0

0 . . . . . . 0 1

0 1 . . . n2 n1

A

x +

0

0......

B

u

A is in companion matrix form.

y =[

1 0 0 . . . 0]

C

x + [0]D

u

Another method: Let 0, 1, . . . , n1 and be constants.

Dny + n1Dn1y + n2Dn2y + . . . + 1Dy + 0y = u

Dny = n1Dn1y n2Dn2y . . . 1Dy 0y + uy = Dn

{n1Dn1y n2Dn2y . . . 1Dy 0y + u}= D1

{ n1y + D1{ n2y + D1{ . . .+D1(0y + u ) . . . }

}}

n

EL 625 Lecture 2 23

z1 = n1z1 + z2z2 = n2z1 + z3

...

zn1 = 1z1 + xnzn = 0z1 + uy = z1

z =

z1

z2...

zn1zn

z =

n1 1 0 . . . . . . 0n2 0 1 0 . . . 0

... ... . . . . . . . . . ...

2 ... 0 1 01 0 . . . . . . 0 10 0 . . . . . . . . . 0

A

z +

0

0...

0

0

B

u

y =[

1 0 0 . . . 0]

C

z + [0]Du

This method gave different A and B matrices . . . - z and x are

related through a similarity transformation.

EL 625 Lecture 2 24

z = Tx

where

T =

1 0 . . . . . . 0

n1 1 0 . . . ...... . . . . . . . . . ...

2 . . . n1 1 0

1 . . . . . . n1 1

It can be checked that A = TAT1, B = TB, C = CT1 and

D = D.

The D matrix does not change under a similarity transforma-

tion.

A non-singular D matrix = the impulse response has an impulse.

EL 625 Lecture 2 25

MIMO systems:

y1 + t2y2 + y1 + ty1 + y2 = tu1 + u2

y2 + ty1 + y2 y1 = tu1 + u2From the first equation,

D2y1 + t2Dy2 + Dy1 + ty1 + y2 = tu1 + Du2

y1 = D2{ t2Dy2 Dy1 +Du2 + tu1 ty1 y2}

= D1{D1(t2Dy2 Dy1 + Du2) + D1{tu1 ty1 y2}

}

D1(t2Dy2) = t2y2 D1(2ty2)Thus,

y1 = D1{u2 y1 t2y2 + D1{2ty2 + tu1 ty1 y2}

}Similarly, from second equation,

D2y2 + tDy1 + y2 y1 = tDu1 + u2

y2 = D2{tDu1 tDy1 + u2 + y1 y2

}= D1

{D1(tDu1 tDy1) + D1{u2 + y1 y2}

}

EL 625 Lecture 2 26

D1(tDu1) = tu1 D1u1D1(tDy1) = ty1 D1y1

y2 = D1{tu1 ty1 + D1{2y1 y2 u1 + u2}

}

State-space realization:

x1 = x1 +x2 t2x3 +u2x2 = tx1 +(2t 1)x3 +tu1x3 = tx1 +x4 +tu1x4 = 2x1 x3 u1 +u2y1 = x1

y2 = x2

EL 625 Lecture 2 27

Simulation Diagram

-

-

-

-

-

-

-

-

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?

AAAAAA

t?

?

AAAAAA

2

?

?

?

AAAAAA

t

6

6

AAAAAA

2t 16

6

AAAAAA

t2

- -

HHHHHH

t

6

6

AAAAAA

t

@@R

6

@@@R

y1

y2

u1

u2

x1

x3

x2

x4

+

+

+

+

++

+

+