chapter 24, dynamic response of discrete-time systems

- CHAPTER 24

Dynamic Response ofDiscrete-Time Systems

In earlier chapters we have seen that the Laplace transform provides a convenientway of analyzing the dynamic behavior of continuous-time systems. The Laplacetransform is also applicable to discrete-time systems but is somewhat awkward touse. Thus we consider a related transform for discrete-time systems, the z-transform. The z-transform has the same utility as the Laplace transform in that its useleads to a compact mathematical description of a dynamic system and permits theuse of algebraic operations for system analysis. By using z-transforms, transferfunctions for discrete-time processes can be defined. This procedure facilitatessubsequent analysis of sampled-data control systems, namely systems where a sampled signal appears. In this chapter we introduce z-transforms and pulse transferfunctions and show how they can be used to calculate transient responses.

24.1 THE z-TRANSFORM

Consider the operation of an ideal, periodic sampler as shown in Fig. 24.1. Thesampler converts a continuous signal f(t) into a discrete signal f*(t) at equallyspaced intervals of time. Mathematically it is convenient to consider impulse sampling, where f* (t) is the sampled signal formed by a sequence of impulses or Diracdelta functions:

f*(t) = ~ f(n6.t) 8(t - ntlt)n~O

(24-1)

Recall that the unit impulse 8(t) was defined in Chapter 3 as the limit of a rectangularpulse with infinitesimal width. The area under the pulse has a value of unity. Thus,it follows that if we integrate the sampled signal over a very small time periodincluding the nth sampling instant,

fnc!.t+nc!.t- f*(t) dt = f(n6.t)(24-2)

In practice, impulse sampling is not attainable because the sampler remainsclosed for a small but finite amount of time. However, the time of closure is usuallysmall (i.e., microseconds) compared to the sampling period and, consequently,impulse sampling provides a suitable idealization.

559

560 DYNAMIC RESPONSE OF DISCRETE-TIME SYSTEMS

0123456789

Time, n

f*(t)

o 234 5 6 7 8 9

Time, nFigure 24.1 Sampled data impulse represen

tation of continuous signal f(t).

Next, consider the Laplace transform of Eq. 24-1, pes). The value off(n!::..t)is considered to be a constant in each term of the summation and thus is invariantwhen transformed. Since ~ [oCt)] = 1, it follows from the Real Translation Theorem(3-104) that the Laplace transform of a delayed unit impulse is .s: [oCt - nM]

e-nMs. Thus the Laplace transform of (24-1) is given by

F*(s) = ~ f(nM)e-nMSI/~O

(24-3)

By introducing the change of variable, z ~ eSt;.!, we define F(z), the z-transformof both f*(t) and f(t), as

F(z) ~ Z[f*(t)] ~ f(n!::..t)z-nI/~O

(24-4)

To simplify the notation, denote f(n!::..t) by fll' Then (24-4) can be written as

F(z) = Z[f*(t)] = ~ f,z-n1/=0

(24-5)

In summary, a z-transform can be derived by taking the Laplace transform ofa sampled signal and then making the change of variable, z = eSM. Thus, thez-transform is a special case of the Laplace transform that is especially convenientfor sampled-data systems. Although (24-5) is an infinite series, F(z) can be writtenin closed form if the Laplace transform of f(t) is a rational function [1].

Next we derive the z-transforms of several simple functions.

Step Function. To calculate the z-transform of a unit step input S(t), setfn = 1 for all n 2: O. Note that fo = 1, which implies the sampled value is takenat f(O+). It follows from (24-5) that

F(z) = 1 + Z-l + Z-l + ... (24-6)

For Izi > 1 this infinite series converges, yielding

1F( z) = -~-- z

24.1 The z-Transform 561

(24-7)

Note that Izl > 1 corresponds to es.ll > 1 (or s > 0). This condition on s is thesame as that used to derive the Laplace transform table in Chapter 3.

Exponential Function. For the exponential function f(t) = Ce -at,

F(z) - ~ f(nM)z-1I = ~ Ce-all.l1z-1I11-0 11=0

(24-8)

Since Eq. 24-8 is a power series in e-a.l1z-1 that converges for le-a.l1z-11< 1 (whichimplies that s > - a), then

F( z) = 1C

(24-9)

Properties of the z-transform. Some important properties of the z-transformare summarized below. Further information is available in Refs. 1and 2.

1. Linearity. The z-transform is a linear transformation, which implies that

(24-10)

where al and az are constants. This important property can be derived from thedefinition of the z-transform given in Eq. 24-1.

2. Real Translation Theorem. The z-transform of a function delayed in timeby an integer multiple of the sampling period is given by

Z[f(t - i~t)] = ri F(z) (24-11)

where i is a positive integer, provided thatf(t) = 0 for t < O. F(z) is defined onlyfor positive values of t.

The theorem can be easily proved. From (24-4),

Z[f(t - i~t)] = ~ f(n~t - iilt)Z-1III~O

Now substitute j = n - i(24-12)

Z[f(t - i~t)] ~ f(j~t)Z-j-ij= -i

(24-13)

Since f(jM) = 0 forj < 0, we can write

Z[f(t - i~t)] = Z-i ~ f(jM)rjj=O

(24-14)

= z-iF(z) (24-15)

If the time delay is not an integer multiple of the sampling period, then the modifiedz-transform described below must be employed.

3. Complex Translation Theorem. This theorem helps deal with z-transforms of functions containing exponential terms, which often arise with linear,continuous-time models.

(24-16)


To demonstrate the validity of (24-16), use (24-4) as the starting point:

Z[e-al{(t)] = 2: e-antltf(nb..t)z-nn~O

= 2: f(nM)(zea::'t)-nn~O

(24-17)

(24-18)

(24-19)

We will illustrate the use of this theorem later in Example 24.2.4. Initial Value Theorem. The initial value of a function can be obtained

from its z-transform:

hm f( nM) = lim F( z)/1->0 z->oo

(24-20)

This result follows directly from Eq. 24-4, with the condition that Izi > 1.5. Final Value Theorem. The final or large-time value of a function can be

found from its z-transform, providing that a finite final value does exist:

lim f(nb..t) = hm (1 - z-l)F(z)n-'?X z-)1

(24-21)

Note that (24-21) is analogous to the final value theorem for Laplace transformswith stable poles (ct. Eq. 3-94).

To prove this theorem, substitute the definition of F(z) into the right side of(24-21):

(1 - Z-I) 2: f(nb..t)z-nn~O

= {f(0) + [J(M) - f(O)] Z-1

+ [J(2b..t) - f(b..t)]Z-2 + ... }

(24-22)

(24-23)

Taking the limit as z --7 1, all of the terms in the infinite series except the last onecancel, yielding

lim f(nb..t) = hm (1 - z-I)F(z)/1->00 z-> 1

(24-24)

6. Modified z-Transform. The modified z-transform is a special version ofthe z-transform that was developed to analyze continuous systems containing fractional time delays, namely those that are not an integer multiple of the samplingperiod. Suppose that a time delay e is expressed as

e = (N + CT)b..t (24-25)

where 0 < CT < 1 and N is a positive integer. The sampled values of the delayedfunction are clearly not the same as those of the function with CT = O. Using thereal translation theorem, the z-transform of the delayed function f(t - e) is

Z[J(t - e)] = 2: f(nb..t - Nb..t - CTM)z-nn~O

Defining m = 1 - CT and k = n - N - 1 yields

Z[J(t - e)] = 2: f(kb..t + mb..t)z-hV-lk~ -N-1

(24-26)

(24-27)


Since f(nt:.t) = 0 for n < 0, the lower limit can be changed to zero. In additionZ-N-l can be factored out:

= rN-1 2: f(k!::.t + mt:.t)z-kk~O

Thus, it is convenient to define the modified z-transform by

F(z, m) ~ Z-N-:-I 2: f(k!::.t + mt:.t)z-kk~O

(24-28)

(24-29)"

where m is the modified z-transform variable. The modified z-transforms of somecommon functions have been tabulated by Ogata [1]. The theorems developedabove (initial value, complex translation, etc.) can be extended to modified ztransforms. Unlike the z-transform, the modified z-transform contains informationabout the values of the function between samples (0 < IT < 1). However, thisinformation is available only if we know the continuous functionf(t). Applicationsof the modified z-transform have been discussed by Smith [3] and Deshpande andAsh [4].

Having introduced the properties of the z-transform, we now illustrate howthey can be used to calculate z-transforms through a series of examples.

EXAMPLE 24.1

Derive the z-transform F(z) for the function, f(t) = t.

