digital controls & digital filters lectures 17 & 18 17-18.pdfstep response of the digital control...

Time-Domain Analysis of Closed-Loop Digital Control Systems

Digital Controls & Digital FiltersLectures 17 & 18

M.R. Azimi, Professor

Department of Electrical and Computer EngineeringColorado State University

Spring 2017

M.R. Azimi Digital Control & Digital Filters


Sampling Period & System PerformanceChoice of sampling period T determines:

a accuracy in representing c(n) or c∗(t) with respect to c(t),

b stability,

c response characteristics and overall performance.

Comparison of time response of analog and digital control systems:

Consider the simplified space-vehicle (analog) system

where Kp = 1.65× 106, Kr = 3.17× 105, Jv = 41822.

The loop transfer function is:

G(s) = C(s)E(s) =Kp

s(Jvs+Kr):Type I system =⇒ ess = 0 to unit step input.



Sampling Period & System Performance-Cont.

The closed-loop transfer function

T (s) = C(s)R(s) =G(s)

1+G(s) =Kp

Jvs2+Krs+Kp= 39.45

s2+8.87︸︷︷︸2ηωn

s+39.45︸︷︷︸ω2n

Characteristic Equation (CE) is :ρ(s) = s2 + 2ηωns+ ω

2n = 0

Hence ωn = 6.3 rad/s and η = 0.706i.e. Underdamped case.

Recall:ωn : Natural undamped frequencyη : Damping ratioα = ηωn : Damping factor0 < η < 1 underdampedη = 1 critically dampedη = 0 undampedη > 1 overdampedη < 0 negative damped/unstable




Digital Control Version:

The OSFG is:

C(s) = A∗(s)Gho(s)1

Jvs2

A(s) = (R(s)−A∗(s)Gho(s) 1Jvs2 )Kp −KrGho(s)

JvsA∗(s)

Star these Eqs.

C∗(s) = A∗(s)[Gho(s)Jvs2

]∗A∗(s) =

(R∗(s)−A∗(s)

[Gho(s)Jvs2

]∗)Kp −Kr

[Gho(s)Jvs

]∗A∗(s)




Apply Mason’s gain:

C∗(s)R∗(s) =

Kp[Gho(s)

Jvs2

]∗1+Kp

[Gho(s)

Jvs2

]∗+Kv

[Gho(s)

Jvs

]∗Z[Gho(s)Jvs2

]= (1− z−1)Z

[1

Jvs3

]Z[Gho(s)Jvs

]= (1− z−1)Z

[1

Jvs2

]T (z) = C(z)R(z) =

KpT2(z+1)

2Jvz2+(2KrT−4Jv)z+(2Jv−2KrT ) =1.65×106T 2(z+1)

az2+bz+c

where a = 83644b = 6.34× 105T + 1.65× 106T 2 − 167288c = −6.34× 105T + 1.65× 106T 2 − 83644




Step response of the digital control system for different choices of T is shownbelow. As can be seen the performance is directly dependent on choice of T .

Rule of thumb:Choose T < τmax5τmax :Largest time constant of analog system.

In this case, τ = 1ηωn =1

4.43 = .225

Thus T < τ/5 = 0.045 Sec.



Mapping Between s- and z-Planes

Describes behavior of digital system v.s. analog one.

Map important Loci from s- to z-planes.

The s-plane is divided into a Primary Strip (extends from −ωs2

to +ωs2

) and an infinitenumber of periodic Complementary Strips as shown.If we consider Primary Strip, path 1©- 2©- 3©- 4©- 5©- 6© is mapped into the unit circleusing the mapping z = eTs where s = σ ± jω. To see this, write

z = esT = eσT e±jωT = re±jΩ

1© s = 0 =⇒ z = 12© s = jωs/2 =⇒ z = ejωsT/2 = ejπ since ωs = 2πT3© s = −∞+ jωs/2 =⇒ r = 0,Ω = jπ (i.e. at zero with phase π)



Mapping Between s- and z-Planes-Cont.

