Ifthe system description is available in difference-equation form. and if the input signal elementary, then the first three steps of this process require little effort or computation. The final step, however, tedious if done by hand.
and, because we will later be preoccupied with design of transfer functions to give desirable responses, we attach great benefit to gaining intuition for the kind of response to be expected from a transform without actually inverting it or nu-merically evaluating the response. Our approach to this problem is to present a
1. Compute the transfer function of the systemH 2. Compute the transform of the input signal,
3. Form the product. which is the transform of the output signal,U.
4. Invert the transform to obtainu(k
Signal Analysts and D)11amic Response 119
Thus we see that nonlinear models can be approximated as linear state-space models or as transfer function models. The accuracy of the approximation varies with the problem, but is generally useful in designing the control system. The final design of the control system should always be checked via numerical simulation of the nonlinear equations.
from which the transfer function is
We go even further and restrict ourselves to the case of single input and single output and discrete time. We then write the model as
x(k
+
I)=+
ruCk).y(k) = Hx(k)
+
JBut now the notation is overly clumsy. so we drop the and notation and simply call them x. u and y and define the constant matrices
F
=
f" (xo' uo). G= r.
u(xo'H
=
(xo' uo)' J=
h.o uo)'This results in the form we used earlier in Section 2.1.1
4.4
The row vector in Eq. (4.88) is called thegradientof the scalar
I,
with respect to the vector x.Ifr
ISa vector, we define its partial derivatives with respect to the vector x as the matrix (called theJacobean)composed ofrows of gradients. In the subscript notatiun, if we mean to take the partial ofall components, we omit the specific subscript such asIor2but hold its place by the use of a commaaI,
The assumption of small signals can be reflected by taking x and u to be always close to their reference values xo' uo'and these values, furthermore, to be an equilibrium point of Eq,(4.82).where
r(xo' uo)
=
0, (4.85)Now. ifx and u are toXo and uo' they can be written as x= Xo
+
u=Uo
+
and these can be substituted into Eq.(4.84),The fact that and are small is now used to motivate an expansion of Eq. (4.84) about andUoand to suggest that the only terms in the first power of the small quantities Sx and need to be retained. We thus have a vector equation and need the expansion of a vector-valued function of a vector variable,dfd
+
8x)= f(xo+
Uo+
8u). (4,86)Ifwe go back to Eq,(4,82)and do the expansion of the components one at a time. it is tedious but simple to verify that Eq. (4,86) can be written as'7
= f(xo' uo)
+
(xo' uo)Sx+
(xo'uo)Su+ ...
(4.87)where we define the partial derivative of a scalar with respect to the vector x by a subscript notation:
Now, to return to Eq. (4.87), we note that by Eq. (4.85) we chosex u to be an equilibrium point, so the first term on the right of Eq, (4.87) is zero,o because the terms beyond those shown depend on higher powers of the small signals Sx and we are led to the approximation
(xo'
+
u(xo'= h"
+
h,o 118 Chapter4 Discrete Systems AnalysisI'
4.4.1 The Unit Pulse
(4.96) (k 0)
(k < 0)
..
:-(k-' l_e__e__ e e_e
k'"
(a) (b)
Signal Anah'sls and DynamIC Response 121
which is the as using the symbol Ilk) for the unit step function.
we get
Here the transfonn is characterized by a zero at = 0and a pole at = I.The significance of the convergence being restrictedto > Iwill be explored later we consider the inverse transfonn operation. The Laplace transform of the unit step isI! we may thus keep in mind that a pole at = 0for a continuous signal corresponds in some way to a pole at = I for discrete signals. We will explore this further later. In any event. we record that a pole at = I with convergence outside the unit circle. = 1.will correspond to a constant for positive time and zero for negative time.
To emphasize the connection between the time domain and the we sketch in Fig.4.21 the with the unit circle shown and the pole of marked x and the zero markedc.Beside the we sketch the time plot of
4.4.3 Exponential
The one-sided exponential in time e,(k)=
=0
Figure 4.21
(a) Pole andzeroof£I(Z) in the z-plane. Theunit circleis shownfor reference. (b) Plot of e1lk)
(4.94)
(k 0) (k < 0)
E (-) = = I.
therefore we have
of elementary signals with known features and to learn their represen-tation the transfonn or Thus, when an unknown transform we will be by reference to these known to infer the major tures of the tlme-domam signal and thus to detennine whether the unknown is of sufficient interesttowarrant the effort of detailed time-response computation.
To begm this process of attaching a connection between the time domain and the z-transform domain, we compute the transforms of a few elementary signals.
We have already seen that the unit pulse is defined
e,(k)=l (k=O)
= 0 (k 0)
=
This result is much like the continuous case, wherein the Laplace transform of the unit impulse is the constant 1.0.
The quantity E, gives us an instantaneous method to relate sianals to To characterize the systemH(z),consider the signalu(k), is the umt pulse response: then = H(z).
4.4.2 The Unit Step
Consider the unit step function defined by
= I
=0 I(k).
In this case. the z-transfonn is 120 Chapter 4 Discrete Systems
,
,
= =
= 1
= I -(4.95)
(lzl> I).
I -I
-
I18 h.a,ve notation to use rather than for kth Identity dIfferent signals.
>
(4.99)
(4.101) (4.100)
>1').
>
(izi > 1').
(b)
• • •
+
J {
-E =
+ .