Solution

Since f(nt:.t) = nt:.t, F(z) is given by the formula,

F(z) = 2: f(nt:.t)z-n = 2: n!::.tz-n = t:.t 2: nrnn~O n~O n~O

To obtain a closed-form expression for the summation, let

5(z) - 2: nz-n = rl + 2z-2 + 3r3 + 4z-4 +n=O

Multiplying by z -I gives

z-15(z) = Z-2 + 2z-3 + 3z-4 + 4z-5 +

Subtract (24-32) from (24-31):

(24-30)

(24-31)

(24-32)

11 - Z-I - 1 (24-33)

Note that the left side is (1 - z-I)5(z). Solving for 5(z) gives

15(z) = ~rl)2

Hence, the z-transform of f(t) = tis

(24-34)

F(z) = !::.t5(z) (24-35)


EXAMPLE 24.2

Derive the z-transform of cos bt. Then, using the complex translation theorem,find the z-transform of f(t) = e-al cos bt.

Solution

Applying the definition of the z-transform,

F(z) = Z(cos bt) = ~ (cos nM.t)z-nn= 0

(24-36)

To derive a compact formula for the power series, use Euler's relation for thecosine function:

cas nb!::.t = ! (e jnb::'1 + e - jnbM)

where j = v=1. This leads to

1('" '" )F(z) = - ~ ejnbMz-n + ~ e-jnb:o.lz-n

2 n~O n~O

We can then employ a previous result given in (24-8) and (24-9):

1( 1 1)Fz =- . + .

() 2 1 - e]bMrl 1 - e- ]bMZ-l

1 ( 2 - (e jb:o.l + e - jMI) r 1 )=:2 1 - (ejbM + e-jb:o.l)z-1 + Z-2

Using Euler's relation once more to return to trigonometric functions

F 1 - Z -1 cos b!::.t(z) = 1 - 2z-1 cas b!::.t + r2

(24-37)

(24-38)

(24-39)

Note that for b = 2mr/!::.t, F(z) = 1/(1 - Z-I), which is the same expression asthe z-transform of the unit step function. In other words, the sampled cosine hasthe identical appearance of the sampled unit step function Un = 1 for all n), anexample of aliasing (see Fig. 22.5). Hence, the two z-transforms are identical inthis special case.

To obtain the z-transform of the composite function of f(t) = e-at cas bt,

apply the complex translation theorem:

Z(e-at cos bt) = F(zea':'t) (24-40)

Equation 24-40 implies that we substitute zea':'t every place z appears in (24-39).This step gives

Z(e-al cos bt)1 - z-le-a::'1 cas b!::.t

1 - 2z-1e-a::'1 cas b!::.t + z-2e-2a:o.{(24-41)

EXAMPLE 24.3

Find the modified z-transform of e-al using Eq. 24-29. Show that the case m = 0(a = 1) corresponds to a pure time delay of one sampling period, that is, 8 = !::.t.


Solution

Using (24-29) as the definition of the modified z-transform and substituting(k + m) Ilt and N = 0 for the value of t in e -at yields

F(z, m) = rl L e-a(k+m)j.{z-kk~O

e-amj.{Z-1 L e-akj.{z-kk~O

(24-42)

(24-43)-

(24-44)

(a > 0)

Using (24-8) and (24-9) with k as the summation index gives

e -am::H Z-IF(z, m) = -aM1 - e z

When m = 0 (CT = 1), the numerator of (24-44) becomes z -I. This termindicates a one unit time delay (N = 0 and CT = lor N = 1 and CT = 0). Thereforeit is consistent with Property 2 (the Real Translation Theorem) stated in Eq.24-11.

As these examples illustrate, we can readily construct tables of z-transformscorresponding to functionsf(t) and F(s). Table 24.1 provides a representative listof z-transforms; more extensive tables can be found in Ogata [1] and Deshpandeand Ash [4].

EXAMPLE 24.4

Given the transform,

1F(s) - -

s(s + a)

which might represent the step response of a first-order system, determine the finalsteady-state value, lim f(nllt).

J1~X

Solution

From Table 24.1,

1 ( 1F(z) = ~ 1=-Z-I

Then from the Final Value Theorem,

limf(nllt) = !lim [(1 - Z-I) ( 1 _] :aj.{ -I)]n~X a z~1 1 - z 1 - e z

limf(nllt) = !lim [1 - z - _la~t] = !n~X a z~ 1 Z - e a

This result agrees with the continuous-time limit of f(t) as t ~ x obtained fromthe final value theorem for Laplace transforms. Note that if a ::5 O,f(t) and f(nllt)

are unbounded, and application of the Final Value Theorem would provide misleading results. Why?


Table 24.1 z-Transforms (At = Sampling Period)

Time Function f(t)

Set), unit step

1b _ a (e-al - e-bl)

1 ( b- Set) + -- e-alab a - b

__ a_ e-bl)a - b

1b e-al sin bt

e-al cas bt

S(t), unit impulse

f(t - kAt)

Laplace

Transform F(s)

1s

(n - I)!sn

1s + a

1

(s + a)(s + b)

1s(s + a)(s + b)

1

(s + a)2

1

(s + a)2 + b2

s + a

(s + a)2 + b2

F(s)e-kMS

z-transform F(z)

11 - r1I:::.tr1

(1 - r1f

an-1 ( 1 )lim -1 n-I __a~O ( ) aan-1 1 - e-aMz-1

11 - e-acl1z-1

b ~ a C - e~aMrl - 1 - e~b~lrl)

2. 1__ 1_ + bab Ll - rl (a - b) (1 - e-aMz-l)

_ a]

!1t e -aMz-I

(1 - e-acl1r1f

1 r1e-acll sin bl:::.t

b 1 - 2r1e-acll cas b!1t + e-2acllr2

1 - rle-a~1 cas b!1t

1 - 2rle-acll cas b!1t + e-2aMr2

1

F(z)rk

24.2 INVERSION OF z-TRANSFORMS

Once a z-transform has been obtained (by whatever means), we need to be ableto obtain the values of its corresponding time-domain function at the samplinginstants. This is analogous to inverting Laplace transforms back to the time domain.The inversion of a z-transform F(z) to its corresponding time domain functionf(t)is not unique because the inverse z-transform does not yield a continuous timefunction. Instead the values of the function are obtained only at the samplinginstants. We know that a variety of continuous signals can be reconstructed fromf*(t); that is, aliasing prevents the unique identification of the continuous functionof time. On the other hand, the transformation from F(z) to f*(t) (or, equivalently,from F(z) to f(nllt)) is unique. Consequently, we define the inverse z-transformoperator, denoted by Z-1, as follows:

f*(t) = {f(nllt)} = Z-l[F(z)] (24-45)

The inverse z-transform consists of the sampled values f*(t), represented at thenth sampling instant as f(n/1t).

To illustrate the inversion process, consider F(z) = rl/(l - P1r1). If F(z)

24.2 Inversion of z-Transforms 567

above is expanded as an infinite series,

1 ~ = rl(l + PIZ-I + PIZZ-z + ... + PI"r" + ... ) (24-46)- PIZ

Comparing this expression to (24-1), note that the inverse z-transform gives anexpression for the value of the function at the nth sampling instant

(24-47)

(24-48)

In most cases the z-transform to be inverted consists of a ratio of polynomials inz -I. To invert such expressions, three methods are available:

(a) Partial fraction expansion(b) Long division(c) Contour integration

(a) Partial Fraction Expansion

This method is analogous to the procedure for expanding a complicated Laplacetransform F(s) into simpler functions prior to taking the inverse Laplace transform.Note that the z-transform table contains expressions that are functions of Z-I ratherthan z. Consequently, each term in the partial fraction expansion should be in thissame form.

Suppose that F(z) has the following form:

F(z) = VI(z)Vz(z)

where VI (z) = kth-order polynomial in Z-I (excluding a possible time-delay termZ-N, which can be factored)

Vz(z) - mth-order polynomial in Z-I (the denominator is normalized sothat the coefficient of ZO is unity).

Assume that the denominator, Vz(z), can be factored into m distinct real roots(i.e., poles of F(z)), denoted by PI> pz, ... , Pm' Thus,

F(z) = VI(z) (24-49)(1 - PIZ-I)(l - pzZ-I) ... (1 - Pmz-I)

Then choose a partial fraction expansion of the form:

F(z) __ r_l__ + __ r_z__ + ... + __ r_m1 - PIZ-1 1 - pZZ-l 1 - Pmz

(24-50)

Each numerator coefficient r; can be calculated in a manner similar to that usedfor Laplace transforms (e.g., Heaviside's rule with z = lip;). Taking the inversez-transform of (24-50) term by term gives

f(n!::.t) = Z-I( rl _) + Z-l( rz _)1 - PIZ 1 1 - pzz I

+ ... (24-51)

Since the inverse transform of rl/(l - PIZ-l) is rl(PI)", then

f(nt:..t) = rl(Pl)" + rz(pz)" + ... + rm(PnY (24-52)


If there were only a single term in the z-transform, then f(n6.t) = rIPI". For thissimple case, we can examine how the sampled representation will vary as a functionof the sign and magnitude of PI' Figure 24.2 shows the discrete-time responses fordifferent values of PI along with possible continuous-time interpretations for thecase of a first-order z-transform.