4© s = −∞− jωs/2 =⇒ r = 0,Ω = −jπ (i.e. at zero with phase −π)5© s = −jωs/2 =⇒ z = e−jωsT/2 = e−jπ6© s = 0 =⇒ z = 1

Thus, the entire LHS of the s-plane (including the complementary strips) is mapped tothe interior of the unit circle.1. Constant Damping Factor Loci

Recall: the roots of CE ρ(s) = s2 + 2ηωns+ ω2n = 0 are:

s1,2 = − ηωn︸︷︷︸α

±j√

1− η2ωn︸︷︷︸ω

where α = ηωn: Damping factor and ω =√

1− η2ωn: Damped frequencyand η: Damping ratio, ωn: Natural undamped frequency.Thus, for constant damping factor loci α is fixed while ω variesz = eTs = e−αT e±jωT = re±jΩ




2. Constant Frequency Lociω fixed, α variesz = eTs = e−αT e±jωT = re±jΩ

Note: if α = 0 end point of the lines in z-plane are on the unit circle.3. Constant Damping Ratio Lociη fixeds1,2 = −ηωn ± j ωn

√1− η2︸︷︷︸ω

From the plot of s1,2 roots:

sinβ = η and η√1−η2

= tanβ

Thus,s1,2 = −ω tanβ ± jω




Using this equation, the z-transform relation is then,

z = eTs = e−ω tan βT e±jωT = re±jΩ

1© ω = 0 =⇒ r = 1, Ω = 0

2© ω = ωs/4 =⇒ r = e−π2

tan β , Ω = π/2

3© ω = ωs/2 =⇒ r = e−π tan β , Ω = π

4© ω = 3ωs/4 =⇒ r = e−32

tan β , Ω = 3π2

i.e. the trace in the z-domain is a logarithmic spiral except for β = 0 (r = 1 i.e. unitcircle-oscillatory) and β = π/2 (r = 0 i.e. the origin).




Important Remarks:

Recall that the time constant (analog) is,

τ = 1ηωn

Now, using s1,2 = −ηωn ± jω, we can write

z = eTs = e−ηωnT e±jωT = re±jΩ i.e.

r = e−ηωnT and Ω = ωT = ωnT√

1− η2

It can easily be shown that we have the following important relations,

η = − ln r√(ln r)2+Ω2

,

ωn =1r

√(ln r)2 + Ω2 and

τ = − Tln r

The latter gives a relationship between time constant, r, and sampling period.



s- and z-Planes Poles

Several cases of pole locations of 2nd order system in the s-plane and z-plane and

their corresponding time responses are shown.



Transient Response

Example: Consider the control system below. Find η, ωn, and τ for closed loop systemwhen Gp(s) =

1s(s+1)

and for T = 0.5 sec and T = 0.1 sec and comment on the result.

G(z) = (1− z−1)Z[

1s2(s+1)

]= (T−1+e

−T )z+(1−e−T−Te−T )(z−1)(z−e−T )

Then closed-loop transfer function is,

T (z) = G(z)1+G(z)

= (T−1+e−T )z+(1−e−T−Te−T )

z2+(T−2)z+(1−Te−T )

For T = 0.5, CE is

%(z) = z2 + 1.5z + 0.6967 = 0 =⇒ z1,2 = 0.75± j0.3663 = 0.8347∠± 0.454 orr = 0.8347 and Ω = 0.454 Rad

Thus using the equations in previous remark

η = − ln r√(ln r)2+Ω2

= 0.3698

ωn =1r

√(ln r)2 + Ω2 = 0.9773

τ = − Tln r

= 2.767 secM.R. Azimi Digital Control & Digital Filters


Transient Response-Cont.

