Ccip Lessons

8/3/2019 Ccip Lessons

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Implementation of digital control system

C(kT) = ∑m i=0 a(i)r[(k-i)T] + ∑ n j=1 b(j)c[(k-j)T]

The above equation represents a control law in difference equation form. Performance is directly related to

representations of coefficients a(i), b(j); processes involved in arithmetic operations; data input (r ) and output (c

), i/o for multiple control tasks for computer.

Control logic: symbolic logical design and Boolean algebra

Computer architecture for control: control algorithm input, output, calculation, control, information

storage. Speed, power etc. are considered – microcontrollers are specialized.

Software for control: apart from solving difference equations, human command interfacing through control

panels and switching discrete event control like alarms and timers, device calibration, environmental protection,

checks/audits required. System integration in time and space, scheduling are tough. Developing algorithms and

implementing on languages – critical task.

Software engineering in control: defining SDLC:- requirement specs-design-development-integrate-test-

maintain. Use of DFD etc. to depict the system.

Real time design for digital control: calculating time for each digitization step is important. Interrupt latency,

context switching are to be worked out.

Software for direct design method: use of structural programming, numerical techniques

Real time scheduling: complex industrial process, aircraft control etc. require system integration that involves

scheduling various jobs.

Operating system: managing resources of computer system, schedule periodic tasks, coordinate

communication between tasks, integrity of control system etc.

Sampling time selection: response to given control input, external disturbance, sensitivity to plant model etc.

are tuned to desired ones- optimal sampling, multi-rate sampling (more complex)



Introduction to modern control theory

Conventional classical control theory is based on frequency domain. It is only applicable to single-input-single-

output (SISO) systems that are linear time-invariant (LTI).

Modern control theory is applicable to time-varying as well as non-linear multi-input multi-output (MIMO)

systems.

State variable: Smallest set of variables whose knowledge at t=t 0 together with subsequent inputs at t≥t0

completely determines the behaviour of the system for any time t≥ t0

Meaning of linear time invariant system:

Superposition theorem:

r1 (t) c1 (t) and r2 (t) c2 (t)

a1 r1 (t) + a2 r2 (t) a1 c1 (t) + a2 c2 (t)

c(t) = f{r(t)} is the time domain transfer function.

c(kT) = f{ a1 r1 (kT) + a2 r2 (kT)} = a1 f{ r 1 (kT)} + a2 f{r2 (kT)}

Ex. c(t) = r(t)+α. Then, c1= r1+α and c2= r2+α

For r = a1 r1 + a2 r2

c(actual) = a1 r1 + a2 r2 + a1α + a2α.

c(reqd for SPT) = a1 r1 + a2 r2 + α

Hence, the function is non-linear.

c(kT)=r^2(kT) is non-linear.

c(t) = dr(t)/dt i.e. c(kT) = (1/T)(r(kT)-r(kT-T))

Time invariance:

When c(t-τ)=f{r(t-τ)} or c((k-p)T) = f{r((k-p)T)}

System TF followed by time delay Ξ Time delay followed by system TF

Ex. c (t) + c(t) = t r(t)



Here, c (t-τ) + c(t-τ) = (t-τ) r(t-τ) taking first the system TF, then the delay.

Again, c (t-τ) + c(t-τ) = t r(t-τ) taking first the delay, then the system TF.

∴ This is not time invariant.

Ex. c(kT) = kT r(kT)

Ex. c((k+1)T)-c(kT) = r(kT)

Let x1 , x2, x3 , … , xn be the state variables of a system and the state vector is denoted as X(t) in continuous time

domain at time t. Also, u(t) is the input variable at time t.

Then the system can be described in terms of differential equations involving the rate of change of state

variables with respect to time.

ẋ =Ax(t) + Bu(t)

ẍ ẏ ϊ ϊ ύ ϋ

L + Ri + vί c = v and Cύc = i describes the differential equations governing a L-R-C circuit.

Here, = di/dt andί ύc = d vc /dt are the rate of change of current and voltage across the capacitance respectively.

Taking x1 = i(t) and x2 = vc(t) with u(t)=applied voltage v; one can write ẋ=Ax(t) + Bu(t)

One can easily obtain A=[-R/L, -1/L; 1/C, 0] and B=[1/L; 0].

Electric Motor Transfer Function:

Simple harmonic motion

Spring-mass system

Free falling body

State-Space methods in continuous domain

Consider n-th order differential equation describing a system:

y(n) + a1 y(n-1) + …+ an-1ẏ + any = u



Knowledge of y(0), ẏ(0), ÿ(0), …, y(n-1)(0)together with u(t) for t≥ 0 determines completely the future

behaviour of the system. Hence y(t), ẏ(t), ÿ(t), …, y(n-1)(t) is a set of SVs.

Mathematically convenient to express the system as a set of simultaneous first order differential equations

involving x1=y, x2= ẏ , x3=ÿ, … , xn = y(n-1)

(t)i.e.

ẋ1 = x2 ; ẋ2 = x3; ẋn-1= xn and ẋn = - an x1- an-1 x2 - … - a1 xn-1+ u

Now, one can express the system as ẋ =Ax(t) + Bu(t) ; y=C x(t) where

A= [ 0 1 0 0 …. 0

0 0 1 0...... 0

………………...