In the special case when all roots are bounded by 0 :S Pi :S 1 and 6.t is known,then we can express the inverse z-transform for (24-46) as

f(n6.t) = rle-q,,,:>t + r2e-q2,,:>r + ... + rme-qm,,:>r (24-53)

Relating (24-53) to (24-52), note that

111

ql = - 6.tlnpl, q2 = - 6.tlnp2,···, qm = - 6.tlnpm

If any Pi < 0, Eq. 24-52 should be used in place of (24-53).If the denominator of F(z) cannot be factored into real roots (i.e., complex

roots appear), then the partial fraction expansion must contain a second-orderpolynomial in the denominator for each pair of complex roots. When invertingsuch a term, the appropriate quadratic form for damped sines and cosines in Table24.1 should be used.

One other unusual case can arise in using the partial fraction expansion procedure. If the order of the numerator is greater than that of the denominator (i.e.,k > n in Eq. 24-48), then Eq. 24-50 is not strictly applicable. In this case, it ispreferable to use other methods for generating the discrete-time sequence, suchas the long division method discussed below.

EXAMPLE 24.5

Using a partial fraction expansion, find the inverse z-transform ofF(z) = 0.5z-I/[(1 - z-I)(l - 0.5z-I)] for a sampling period 6.t = 1.

Solution

Expanding F(z) into the sum of two fractions yields

F 0.5z-1 rl r2(z) = (1 _ z-I)(l - 0.5z-1) = 1 - Z-I + 1 - 0.5z-1

(24-54)

bmrr

/;b1ill

z-planeImaginary

(1)

trrn-m,

Real

Figure 24.2 Time-domain responses for different locations of the root of F(z).

24.2 Inversion of z-Transforms 569

Using Heaviside's rule (see Section 3.3) for finding rl and rz, multiply F(z) by(1 - Z-I) and set z = 1:

0.5rl = - = 10.5

Next multiply F(z) by (1 - 0.5z-1) and set z = 0.5, that is, Z-1 = 2:

0.5(2)rz = -- = -1

1 - 2

Substituting into (24-54) gives

F( z) = 1

Equation 24-53 with b.t = 1 leads to

11 - 0.5z-1 (24-55)

1ql = - b.t In (1) = 0

1qz = - M In (0.5) = 0.693

so that

f(nb.t) = 1 - e-O.693nM

When Eq. 24-52 is used to check this result (M = 1, rl = 1, PI

pz = 0.5), the same expression results since e-O.693n = (0.5)".

(b) Long Division

(24-56)

1,rz- -1,

Long division provides a second method for obtaining an inverse z-transform. Inmost cases it is considerably easier to use this method to obtain the inverse z

transform than to use partial fraction expansion. However, the result (an infiniteseries) may not be as useful as an analytical expression. Inversion via long divisionis an operation unique to discrete-time systems; no analogous method exists forcontinuous-time systems. From the definition of the z-transform as

F(z) = L f(nb.t)z-II11=0

(24-57)

the coefficients in the power series expansion of F(z) are the values of f(t) at thesampling instants:

F(z) = f(O) + f(M)z-1 + f(2M)z-Z + ...

Let F( z) be a rational function represented by

() bo + b1z-1 + bzz-z + ... + bkz-kFz =--------------ao + alz-1 + azz-z + ... + amz-m

Dividing the denominator into the numerator by long division 1 gives

F(z) = Co + C1Z-1 + czz-z + ...

(24-58)

(24-59)

(24-60)

'The order of division is based on dividing ao into the numerator and its remainders; see Example 24.6.

(24-64)


Referring back to (24-58), we can perform long division and equate the sequences{cn} and Un} to obtain

bo(24-61)

fo = Co = -ao

fj = Cj _ bj _ bO~j

(24-62)ao ao-

fo = C2 = bo _ bOa2 _ bjaj + bOaj2

(24-63)__ 0 ao ao ao-

Some important properties of sampled-data systems can be obtained from longdivision of their z-transforms. For example, for a first-order z-transform,

boF(z) = -1 -

- ajZ

the equivalent sampled signal in the time domain can be found by long division,resulting in the infinite series (cf. (24-46)):

F(z) = boO + ajz-j + aj2z-2 + ... + aI"rn + ... ) (24-65)

Therefore, the sampled data response of this system is given by the sequence{bo, boa), bOaj2,. .. }. This agrees with (24-61) through (24-63), taking into accountthe negative sign before aj in (24-64). Note that if lajl > 1, the magnitude of thesignal grows steadily over time (see Fig. 24.2), but if lajl < 1, the signal will beattenuated over time. This coefficient in a first-order z-transform indicates if asystem producing the signal is stable or unstable.

EXAMPLE 24.6

Repeat Example 24.5 using long division to generate the first five terms.

Solution

Dividing the denominator into the numerator,O.5rl + O.75z-1 + O.875z-3 + O.9375r< + O.9687rS + ...

1 - 1.5z-1 + O.5z-2 IO.5z-1O.5rl - O.75z-1 + O.25r3

O.75z-2 - O.25z-3

O.75z-2 - 1.125z-3 + O.375r<

O.875z-3 - O.375z-<O.875r3 - 1.3125r< + 0.4375rS

O.9375r< - 0.4375r5O.9375r< - 1.4062z-S - 0.4688z-6

O.9687z-S + O.4688r6

Note that f(O) is zero in this expression. Long division does yield the same timesequence as does partial fraction expansion, with considerably less effort. However,the result is in the form of an infinite series.

(24-66)

24.3 The Pulse Transfer Function 571

(c) Contour Integration

A final method for inverting z-transforms utilizes a contour integral:

f(nflt) = f-. ( F(z)zn-l dz'IT) Jr

where the contour r must be appropriately specified. Although the integral can beevaluated using the residue theorem [1, p. 83], this method is seldom used inpractice.

24.3 THE PULSE TRANSFER FUNCTION

In analogy with continuous systems, the analysis and design of sampled-datacontrol systems is facilitated by the use of transfer functions based on z-transforms.The pulse transfer function represents a dynamic relationship and is defined as theratio of the output and input z-transforms, assuming both output and input areinitially at steady state, analogous to continuous-time systems. In addition, boththe input and output signals are sampled at the same rate and synchronously.

Consider the sampled-data system in Fig. 24.3b where X(z) and Y(z) are ztransforms of the sampled input and output signals, x(nM) and y(nflt), respectively. The response of a continuous linear process (Fig. 24.3a) is given by theconvolution integral [1, p. 180],

yet) = L get - T)X*(T) dT(24-67)

where get) is the impulse response of the process (see Chapter 3), T is the dummyvariable of integration, and X*(T) is a series of impulses that can be expressed as

X*(T) = L x(kflt) OCT - kflt)k~O

Substituting (24-68) into (24-67) gives

t x

yet) = 10 get - T) t:o x(kflt) OCT - kflt) dT

(24-68)

(24-69)

As shown by Ogata [1], an impulse oCt) has the important property that, for anarbitrary function h(T),

J: h(T) OCT - kflt) dT = h(kflt)for 0 :S T :S t (24-70)

xes)

x(t)

yes)--y(t)

(a) Continuous input

(b) Sampled input and output

Y*(s)y*(t)

Figure 24.3 Transfer function with (a) continuous input and (b) sampled input.


Thus, Eq. 24-69 reduces to

yet) = ~ get - kM)x(kl:1t)k~O

In particular, for t = nl:1t

y(nM) = ~ g(nl:1t - kl:1t)x(kl:1t)k=O

The z-transform of the output signal is defined by

Y(z) = ~ y(nl:1t)z-1lIl~O

Substitution of y(nl:1t) from (24-72) gives

Y(z) = ~ ~ g(nM - kl:1t)x(kl:1t)r"Il~Ok~O

Let i = n - k:

Y(z) = ~ ~ g(il:1t)Z-(i+k)X(kl:1t)i~ -k k~O

(24-71)

. (24-72)

(24-73)

(24-74)

(24-75)

Since g(il:1t) is zero for i < 0, the lower limit on i can be changed to zero and thesummations separated:

or

Y(z) = G(z)X(z)

where G(z) is defined asx

G(z) ~ ~ g(il:1t)Z-ii~O

(24-76)

(24-77)

(24-78)

We will refer to G(z) as the pulse transfer function of the system. It relates thediscrete-time input and output signals in the same manner as a transfer functionin the s-domain relates continuous signals (see Fig. 24.4).

