-
CHAPTER
24
Dynamic Response of
Discrete-Time Systems
In earlier chapters we have seen that the Laplace transform provides a convenient
way of analyzing the dynamic behavior of continuous-time systems. The Laplace
transform is also applicable to discrete-time systems but is somewhat awkward to
use. 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 use
leads to a compact mathematical description of a dynamic system and permits the
use of algebraic operations for system analysis. By using z-transforms, transfer
functions for discrete-time processes can be defined. This procedure facilitates
subsequent analysis of sampled-data control systems, namely systems where a sampled signal appears. In this chapter we introduce z-transforms and pulse transfer
functions 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. The
sampler converts a continuous signal f(t) into a discrete signal f*(t) at equally
spaced 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 Dirac
delta functions:
f*(t)
= ~
f(n6.t)
8(t - ntlt)
(24-1)
n~O
Recall that the unit impulse 8(t) was defined in Chapter 3 as the limit of a rectangular
pulse 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 period
including 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 remains
closed for a small but finite amount of time. However, the time of closure is usually
small (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
Figure 24.1 Sampled data impulse represen-
Time, n
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 invariant
when 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)
= ~
(24-3)
f(nM)e-nMS
I/~O
By introducing the change of variable,
of both f*(t) and f(t), as
F(z)
z ~ eSt;.!,
~ Z[f*(t)]
~
we define
f(n!::..t)z-n
F(z),
the z-transform
(24-4)
I/~O
To simplify the notation, denote
F(z)
=
f(n!::..t)
Z[f*(t)]
by fll' Then (24-4) can be written as
= ~
f,z-n
(24-5)
1/=0
In summary, a z-transform can be derived by taking the Laplace transform of
a sampled signal and then making the change of variable, z = eSM. Thus, the
z-transform is a special case of the Laplace transform that is especially convenient
for sampled-data systems. Although (24-5) is an infinite series, F(z) can be written
in 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), set
= 1 for all n 2: O. Note that fo = 1, which implies the sampled value is taken
fn
at f(O+).
It follows from (24-5) that
F(z)
=
1
+
Z-l
+
Z-l
+ ...
(24-6)
24.1 The z-Transform
561
For Izi > 1 this infinite series converges, yielding
1
= -~-- z
z)
F(
(24-7)
Note that Izl > 1 corresponds to es.ll > 1 (or s > 0). This condition on s is the
same as that used to derive the Laplace transform table in Chapter 3.
For the exponential function
Exponential Function.
F(z)
-
~
11-0
= ~
f(nM)z-1I
11=0
f(t)
= Ce -at,
Ce-all.l1z-1I
(24-8)
Since Eq. 24-8 is a power series in e-a.l1z-1 that converges for le-a.l1z-11< 1 (which
implies that s > - a), then
C
F(
z) =
(24-9)
1
Some important properties of the z-transform
are summarized below. Further information is available in Refs. 1 and 2.
Properties of the z-transform.
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 the
definition of the z-transform given in Eq. 24-1.
2. Real Translation Theorem. The z-transform of a function delayed in time
by an integer multiple of the sampling period is given by
Z[f(t
-
i~t)]
=
ri
(24-11)
F(z)
i
where is a positive integer, provided thatf(t)
= 0 for
for positive values of t.
The theorem can be easily proved. From (24-4),
Z[f(t
-
i~t)]
= ~
f(n~t
-
t
< O. F(z)
iilt)Z-1I
is defined only
(24-12)
II~O
Now substitute
j
= n -
i
Z[f(t
-
i~t)]
~
j=
Since f(jM)
-i
f(j~t)Z-j-i
(24-13)
= 0 forj < 0, we can write
Z[f(t
-
i~t)]
=
Z-i ~
f(jM)rj
(24-14)
j=O
=
z-iF(z)
(24-15)
If the time delay is not an integer multiple of the sampling period, then the modified
z-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)
562
DYNAMIC
To demonstrate
RESPONSE
OF DISCRETE-TIME
SYSTEMS
the validity of (24-16), use (24-4) as the starting
Z[e-al{(t)]
point:
= 2:
e-antltf(nb..t)z-n
(24-17)
= 2:
f(nM)(zea::'t)-n
(24-18)
n~O
n~O
(24-19)
We will illustrate
the use of this theorem later in Example 24.2.
The initial value of a function
from its z-transform:
4.
Initial Value Theorem.
hm
=
f( nM)
/1->0
can be obtained
lim F( z)
(24-20)
z->oo
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)
(24-21)
z-)1
n-'?X
Note that (24-21) is analogous to the final value theorem for Laplace transforms
with stable poles (ct. Eq. 3-94).
To prove this theorem, substitute the definition of F(z) into the right side of
(24-21):
2:
(1 - Z-I)
(24-22)
f(nb..t)z-n
n~O
=
+
{f(0)
+
[J(M)
[J(2b..t)
-
f(O)]
- f(b..t)]Z-2
Z-1
+ ... }
(24-23)
Taking the limit as z --7 1, all of the terms in the infinite series except the last one
cancel, yielding
= hm (1 -
lim f(nb..t)
z->
/1->00
1
(24-24)
z-I)F(z)
6. Modified z-Transform. The modified z-transform is a special version of
the z-transform that was developed to analyze continuous systems containing fractional time delays, namely those that are not an integer multiple of the sampling
period. Suppose that a time delay e is expressed as
e
=
(N
+
(24-25)
CT)b..t
where 0 < CT < 1 and N is a positive integer. The sampled values of the delayed
function are clearly not the same as those of the function with CT = O. Using the
real 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-n
(24-26)
n~O
Defining m
= 1 -
CT
Z[J(t
and k
-
= n - N - 1 yields
e)]
=
2:
k~ -N-1
f(kb..t
+
mb..t)z-hV-l
(24-27)
24.1 The z-Transform
563
Since f(nt:.t) = 0 for n < 0, the lower limit can be changed to zero. In addition
can be factored out:
Z-N-l
=
2: f(k!::.t + mt:.t)z-k
rN-1
(24-28)
k~O
Thus, it is convenient to define the modified z-transform by
F(z, m) ~
2: f(k!::.t + mt:.t)z-k
Z-N-:-I
(24-29)"
k~O
where m is the modified z-transform variable. The modified z-transforms of some
common functions have been tabulated by Ogata [1]. The theorems developed
above (initial value, complex translation, etc.) can be extended to modified ztransforms. Unlike the z-transform, the modified z-transform contains information
about the values of the function between samples (0 < IT < 1). However, this
information is available only if we know the continuous functionf(t). Applications
of the modified z-transform have been discussed by Smith [3] and Deshpande and
Ash [4].
Having introduced the properties of the z-transform, we now illustrate how
they 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,
2: f(nt:.t)z-n
n~O
F(z) =
=
2: n!::.tz-n = t:.t 2: nrn
n~O
n~O
(24-30)
To obtain a closed-form expression for the summation, let
5(z) -
2: nz-n =
n=O
rl
+ 2z-2 +
3r3
+ 4z-4 +
(24-31)
Multiplying by z -I gives
z-15(z)
=
+ 2z-3 + 3z-4 + 4z-5 +
Z-2
(24-32)
Subtract (24-32) from (24-31):
1
1 - Z-I - 1
Note that the left side is (1 - z-I)5(z).