For T = 0.1, CE is

%(z) = z2 + 1.9z + 0.9095 = 0 =⇒ z1,2 = 0.95± j0.0837 then r = 0.8347 andΩ = 0.0878 Rad

Thus we get

η = − ln r√(ln r)2+Ω2

= 0.475

ωn =1r

√(ln r)2 + Ω2 = 0.998

τ = − Tln r

= 2.11 sec

Note that the closed-loop transfer function for analog system alone (without samplerand hold) is

T (s) = 1s2+s+1

=⇒ ωn = 1 and η = 0.5 and τ = 1ηωn = 2 secThus, as the sampling period decreases these parameters become closer to those of theclosed-loop analog system without sampler and hold.



Steady-State Error Analysis

An important consideration in a control system is its ability to follow certain inputs(step, ramp, etc.) with minimum error. Consider a typical closed-loop system.

e(t) = r(t)− b(t)G(s) = 1−e

−Ts

sGp(s)

We look at steady state behavior of e∗(t) instead of e(t)e∗ss = lim

t→∞e∗(t) = lim

n→∞e(nT )

Using FVT: e∗ss = limn→∞

e(nT ) = limz→1

(1− z−1)E(z)Note:Condition to use FVT, is that (1− z−1)E(z) should not have any pole on or outsidethe unit circle.E(z) = R(z)

1+GH(z)

GH(z) = Z[Gho(s)Gp(s)H(s)] = (1− z−1)Z[Gp(s)H(s)

s

]e∗ss = lim

z→1(1− z−1) R(z)

1+GH(z)



Steady-State Error Analysis-Cont.

Case 1: Step Input

Let r(t) = Rus(t) =⇒ R(z) = Rzz−1 and e∗ss = lim

z→1

(1−z−1) Rzz−1

1+GH(z)= R

1+Kp

where Kp = limz→1

GH(z): Step Error Constant

For e∗ss = 0 =⇒ Kp =∞ =⇒ GH(z) must have at least one pole at z = 1

Case 2: Ramp Input

Let r(t) = Rtus(t) =⇒ R(z) = RTz(z−1)2

e∗ss = limz→1

(1− z−1)RTz

(z−1)2

1+GH(z)

e∗ss = limz→1

RT

(z−1)(1+GH(z)) = limz→1RKv

Kv =1T

limz→1

(z − 1)GH(z): Ramp Error Constant

For e∗ss = 0 =⇒ Kv = 0 =⇒ GH(z) must have at least two poles at z = 1




Example:Consider the regulator (maintains output in presence of disturbance) systembelow.

Find c∗ss for (a) D(z) = 1 and (b) D(z) = 1 +0.1zz−1

Inputs: F, E∗

Non-inputs: C and E

C(s) = F (s)s+1

+ E∗(s)D∗(s)( 1−e−Ts

s) 1s+1

E(s) = −F (s)s+1− E∗(s)D∗(s)( 1−e

−Ts

s) 1s+1




Star these equations

E∗(s) = −[F (s)s+1

]∗− E∗(s)D∗(s)(1− e−Ts)

[1

s(s+1)

]∗C∗(s) = −E∗(s)

Thus,

C(z) =Z[F (s)s+1

]1+D(z)(1−z−1)Z

[1

s(s+1)

]Here F (s) = 1/s thus Z

[1

s(s+1)

]= z

z−1 −z

z−0.9048

(a) No digital controller D(z) = 1C(z) = 0.095z

(z−0.8096)(z−1)c∗ss = lim

n→∞c(n) = lim

z→1(1− z−1)C(z) = 0.095

1−0.8096 = 0.5




(b): With digital controller D(z) = 1 + 0.1zz−1 =

1.1z−1z−1

D(z)(1− z−1)Z[

1s(s+1)

]= 0.095(1.1z−1)

(z−0.9048)(z−1)

Thus,C(z) = 0.095z

z2−1.8z+0.8096

with poles at s1 = 0.92 and s1 = 0.88 hence FVT can be used.

c∗ss = limz→1

(1− z−1)C(z) = 0



digital controls & digital filters lectures 17 & 18 17-18.pdfstep response of the digital control...

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