Other derivations of (24-77) are available [1]. Note that G(z) can be calculateddirectly after get), the impulse response function, has been determined (d. Eq.24-78).

EXAMPLE 24.7

Find the pulse transfer function for a first-order continuous process,G(s) = K/(TS + 1).

~~)~Figure 24.4 Pulse transfer function.

24.4 Relating Pulse Transfer Functions to Difference Equations 573

Solution

The continuous time impulse response get) is found from

K _get) = ,\:,-I[G(S)] = - e-t/, (24-79)

T

The z-transform is

(24-80}

This result can also be obtained from Table 24.1.

EXAMPLE 24.8

A second-order discrete process has the pulse transfer function

G(z) = -O.3225z-1 + O.5712z-2 (24-81)

Determine its discrete-time response when forced by a sampled unit step input.

Solution

For a unit step input,

X(z) = 1 + Z-l + Z-2 + Z-3 + ...1

1 - Z-l

Here we use the closed-form expression to minimize the number of terms. Equation24-77 is used to develop the expression for the step response Y(z). We could applythe partial fraction expansion method to find y(ntlt), but this would be relativelytime-consuming due to the need to factor the denominator and use the Heavisideexpansion. Therefore, long division is employed to determine the response. Substitution gives

_ (_ -O.3225r1 + O.5712z-2 1 ? 2)Y(z) - G(z)X z) - 1 r. r.""",,, __ 1 , r. """,,'H _, , _1 (_4-8

Applying the long division procedure yields

Y(z) = -O.3225z-1 - O.0665z-2 + O.2568z-3 + O.5136z-4

+ O.6918z-5 + O.8082z-6

+ O.8820z-7 + O.9277z-8 + ... (24-83)

We note in (24-83) that y(ntlt) is steadily increasing and may be approaching asteady state value. Using the Final Value Theorem. we can calculate the value ofy(ntlt), as n -,> x. Returning to (24-82), multiply by (1 - Z-l) and set z = 1. Theultimate (steady-state) value for the response is thus 1. Since at steady state bothinput and output values are 1, the gain of the pulse transfer function is also 1. Thiscan be verified by evaluating G(Z)IZ~I'

24.4 RELATING PULSE TRANSFER FUNCTIONSTO DIFFERENCE EQUATIONS

A pulse transfer function, representing a dynamic relation between an input andan output, has a unique correspondence with a difference equation. To demonstrate


this, consider a general difference equation given by

aOYIl + alYn~1 + ... + amY,,-m = boxn + b1x"_1 + ... + bkXIl~k (24-84)

where {aJ and {bJ are sets of constant coefficients and k and m are positive integers.Before taking the z-transform of (24-84), we recall the real translation theorem inEq.24-11:

Z(YIl-J = Z[y(n~t - i~t)] = riy(z)

Taking the z-transform of both sides of Eq. 24-84 gives

aoY(z) + alz~1y(z) + ... + amz-my(z)

boX(z) + b1rl X(z) + ... + bkZ~k X(z)

Collecting terms, we solve for Y(z):

() bo + b1z-1 + ... + bkz-k ()Y z = ----------X zao + alz-1 + ... + amz-m

(24-85)

(24-86)

(24-87)

(24-88)

The pulse transfer function of the discrete-time process is therefore given by

Y(z) bo + b1rl + ... + bkz-kG(z) -- --------

X(z) ao + alz~1 + ... + amz-m

For most processes bo is zero, indicating that the input does not instantaneouslyaffect the output. However, for proportional controllers and processes which aremodeled simply by steady-state gains, all coefficients except ao and bo are zero.Also, the leading coefficient in the denominator can be set to unity by dividingboth numerator and denominator by ao.

Physical Realizability

In Chapter 4 we addressed the notion of physical realizability for continuous-timetransfer functions. An analogous condition can be stated for a pulse transfer function, namely a discrete-time model cannot have an output signal that depends uponfut~re inputs. Otherwise the model is not physically realizable. Consider the ratioof polynomials given in Eq. 24-88. The transfer function will be physically realizableas long as ao #- 0, assuming that G(z) has been reduced so that common factorsin numerator and denominator have been cancelled. To show this property, examineEq. 28-84. If ao = 0, the difference equation is

This equation requires a future input x" to influence YIl-1, which is physicallyimpossible (unrealizable). Another way to test physical realizability is to use longdivision. When the denominator in (24-88) is divided into the numerator, no termswith positive powers of z should occur for a physically realizable system.

EXAMPLE 24.9

Check the transfer function

Y(z)

X(z)

for physical realizability.

1 + 2z-1 + 3Z~2

5z-1 + 2z-2(24-90)


Solution

Note that the leading coefficient in the denominator is a power of Z-I, violatingphysical realizability (ao = 0 in (24-88)). The corresponding difference equation is

5Yn-l + 2Yn-2 = x" + 2X"-1 + 3X"_2

An equivalent difference equation is

5y" + 2Yn-l = Xn+l + 2x" + 3X"-1

(24-91)

(24-92)

This equation requires knowledge of the future input XI1+1 to generate the currentoutput value Ytl" A similar conclusion can be deduced using long division, whichyields positive powers of z. This exercise is left for the reader.

The Zero-Order Hold

Most sampled-data or computer control systems require a device to convert a digitaloutput signal from the controller to an analog signal, which can then be utilizedby a final control element such as a valve to manipulate the process. In many casesthe final control element requires a continuous signal as input rather than a digitalinput to set its position (although in the specific case of a stepping motor, a continuous signal is generated from the digital input). The device usually employedfor this purpose in process control is the digital-to-analog converter (DAC) whichfunctions as a zero-order hold (ZO H), although there are other types of holdsavailable (see Chapter 22). The zero-order hold converts the digital signal fromthe controller into a continuous staircase function. This device ordinarily must

appear in conjunction with a continuous process for digital control to be carried out.The process transfer function, when converted to a pulse transfer function,

must incorporate the zero-order hold. The digital controller output signal may bethought of as an idealized sequence of impulses through the ZOH and then throughthe process. We know from Laplace transform theory that if H(s) is the ZOHtransfer function and G(s) is the process transfer function (including the finalcontrol element), then the overall transfer function is H(s)G(s). Although Table24.1 gives the corresponding discrete-time form for a first- or second-order transferfunction (see Example 24.7 for the first-order case), this table is based on an inputthat can be represented as a series of impulses. Thus, it cannot be used directlyfor the type of input that is characteristic of a ZOH (DAC) output, which is astaircase function. However, by taking the hold device into account, we can derivepulse transfer functions that are appropriate for process control calculations.

First we derive the appropriate expression for H(s). Figure 24.5 shows theresponse of the hold device to an impulse input of unit strength. The hold yieldsa constant output value over the sampling period I:::.t, which is the same as thestrength of the impulse. The impulse response of the ZOH over the interval

hW :Clo !:J.t

TimeFigure 24.5 Response of a zero-order hold element to an impulse of unit

magnitude.


t = 0 to t = 6.t can be written as the difference of two unit step functions, Set)and S(t-6.t):

h(t) = Set) - S(t-M)

From Eq. 3-22, the Laplace transform of h(t) is

1 e -s6.1 1 - e -s6.tH(s) = - - - = ---s s s

(24-93)

(24-94 )

For processes with a piecewise constant input, the difference equation model mustbe based on the product H(s)G(s) rather than G(s) alone. Then Table 24.1 canbe used to convert H(s)G(s) to a z-transform.

EXAMPLE 24.10

For a first-order transfer function with gain equal to one,

1G(s) = TS + 1

Show by transform techniques that a zero-order hold placed ahead of this processwill yield the same difference equation model for the combined system (ZOH plusfirst-order process) as was derived in Eq. 23-19 for piecewise constant inputs to afirst-order process.