(24-33)
Solving for 5(z) gives
1
5(z)
(24-34)
= ~rl)2
Hence, the z-transform of f(t)
F(z)
= tis
= !::.t5(z)
(24-35)
564
DYNAMIC RESPONSE OF DISCRETE-TIME SYSTEMS
EXAMPLE
24.2
Derive the z-transform of cos bt. Then, using the complex
find the z-transform of f(t) = e-al cos bt.
translation
theorem,
Solution
Applying
the definition
of the z-transform,
= Z(cos
F(z)
To derive a compact
cosine function:
formula
= v=1.
= ~
(cos nM.t)z-n
n=
=
!
series, use Euler's
relation
(e jnb::'1 + e - jnbM)
for the
(24-37)
This leads to
F(z)
= -
~
21('"n~O
We can then employ a previous
Fz
()
=-
+ ~
ejnbMz-n
'"
e-jnb:o.lz-n
relation
.
+
1-
.
e- 1)]bMZ-l
r+
once more to return to trigonometric
(z)
=
(24-38)
result given in (24-8) and (24-9):
21( 1 - e]bMrl
1
F
)
n~O
- (e jb:o.l
+ e - jMI)
=:21 ( 1 - 2 (ejbM
+ e-jb:o.l)z-1
Using Euler's
(24-36)
0
for the power
cas nb!::.t
where j
bt)
1 - Z -1 cos b!::.t
1 - 2z-1 cas b!::.t +
1 Z-2 )
functions
(24-39)
r2
Note that for b = 2mr/!::.t, F(z) = 1/(1 - Z-I), which is the same expression as
the z-transform of the unit step function. In other words, the sampled cosine has
the identical appearance of the sampled unit step function Un = 1 for all n), an
example of aliasing (see Fig. 22.5). Hence, the two z-transforms are identical in
this 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)
Equation 24-40 implies that we substitute
This step gives
=
zea':'t
every place z appears
1 - z-le-a::'1
Z(e-al
EXAMPLE
cos bt)
1 - 2z-1e-a::'1
(24-40)
F(zea':'t)
in (24-39).
cas b!::.t
cas b!::.t
+
z-2e-2a:o.{
(24-41)
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.
24.1 The z-Transform
565
Solution
Using
(k
+
(24-29) as the definition of the modified z-transform
m) Ilt and N = 0 for the value of t in e -at yields
F(z,
m)
=
and substituting
rl L e-a(k+m)j.{z-k
(24-42)
k~O
e-amj.{Z-1
L e-akj.{z-k
(24-43)-
k~O
Using (24-8) and (24-9) with k as the summation
index gives
e -am::H
F(z,
= 1-
m)
Z-I
(24-44)
e -aM z
When m = 0 (CT = 1), the numerator of (24-44) becomes z -I. This term
indicates a one unit time delay (N = 0 and CT = lor N = 1 and CT = 0). Therefore
it 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-transforms
and F(s). Table 24.1 provides a representative
list
corresponding to functionsf(t)
of z-transforms; more extensive tables can be found in Ogata [1] and Deshpande
and Ash [4].
EXAMPLE 24.4
Given the transform,
1
F(s)
-
s(s
+
a)
(a
> 0)
which might represent the step response of a first-order
steady-state value, lim f(nllt).
system, determine
the final
J1~X
Solution
From Table 24.1,
F(z)
= ~1 ( 1=-1Z-I
Then from the Final Value Theorem,
!
!
n~X
limf(nllt)
=
limf(nllt)
z~ 1
= a lim
n~X
a lim
z~1
[(1 -
[1 -
Z-I)
Z z-
(
1 - 1z _]
1-
e :aj.{ z
-I)]
!
-e _la~t] = a
This result agrees with the continuous-time
limit of f(t) as t ~ x obtained from
the 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?
566
DYNAMIC RESPONSE OF DISCRETE-TIME SYSTEMS
Table 24.1
z-Transforms (At = Sampling Period)
Time Function f(t)
Laplace
Transform F(s)
1
s
unit step
Set),
z-transform
F(z)
1
r1
1-
I:::.tr1
(1 - r1f
(n -
I)!
lim -1 n-I __
sn
a~O
(
aan-1
an-1
)
1
s +
1
ab
1
(
-
e-bl)
+ -a -
Set)
b b
+
(s
e-al
1
b e-al sin
a)(s
+
b)
a)(s
b ~ a
bt
C-
+
b
(a - b) (1 - e-aMz-l)
!1t e -aMz-I
(1 - e-acl1r1f
1
(s + a)2 + b2
b
1-
2r1e-acll
1b2
sin bl:::.t
r1e-acll
1-
cas b!1t
rle-a~1
2rle-acll
+
e-2acllr2
cas b!1t
cas b!1t
+
e-2aMr2
1
unit impulse
INVERSION
e~b~lrl)
a]
1
+ a)2
s + a
+ a)2 +
1-
b)
1
(s
- kAt)
-
_
e-al cas bt
24.2
)
e-acl1z-1
e~aMrl
2.
1__ 1_ +
ab Ll - rl
e-bl)
(s
f(t
e-aMz-1
1
1
1-
a
1
s(s +
__a -a_b
S(t),
1-
1
b _ a (e-al
-
(
F(z)rk
F(s)e-kMS
OF z-TRANSFORMS
Once a z-transform has been obtained (by whatever means), we need to be able
to obtain the values of its corresponding
time-domain
function at the sampling
instants. 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 time
function. Instead the values of the function are obtained only at the sampling
instants. We know that a variety of continuous signals can be reconstructed
from
f*(t); that is, aliasing prevents the unique identification of the continuous function
of 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-transform
operator, denoted by Z-1, as follows:
f*(t)
= {f(nllt)}
= Z-l[F(z)]
The inverse z-transform consists of the sampled
nth sampling instant as f(n/1t).
To illustrate the inversion process, consider
(24-45)
values f*(t),
F(z)
represented
= rl/(l - P1r1).
at the
If F(z)
24.2
Inversion of z-Transforms
567
above is expanded as an infinite series,
1
-
=
~PIZ
+ PIZ-I +
rl(l
PIZZ-z
+ ...
+
+ ... )
PI"r"
(24-46)
Comparing this expression to (24-1), note that the inverse z-transform gives an
expression for the value of the function at the nth sampling instant
(24-47)
In most cases the z-transform to be inverted consists of a ratio of polynomials in
z -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 Laplace
transform 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 rather
than z. Consequently, each term in the partial fraction expansion should be in this
same form.