-an -an-1 …..… - a1 ]

B=[ 0

0

…

1]

C = [1 0 0 … 0]

Consider a system: y(3) + 6ÿ +11ẏ + 6y = 6u

A= [0 1 0; 0 0 1; -6 -11 -6]B = [0; 0; 6], C=[1 0 0]

Non-uniqueness of state variables

Write the above system in s-domain transfer function (TF): Y(s)/U(s) = 6/(s3 + 6s2 + 11s + 6)

Expanding partial fractions, above TF = 3/(s+1) – 6/(s+2) + 3/(s+3) = X 1(s) + X2(s) + X3(s)

Hence, ẋ1= - x1 + 3 u ; ẋ2 = -2x2 - 6 u; ẋ3 = -3 x3 + 3 u

and y = x1 + x2 + x3

so that A = [-1 0 0; 0 -2 0; 0 0 -3], B = [3; -6; 3], C = [1 1 1]

Multiple forcing function:

y(n) + a1 y(n-1) + …+ an-1ẏ + any = b0 u(n) + b1 u(n-1) + … bn-2 ü + bn-1 ύ + bnu



Now, for unique solution, derivatives of u must be eliminated from state equations. Define:

x1=y- β0u ; x2= ẏ- β0ύ - β1u = ẋ1- β1u ; x3= ÿ - β0ü - β1ύ - β2 u = ẋ2 - β2 u ; … , xn = ẋn-1 - βn-1 u

Solving, we get β0 = b0; β1 = b1- a1 β0 ; β2 = b2- a1 β1- a2 β0;

βn = bn- a1 βn-1- … - an-1 β1 - an β0

Hence, A = same as earlier; C = same as earlier

B = [β1 ; β2 ; … ; βn-1 ; βn ] ; D = β0 = b0

Derivatives only affect the B matrix. The Transfer function has introduced zeroes.

Y(s)/U(s) = ( b0 sn + b1 sn-1 + … + bn-1 s + bn )/( sn + a1 sn-1 + … + an-1 s + an )

Ex. y(3) + 18 ÿ + 192 ẏ + 640y = 160 ύ + 640 u

Y(s)/U(s) = {4(s+4)/(s+16)}*{40/s(s+2)}

A= [0 1 0; 0 0 1;-640 -192 -18]; B = [0; 160; -2240]; C=[1 0 0]; D=[640]

Solution of the homogeneous state equation:

ẋ = Ax solves to x = exp (At) x(0) = x(0){ I + At + …. + (1/k!)A k tk + … } converges for all t.

exp(-At) is inverse of exp(At). Derivative of exp(At) = A exp(At)

Laplace transform approach:

X(s) = (sI-A) -1 x(0)

Then, x(t) = L-1 {(sI-A) -1 }

(sI-A) -1 = I/s + A/s2 + A2 /s3 + ….

L-1 {(sI-A) -1 }= I + At + (1/2!)A2t2 + (1/3!)A3 t3 + …. = exp(At)

State transition matrix Φ(t) Phi(t):

x(t) = Φ(t)x(0)i.e. Φ(t) = exp(At)

Φ(t1 + t2)= exp(A(t1+ t2))= exp(A(t1)) exp(A(t2))= Φ(t1 ) Φ(t2)

Φ(t)contains the Eigenvalues exp(λit) for non-repetitive roots.

Φ(t)contains texp(λ jt), …, tm exp(λ jt) for m-times repeating roots.

Also, (Φ(t) ) n= Φ(nt) and Φ(t2 - t1) Φ(t1 - t0) = Φ(t2 - t0) = Φ(t1 - t0) Φ(t2 - t1)



This shows the power of state transition matrix as it simply transforms the initial condition.

Ex. A = [0 1; -2 -3] Then sI-A = [s, -1; 2, s+3]

{(sI-A) -1 } = {1/(s+1)(s+2)} [ s+3, 1; -2, s]

L-1 {(sI-A) -1 } = [ 2exp(-t) –exp(-2t), exp(-t) –exp(-2t); -2exp(-t)+2exp(-2t), -exp(-t)+2exp(-2t)]

Φ-1 (t) = Φ(-t) gives an easy way of inverting the state transition matrix.

For non-homogeneous state equation:

X(s) = (sI-A) -1 x(0) + (sI-A) -1 B U(s)

While taking inverse Laplace transform, convolution integral is required.

x(t) = Φ(t)x(0) + ∫ t0 exp(A(t-τ)Bu(τ)d(τ)

In the above example, take B=[0;1] with u(t)=1 (unit step) when x(0)=0 to get

Upon simplification, x(t) = [½ – exp(-t) + ½ exp(-2t); exp(-t)-exp(-2t)]

Relation between time and frequency domain:

ẋ 1=- 5 x1 -x2 +2u; ẋ2 = 3x1 -x2 +5u ; and y = x1 + 2x2

From this, we get G(s)=(12s+59)/(s+2)(s+4)

State-Space methods in discrete domain

1. Phase variable approach

Ex. y(k+2)+y(k+1)+0.16y(k)=u(k+1)+2u(k)

State variables: x1 (k)= y(k) and x2(k) = x1 (k+1) - u(k)

To consider x(k+1) = Gx(k)+Hu(k) and y(k)=Cx(k)+Du(k)

G = [0 1;-0.16 -1]; H = [1 1]; C = [1 0]; D = 0

T ẋ = x(k+1)-x(k) as T 0 is the connection between continuous and discrete domain

Consider n-th order difference equation describing a system:

y(k+n) + a1 y(k+n-1) + …+ an-1y(k+1) + any(k) = b0u(k+n)+ b1u(k+n-1)+…+ bn-1u(k+1)+ bnu(k)

Choose x1(k) = y(k)- h0 u(k)

x2(k) = x1(k+1) - h1 u(k)

x3(k) = x2(k+1) - h2 u(k)



xn(k) = xn-1(k+1) - hn-1 u(k)

h0 = b0; h1 = b1- a1 h0 ; h2 = b2- a1 h1- a2 h0;

hn = bn- a1 hn-1- … - an-1 h1 - an h0

G = [ 0 1 0 0 …. 0

0 0 1 0...... 0

………………...