Solution

First form the product H(s)G(s):

H(s)G(s)1 - e -s6.1 1

s TS + 1 (24-95)

(24-96)

To convert this expression to its equivalent z-transform, we use a partial fractionexpansIOn:

( 1 1 -s6.t C 1)H(s)G s) = ~ - s-+-l/-T - e ~ - s-+-l/-T

HG(z) is defined as the z-transform of the combined ZOH plus process. Z is usedas a shorthand way to denote the z-transform of the time-domain function givenby the inverse Laplace transform,

HG(z) = Z [H(s)G(s)] = Z{cS:'-1[H(s)G(s)]}

- Z C) Z (_1 ) Z [-S6.tC u_1 )]~ - s + 1/T - e ~ - s + liT(24-97)

(24-98)

Table 24.1 is then used to convert each term in Eq. 24-97 into its equivalentz-transform. Since Z [e-S6.tp(s)] = r1 Z [pes)], Eq. 24-97 becomes

HG(z) = C _1 Z-1 - 1 _ e ~6.tI'Z-1)

-1( 1 1)- z 1 _ Z-1 - 1 _ e-6.II'z-1


or

-I [ z-l(l - e-M/T) ]HG(z) = (1 - z ) (1 _ z-I)(l _ e-Llt/Tz-I)

Z-I (1 - e-LltIT)

= 1 - e-M/Tz-1

Defining al ~ e-M/T, (24-99) can be written as

Y(z) = HG(z) = (1 - al)z-lX(z) 1 - alz-I

In difference equation form, (24-100) becomes

y" - alY,,-1 = (1 - al) X,,_I

or

y" = aIY,,_1 + (1 - al) X,,_I

the same result as given in Eq. 23-19.

(24-99)

(24-100)

(24-101)

(24-102)

Several comments should be made about the procedure for transformingH(s)G(s). First recognize that the combined transfer function H(s)G(s) couldhave been converted to a series of impulse terms as in Example 24.7 by using thetime-domain impulse response followed by transformation to the z-domain. Theapproach taken in Example 24.10, however, is more direct, because Table 24.1provides the relation between sand z, thus bypassing the need to convert fromthe s-domain to the t-domain (and then to the z-domain). We can also generalizethe results in (24-97) to (24-99) as

HG(z) = Z(H(s)G(s)) = (1 - Z-I)Z(G;S)) (24-103)

In the second term, the operator Z indicates that the z-transform equivalent ofG(s)/s can be determined using Table 24.1. Therefore, the calculation of HG(z)

first requires partial fraction expansion of G(s)/ s, followed by transformation toits z-transform. Finally multiplication by (1 - Z-I) yields HG(z).

Equation 24-103 can also be employed to illustrate an important property ofthe zero-order hold; in general, for dynamic systems,

HG(z) ¥ H(z)G(z)

First calculate the z-transform of H(s):

H(z) = Z e se-s;!.t) = (1 - Z-I)Z G)1 - Z-I

H(z) = ~- = 11 - Z-I

(24-104 )

(24-105)

(24-106)

This result is consistent with the fact that the gain of H(s) is unity. Using

H(z) = 1, Eq. 24-104 becomes

HG(z) ¥ G(z) (24-107)


Substituting (24-103) for HG(z), the above expression is

(1 - rl)Z [G;S)] ¥ Z[G(s)](24-108)

The inequality in 24-108 is valid, in general, except for the case G(s) - K. Oneother interesting characteristic of the zero-order hold is that

lim HG(z) = G(s)clt~O

(24-109)

This result seems intuitively correct in view of Fig. 22.2b. For example, if wesubstitute z = esclt in Eq. 24-99 and apply L'Hospital's rule for M -7 0, thez-transforrn reduces to the original first-order transfer function G(s) = 1/(TS + 1).In contrast, the limit of G(z) as /1t -7 0 does not equal G(s) because of thediscontinuous nature of the pulse transfer function G(z) [1]; for example, see Eq.24-80.

EXAMPLE 24.11

Derive the difference equation that corresponds to an integrating element,G(s) = Y(s)IX(s) = lis, using the ZOH and Eq. 24-103.

Solution

First determine G(s)/s, which is lIs2. The z-transform is

/1t rlZ(1/s2) = (1 _ Z-I)2

Multiply (24-110) by (1 - Z-I), yielding HG(z):

HG(z) = Mrl1 _ Z

The corresponding difference equation is

Yn - Yn-I - /1t Xn-I

(24-110)

(24-111)

(24-112)

The infinite series version of (24-112) can be obtained by long division ofHG(z) in (24-111) and conversion to discrete-time form:

n

Yll /1t ~ Xk-Ik=1

(24-113)

(24-114)

Equations 24-112 and 24-113 describe the two equivalent forms of the integratingelement in discrete time.

Higher-Order Systems

As discussed in Chapters 6 and 7, many processes can be approximated by a secondorder transfer function with time delay 8:

G(s) = yes) = Ke-esXes) (TIS + 1)(T2S + 1)

Assume that 8 is an integer multiple of the sampling period (8 = N/1t), and x(t)is a piecewise constant input (i.e., G(s) is placed in series with a ZOH). The

24.5 Effect of Pole and Zero Locations 579

following difference equation results:

(24-115)

The relations between (a), az, b), bz) and (K, T), TZ) have been previously derivedin Eqs. 23-21 through 23-25, based on the analytical solution of the differentialequation. The pulse transfer function for (24-115) is

G(z) = Y(z) = (b1r1 +_~zz-Z)z=~ _ (b1 + bz_~-I)rN=~ (24-116)X(z) l+alz. +azz- l+alz +azz-

Note that (24-116) is a general expression for a second-order discrete-timemodel with a time delay of N sampling periods (the apparent time delay is onesampling period longer (i.e., N + 1) in (24-116) because the output cannot respondinstantly). Neuman and Baradello [6] have derived difference equations incorporating the zero-order hold for a variety of linear process models. Table 24.2 givespulse transfer functions for a number of transfer functions with the zero-order hold.Note that the transfer functions are given in pole-zero form (rather than usingtime constants). Only overdamped and integrating systems are considered.

24.5 EFFECT OF POLE AND ZERO LOCATIONS

For both continuous-time and discrete-time systems, the nature of the dynamic

response is influenced by the location of the poles and zeros of the transfer function(see Section 6.1). In fact, a pole Pi of a continuous system maps into a pole indiscrete time as ePiM (recall the transformation z = eSM). A process with twotime constants (T), TZ) has two negative continuous-time poles (-lIT), -l/Tz);

therefore, after conversion to discrete-time, two positive poles in the z-plane(e -MITl, e-M/TZ) occur. This result is consistent with the expressions for al and azin a second-order difference equation (see Eqs. 23-22 and 23-23).

It is possible to categorize discrete-time responses for first- and second-ordercontinuous-time processes with no zeros into eight different patterns [1], as shownin Fig. 24.6a-h. All responses exhibit an apparent one-unit time delay as notedabove. For first-order systems, recall that in Fig. 24.2 we identified several possibleresponses depending on the pole location. The appropriate first-order differenceequation is

Yll - PIYIl-l - XIl-l (24-117)

Table 24.3 gives the response for the case of a unit impulse input at n = 0(xo = 1 and Yo = 0). The analytical solution is Yll = (Pl)ll· It is clear that a negativepole near the unit circle has a pronounced effect on the response. The alternationin sign of Y is referred to as ringing of the output signal (see Chapter 26 for adiscussion of ringing in control systems). Negative poles nearer the origin, althoughthey produce a change in sign, are heavily damped and their results are not sonoticeable. Positive poles do not cause the output to change in sign.

In Fig. 24.6 case c corresponds to a second-order overdamped system withtwo positive poles on the real axis. It has similar properties to case b. Cases d ande are noteworthy, because the two complex poles in continuous time map into asingle pole in discrete time. In case e, the pole is located at s = ±w)2j in continuoustime, indicating an undamped oscillation. Since Ws is the sampling frequency,and Ws = 2'IT/6.t, the discrete-time pole (eSM) is given by e±"ITJ. Via Euler's identity, the value of the single pole in discrete time is established as e±"ITj = COS'IT ±j sin 'IT = -1. It is important to remember that a first-order discrete-time system

Table 24.2 Pulse Transfer Functions with Zero-Order Hold

HG(z) = Z[H(s)G(s)]Transfer Function G(s)

K

K

.I + r

K

(.I + r)(s + 1')

K

.1(.1 + r)

K(s + 'I)

(.I + r)(s + 1')

K

(.I + r)(.\' + 1')(.1 + v)

~1 + air I

blrl

1 + alz-I

bIZ-I + b1z-2

+ (l\Z-1 + (l2Z-2

blz-I + b,Z-2

+ (1\z-} + (12Z-2

h]z- [ + h2z-2

+ ([IZ-! + (l2Z-2

h1z-' + [,22-2 + h3z-J

I + (ltZ-! + (l2Z-'2 + (lJz-J

al = -Ibl = Kt.!