Suppose that F(z) has the following form:
=
F(z)
(24-48)
VI(z)
Vz(z)
where VI (z) = kth-order polynomial in Z-I (excluding a possible time-delay term
Z-N, which can be factored)
Vz(z)
- mth-order polynomial in Z-I (the denominator is normalized so
that 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)
=
(1 -
VI(z)
- pzZ-I)
PIZ-I)(l
...
(1 -
(24-49)
Pmz-I)
Then choose a partial fraction expansion of the form:
F(z)
__
r_l__
1 - PIZ-1
+ __
1 -
r_z__
pZZ-l
+ ...
+ __
1 -
r_m
Pmz
(24-50)
Each numerator coefficient r; can be calculated in a manner similar to that used
for Laplace transforms (e.g., Heaviside's rule with z = lip;). Taking the inverse
z-transform of (24-50) term by term gives
f(n!::.t)
= Z-I( 1 -
rlPIZ _)1
+ Z-l( 1 -
rz
pzz _)I
+ ...
Since the inverse transform of
f(nt:..t)
=
rl/(l
rl(Pl)"
- PIZ-l)
+
rz(pz)"
is
(24-51)
rl(PI)",
+ ...
+
then
rm(PnY
(24-52)
568
DYNAMIC
RESPONSE
OF DISCRETE-TIME
If there were only a single term
simple case, we can examine how
of the sign and magnitude of PI'
different values of PI along with
case of a first-order z-transform.
SYSTEMS
in the z-transform, then f(n6.t)
= rIPI". For this
the sampled representation
will vary as a function
Figure 24.2 shows the discrete-time responses for
possible continuous-time
interpretations
for the
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)
Relating
rle-q,,,:>t
+
r2e-q2,,:>r
+ ... +
(24-53)
rme-qm,,:>r
(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-order
polynomial in the denominator
for each pair of complex roots. When inverting
such a term, the appropriate quadratic form for damped sines and cosines in Table
24.1 should be used.
One other unusual case can arise in using the partial fraction expansion
cedure. If the order of the numerator is greater than that of the denominator
k > n in Eq. 24-48), then Eq. 24-50 is not strictly applicable. In this case,
preferable to use other methods for generating the discrete-time sequence,
as the long division method discussed below.
EXAMPLE
24.5
Using
partial
F(z)
a
fraction
- z-I)(l
= 0.5z-I/[(1
expansion,
find the inverse
z-transform
- 0.5z-I)] for a sampling period 6.t = 1.
pro(i.e.,
it is
such
of
Solution
Expanding
F(z)
into the sum of two fractions
rl
0.5z-1
F
(z)
= (1 _ z-I)(l
- 0.5z-1)
r2
= 1 - Z-I + 1 - 0.5z-1
Imaginary
z-plane
yields
bmrr
/;b1ill
(1)
Real
trrn-m,
Figure 24.2
Time-domain
responses for different locations of the root of F(z).
(24-54)
24.2
569
Inversion of z-Transforms
Using Heaviside's rule (see Section 3.3) for finding rl and rz, multiply
(1 - Z-I) and set z = 1:
F(z)
by
0.5
= -0.5 = 1
rl
Next multiply F(z)
by (1 - 0.5z-1)
rz
Substituting
and set z
= 0.5, that is,
0.5(2)
-1
= -=
1 - 2
Z-1
= 2:
into (24-54) gives
1
Equation
1-
= 1
F( z)
(24-55)
0.5z-1
= 1 leads to
1
ql = - b.t In (1) = 0
24-53 with b.t
1
= - M
qz
= 0.693
In (0.5)
so that
= 1-
f(nb.t)
(24-56)
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)
1,rz-
-1,
Long Division
Long division provides a second method for obtaining an inverse z-transform. In
most cases it is considerably easier to use this method to obtain the inverse ztransform than to use partial fraction expansion. However, the result (an infinite
series) may not be as useful as an analytical expression. Inversion via long division
is an operation unique to discrete-time systems; no analogous method exists for
continuous-time
systems. From the definition of the z-transform as
F(z)
=
L f(nb.t)z-II
(24-57)
11=0
the coefficients in the power series expansion
sampling instants:
= f(O) +
at the
(24-58)
=-------------+ bzz-z + ... + bkz-k
bo +
ao + alz-1
+ azz-z + ... + amz-m
(24-59)
Let F( z) be a rational function
f(M)z-1
+
are the values of f(t)
+ ...
F(z)
Fz
()
of F(z)
represented
f(2M)z-Z
by
b1z-1
Dividing the denominator
into the numerator
Co
+
'The order of division is based on dividing
ao
F(z)
=
C1Z-1
by long division
+ czz-z + ...
1
gives
(24-60)
into the numerator and its remainders; see Example 24.6.
570
DYNAMIC
RESPONSE
OF DISCRETE-TIME
SYSTEMS
Referring back to (24-58), we can perform long division and equate the sequences
and Un} to obtain
{cn}
bo
ao
0
fjfo
=
=
Cj
C2 _
ao
ao
+
bj __ bO~j
bo
bOa2 _ bjaj
ao-
(24-61)
(24-62)
(24-63)
bOaj2
Some important properties of sampled-data systems can be obtained from long
division of their z-transforms. For example, for a first-order z-transform,
=
F(z)
bo
-1
(24-64)
-
-
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 account
the negative sign before aj in (24-64). Note that if lajl > 1, the magnitude of the
signal grows steadily over time (see Fig. 24.2), but if lajl < 1, the signal will be
attenuated over time. This coefficient in a first-order z-transform indicates if a
system 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-1
O.5rl - O.75z-1 + O.25r3
O.75z-2
O.75z-2
-
O.25z-3
1.125z-3
O.875z-3
O.875r3
+ O.375r<
-
O.375z-<
1.3125r< + 0.4375rS
O.9375r< - 0.4375r5
O.9375r< - 1.4062z-S
O.9687z-S
-
0.4688z-6
+
O.4688r6
Note that f(O) is zero in this expression. Long division does yield the same time
sequence as does partial fraction expansion, with considerably less effort. However,
the result is in the form of an infinite series.
24.3 The Pulse Transfer Function
571
(c) Contour Integration
A final method for inverting z-transforms utilizes a contour integral:
f-. Jr( F(z)zn-l
=
f(nflt)
'IT)
(24-66)
dz
r
where the contour must be appropriately specified. Although the integral can be
evaluated using the residue theorem [1, p. 83], this method is seldom used in
practice.
24.3 THE PULSE TRANSFER FUNCTION
In analogy with continuous systems, the analysis and design of sampled-data
control 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 the
ratio of the output and input z-transforms, assuming both output and input are
initially at steady state, analogous to continuous-time systems. In addition, both
the 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 the
convolution integral [1, p. 180],
yet)
= L get -
(24-67)
T)X*(T)
dT
where get) is the impulse response of the process (see Chapter 3), T is the dummy
variable of integration, and X*(T) is a series of impulses that can be expressed as
X*(T)
=
L x(kflt)
OCT -
(24-68)
kflt)
k~O
Substituting (24-68) into (24-67) gives
x
t
yet)
=
10
get -
As shown by Ogata [1], an impulse
arbitrary function h( T),
J: h(T)
OCT -
xes)
t:o x(kflt)
T)
kflt)
oCt)
dT
=
OCT -
kflt)
(24-69)
dT
has the important property that, for an
for 0 :S T :S t
h(kflt)
(24-70)
yes)
---
x(t)
y(t)
(a) Continuous
input
Y*(s)
y*(t)
(b)
Sampled input and output
Figure 24.3
Transfer function with (a) continuous input and (b) sampled input.