-an -an-1 …..… - a1 ]

HT = [h1; h2 ; … ; hn-1 ; hn] ;

C = [1 0 0 … 0]

D = h0 = b0

2. Canonical variable

G(z)=Y(z)/U(z)

Use partial fraction method to get G(z) = const KD + for distinct poles Ai /z-pi

Xi (z) = U(z) /z-pi

z Xi (z)- pi Xi (z) = U(z)

xi (k+1)= pi xi (k)+u(k)

F = [p1 0 … 0 ; 0 p2 0 … 0; …; 0 0 … pn ]

HT = [1 1 … 1]; C = [A1 A2 … An]; D = KD

If the pole p1 repeats q times, the partial fraction expansion will contain

A1,q /(z-p1)q , A1,q-1 /(z-p1)q-1 ,…, A12 /(z-p1)2 , A11 /(z-p1)

A2 /(z-p2), A3 /(z-p3), …, An-q+1 /(z-pn-q+1)

X1 (z) = U(z)/(z- p1), X2 (z) = U(z)/(z-p1) 2 , …,Xq-1(z) = U(z)/(z-p1)q-1 , Xq(z) = U(z)/ (z-p1)q

Xq+1(z) = U(z)/(z-p2), Xq+1(z) = U(z)/(z-p3), …, Xn(z) = U(z)/(z-pn-q+1)

p1 0 0 ……….. ………0

1 p1 0 0 ……………... 0

F = 0 1 p1 0 0 …….…….. 0

0 0 ……1 p1 0 0 …….. 0

0 0 ………0 p2 0 0 ….. 0



0 0 ………0 0 p3 0 ….. 0

………………………….

0 0 ………0 0 0 0 pn-q+1

HT

= [1 0 0 … 0 1 1 … 1]; C = [A11 A12 … A1q A2 … An] q

When s-domain Eigenvalues are λi, Z-domain gives exp(λiT) = pi

3. Physical variable

A cascade block is Ai (z - zi)/ (z- pi) or A j / (z- p j)

For complex conjugate poles, a quadratic expression gets involved.

AiA

i+1 / (z2- 2 z +σ

ω2)

Ex. G(s) = Gzo(s) Gx (s) = KG (1-εsT)/s2(s+2)

G(z) = KG { (T+.5ε -2T -.5)z +.5-.5 ε -2T -T ε -2T }/{2(z-1)(z- ε -2T )}

Taking T = 0.1 sec, KG =2;

G(z) = (0.009365z+0.008762)/(z-1)(z-0.818731)

F = [0 1; -0.81873 1.81873]; HT = [0 0.009365]; C = [0.935611 1]

Then x(0)=0, u(0)=1, u(0.1)=2 gives y(0.1)=0.009365, y(0.2)=0.0445244

Again G(z) = 0.1/(z-1) + 0.0906349/(z-0.818731) gives

F = [1 0; 0 0.818731]; HT = [1 1]; C = [0.1 -0.090635]

G(s) = ½{1/s – 1/(s+2)} gives λ1=0, λ2=2 and G(z) reduces to same expressions.

Now, take 0.009365/(z-1) and (z+0.9356113)/(z-0.81873) as the two physical state variables.

Then, F=[1 0; 1.935611 0.818731]; HT = [0.009365 0.009356]; C = [0 1]

Solution of discrete time state equations

x(1) = Gx(0)+Hu(0); x(2) = Gx(1)+Hu(1) = G2 x(0)+GHu(0)+Hu(1)

Then, x(k) = Gk x(0)+ ∑Gk-j-1 Hu(j)

Consider state transition matrix Φ(k) = Gk

Φ(k+1)= G Φ (k) and Φ(0)=I, so that

x(k) = Φ (k) x(0)+ ∑ Φ (k-j-1)Hu(j) = Φ (k) x(0)+ ∑ Φ (j)Hu(k-j-1)



y(k) = C Φ (k) x(0) + C ∑Φ(j)Hu(k-j-1) + Du(k)

Getting to State Variable representation from Z-domain TF

G(z) = Y(z)/U(z) = {Y(z)/X(z)}{(X(z)/U(z)}

= ( zw

+ f w-1 zw-1

+ … + f 1 z + f 0 )* KD /( zn

+ dn-1 zn-1

+ … + d1 z + d0 )

KDU(z) = zn X(z) + dn-1 zn-1 X(z) + … + d1 z X(z) + d0 X(z)

Y(z) = f 0 X(z)+ f 1 z X(z) + … + f w-1 zw-1 X(z) + zw X(z)

Since Z-1 {zn E(z)} = e[(k+n)T]; taking Z-1 { } of the above expressions,

We get x(k+n) + dn-1 x(k+n-1) + … + d1 x(k+1) + d0 x(k) = KD u(k)

And y(k) = f 0 x(k)+ f 1 x(k+1) + … + f w-1 x(k+w-1) + x(k+w)

Phase variables:

x1 (k) = x(k); x2 (k) = x1 (k+1); …; xn (k) = x1 (k+n-1)

X2 (z) = zX(z); X3 (z) = z X2 (z) = z2 X(z) etc.

This gives F = [0 1 0 … 0; 0 0 1 0…0; …; -d0 -d1 … -dn-1]

HT = [0 0 … 0 KD]; C = [f 0 f 1 … f w-1 1 0 … 0]



Sample and Hold transfer function

e*p(t)= p(t) e(t) = signal e(t) modulated by pulse train p(t)

e*p(t)= (1/ ) ∑ e(kT) {u(t-kT) – u(t- kT- )} from k=0 toγ γ ∝

where e*p(t) = e(kT/ ) for kTγ ≤ t < kT+ andγ

e*p(t) = 0 for kT + γ ≤ t < (k+1)T

E*p(s) = ∑ e(kT / ){exp(-kTs)/s – exp(-(kT+ )s)/s}γ γ

= ∑ e(kT)exp(-kTs) {(1-exp(- s))/ s}γ γ ≈ ∑ e(kT)exp(-kTs)

gzo(t) = u(t) – u(t-T)

Gzo(s) = 1- ε -sT / s = G*z(s) Go(s) with Go(s) being 1/s.