al = -exp( -rt.!)K

bl = - [I - exp( -rt.!)]r

al = - {exp( -rt.!) + exp( -pt.!)}

a, = exp[ - (r + I' )t.!]bl = [Klrp(r - p)][(r - 1')- rexp(-pt.!) + pexp(-rt.!)]

b, =IKlrp(r - p)I{(r - 1') exp[-(r + p)t.!] + I' exp(-pM)-rexp(-rt.!)}

al = -{I + exp(-rt.!)}a, = exp( - rM)hi = -(Klr')[l - rt.t - exp(-rM)]

h, = (Klr')!l - exp( -rt.!) - rM exp( -rt.!)]

al = -{exp( -I'M) + exp( -rM))

a, = exp[ -(r + p)M)1K

bl = --{exp(-pt.!) - exp(-rM) + ('1/1')[1 - exp(-pt.!)1 - (qlr)!1 - exp(-rM)J)I' - r

b, = K{(qlrp) exp[ -(r + p)t.!1 + [(I' - q)/p(r - 1')1 exp( -rt.!) + [('I - r)/r(r - 1')] exp( -pt.I)}

al = -{exp( -rt.t) + exp( -I'M) + exp( -vt.!)}a, = exp[-(r + p)t.I] + exp[-(p + V)t.I) + exp[-(r '" v)t.11

a, = -exp[-(r + I' + v)t.11

bl = [KI(rpv»]( - '1[ exp( - rt.l) + exp( - pt.!) + exp( - Vt.I)]+ {[pv(q - r)l/l(p - r)(v - r)IHI + exp(-pt.l) + exp(-ut.I)1+ {[rv(q - p)I/[(r - p)(v - p)IHI + exp(-rt.l) + exp(- vt.I)1

+ {[rp(ll - v)]I[(r - v)(p - v)[HI + exp( -rt.l) + exp( -pt.!)])b, = [-KI(rpvJl(-q exp[ -(r + p)t.I] + expl-(p + v)t.tl + exp[ -(r + v)t.!J)

+ {[pv(q - r)JI[(p - r)(v - r)IH exp( -pt.l) + exp( -wt.l) + expl-(p + v)t.!J}

<.nCOo

o-<z»s::oJJm(J)-0oZ(J)mo"oen()JJm--im::Js::m(J)-<(J)--ims::(J)

K(s + 'I)

(s + rHs + pHs + v)

K(s + 'I)

.1'(.1' + r)(,,' + 1')

b,2-' + b,Z-2 + b,z-'1 + a,z-' + a,r' + a3z-'

b,r' + b,r2 + b,z-'I + a\z-I +U2Z-2 + uJz-J

+ {[ru(q - p)]/[(r - p )(u - p)]}{ cxp( - rl>t) + cxp( - vl>t) + cxp[- (r + v)l>t]}

+ {[rp(q - v)J/[(r - vHp - v)]}{ cxp( -rl>t) + cxp( -pl>t) + cxp[-(r + p)l>t]})

b, = [KI(rpu)]( - 'I cxp[- (I' + P + v)l>t]

+ {[pv(q - r)]/[(p - r)(u - r)]} cxp[ -(p + v)l>tJ

+ {[rv(q - p)]/[(r - pHu - p)]} cxp[-(r + u)MJ

+ {[rp(q - u)]/[(r - v)(p - v)]} cxp[-(r + p)l>t»

a, = - {cxp( - I'M) + cxp( - pl>t) + cxp( - vl>t)}

a, = cxp[-(r + p)l>t] + cxp[-(p + u)l>tJ + cxpl-(r + v)l>tJ

a3 = -cxp[-(r + p + v)M)]

b, = [KI(rpv)](-q[ cxp(-rl>t) + cxp(-pl>t) + cxp(-uM)]

+ {[pv(q - rW[(p - r)(v - r)]}[l + cxp( -pl>t) + cxp( -vM)]

+ {[ru(q - p)]/[(r - p)(u - p)IHl + cxp( -rl>t) + cxp( -vl>t)]

+ {[rp(q - u)]/[(r - v)(p - v)]}[l + cxp( -rl>t) + cxp( -pl>t)j)

b, = [-KI(rpv)]{-q(cxpl-(r + p)l>t] + cxp[-(p + v)l>t) + cxp[-(r + v)l>t])

+ {[pu(q - 1')]/[(1' - r)(v - r)]}{ cxp( -I'M) + cxp( -vl>t) + cxp[ -(I' + v)l>t])

+ {[rv(q - 1')]/[(1' - p)(v - p)J}{cxp«-rl>t) + cxp(-vl>t) + cxp[-(r + v)l>t])

+ {[rp(q - v)]/[(r - v)(p - v)]H cxp( -rl>t) + cxp( -pl>t) + cxp[- (r + p)M]})

b, = [KI(rpv)I(- qcxp[-(r + I' + v)M]

+ {[pu(q - r)]/[(p - r)(u - r)]} cxp[ -(I' + v)l>t]

+ {[rv(q - 1')]/[1' - p)(v - p)]} cxp[-(r + v)l>t]

+ {[rp(q - v)]/[(r - v)(p - v)]} cxp[ -(I' + p)l>tj)

a, = -{I + cxp(-rl>t) + cxp(-pl>t)}

"2 = cxp( -rl>t) + cxp( -pl>t) + cxp[-(r + p)l>tl

11, = -cxp[- (r + p)l>t]

b, = (Klrp){ql>t - (I - 'III' - '1/1')11 + cxp( -rl>t) + cxp( -pl>t)J

+ ([1'('1 - 1')1/[1'(1' - 1')1)[2 + cxp( -rl>t)]

+ ([1'('1 - 1')1/[1'(1' - 1')])[2 + cxp(-pM)])b2 = (- Kilp )(ql>t[ cxp( - rl>t) + cxp( - pl>t)]

- (I - 'III' - qlr){cxp( -rl>t) + cxp( -pl>t) + cxpl-(r + p)l>t]}

+ {[p(q - 1')1/[1'(1' - p)IH 1 + 2 cxp( -pl>t)1

+ {[r(q - 1')]/11'(1' - r)IHI + 2 cxp(-rl>t)j)

1>, = (Klrp({[(ql>t - 1)1'1' + '1(1' + p)l/(rp)}cxp[-(r + p)l>t]

+ {[r(q - 1')[/11'(1' - r)]}cxp(-rl>t)

+ {[p(q - 1')1/[1'(1' - p)]} cxp( -pl>t»

'""'":".

JJCD

§IS'

(Q-0cenCD

-I03::J(f)

~"C::J

"0'::J(f)

oo:::;;CD

(i)::J()CD

m..0c~0'::J(f)

U1CO....•.


s-plane y(l) z-plane y*(1)

First-order

systems(s-domain)

(a)

(b)

Second-order

systems(s-domain)

(c)

(d)

(e)

(f)

(g)

(h)

-+c,$Will,

-+b,-$lm,

-+~,-$-~,

=t~,-$~,--- -

-ws/2

=f W'$W'-ws/2 =t ~,$~,-wsf2 :f' ~, -$~,-ws/2 rr w'-$W'-ws/2

Figure 24.6 Effect of pole locations on impulse response.

K(s + 'I)

(s + rHs + p)(s + v)

K(s + 'I)

s(s + rHs + 1')

b12-1 + b,Z-2 + b,z-'

1 + (11Z-] + (12Z-2 + (1:1Z-3

b12-1 + b22-2 + b,z-'

1 + (1!Z-1 +(/2Z-2 + U3Z-J

+ {[ru(q - p)]/[(r - p)(u - p)J}{cxp(-rt.l) + cxp(-uM) + exp[-(r + u)M]}

+ {[rp(q - u)]/[(r - u)(p - u)]}{ cxp( -rt.l) + cxp( -pM) + cxp[ -(r + p)M]})b, = [KI(rpv)](- qcxp[-(r + p + u)t.l]

+ {[pv(q - r)]/[(p - r)(v - r)J} cxp[-(p + v)M]

+ {[rv(q - p)]/[(r - p)(v - p)J} cxp[ -(r + u)t.l]+ {[rp(q - v)]I[(r - vHp - v)]} cxp[-(r + p)M))

a1 = -{cxp( -rt.l) + cxp( -pt.l) + cxp( -vt.l))

a, = cxp[-(r + p)M] + cxp[-(p + u)t.l] + cxp[ -(r + U)t.l]

a, = -cxp[-(r + I' + V)t.l)]b1 = [KI(rpu)]( - q[ cxp( - rM) + cxp( - pt.l) + cxp( - Ut.l)]