572
DYNAMIC
RESPONSE
OF DISCRETE-TIME
SYSTEMS
Thus, Eq. 24-69 reduces to
yet)
= ~
(24-71)
get - kM)x(kl:1t)
k~O
In particular, for t
=
nl:1t
y(nM)
= ~ g(nl:1t - kl:1t)x(kl:1t)
. (24-72)
k=O
The z-transform of the output signal is defined by
= ~ y(nl:1t)z-1l
Y(z)
(24-73)
Il~O
Substitution of y(nl:1t) from (24-72) gives
= ~ ~ g(nM
Y(z)
Let i
=n -
Il~Ok~O
(24-74)
- kl:1t)x(kl:1t)r"
k:
Y(z)
Since g(il:1t) is zero for i
summations separated:
= ~
i~
~ g(il:1t)Z-(i+k)X(kl:1t)
(24-75)
-k k~O
< 0, the
lower limit on i can be changed to zero and the
(24-76)
or
Y(z)
=
(24-77)
G(z)X(z)
where G(z) is defined as
x
G(z)
(24-78)
~ ~ g(il:1t)Z-i
i~O
We will refer to G(z) as the pulse transfer function of the system. It relates the
discrete-time input and output signals in the same manner as a transfer function
in the s-domain relates continuous signals (see Fig. 24.4).
Other derivations of (24-77) are available [1]. Note that G(z) can be calculated
directly after get), the impulse response function, has been determined (d. Eq.
24-78).
EXAMPLE 24.7
Find
G(s)
the
=
pulse
K/(TS
transfer
function
for
a first-order
+ 1).
~
~~)
Figure 24.4
Pulse transfer function.
continuous
process,
24.4
Relating Pulse Transfer Functions to Difference
Equations
573
Solution
The continuous time impulse response
is found from
get)
K
_
= ,\:,-I[G(S)] = -T
get)
(24-79)
e-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
(24-81)
O.5712z-2
Determine its discrete-time response when forced by a sampled unit step input.
Solution
For a unit step input,
X(z)
= 1 + Z-l +
1
Z-2
+
Z-3
+ ...
1 - Z-l
Here we use the closed-form expression to minimize the number of terms. Equation
24-77 is used to develop the expression for the step response Y( z). We could apply
the partial fraction expansion method to find y(ntlt), but this would be relatively
time-consuming due to the need to factor the denominator and use the Heaviside
expansion. Therefore, long division is employed to determine the response. Substitution gives
Y(z)
_- G(z)X (_z) -
1
-O.3225r1
r. r.""",,,
__
+
1
,
O.5712z-2
r. """,,'H
_,
,
1
_1
(_4-8
? 2)
Applying the long division procedure yields
Y(z)
=
-O.3225z-1
O.0665z-2
-
+
O.6918z-5
+
O.8082z-6
+
O.8820z-7
+
O.9277z-8
+ O.2568z-3 + O.5136z-4
+ ...
(24-83)
We note in (24-83) that y(ntlt) is steadily increasing and may be approaching a
steady state value. Using the Final Value Theorem. we can calculate the value of
y(ntlt), as n -,> x. Returning to (24-82), multiply by (1 - Z-l) and set z = 1. The
ultimate (steady-state) value for the response is thus 1. Since at steady state both
input and output values are 1, the gain of the pulse transfer function is also 1. This
can be verified by evaluating G(Z)IZ~I'
24.4
RELATING PULSE TRANSFER
TO DIFFERENCE EQUATIONS
FUNCTIONS
A pulse transfer function, representing a dynamic relation between an input and
an output, has a unique correspondence with a difference equation. To demonstrate
574
DYNAMIC RESPONSE OF DISCRETE-TIME SYSTEMS
this, consider
aOYIl
+
a general
difference
+ ... +
alYn~1
equation
amY,,-m
=
given by
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 in
Eq.24-11:
=
Z(YIl-J
Taking the z-transform
aoY(z)
+
Collecting
alz~1y(z)
Z[y(n~t
-
=
i~t)]
(24-85)
riy(z)
of both sides of Eq. 24-84 gives
+ ... +
amz-my(z)
boX(z)
+ b1rl
X(z)
+ ... +
bkZ~k X(z)
(24-86)
terms, we solve for Y(z):
Y ()z
The pulse transfer
=
function
G(z)
--
----------X
+ b1z-1 + ...
+
+ ...
bo
ao
alz-1
of the discrete-time
Y(z)
X(z)
bo
+ bkz-k
+ amz-m
()z
process is therefore
+ b1rl + ... + bkz-k
+ alz~1 + ... + amz-m
-------ao
For most processes bo is zero, indicating that the input does
affect the output. However, for proportional
controllers and
modeled simply by steady-state gains, all coefficients except
Also, the leading coefficient in the denominator
can be set
both numerator and denominator
by ao.
(24-87)
given by
(24-88)
not instantaneously
processes which are
ao and bo are zero.
to unity by dividing
Physical Realizability
In Chapter 4 we addressed the notion of physical realizability for continuous-time
transfer functions. An analogous condition can be stated for a pulse transfer function, namely a discrete-time model cannot have an output signal that depends upon
fut~re inputs. Otherwise the model is not physically realizable. Consider the ratio
of polynomials given in Eq. 24-88. The transfer function will be physically realizable
as long as ao #- 0, assuming that G(z) has been reduced so that common factors
in numerator and denominator have been cancelled. To show this property, examine
Eq. 28-84. If ao = 0, the difference equation is
This equation requires a future input x"
impossible (unrealizable).
Another way to
division. When the denominator in (24-88)
with positive powers of z should occur for
EXAMPLE
to influence YIl-1, which is physically
test physical realizability is to use long
is divided into the numerator, no terms
a physically realizable system.
24.9
Check the transfer
function
Y(z)
X(z)
for physical realizability.
1 + 2z-1 +
5z-1
+
3Z~2
2z-2
(24-90)
24.4
Relating Pulse Transfer Functions to Difference
Equations
575
Solution
Note that the leading coefficient in the denominator is a power of Z-I, violating
physical realizability (ao = 0 in (24-88)). The corresponding difference equation is
5Yn-l
+
=
2Yn-2
x"
+
2X"-1
+
3X"_2
(24-91)
An equivalent difference equation is
5y"
+
2Yn-l
=
Xn+l
+
2x"
+
3X"-1
(24-92)
This equation requires knowledge of the future input XI1+1 to generate the current
output value Ytl" A similar conclusion can be deduced using long division, which
yields 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 digital
output signal from the controller to an analog signal, which can then be utilized
by a final control element such as a valve to manipulate the process. In many cases
the final control element requires a continuous signal as input rather than a digital
input 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 employed
for this purpose in process control is the digital-to-analog converter (DAC) which
functions as a zero-order hold (ZO H), although there are other types of holds
available (see Chapter 22). The zero-order hold converts the digital signal from
the 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 be
thought of as an idealized sequence of impulses through the ZOH and then through
the process. We know from Laplace transform theory that if H(s) is the ZOH
transfer function and G(s) is the process transfer function (including the final
control element), then the overall transfer function is H(s)G(s).