Gzo(jω) = 1- ε -jωT / jω = 2ε -jωT/2 (ε jωT/2 - ε -jωT/2 )/ 2jω = 2ε -jωT/2.sin(ωT/2) / ω

Now, using T = 2π / ωs where ωs = 2πf s ;

Gzo(jω) = (2π / ωs ) {sin(πω / ωs)/(πω / ωs)}ε -j π ω / ωs

= Gzo(jω) ∠ {-(πω / ωs) + m π}

Zero order polynomial passes without distortion.

Taking z= εsT = exp(sT) or s=(1/T) ln z;

We get E(z) = E*(s) evaluated for z = ∑ e(kT) z-k

This gives a series in z, which is algebraically more convenient.

L-1(E*p (s)) = L-1( ∑ e(kT)exp(-kTs)) ≈ ∑ e(kT) δ(t-kT)

e*p(t) = e(t) δT(t) where δT(t) = ∑ δ(t-kT)

Now, s = σ + jω means Re z = εσT cos (ωT) and Im z = εσT sin (ωT)

G*(s) = (1/T) ∑ G(s+jnωs); G* (s) = C*(s)/ R*(s); G(z)=C(z)/R(z)

Mapping from s-domain to z-domain:

Interval [a,b): z = 1∠ ωT, 0 ≤ ∠ ωT ≤ (180°)-

Interval [b,c): z = z∠ (180°)-, 1≥ z > 0

Interval [c,d]: z = z∠ ±ωT, 0+≥ z > 0, (180°)-≥ ∠ ±ωT ≥ (-180°)-



Interval (d,e]: z = z∠(-180°) -, 1≥ z > 0

Interval (e,a): z = 1∠ -ωT, (-180°) - < ∠ -ωT < 0°

A: origin σ=0, jω=0; Re z =1, Im z =0 Circular arc of unit radius in +ve ImZ from A to B

B: 0, + jωs /2; Re z = -1, Im z ≈ 0 Straight line || to ReZ axis back to ImZ axis B to C

C: - ∝, + jωs /2; Re z = 0, Im z ≈ 0 Circular arc of zero radius crossing ReZ axis C to D

D: - ∝, - jωs /2; Re z = 0, Im z ≈ 0- Straight line || to ReZ axis away from ImZ axis D to E

E: 0, - jωs /2; Re z = -1, Im z≈ 0- Circular arc of unit radius in –ve ImZ back from E to A

Mapping of complementary strips similar to that of primary strip because of the trigonometric identity

involving ωs /2 (Nyquist criteria for signal frequency) and the sampling frequency multiples nωs

Z-transform and related problems

Unit step: Z(1(t)) = ∑ 1(kT) z-k = 1+ z-1 + z-2 +…. = z/(z-1)

Exponential decay: x(t)=0 for t<0; exp(-at) for t ≥ 0

Z(exp(-at)) = ∑ ε-akT z-k = 1+ ε-aT z -1 + ε-2aT z -2 + … = z/(z-ε-aT )

Sinusoidal function: Z{ sin(ωt) } = Z{ε jωt - ε

-jωt /2j }

= (z/(z-ε jωT )– z/(z-ε-jωT))/2j = z sin ωT / (z2 – 2z cos(ωT) +1)

Given Laplace transform:

X(s) = 1/(s(s+1)) = 1/s – 1/(s+1) implies X(z) = z/(z-1) –z/(z-ε-T)

Differentiation:

Z{x(k+1)} = ∑ x(k+1) z -k = z ∑ x(k+1) z –(k+1) = z X(z) – z x(0)

Z-transform theorem:

Linearity:

Z{ag(t)} = aG(z) and Z{g1(t) ± g2(t)} = G1(z) ± G2(z)



Real translation:

Z{g(t-nT)}= z –n G(z) and Z{g(t+nT)}= z n G(z)

g(t-nT) = ∑ g(kT-nT) z -k = z -n ∑ g(kT-nT) z -(k-n) = z –n G(z)

To tackle g(t+nT), take g(kT) = 0 for 0≤ k ≤ n-1

Z{g(t+nT)}= g(nT) + g(nT+T) z-1

+ g(nT+2T) z-2

+ …

= z n (g(0) + g(T) z -1 +…+ g(nT) z -n + g(nT+T) z –(n+1)+ g(nT+2T) z –(n+2) + …)

= z n G(z)

Scale change:

Z{ εat g(t)} = G(ε-aT z)

Z{ ε-at g(t)} = G(εaT z)

Z{ εat g(t)} = ∑ {εanT g(nT) z -n }

Replacing z= ε

sT

and substituting z 1= ε

(s-a)T

= z ε

-aT

Z{ εat g(t)} = G(z1) = G(ε-aT z)

Change of sign proves the other result.

Final value theorem:

limz1 { (z-1)/z G(z)} = lim t∝ { (g*(t)}

provided there is no pole on or outside unit circle

Consider two sequences:

∑ n f(kT) z -k } = f(0) + f(T) z -1 +…+ f(nT) z –n

∑ n f((k-1)T) z -k } = f(0) z -1 + f(T) z -2 +…+ f((n-1)T) z –n

= z -1 ( f(0) + f(T) z -1 +…+ f((n-1)T) z –n+1) = z -1∑ n-1 f(kT) z -k

Now, f(nT) = ∑n f(kT) - ∑ n-1 f(kT) = limz1 {∑ n f(kT) z -k - z -1∑ n-1 f(kT) z -k }

In the limit, lim t∝ f(nT) = limz1 {F(z) - z -1 F(z)}

by definition of the Z-transform as the sum of the series in z that runs to infinity.