+ {[pu(q - r)]/[(p - r)(u - r)]}[! + cxp( -pt.l) + cxp( -vt.l)J

+ {[rv(q - p)I/[(r - p)(v - p)]}[1 + cxp( -rt.l) + cxp( -Vt.l)]+ {[rp(q - v)]/[(r - vHp - v)]}[! + cxp( -rt.l) + cxp( -pt.l)])

b2 = [-KI(rpv)J{ -q( cxp[ -(r + p)t.l] + cxp[-(p + u)M) + cxp[ -(r + v)MJ)

+ {[pu(q - r)]I[(p - r)(u - r)]H cxp( -pt.l) + cxp( -ut.l) + cxp[ -(p + U)t.lJ)

+ {[ru(q - p)J/[(r - p)(u - p)]}{ cxp« -rt.l) + cxp( -Vt.l) + cxp[-(r + U)t.lJ}

+ {[rp(q - v)]I[(r - u)(P - v)]H cxp( -rt.l) + cxp( -pt./) + cxp[ -(r + p)M]})b, = [KI(rpv)I(- 'I cxp[ -(r + I' + U)t.l]

+ {[pU(11 - r)]/[(p - r)(u - r)]} cxp[ -(I' + u)t.ll

+ {[rv(q - p)l/[r - p)(v - p)]} cxp[ -(r + v)t.lj

+ {[rp(q - u)]/[(r - u)(p - v)]} cxpl-(r + p)t.lj)

a1 = -{I + cxp( -rt./) + cxp( -pt.l)}

"2 = cxp( - rt.l) + cxp( - pt.l) + cxp[- (r + I' )t.l]

", = -cxp[-(r + p)t.l]bl = (Klrp){qt./ - (I - 'III' - qlr)[ I + cxp( -rt.l) + cxp( -pM)1

+ (lr(q - 1')1/11'(1' - r)J)12 + cxp( -rt./)]+ (11'('1 - r)l/[r(r - 1')1)[2 + cxp( -pt.l)]}

b2 = (- Klrp )(qt.ll cxp( - rt./) + cxp( - pt.l) I- (1 - 'III' - qlr){cxp( -rM) + cxp( -pt./) + cxp[-(r + p)t./]}

+ {[p(q - r)l/lr(r - p)]}11 + 2cxp(-pt./)]+ {Ir(q - 1')1/11'(1' - r)]}[ I + 2 cxp( -rt./)J)

b, = (Klrp({[(qt./ - I)rp + q(r + p)l/(rp)} cxp[ -(r + p)t./]+ {[r(q - 1')]/11'(1' - r)J} cxp( -rt./)

+ {[p(q - r)]I[r(r - p)]} cxp( -pt./))

I\)"':".

:IJ<1>

![S'to-uc:en<1>

-iOJ::J(J)

~T1c:::J

>1o'::J(J)

oo=0<1>

ro::J()<1>

m.Dc:~o'::J(J)

01CO•...•.


s-plane y(t) z-plane y*(t)

First-order

systems(s-domain)

(a)

(b)

Second-order

systems(s-domain)

(c)

(d)

(e)

(f)

(g)

(h)

-+ c,-$lllilL

-+b,-$~,

-+~,-$-~,

=t~,-$~,-----Jo( -

-ws/2

~

W'-$W'-ws/2 =t ~,-$~,-wsl2 :f' ~, -$~,- wsl2

iTW'-$W'

-wsl2

Figure 24.6 Effect of pole locations on impulse response.

(24-118)

24.6 Conversion Between Laplace and z-Transforms 583

Table 24.3 First-order Difference Output as a Function of Pole Location (unitimpulse input)

Yn

n

XnPI = -0.9PI = -0.3PI = 0.8

0

1 0 001

0 1 112

0 -0.9 -0.30.83

0 +0.81 +0.090.644

0 -0.729- 0.0270.5125

0+ 0.656+ 0.00810.4106

0 - 0.590-0.00240.328

can oscillate. Such behavior is not possible with a first-order continuous-time system.

Oscillation can also occur for second-order discrete-time systems, if the poleshave imaginary (complex conjugate) values (see cases f-h in Fig. 24.6). When apositive or negative zero occurs in the discrete-time model, the degree of oscillationas well as its frequency can be affected. Unfortunately it is not possible to categorizethis case easily, so we refer the reader to some examples presented by Franklinand Powell [5, pp. 32-35].

The mapping of zeros from continuous time to discrete time is unpredictable,mainly due to sampling effects. Consider the second-order difference equationresulting from a second-order continuous-time transfer function with no zero. In

Eq. 24-116, the poles are found from factoring the denominator polynomial intotwo roots. The zero of the discrete-time transfer function is - bz/ b1, which is fairlycomplicated when expressed in terms of TJ and TZ (see Eqs. 23-24 and 23-25).Therefore, there is no apparent simple relation between continuous- and discretetime zeros. In addition, the sampling period can have a profound influence on thesampled response [7]. For example, an inverse response in continuous time (seeFig. 6.3) may not be observed at the sampling instants if the sampling rate is tooslow.

24.6 CONVERSION BETWEEN LAPLACE AND z-TRANSFORMS

We have previously seen that Table 24.1 can be used to convert Laplace transformsto z-transforms and vice versa. However, implicit in this approach is the requirement that partial fraction expansion be performed to obtain the correct conversion.

An alternative approach that avoids partial fraction expansion yields an approximate result merely by performing a variable substitution. No zero-order hold isexplicitly considered in this approach.

As discussed earlier, the transform variable z was defined by z = eS::'t orZ-1 = e-s!lt. To obtain an approximate relation expressing s in terms of a ratio ofpolynomials in z, we can use the Pade approximation for e -s!lt

-stlt _ 2 - stite =---2 + stit

Equating to r J gives

1 2 - stitz- :::= ---

2 + stit (24-119)


or

(24-120)

The approximation suggests that a Laplace transform can be converted to az-transform by substituting (24-120) for s. Such an approach is known as Tustin'smethod [1]. A less accurate expression for s can be derived using the power series

s26.t2e-s':'t = 1 - s6.t + -- - ... (24-121)2

Retaining only the first two terms, we have

Z-I = e-s':'t == 1 - s6.t

or

s= (24-122)

which is equivalent to the backward difference formula we have used in Chapter23. When the algebra involved in the substitution is not too complicated, (24-120)should be used instead of (24-122) to improve accuracy.

Ogata [1] has listed more accurate formulas for algebraic substitution into atransfer function G(s). Approximate substitution is a procedure that should alwaysbe used with care. Exact conversion, especially of the process model (up to thirdorder), is recommended. A bilinear transformation similar to (24-120) in form issometimes used for stability analysis; its use will be discussed in Chapter 25.

EXAMPLE 24.12

Find an expression for the pulse transfer function of an ideal PID controller,

Ge(s) = Ke( 1 + ~ + TDS) (24-123)TIS

using the approximation in (24-122). Compare your result with the velocity formof the PID algorithm given in Eq. 8-18.

Solution

Substituting (24-122) into (24-123) gives

Ke(ao + alrl + a2z-2)1 - Z-I (24-124)

(24-125)

If e" is the error signal and p" is the output from the controller, then Gc(z) =P(z)/E(z) and

(24-126)

Summary 585

Using the real translation theorem and converting the controller equation into adiscrete-time form gives:

(24-127)

Substituting for aQ, al> and az and collecting terms with respect to the controllersettings Kn Tf, and TD gives

[ IlT TD ]Pn - Pn-I = Kc (en - en-I) + ---:-en + A(en - 2en_1 + en-z) (24-128) .-II ~t

Note that this equation is identical to Eq. 8-18 which was derived using a finitedifference approximation.

SUMMARY

In this chapter we have introduced the z-transform and its properties, in much thesame fashion as was done for Laplace transforms and continuous-time linear systemsin Chapters 3 and 4. Operational use of the z-transform with linear process modelshas been emphasized here, since z-transforms are a convenient medium to analyze

Table 24.4 Discrete/Continuous Conversions for Linear Systems

Conversion Method (section or equation)

(A) Impulse response ~ z-transform

Conversion table (§24.1)

(B) Laplace transform ~ z-transform

(1) Conversion table (§24.1, §24.2)

(2) Approximate substitution (§24.6)(C) z-transform ~ difference equation

Translation theorem (§24.2)

(D) Difference equation ~ z-transform

Translation theorem (§24.2)

(E) Laplace transform ~ difference

Zero-order hold transfer function,

equation (piecewise constant input)Eq. 24-104 ~ partial fraction

expansion ~ Table 24.1 ~ thenuse (C) (§24.2)(F) Differential equation ~

(1) Convert to Laplace transform;difference equation

then use (E) (§24.4)

(2) Analytical solution for piecewiseconstant input, Table 24.2 or Eq.23-21(3) Finite difference approximation(§23.1)(0) Data ~ difference equation

(1) Linear regression (§23.3)(2) Nonlinear regression in continuoustime (§7.1) ~ analytical solution(§23.2)

(H) z-transform ~ response

(1) Use (C) followed by integration ofdifference equation (§23.1)(2) Power series in rk by longdivision (§24.3)(3) Contour integration, Eq. 24-66


digital feedback control systems (Chapters 25 and 26). In Chapters 22-24 we havepresented a rather diverse set of techniques for converting continuous models todiscrete models and vice versa. Hence, a suitable epilog to this chapter would beto provide a summary of the possible avenues for interconversion. Table 24.4 givesa list of different approaches, the steps involved, and the pertinent sections orequations where the specifics are demonstrated.