Although Table
24.1 gives the corresponding discrete-time form for a first- or second-order transfer
function (see Example 24.7 for the first-order case), this table is based on an input
that can be represented as a series of impulses. Thus, it cannot be used directly
for the type of input that is characteristic of a ZOH (DAC) output, which is a
staircase function. However, by taking the hold device into account, we can derive
pulse transfer functions that are appropriate for process control calculations.
First we derive the appropriate expression for H(s). Figure 24.5 shows the
response of the hold device to an impulse input of unit strength. The hold yields
a constant output value over the sampling period I:::.t, which is the same as the
strength of the impulse. The impulse response of the ZOH over the interval
hW
:Cl
o
!:J.t
Time
Figure 24.5
Response of a zero-order hold element to an impulse of unit
magnitude.
576
DYNAMIC
RESPONSE
OF DISCRETE-TIME
t = 0 to t = 6.t can be written
and S(t-6.t):
SYSTEMS
as the difference
h(t)
=
Set)
of two unit step functions,
S(t-M)
-
Set)
(24-93)
of h(t) is
From Eq. 3-22, the Laplace transform
1 e -s6.1 1 - e -s6.t
H(s) = - - s
s = --- s
(24-94 )
For processes with a piecewise constant input, the difference equation model must
be based on the product H(s)G(s)
rather than G(s) alone. Then Table 24.1 can
be 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,
=
G(s)
1
+ 1
TS
Show by transform techniques that a zero-order hold placed ahead of this process
will yield the same difference equation model for the combined system (ZOH plus
first-order process) as was derived in Eq. 23-19 for piecewise constant inputs to a
first-order process.
Solution
First form the product
H(s)G(s):
1 - e -s6.1
H(s)G(s)
To convert this expression
expansIOn:
H(s)G ( s)
s
to its equivalent
= ~1 -
s-+-l/-T
1
TS
1
+ 1
z-transform,
(24-95)
we use a partial fraction
(24-96)
- e -s6.t C
~ - s-+-l/-T
1)
HG(z) is defined as the z-transform of the combined ZOH plus process. Z is used
as a shorthand way to denote the z-transform of the time-domain function given
by the inverse Laplace transform,
HG(z)
= Z [H(s)G(s)]
- Z
~
C)
= Z{cS:'-1[H(s)G(s)]}
(_1 ) -
- Z s + 1/T
Table 24.1 is then used to convert
z-transform. Since Z [e-S6.tp(s)] =
HG(z)
e
~ - su_1
+ liT
Z [-S6.tC
)]
(24-97)
each term in Eq. 24-97 into its equivalent
Z [pes)], Eq. 24-97 becomes
r1
= C _1 Z-1 - 1 _ e ~6.tI'Z-1)
- z -1( 1 _1Z-1 - 1 _ e-6.II'z-1
1)
(24-98)
24.4
Relating Pulse Transfer Functions to Difference
577
Equations
or
= (1 - z -I ) [ (1 _ z-l(l
z-I)(l - _e-M/T)
e-Llt/Tz-I)
HG(z)
]
Z-I (1 -
e-LltIT)
e-M/Tz-1
= 1Defining
al ~ e-M/T,
Y(z)
X(z)
In difference
=
(24-99)
(24-99) can be written
= (1 -
HG(z)
equation
as
(24-100)
al)z-l
alz-I
1-
form, (24-100) becomes
y" -
alY,,-1
= (1 - al)
(24-101)
X,,_I
or
y"
=
+ (1 - al)
aIY,,_1
X,,_I
(24-102)
the same result as given in Eq. 23-19.
Several
comments should be made about the procedure for transforming
First recognize that the combined transfer function H(s)G(s)
could
have been converted to a series of impulse terms as in Example 24.7 by using the
time-domain impulse response followed by transformation
to the z-domain. The
approach taken in Example 24.10, however, is more direct, because Table 24.1
provides the relation between sand z, thus bypassing the need to convert from
the s-domain to the t-domain (and then to the z-domain). We can also generalize
the results in (24-97) to (24-99) as
H(s)G(s).
HG(z)
=
= (1 -
Z(H(s)G(s))
Z-I)Z(G;S))
(24-103)
In the second term, the operator Z indicates that the z-transform equivalent of
G(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
to
its z-transform. Finally multiplication by (1 - Z-I) yields HG(z).
Equation 24-103 can also be employed to illustrate an important property of
the zero-order hold; in general, for dynamic systems,
¥
HG(z)
First calculate
the z-transform
of H(s):
e
H(z)
=
H(z)
1 - Z-I
= 1 - ~Z-I = 1
Z
(24-104 )
H(z)G(z)
se-s;!.t)
= (1 -
¥
G)
(24-105)
(24-106)
This result is consistent with the fact that
H(z)
= 1, Eq. 24-104 becomes
HG(z)
Z-I)Z
the gain of H(s)
G(z)
is unity.
Using
(24-107)
578
DYNAMIC RESPONSE OF DISCRETE-TIME SYSTEMS
Substituting (24-103) for
the above expression is
HG(z),
rl)Z [G;S)]
(1 -
¥
Z[G(s)]
(24-108)
The inequality in 24-108 is valid, in general, except for the case G(s) - K. One
other interesting characteristic of the zero-order hold is that
lim
=
HG(z)
(24-109)
G(s)
clt~O
This result seems intuitively correct in view of Fig. 22.2b. For example, if we
substitute z = esclt in Eq. 24-99 and apply L'Hospital's rule for M -7 0, the
z-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 the
discontinuous 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
which is
G(s)/s,
lIs2.
The z-transform is
/1t
Z(1/s2)
rl
= (1 _
Multiply (24-110) by (1 - Z-I), yielding
(24-110)
Z-I)2
HG(z):
= 1M_ rlZ
HG(z)
(24-111)
The corresponding difference equation is
Yn
-
Yn-I
-
(24-112)
/1t Xn-I
The infinite series version of (24-112) can be obtained by long division of
in (24-111) and conversion to discrete-time form:
HG(z)
n
Yll
(24-113)
/1t ~ Xk-I
k=1
Equations 24-112 and 24-113 describe the two equivalent forms of the integrating
element 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)
Xes)
=
(TIS
+
Ke-es
1)(T2S
+ 1)
(24-114)
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
following difference
equation
Effect of Pole and Zero Locations
579
results:
(24-115)
The relations between (a), az, b), bz) and (K, T), TZ) have been previously derived
in Eqs. 23-21 through 23-25, based on the analytical solution of the differential
equation. The pulse transfer function for (24-115) is
G(z)
= Y(z)
X(z)
= (b1r1
l+alz.