Ex: g(t)=b(1-ε-at )

G(z)=bz(1-ε-aT )/(z-1) (z-ε-aT )

Both values tend to b in the limit.

Initial value theorem:



lim z∝ { G(z)} = lim t0 { (g(t)}

G(z) = g(0)+ g(T) z -1 +…+ g(nT) z –n = g(0) in the limit as zinfinity

Now, z G(z) = Z {g(t+T)} from the result Z{x(k+1)} = z X(z) – z x(0)

Then, limz∝ { z G(z)} = lim tT { g(t)}

In general, limz∝ { zn G(z)} = lim tnT { g(t)}

Inverse Z-transform:

e(kT) = (1/2π j) E(z) z∫◦ k-1 dz

where integration contour encompasses all singularities of E(z).

e(kT) = ∑ i residues of E(z) at singularity z i

Z-transform contains information only about e(t) at sampling instants.

Divide Num(z) by Den(z) and expand as a power series in z -1

Example: Num(z) = 11 z 2 - 15 z + 6; Den(z) = z 3 - 4z 2 + 5z – 2

Quotient is 11z -1 + 29 z -2 +67z -3 + 145 z -4 + ….

Stability analysis

Mapping the stable region of s-plane i.e. the left-half plane into the inside of the unit circle around origin of z-

plane fulfils the basic stability criterion.

Bilinear transformation extends the discrete system to use Routh criterion and s-plane frequency domain

analysis.

Consider z = (w+1)/(-w+1) or w = (z-1)/(z +1) as transformation.

Then as T 0; putting z=εsT ; (2/T) w s as higher order terms of T vanish in the limit.

So, (2/T)(z-1/z+1) can be taken as w and may be used approximately like s.

Then, w = (2/T) (εsT – 1)/ (εsT + 1) = (2/T) (εsT/2 – ε-sT/2 )/ (εsT/2 + ε-sT/2 ) = (2/T) tanh (sT/2)

For s-plane imaginary axis, substitute s = jωsp , we get w = σwp + j ωwp



Then, w = (2/T) tanh ( jωsp T /2) = j(2/T) tan (ωspT/2)

If ωspT/2 is small, ωwp ≈ ωsp for ωspT/2 < 0.297

On real axis of s-plane, w = (2/T) tanh (σsp T/2) = σwp

Expand tanh(α) = (α +α 3 / 3! + ..)/(1+α 2 / 2! + ..)

For α 2 << 2; σsp ≈ σwp

Hence, for valid approximations, w s.≈

When not valid, pre-warping of s-plane poles and zeroes into w-plane through actual mapping has to be done.

The two planes are similar from stability analysis view point.

Applying Routh criterion to examine if any right half poles are present: Form the Routh array and no of rootswith positive real part is equal to the number of changes of sign in the first column. The necessary and sufficient

condition for all roots to lie in the left half s-plane is that all coefficients of the polynomial are positive and all

terms in the first column of the array have positive signs.

a0 sn + a1 sn-1 + … + an-1 s + an = 0

sn : a0 a2 a4 a6 ……..

sn-1 : a1 a3 a5 a7 ……..

sn-2 : b1 b2 b3 b4 ……..

sn-3 : c1 c2 c3 c4 ……..

sn-4 : d1 d2 d3 d4 ……..

s2 : e1 e2

s1 : f 1

s0 : g1

b1 = a1a2 -a0a3 / a1

b2 = a1a4 -a0a5 / a1

b3 = a1a6 -a0a7 / a1

…

c1 = b1a3 -a1b2 / b1

c2 = b1a5 -a1b3 / b1



...

d1 = c1b2 -b1c2 / c1

d2 = c1b3 -b1c3 / c1

etc.

Since sign only matters, simplifications are allowed.

Ex. Q(z) = z3 - 4 z2 + 5z – 2 is characteristic equation with T = 2 sec.

Then, Q(w) = 12w3 - 4 w2

Upon substitution of z = (2/T) (w+1)/(-w+1) with denominator (1-w) 3

Routh table gives:

w3 : 12 0

w2: - 4 0

w1 : 0

w0 : 0

One change of sign can be seen in the first column => One root in the right half w-plane

Two zeroes in first column => Two roots on imaginary axis

In fact, Q(z) = (z-2) (z-1) 2

Ex. Q(z) = z3 – 0.8 z2 - 0.03z – 0.17 with T=2 sec.

Q(w) = w(1.94w2

+ 3.32 w + 2.74)

w2: 1.94 2.74

w1 : 3.32

w0 : 2.74

The root at w=0 => A root of Q(z) at z=1

Hence, step input will give ramp output.

System will be clearly stable with no change of sign.

Pseudo-continuous time (PCT) control system

Sampling introduces an attenuation factor of (1/T)

Pade Approximation:

F(x) A≈ m(x)/Bm(x) is a rational fraction approximation of a differentiable power series of the form

c0 + c1 x1 + c2 x2 + …where Am(x), Bm(x) are polynomials in x.



First order Pade approximation of εx=1+x+ x2 /2!+ …yields

F(x) = (a0 + a1 x )/ (b0 + b1 x) = (1+x/2)/(1-x/2) by solving for a,b by comparing coefficients.

Now, Gzoh(s) = 1- ε -sT / s 2T/(Ts+2) using Pade approximation of first order.≈

Together with the sampler (1/T), the overall addition to the system TF is 2/(Ts+2)

As T0, ωs∞; 2/(Ts+2) 1

Hence, pseudo-continuous nature is observed.

Note that the second order Pade approximation of εx (1+x/2 + x≈ 2 /12)/ (1-x/2 + x2 /12)

So, poles of higher order would be introduced into the overall TF.