REFERENCES

1. Ogata, K., Discrete Time Control Systems, Prentice-Hall, Englewood Cliffs, NJ, 1987.2. Corripio, A. B., Module 3.3 in AIChemI Modular Instruction, Series A, Vol. 3, AIChE, New York

(1983).3. Smith. C. L., Digital Computer Process Control, InText, Scranton, PA, 1972.4. Deshpande. P. B., and R. H. Ash, Elements of Computer Process COllfrol, Instrum. Soc. of America,

Research Triangle Park, NC, 1981.5. Franklin, G. E, and J. D. Powell, Digital Control of Dynamic Systems, Addison-Wesley, Reading.

MA. 1980.6. Neuman, C. P., and C. S. Baradello. Digital Transfer Functions for Microcomputer Control. IEEE

Trans. Systems, Man, Cybernetics SMC-9 (12), 856 (1979).7. Astrom, K. J., and B. Wittenmark, Computer-Controlled Systems, Prentice-Hall. Englewood Cliffs,

NJ. 1984.

EXERCISES

24.1. What is the z-transform F(z) of the triangular pulse in the figure if the samplingperiod has the following values:

(a) I1t = 5 s(b) I1t = 10s

f

2

1

00

Time (s)

I40

24.2. A temperature sensor has the transfer function,

T~,(S )

T' (s)

1lOs + 1

where T;/1 is the measured temperature and T' is the temperature (both in deviationvariables). The temperature measurement is sampled every five seconds and sent toa digital controller. Suppose that the actual temperature changes in the followingmanner,

{ 350 of

T(t) = 370 of350 of

for 0 :s:; t < 4 sfor 4 :s:; t < 12 sfor t 2: 12 s

(a) What is the z-transform of this signal, T(z)?(b) Derive an expression for the z-transform of the measured temperature T", (z).(c) The digital controller sounds an alarm if the sampled value of T", exceeds 360 oF.

Does the alarm sound?

(d) What is the maximum value of the measured temperature T",(t)?

Exercises 587

24.3. Suppose that1 - 0.2rl

F(z) = (1 + 0.6r1)(1 - 0.3rl)(1 - rl)(a) Calculate the corresponding time-domain response r (t).

(b) As a check, use the final value theorem to determine the steady-state value ofr(t).

24.4. Determine the inverse transform of

z(z + 1)

(z - 1) (Z2 - Z + 1)

by the following methods:

(a) Partial fraction expansion.(b) Long division.

24.S. Calculate the z-transform of the rectangular pulse shown in the drawing. Assumethat the sampling period is t:.t = 2 min. The pulse is f = 3 for 2 s:: t < 6.

6.-

f

3

oLL0

246

Time (min)

18

24.6. The pulse transfer function of a process is given by

Y(z) 5(z + 0.6)X(z) Z2 - Z + 0.41

(a) Calculate the response Y(nLlt) to a unit step change in x using the partial fractionmethod.

(b) Check your answer in part (a) by using long division.(c) What is the steady-state value of y?

24.7. The desired temperature trajectory T(t) for a batch reactor is shown in the drawing.

(a) Derive an expression for the Laplace transform of the temperature trajectory,T(s).

(b) Determine the corresponding z-transform T( z) for sampling periods of t:.t = 4and 8 min.

80r /25~

T(OC)

o 20

Time (min)

J

40

24.8. The dynamic behavior of a temperature sensor and transmitter can be described bythe first-order transfer function,

T;"(s) _ e-2s

T' (s) - 8s + 1


where the time constant and time delay are in secondsT = actual temperatureTm = measured temperature

If the actual temperature changes as follows (t in seconds):

{ 70 °C for t < a

T = 85°C for a :S t < 1070 °C for t ~ 10

(a) What is the maximum value of the measured temperature Tm?

(b) If samples of the measured temperature are automatically logged in a digitalcomputer every two minutes beginning at t = 0, what is the maximum value ofthe logged temperature?

24.9. The transfer function for a process model and a zero-order hold can be written as

H G (1 - e-s~t) 3.8e-2s(s) pes) = s (lOs + 1)(5s + 1)

Derive an expression for the pulse transfer function of H(s) G p(s) when !:::,.t = 2.

24.10. The pulse transfer function of a process is given by

Y(z) 2.7r1(z + 3)X(z) Z2 - 0.5z + 0.06

(a) Calculate the response y(n!:::"t) to a unit step change in x using the partial fractionmethod.

(b) Check your answer in part (a) by using long division.(c) What is the steady-state value of y?

24.11. A gas chromatograph is used to provide composition measurements in a feedbackcontrol loop. The open-loop transfer function is given by

G(s) = GcHGpGm (Gu = 1)

and is

G(s) = B(s) = 2(1 + ~)C - e-SM)( 10 )e-2SE(s) 8s 5 125 + 1

(a) Suppose that a sampling period of At = 1 min is selected. Calculate HG(z), thepulse transfer function of G(s) with ZOH.

(b) If a unit step change in the controller error signal e(t) is made, calculate thesampled open-loop response b(n!:::"t) using HG(z).

24.12. Determine the pulse transfer function with zero-order hold for the second-order

process Gp(s) = K/[(5s + 1)(3s + 1)] using partial fraction expansion in thes-domain. Check your results with those in Section 23.3. Note that At is unspecifiedhere.

24.13. FindHG(z)ifG(s) = (1- 9s)/[(3s + 1)(15s + 1)] for At = 4 (use partial fractionexpansion). What is the corresponding difference equation? Do you detect inverse

response in the output Y n for a step change in the input at this sampling period?

24.14. Verify the z-transform in Table 24.1 for J(t) = t2. What is the z-transform forJ(t) = 1 - e -at?

24.15. Find the response Yn for the difference equation

Yn - Yn-l + 0.21 Yn-2 = Xn-2

Let Xo = 1, Xn = a for n ~ 1. Use long division as well as direct integration to checkthe results.

Exercises 589

24.16. Use long division to calculate the first eight coefficients of the z-transform given by

0.8r1

F(z) = (1 _ 0.8r1)Z

24.17. Derive the pulse transfer function for an analog lead-lag device cascaded with a zeroorder hold. The lead-lag device has the transfer function (TIS + 1)1(TZS + 1). Check

the steady-state gain of the pulse transfer function.

24.18. Determine the sampled function f(n6.t) corresponding to the z-transform

F( ) 0.5r1z = 1 _ 1.5r1 + 0.5z-2

Use partial fraction expansion (6.t = 1) and compare the results with the long divisionmethod for the first six sampled values (n = 0, 1, ... ,5).

24.19. For G(s) = 1/[(s + 1)(s + 2)], obtain G(z) for IJ.t = 1. Determine the responseto a unit step change in the input. Repeat using Tustin's method (approximatez-transform) and compare the step responses for the first five samples.

24.20. To determine the effects of pole and zero locations, calculate and sketch the unit

step responses of the pulse transfer functions shown below for the first six samplinginstants, n = 0 to n = 5. What conclusions can you make concerning the effect of

pole and zero locations?

1(a) 1=-r1 1(b) 1 + 0.7r1 1(c) 1 _ 0.7r1

1

(d) (1 + 0.7z-1)(1 - 0.3r1)1 - 0.5r1

(e) (1 + 0.7r1)(1 - 0.3r1)

f 1 - 0.2rl( ) (1 + 0.6r1)(1 - 0.3r1)

24.21. For the transfer functions shown below, determine the corresponding pulse transfer

function HGp(z) for the system and a zero-order hold.

1(a) Gp(s) = (s + 1)3

6(1 - s)(b) Gp(s) = (s + 2)(s + 3)

For sampling periods of 6.t = 1 and IJ.t = 2, determine whether any poles or zeros

of HG p(z) lie outside the unit circle for either process. Discuss the significance ofthese results.

chapter 24, dynamic response of discrete-time systems

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