+ _~zz-Z)z=~
+azz-
_
(b1
+ bz_~-I)rN=~
l+alz
+azz-
(24-116)
Note that (24-116) is a general expression for a second-order
discrete-time
model with a time delay of N sampling periods (the apparent time delay is one
sampling period longer (i.e., N + 1) in (24-116) because the output cannot respond
instantly). Neuman and Baradello [6] have derived difference equations incorporating the zero-order hold for a variety of linear process models. Table 24.2 gives
pulse 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 using
time 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 in
discrete time as ePiM (recall the transformation
z = eSM).
A process with two
time 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 az
in a second-order difference equation (see Eqs. 23-22 and 23-23).
It is possible to categorize discrete-time responses for first- and second-order
continuous-time
processes with no zeros into eight different patterns [1], as shown
in Fig. 24.6a-h. All responses exhibit an apparent one-unit time delay as noted
above. For first-order systems, recall that in Fig. 24.2 we identified several possible
responses depending on the pole location. The appropriate first-order difference
equation is
Yll -
PIYIl-l
-
XIl-l
(24-117)
24.3 gives the response for the case of a unit impulse input at n = 0
Yo = 0). The analytical solution is Yll = (Pl)ll·
It is clear that a negative
pole near the unit circle has a pronounced effect on the response. The alternation
in sign of Y is referred to as ringing of the output signal (see Chapter 26 for a
discussion of ringing in control systems). Negative poles nearer the origin, although
they produce a change in sign, are heavily damped and their results are not so
noticeable. Positive poles do not cause the output to change in sign.
In Fig. 24.6 case c corresponds to a second-order overdamped
system with
two positive poles on the real axis. It has similar properties to case b. Cases d and
e are noteworthy, because the two complex poles in continuous time map into a
single pole in discrete time. In case e, the pole is located at s = ±w)2j in continuous
time, 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
(xo
= 1 and
<.n
CO
o
Table 24.2
Pulse Transfer Functions with Zero-Order Hold
Transfer Function
G(s)
HG(z)
~
K
1
K
r
1
K
(.I
+
+
bl
+ 1')
+
K
+
+
r)
(1\z-}
=
=
-I
o
Kt.!
m
JJ
o-0Z
al
= -exp( -rt.!)
bl
= -r [I - exp( -rt.!)]
(J)
m
K
+
b1z-2
+
(l\Z-1
blz-I
z»
s::
(J)
alz-I
bIZ-I
r)(s
.1(.1
air I
o
-<
Z[H(s)G(s)]
al
+
blrl
+
.I
=
+
(l2Z-2
"o
o
= - {exp( -rt.!)
+ exp( -pt.!)}
a, = exp[ - (r + I' )t.!]
bl
= [Klrp(r - p)][(r - 1')- rexp(-pt.!)
+ pexp(-rt.!)]
+ p)t.!] + I' exp(-pM)-rexp(-rt.!)}
b, =IKlrp(r
- p)I{(r - 1') exp[-(r
al
en
()
JJ
m
--i
m
::J
s::
b,Z-2
+
(12Z-2
m
= -{I + exp(-rt.!)}
a, = exp( - rM)
al
hi
h,
=
=
-(Klr')[l
(Klr')!l
(J)
- rt.t -
-<
(J)
--i
exp(-rM)]
exp( -rt.!)
-
m
rM exp( -rt.!)]
s::
(J)
K(s
(.I
+
+
r)(s
'I)
h]z- [ + h2z-2
+ 1')
+
([IZ-!
+
(l2Z-2
= -{exp( -I'M)
+ exp( -rM))
a, = exp[ -(r + p)M)1
al
K
bl
= --{exp(-pt.!)
I' -
r
b, = K{(qlrp)
K
(.I
+
r)(.\'
+
h1z-'
1')(.1
+
v)
I
+
(ltZ-!
+ [,22-2 + h3z-J
+ (l2Z-'2 + (lJz-J
exp[ -(r
- exp(-rM)
+ ('1/1')[1 - exp(-pt.!)1
+ p)t.!1 + [(I' - q)/p(r
- 1')1 exp( -rt.!)
- (qlr)!1
-
exp(-rM)J)
+ [('I - r)/r(r
- 1')] exp( -pt.I)}
= -{exp( -rt.t)
+ exp( -I'M)
+ exp( -vt.!)}
+ 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)[H + 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}
al
a, = exp[-(r
I
b,
K(s
(s
+
.1'(.1'
rHs
+ 'I)
+ pHs +
K(s
+
+
r)(,,'
1
b,2-' + b,Z-2 + b,z-'
+ a,z-' + a,r' + a3z-'
b,r'
'I)
+
v)
1')
I + a\z-I
+ b,r2 + b,z-'
+U2Z-2
+
uJz-J
=
+
{[ru(q
-
p)]/[(r
-
p )(u
-
p)]}{
+
{[rp(q
-
v)J/[(r
-
vHp
-
v)]}{
cxp[-
[KI(rpu)](
-
'I
+
+
+
{[pv(q
-
r)]/[(p
{[rv(q
-
p)]/[(r
{[rp(q
-
u)]/[(r
+
(I'
-
+
P
cxp( - rl>t)
cxp( -rl>t)
+ cxp( - vl>t) + cxp[- (r +
+ cxp( -pl>t) + cxp[-(r +
v)l>t]}
p)l>t]})
v)l>t]
r)(u
-
r)]}
-
pHu
-
p)]}
-
v)(p
-
v)]}
cxp[ -(p
cxp[-(r
cxp[-(r
+ v)l>tJ
+ u)MJ
+ p)l>t»
a, = - {cxp( - I'M) + cxp( - pl>t) + cxp( - vl>t)}
= cxp[-(r + p)l>t] + cxp[-(p + u)l>tJ + cxpl-(r + v)l>tJ
+ p + v)M)]
a3 = -cxp[-(r
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)
+ p)l>t] + cxp[-(p + v)l>t) + cxp[-(r + v)l>t])
b, = [-KI(rpv)]{-q(cxpl-(r
+ {[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,
a,
"2
11,
b,
b2
= -{I + cxp(-rl>t)
+ cxp(-pl>t)}
= cxp( -rl>t) + cxp( -pl>t) + cxp[-(r + p)l>tl
= -cxp[- (r + p)l>t]
= (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)])
= (- 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
+
1>,
=
{[r(q
-
1')]/11'(1'
(Klrp({[(ql>t
+
+
{[r(q
{[p(q
-
+
1)1'1'
1')[/11'(1'
-
1')1/[1'(1'
r)IHI +
-
-
'1(1'
+
r)]}cxp(-rl>t)
p)]}
cxp( -pl>t»
"'"
:".