The main advantage is that all the analysis of continuous time system applies to the PCT.Interestingly, the role of sampling time T is evident as pole introduced contains it.



From the state equations describing the system, one can write:

x(k+1) = F x(k) + H u(k)

z X(z) = F X(z) + H U(z)

X(z) = (z I – F)-1 H U(z)

x(1) = F x(0) + H u(0)

x(2) = F x(1) + H u(1) = F (F x(0) + H u(0) ) + H u(1)

= F2 x(0) + FH u(0) + H u(1)

x(n) = Fn x(0) + Fn-1 Hu(0) + Fn-2 H u(1) + … + H u(n-1)

= Fn x(0) + uΩ

= [H FH FΩ 2H … Fn-2 H Fn-1 H] is the state controllability matrix for linear time invariant systems.

y(k) = Cx(k) + Du(k)

y(k+1)= Cx(k+1) + Du(k+1) = C(Fx(k)+Hu(k))+Du(k+1)

= CF x(k) + CH u(k) + Du(k+1)

y(k+2) = CF2 x(k) + CFH u(k) + CH u(k+1) + D u(k+2)

y(k+N) = CF N x(k) + uΓ

= [CFΓ N -1 H CFN-2 H … CFH CH D] is the output controllability matrix for linear time invariant systems N

steps ahead.

Controllability criterion

No of state variables: n

No of input variables: r

No of output variables: p

State matrix F: n x n

Control (input) matrix H: n x r

Output matrix C: p x n

Feedforward matrix D: p x r



The LTI system is completely state controllable iff

• The nXNr controllability matrix is of rank n.Ω

• The nxn matrix ΩΩ` (Gramian matrix) is non-singular.

• The rows of the matrix (z I – F)-1 H are linearly independent.

Proof: x(N) can be generated as linear combination of F i H iff has rank n. Therefore, each of u(i)’s areΩ

allowed to play a role in controlling the system from x(0) to x(N); which is the essence of controllability.

The LTI system is completely output controllable iff the matrix is of rank p.Γ

Feedback control law u(k) = -Kx(k) gives x(k+1) = [F-HK] x(k)

Q(z) = [zI-F+HK] = (z-p1) (z-p2) … (z-pn)

The desired poles are indicated in the design problem. The characteristic equation described above can then be

solved to find the required feedback gains K that ensure the desired pole placement.

Ackerman’s formula: K = [0 0 … 1] Ω -1 Q(F)

Every arbitrary initial state Final state using unconstrained control sequences

Strongly controllable: Each vector (discrete state or output) is controllable by a separate control, when all other

control inputs are zero.

Weakly controllable: Each vector is not strongly controllable.

Complete state controllability: Can find u(k i) for every initial state x(k 0) moves to any final state x(k N)

Complete output controllability: Every y(k 0) moves to any final y(k N)

Complete controllability: Control on every initial condition is implied.

Total controllability: additional control on k 0 and k N combination. For LTI systems both are same.

Reachability: Every state x(k N) can be reached from any x(k 0)

Reachable states: Subspace of state space and unconstrained controllers u(k i) with i = 0, 1, … , N-1; every state in

subspace can be reached.

Ex. F=[a b; c d] H=[g; h] D = [v]

n=2; r=1; p=1; N=2 (= n)

= [g ag+bh; h cg+dh]Ω

Det { } = cgΩ 2+dgh-agh-dh2 ≠0 is required for full rank.



Say, a=0, b=1, c= -1, d =2, g = 1, h= -1.

Then = [ 1, -1; -1, 1]Ω

Det{ } = 0 implies system is uncontrollable.Ω

Subspace [1; -1] is reachable.

For time-varying systems, continued product { F (k Π i) } has to be considered.

= [Θ θ 0 θ1 … θN-1 ] where θi = F(k N-1) F(k N-2) … F(k i+1) H(k i) for i=0,1,…,N-1

is nxNr matrix of rank n in case of time varying systems.Θ

= [Ρ ρ 0 ρ1 … ρN-1] where ρi = C(k N)F(k N-1)F(k N-2) … F(k i+1) H(k i)

is matrix of rank p in case of time varying systems.Ρ

Observability criterion

All state variables are involved in feedback. If every initial state x(0) can be determined from the observation of y(t)

over a finite time interval – helps to reconstruct un-measurable state variables from the measured ones. Since u(kT)

are known, we use unforced system in the analysis.

Then all state variables can participate in feedback control.

x(k+1) = F x(k) and y(k) = C x(k) F n x n and C m x n

x(k) = Fk x(0) and y(k) = C Fk x(0)

Complete observability means given y(0), y(T), …, y(NT)

To determine x1(0), x2(0), …, xn(0)

y(0)= C x(0)

y(1)= CF x(0)

….

y(n-1) = C F n-1 x(0)

Arranging, y = Υx where Υ is the observability matrix nm x n which must have the full rank n so that the n

simultaneous equations may be solved

Ex. F=[a b; c d], C=[e; f], H=[g; h] D = [v]

n=2; r=1; p=1; N=2 (= n)

Υ = [e ae+cf; f be+df]

Det { Υ} = be2+def-aef-cf 2 ≠0 is required for full rank.

In other words,

x1(1) = a x1(0) + b x2(0)

x2(1) = c x1(0) + d x2(0)

y(0) = e x1(0) + f x2(0)



y(1) = (ae+cf) x1(0) + (bc+df) x2(0)

We must be able to solve for x1(0) and x2(0) from these two simultaneous equations.

So the equations must be linearly independent.

Controllable Observable

Yes Yes Controller can be designed only in this case

Yes No

No Yes

No No

Controller and observer design

Since, x(k+1) = F x(k) + H u(k)

Taking u(k) = -K x(k) we have x(k+1) = (F-HK) x(k)

Taking Z-transform, z X(z) =(F-HK) X(z)

Simplifying, (zI-(F-HK) )-1 is the new state transition matrix.