JJ
CD
§I
S'
(Q
-0
c
en
CD
-I
03
::J
(f)
"
~
C
::J
"-
0'
::J
(f)
o
o
:::;;
CD
(i)
::J
()
CD
m
2 cxp(-rl>t)j)
p)l/(rp)}cxp[-(r
'"
+
..0
p)l>t]
c
~
0'
::J
(f)
U1
CO
....•.
582
DYNAMIC RESPONSE OF DISCRETE-TIME SYSTEMS
$
b,
c,
W
ill,
lm,
-$
W
'
W'
-$w'
-+
=f
=t:f'rr
s-plane
First-order
-ws/2
systems
(s-domain)
(a)
y(l)
z-plane
,
~,
~,
~, ~
~,
(b)
Second-order
systems
(s-domain)
(c)
(d)
(e)
(f)
(g)
(h)
Figure 24.6
Effect of pole locations on impulse response.
y*(1)
+ {[ru(q - p)]/[(r
+ {[rp(q - u)]/[(r
b,
=
[KI(rpv)](-
qcxp[-(r
+ {[pv(q - r)]/[(p
+ {[rv(q - p)]/[(r
+ {[rp(q - v)]I[(r
K(s
(s
+
rHs
+ 'I)
+ p)(s +
b12-1
v)
1
+
+ b,Z-2 + b,z-'
+
(11Z-]
(12Z-2
+
(1:1Z-3
a1
a,
a,
b1
b2
b,
- p)(u
-
r)(v
+ cxp(-uM)
+ cxp( -pM)
- p)J}{cxp(-rt.l)
u)(p +p
-
u)]}{ cxp( -rt.l)
+ u)t.l]
r)J} cxp[-(p
- p)(v - p)J} cxp[ -(r
- vHp - v)]} cxp[-(r
+ exp[-(r
+ cxp[ -(r
+
+
u)M]}
p)M]})
+ v)M]
+ u)t.l]
+ p)M))
= -{cxp( -rt.l) + cxp( -pt.l) + cxp( -vt.l))
= cxp[-(r + p)M] + cxp[-(p
+ u)t.l] + cxp[ -(r + U)t.l]
= -cxp[-(r
+ I' + V)t.l)]
= [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)])
= [-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]})
= [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)
I\)
"'
:".
:IJ
<1>
![
S'
to
-u
c:
en
<1>
-i
OJ
::J
(J)
K(s
s(s
+
+
rHs
'I)
+
b12-1
1')
1
+
+
(1!Z-1
b22-2
+(/2Z-2
+ b,z-'
+
U3Z-J
= -{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(
+ (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(q - r)l/lr(r - p)]}11 + 2cxp(-pt./)]
~
a1
+ {Ir(q - 1')1/11'(1' - r)]}[ I + 2 cxp( -rt./)J)
- I)rp + q(r + p)l/(rp)}
cxp[ -(r
b, = (Klrp({[(qt./
+ {[r(q - 1')]/11'(1' - r)J} cxp( -rt./)
+ {[p(q - r)]I[r(r - p)]} cxp( -pt./))
T1
c:
::J
>1
o'
-pM)1
::J
(J)
o
o
=0
<1>
+
p)t./]}
ro
::J
()
<1>
m
+
p)t./]
.D
c:
~
o'
::J
(J)
01
CO
•...•.
582
DYNAMIC RESPONSE OF DISCRETE-TIME SYSTEMS
c,
b,
lllilL
-$
W'
-$W'
=t:f'iT-+
s-plane
First-order
y(t)
-ws/2
systems
(s-domain)
(a)
z-plane
~,
~,
~,
~, ~,
~
(b)
Second-order
systems
(s-domain)
(c)
(d)
(e)
(f)
(g)
(h)
Figure 24.6
Effect of pole locations on impulse response.
y*(t)
24.6
Table 24.3
Conversion
First-order Difference Output as a Function of Pole Location (unit
impulse input)
Xn
00+
1PI-0.9
00.656
-0.9
+0.81
-0.729- =
10.590
0101-0.3
+0.09
-+PI
0.0024
0.8
0.64
0.410
0.512
0.027
0.328
0.0081
= 0.8
-0.3
can oscillate.
tem.
583
Between Laplace and z-Transforms
Such behavior
Yn
is not possible with a first-order
continuous-time
sys-
Oscillation can also occur for second-order discrete-time systems, if the poles
have imaginary (complex conjugate) values (see cases f-h in Fig. 24.6). When a
positive or negative zero occurs in the discrete-time model, the degree of oscillation
as well as its frequency can be affected. Unfortunately it is not possible to categorize
this case easily, so we refer the reader to some examples presented by Franklin
and 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 equation
resulting from a second-order continuous-time
transfer function with no zero. In
Eq. 24-116, the poles are found from factoring the denominator
polynomial into
two roots. The zero of the discrete-time transfer function is - bz/ b1, which is fairly
complicated 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 the
sampled response [7]. For example, an inverse response in continuous time (see
Fig. 6.3) may not be observed at the sampling instants if the sampling rate is too
slow.
24.6
CONVERSION
BETWEEN
LAPLACE
AND z-TRANSFORMS
We have previously seen that Table 24.1 can be used to convert Laplace transforms
to 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 is
explicitly considered in this approach.
As discussed earlier, the transform variable z was defined by z = eS::'t or
Z-1 = e-s!lt. To obtain an approximate
relation expressing s in terms of a ratio of
polynomials in z, we can use the Pade approximation
for e -s!lt
_ 2 - stit
e -stlt =---
2 + stit
Equating
to
r
J
(24-118)
gives
z- 1
:::=
2 -
---
2
stit
+ stit
(24-119)
584
DYNAMIC
RESPONSE
OF DISCRETE-TIME
SYSTEMS
or
(24-120)
The approximation suggests that a Laplace transform can be converted to a
z-transform by substituting (24-120) for s. Such an approach is known as Tustin's
method [1]. A less accurate expression for s can be derived using the power series
e-s':'t
= 1-
s26.t2
s6.t
+ -- 2
- ...
(24-121)
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 Chapter
23. 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 a
transfer function G(s). Approximate substitution is a procedure that should always
be used with care. Exact conversion, especially of the process model (up to third
order), is recommended. A bilinear transformation similar to (24-120) in form is
sometimes 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)
TIS +
= Ke( 1 + ~
TDS)
(24-123)
using the approximation in (24-122). Compare your result with the velocity form
of 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
P(z)/E(z)
the error signal and
and
p"
is the output from the controller, then
Gc(z)
=
(24-126)
Summary
Using the real translation
discrete-time form gives:
theorem
and converting
the controller
equation
585
into a
(24-127)
Substituting for aQ,
settings Kn Tf, and
Pn - Pn-I
=
al>
TD
and
gives
az
Kc [ (en -
and collecting
en-I)
Note that this equation is identical
difference approximation.