Without controller it is (zI-F) -1

Characteristic equation Q(z) is obtained as determinant

So, the Eigenvalues have reduced through controller

Eigenvalues provide the new poles of the system.

From pole placement, one can solve for the controller gains K.

Ex. F = [1 0; 0 a] H = [1; -1] with a=0.815

Desired poles are given by Q(z) = (z-0.98)(z-0.85)

Then, K = [0.0162, 0.0312] can be obtained.

Desired closed-loop transfer function is recast in the selection of the desired pole locations of the closed-loop n-th

order characteristic equation Q(z). The unknown controller gains for this pole placement through feedback have

been obtained above.

Now consider a continuous system where output is observed, feedback is computed. Due to the computational delay,

the system states propagate by this time. To manage this, we need new mathematics.

Output at time k+1, given output at time k is denoted as (k+1|k)η

(k+1|k) = G (k|k-1)+Kη η oy y(k) + Kou u(k)

Here G , Koy , Kou are unknown matrices of proper dimension.



(k+1|k) is a sequential reconstruction of of the state at time k+1η

given linear equations combining y and u at time k

Form the state at time k and value of forcing function at time k, we are predicting the state at time k+1

Error associated with prediction is e(k+1|k) = x(k+1) - (k+1|k)ηPutting x(k+1) = F x(k) + H u(k) and y(k) = C x(k) + D u(k)

We get e(k+1 | k) = (F - Koy C) x(k) – G (k | k-1) + (H- Kη oy D - Kou ) u(k)

Taking G = F - Koy C and Kou = H- Koy D, we get

Recursive relation e(k+1 | k) = G e(k | k-1)

Starting from an arbitrary initial error, whether the error converges depends on Gk

Since e(k|k-1) = G k [x(0) – y(0)]

Matrices F and C depend on system dynamics.

Using appropriate Koy , we can adjust G such that Gk 0 as k ∞

Here y(0) is a user-supplied initial condition.

The computation of the predictor output takes a finite time. This delay is to be managed.

Substituting G = F - Koy C and Kou = H- Koy D, we get

(k+1 | k) = F (k | k-1) + H u(k) + Kη η oy [y(k) – C (k | k-1) – D u(k)]η

F, H, C, D are obtained from system dynamics.

Koy and Kou are to be designed for error stability – this is essence of observer design.

Important:

Placement of Eigenvalues of G

Poles of the characteristic equation of G inside UC

Pole and zero cannot coincide, then order of system decreases and rank falls below n.

Specific selection of Eigenvalues depends on desired sensitivity of error.

Too much error sensitivity means faster reach to zero.

But this results in higher overshoot and settling time.

When objective is to bring the plant states to origin, F-HK are the regulator poles.

If both observer and regulator poles are assigned to the origin,

Observer reconstructs plant states in n steps,

Then regulator would be driving the states to origin in n steps.



This is output feedback deadbeat control system.

Ex. Position control of a motor system

Given ẋ(t) = [0 1; 0 –a] x(t) + [0; b] u(t) and y(t) = [1 0] x(t)

Let a = 4.6 s-1 , b = 0.8 rad/volt s2 and T = 0.1 sec, delay = 0.02 sec.

= T – delay. Taking deadbeat controller, K = [1.16 7.14]τ

(t, tΦ 0) = [1, (1/a)*(1-exp(-a(t-t0)); 0, exp(-a(t- t0))]

F = [1 (1/a)*(1-exp(-aT); 0 exp(-aT)]

H = [(b/a)*(T-1/a + exp(-aT)/a; (b/a)*(1-exp(-aT))]

Cdiscrete = [1, (1/a)*(1-exp(-a ))]τ

Ddiscrete = ((b/a)*( – (1/a) – (exp(-a ))/a)τ τ

When = 0; observer form is (k | k-1)τ η

When = T; observer form is (k | k)τ η

Discrete system equation is x(k+1) = [1 0.08; 0 0.63]x(k) + [0.0034; 0.0631]u(k)

y η (k) = [1 0.066] (k)+ 0.0024 u(k)η

When = 0, u(k) plays no role.τ

y η (k) = [1 0] (k | k-1)η

Since the system is completely controllable, matrix F - Koy Cd with gain matrix [k 1 k 2]

Becomes [1- k 1 , 0.08-0.0661 k 1 ; - k 2 , 0.631 – 0.0661 k 2]

Setting the coefficients of characteristic equation to zero

For deadbeat controller gives K = [1.16 7.14]



More details on discretization:

Concepts of estimation, prediction and smoothing

Introductory ideas of optimal, adaptive and stochastic control

1. Implementation of control system in industry

2. Basic concepts of microcontroller based implementation

3. Basic concepts of DSP based implementation

4. Basic concepts of PLC based implementation



Figure 1: Discrete Mult -Input Multi-Output System

x(k+1) = F x(k) + H u(k)

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



Figure 2: MIMO system controller and observer

System shown with input u and output y. Observer uses both y and u to reconstruct system states.

Controller uses the estimated system states h to compute the control variables for next iteration.



Figure 3: Regulator designed for the MIMO system



Regulator design needs to track the zero inout set point. In optimal control, we need to find the admissible control

sequence that minimizes or maximizes a general performance index constrained by the linear time varying system

equation

x(k+1) = F(k) x(k) + H(k) u(k)

J = S{x(N),N} + S M{x(k), u(k), k}

for the entire zone of k=0 to (N-1)T.

S defines the terminal condition on the state x. S and M are bounded scalar functions that are at least first order

differentiable with respect to their arguments. This ensures a solution to the constrained optimization problem for the

cost function. Solve using Euler-Lagrange approach.

Jo = J – l' (k+1){x(k+1) - F(k) x(k) - H(k) u(k)} after augmenting the performance index with Lagrange multiplier.

Scalar Hamiltonian function involving first and last terms is maximum (min) along the optimal trajectory under the

discrete maximum (min) principle. Taking partial derivatives with respect to the independent variables and setting them

to zero results in equations for the optimal trajectory.

H[x(k), l(k+1), u(k), k] =

M{x(k), u(k), k} + l' (k+1) {F(k) x(k) + H(k) u(k)}

A common quadratic cost function is defined as follows:

Here, the rXr matrix R is symmetric and positive definite.

The nXn matrix Q symmetric and positive semi-definite.

J = (½) [x'(k) Q(k) x(k) + u'(k)R(k)u(k)] + ½ x'(N)S(N)x(N)

It represents the cost function in the form of general energy constraint on the system state and controller. The problem is

known as linear quadratic regulator problem (LQR)

Taking l(k) = P(k) x(k) as a solution after putting appropriate boundary conditions,

u(k) = -R-1 (k)H'(k)l(k+1) and upon substituting this along with assumed form of l;

x(k+1) = [I+H(k)R-1 (k)H'(k)P(k+1)]-1 F(k) x(k)

or, F'(k) P(k+1) x(k+1) = [P(k) – Q(k)] x(k)

Finally,

P(k) = Q(k) + F'(k)P(k+1) [I + H(k)R-1 (k)H'(k)P(k+1)]-1 F(k)

The controller equation can now be obtained as:

u(k) = -[R+H'(k)P(k+1)H(k)]-1 H'(k)P(k+1)F(k)x(k) = Kc(k) x(k)

Other forms of P(k) involving Kc can also be derived.

P(k)=[F HKc(k)]' P(k+1)[F HKc(k)] + Q + Kc'(k)RKc(k)

Kc is called the optimal feedback gain matrix or Kalman gain.

Eqn for P(k) is called the Riccati equation and

P(k) is called the Riccati gain matrix.

The unknown nXn elements of symmetric matrix P(k)can be solved from n(n-1) equations.

Stability of state-space systems:



Let V(x) = x'Px be a Lyapunov function

such that V(x) is positive definite,

V(0)=0 and V(x) is continuous in x.

V(Fx)-V(x) = (Fx)'PFx-x'Px =x'[F'PF-P]x = -x'Qx

Here F'PF-P = -Q is a positive definite matrix

that must exist for asymptotic stability.

In the current problem, in presence of controller,

DV(x) = x'(k+1)P(k+1) x(k+1) - x'(k)P(k)x(k)

= x'(k)[F-HKc]'P(k+1)[F-HKc]x(k) – x'(k)P(k)x(k)

= -(Q+Kc'RKc) which is negative definite.

Now DV(x) is negative definite implies that the system is asymptotically stable. (Lyapunov's second method)

Diagonal elements of R and Q weighting matrices determine the pole placement in the z domain.



Figure 4: Observer reconstructing the states for the Regulator

Here, we show where the observer comes into the picture. In conjunction with Figure 5, it will be obvious that in

presence of noise or disturbance, it will not be possible for the observer to accurately reconstruct system states, thereby

necessitating the designof an estimator that would filter the noise.

x(k+1) = F x(k) + H u(k) + G w(k)

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

w(k) and v(k) are discrete time random disturbance vector sequences of dimension m and r respectively. They are calledprocess (or plant) and measurement (or instrument) noise respectively. They are statistically independent, stationary,

zero mean, white Gaussian noise with given covariance. G matrix models the additive plant disturbance on each state.



Figure 5: System and measurement noise contaminating the MIMO system



Formulation of the estimator problem:

J=(½) E[{x(k+k) – x^(k+k | k)}'{x(k+k) – x^(k+k | k)}]

where x(k+k) –x^(k+k | k) represents the estimation error and J gives the least square norm.

Here, x^(k+k | k) is the best estimate for measurements [y(k N), y(k N-1), ..., y(k 0)] and control inputs [u(k i)]

If k is zero, then it is called filtering problem.

If k > 0, then it is called prediction problem.

If k < 0, then it is called smoothing problem.

x^(k+1|k) = Fx^(k|k-1) + Kf (k) [y(k) - Cx^(k|k-1)]

Kf (k) is called the Kalman estimator gain. It can be generated from the Riccati equations.

P(k | k) = P(k | k-1) –

P(k| k-1)C'[V+CP(k | k-1)C']-1 CP(k | k-1)

P(k+1 | k) = F P(k | k) F' + GWG'

Kf (k) = P(k | k) C' V -1

Each application requires detailed study of covariance matrix W, V and Kalman gain matrix K f .



Figure 6: Filter or estimator employed to replace the observer



Adaptive controller schemes

Self tuning regulator:

Adjust the system parameters using the following algorithm:

1. Recursively identify/ estimate the value of selected parameters based upon measurements y(i) and input u(i) for

i =0, 1 ..., k-1

2. Update discrete controller difference equation coefficients. In case of state-space models, update appropriate

matrix values.

3. Generate control value u(k)

4. Repeat the process

Model reference adaptive control system:

This is similar to STR, differs in the technique for adjusting the controller. A given apriori model for which the

controller was designed is used as reference. It is driven with the identical input to the plant. The two outputs are

compared and the error is driven to zero by using it to modify the controller dynamics.

Gain scheduling:

When mode of the process changes, different gains are utilized depending on the performance requirements of each

mode. A particular mode plant model may be of different order due to exciting of additional dynamics.

In nonlinear systems, extensive simulation is needed to examine stability and desired performance. Parameter sensitivity

is a major challenge to successful implementation of adaptive control.



Figure 7: Adaptive control scheme (STR)

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