+
---:-en
-II
IlT
+
terms with respect
A(en
~t
TD
-
2en_1
+
to the controller
en-z) ]
to Eq. 8-18 which was derived
(24-128) .
using a finite-
SUMMARY
In this chapter we have introduced the z-transform and its properties, in much the
same fashion as was done for Laplace transforms and continuous-time
linear systems
in Chapters 3 and 4. Operational use of the z-transform with linear process models
has been emphasized here, since z-transforms are a convenient medium to analyze
Table 24.4
Discrete/Continuous
Conversions for Linear Systems
hold
function,
Method (section
or (C)
equation)
Conversion
table transfer
(§24.1,
(§24.1)
(§24.2)
§24.2) of
then
use
(E)regression
Translation
theorem
(1) Zero-order
Linear
Use
Convert
24-104
~ followed
to(§24.4)
partial
Laplace
fraction
by
(§23.3)
transform;
integration
Eq.
Conversion
23-21
Finite
difference
approximation
24.2
or
Eq.
constant
input,
Table
(3)
use
Analytical
(§23.2)
Contour
time
Nonlinear
Approximate
(§23.1)
(C)series
(§24.2)
(§7.1)
integration,
regression
~rksubstitution
analytical
for
24.1
Eq.
piecewise
(§23.1)
~in24-66
continuous
then
solution
(§24.6)
expansion
difference
~solution
Table
equation
(2) (2)
Power
in
by
long
division
(§24.3)
586
DYNAMIC
RESPONSE
OF DISCRETE-TIME
SYSTEMS
digital feedback control systems (Chapters 25 and 26). In Chapters 22-24 we have
presented a rather diverse set of techniques for converting continuous models to
discrete models and vice versa. Hence, a suitable epilog to this chapter would be
to provide a summary of the possible avenues for interconversion. Table 24.4 gives
a list of different approaches, the steps involved, and the pertinent sections or
equations 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
values:
period has the following
pulse in the figure if the sampling
(a)
I1t = 5 s
(b) I1t = 10s
2
f
1
I
40
00
Time (s)
24.2. A temperature
sensor has the transfer
function,
T~,(S )
T' (s)
1
lOs +
1
where T;/1 is the measured temperature
and T' is the temperature
(both in deviation
variables). The temperature
measurement
is sampled every five seconds and sent to
a digital controller.
Suppose that the actual temperature
changes in the following
manner,
T(t)
=
370 of
{ 350
350 of
of
for 0 :s:; t < 4 s
for 4 :s:; t < 12 s
for t 2: 12 s
(a) What is the z-transform of this signal, T(z)?
T", (z).
(b) Derive an expression for the z-transform of the measured temperature
(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)?
587
Exercises
24.3. Suppose that
1 - 0.2rl
= (1 + 0.6r1)(1 - 0.3rl)(1
F(z)
-
rl)
r
(a) Calculate the corresponding time-domain response
(t).
(b) As a check, use the final value theorem to determine the steady-state value of
r(t).
24.4. Determine the inverse transform of
z(z
(z - 1) (Z2
+
1)
-
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. Assume
that the sampling period is t:.t = 2 min. The pulse is f = 3 for 2 s:: t < 6.
6.3 46 2
Time (min)
f
oLL
0
18
24.6. The pulse transfer function of a process is given by
Y(z)
X(z)
+ 0.6)
+ 0.41
5(z
Z2 -
Z
(a) Calculate the response Y(nLlt) to a unit step change in x using the partial fraction
method.
(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
and 8 min.
t:.t
=4
T(OC)
80r
25~
/
J
o
40
20
Time (min)
24.8. The dynamic behavior of a temperature sensor and transmitter can be described by
the first-order transfer function,
T;"(s) _
e-2s
T' (s) - 8s
+ 1
588
DYNAMIC RESPONSE OF DISCRETE-TIME SYSTEMS
where
the time constant and time delay are in seconds
T = actual temperature
T m = measured temperature
If the actual temperature
changes as follows (t in seconds):
T =
for a :S t < 10
for
< a10
for tt ~
85°C
70
{ 70 °C
°C
(a) What is the maximum value of the measured temperature
Tm?
(b) If samples of the measured temperature
are automatically
logged in a digital
computer every two minutes beginning at t = 0, what is the maximum value of
the logged temperature?
24.9. The transfer function for a process model and a zero-order
H (s)
Derive an expression
Gpes) = (1 -
(lOs +
1)(5s
3.8e-2s
se-s~t)
for the pulse transfer
hold can be written
+
1)
of H(s) G p(s) when
function
as
!:::,.t
= 2.
24.10. The pulse transfer function of a process is given by
2.7r1(z
Y(z)
X(z)
(a) Calculate
method.
the response
0.5z
Z2 -
+ 3)
+ 0.06
y(n!:::"t) to a unit step change in x using the partial fraction
(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
control
is used to provide composition measurements
loop. The open-loop transfer function is given by
G(s)
=
GcHGpGm
G(s)
=
E(s)
B(s)
(Gu
in a feedback
= 1)
and is
~)C - 5e-SM)(
12510+ 1)e-2S
= 2(1 + 8s
(a) Suppose that a sampling period of At = 1 min is selected. Calculate HG(z),
pulse transfer function of G(s) with ZOH.
(b) If a unit step change in the controller error signal e(t) is made, calculate
sampled open-loop response b(n!:::"t) using HG(z).
the
the
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 the
s-domain. Check your results with those in Section 23.3. Note that At is unspecified
here.
24.13.
FindHG(z)ifG(s)
= (1- 9s)/[(3s + 1)(15s + 1)] for At = 4 (use partial fraction
expansion). 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
J(t)
= 1-
in Table
24.15. Find the response
Yn
for the difference
Yn -
Let Xo = 1,
the results.
24.1 for J(t)
=
t2.
What
is the z-transform
for
e -at?
Xn
= a for
n ~
Yn-l
equation
+
0.21 Yn-2 = Xn-2
1. Use long division as well as direct integration
to check
Exercises
24.16. Use long division to calculate the first eight coefficients of the z-transform
F(z)
589
given by
0.8r1
= (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
the steady-state gain of the pulse transfer function.
24.18. Determine
the sampled
function f(n6.t)
corresponding
(TIS
+
1)1(TZS
+
1). Check
to the z-transform
F( z ) = 1 _ 1.5r1
0.5r1+ 0.5z-2
Use partial fraction expansion (6.t = 1) and compare the results with the long division
method 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 response
to a unit step change in the input. Repeat using Tustin's method (approximate
z-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 sampling
instants, n = 0 to n = 5. What conclusions can you make concerning the effect of
pole and zero locations?
1
1
r1
(b)
(c)(a)11=_+ 0.7r1
1
(d) (1 + 0.7z-1)(1 - 0.3r1)
1 - 0.5r1
(e) (1 + 0.7r1)(1
f
( ) (1 +
- 0.3r1)
1 - 0.2rl
- 0.3r1)
0.6r1)(1
24.21. For the transfer functions shown below, determine
function
HGp(z)
for the system and a zero-order
the corresponding
pulse transfer
hold.
1
Gp(s)
= (s +
(b) Gp(s)
= (s +
(a)
1)3
6(1 2)(s
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 of
these results.