2
-4.4 SignalAnalysisand DynamicResponse 123
• • •
•
The pole-zero pattern of and the time plot of are shown in Fig. 4.23 forI'
=
0.7 and=
We note in passing that if
=
O. then reduces toe3and, withI'=
I, to so that three of our signals are special cases of By exploiting the features of we can draw a number of conclusions about the relation between pole locations in the z-plane and the time-domain signals to which the poles correspond. We collect these for later reference.and thus that
The signales(k)grows without bound ask gets large if and only ifr > Land a system with this pulse response is BIBO stable if and only if 11'1 < I. The boundary of stability is the unit circle. To complete the argument given before for e
4(k) = cos I(k), we see immediately that the other half is found by replacing by
-e
inEq.(4.99)(I)
Figure4.23
(a) Poles and zeros ofE
4(z)for
=
45, r=
0.7Inthe z-plane. (b) Plot ofe4(k) (4.98)(b)
• •
• •
(a)
The pole of is at = r. From Eq. (4.96) we know thate.(k) without bound if11'1 > 1. From Eq. (4.97) we conclude that a that converges for and has a real poleolltsidethe circle I= I corresponds to a growIng signal.Ifsuch a signal were the unit-pulse response of our system such as digital control program. we would say the program wasunstable we saw Eq. (4.39). We in Fig. 4.22 the z-plane and the corresponding time history of as tor the stable value.r= 0.6.
=
=
Our next example considers the modulated sinusoid = cos(kli)]l(k).
where we assumeI'>O. Actually. we can decomposee (k)into the sum of two complex exponentials as
=
+
2 e I(k).and the z-transform is Iinear.19we need only compute the transform of each SIngle complex exponential and add the results later. We thus take first
and compute
19 not formally. The demonstration. the definition of
and Section . .
4.4.4 General Sinusoid
122 Chapter 4 Analysis
Figure4.22 (a) Pole and zero of Inthe z-plane.(b)Plot of e1(k)
aT= 0.3567,
= e cosbl I(I) 1maxis
4
with
4.4.5 Correspondence with Continuous Signals
From the calculation of these few we have established that the duration of a time signal related to the radius of the pole
number of per cycle related to the. angle. Another set of relationships can be established by consldenng the Signalstobe samples from
. . hL I f m E( WIth thiS devlCe we can
continuous Signal,e(l), ap ace trans or " . . _ exploit our kno\-\'ledge of s-plane features by transfemng toeqUlvalent plane properties. For the specific numbers represented the tllustratlon ofe e take the continuous signal
+.+
SIgnal and Response 125plane of constant
T=sampling period
figure4.24 . I I
ftheunit with angle marked numbersofsamp es per eyee
0
-43 21 9 5 0.9
0.8 0.6 0.4
(a) r> I corresponds to a growing signal that will not decay at all.
(b) r= I corresponds to a with constant amplitude (which isnor BI80 stable as a response).
(c) Forr< I.the closerris to0 the shorter the settling time. The corre-sponding BIBO stable. We can compute the settling time in samples,N, in terms of the pole radius,r.
N - - -360 samples/cycle.
(d) A pole at r= 0 corresponds to a transient of finite duration.
Pole Radius Response Duration N
For
=
we haveN=
8, and the plot of given in Fig. 4.23(b) the eight samples in the cycle very clearly.Asketch of the unit circle with several points correspondingtovarious numbers of samples per cycle marked drawn in Fig. 4.24 along with other contours that will be explained in the next section. The sampling frequency in Hertz is1fT,and the signal frequencyf
= 1/NTso thatN =f
and1/Nis anormalized frequency. Since = N, the normalized signal frequency in radians/sample. T the frequency in radians/second.1. The settling time of a transient, defined as the time required for the signal to decay to one percent of maximum value, is set mainly by the value of the radius, of the poles.
2. The number of samples per oscillation of a sinusoidal signal determined If we require cos(fJk) = cos(8(k
+
N», we find that a pcriod of rad containsN samples, whereA compilation of signal versus their pole location in the
shown in Fig. 4.25. It demonstrates visually the features just summarized for the general sinusoid, which encompasses all possible signals.
Chapter 4 Discrete Systems Analysts
(b)
e
= bT(a)
Signal Analysis and Dynamic Response 127
If we know what it means to have a pole in a certain place in the s-plane, then Eq. (4.103) shows us where to look in the to find a representation of discrete samples having the same time features. Itis useful to sketch several major features from the s-plane to the according to Eq. (4.103)tohelp fix these ideas. Such a sketch is shown in Fig. 4.26.
Each feature should be traced in the mind to obtain a good grasp of the relation. These features are given in Table 4.2. We note in passing that the map
20 Unlessa of is repeated. to discrete ofa signal corresponding to higher-order pole. The result readily shown a polynomial in multipl)'ing
(4.103) From Eq. (4.101). the of has poles at
If is a ratio of polynomials in which will be the case ife(k)is generated by a linear difference equation with constant coefficients, then by partial fraction expansion, can be expressed as a sum of elementary terms like and In all such cases. the discrete signal can be generated by samples from continuous signals where the relation between the s-plane poles and the corresponding plane poles is given by
but because equals it follows that
Figure 4.26 Corresponding in the s-plane and the z-plane accord ing to
I · '
...
,..
...
0.5
...
. . .. . . . .
And. taking samples one second apart(T = I). we have
v(kT)= cos- I(k)
4
= cos- I(k) 4
=
The poles of the Laplace transform of (in the s-plane) are at
=
+
jb. -(/ - jb.,
- -
-V •••••
e • • . .
126 Chapter
+
Discrete SystemsFigure 4.25
associated with locations the z-plane
-+-+ Signal 129
Note that if = PI' the members of the one pole-zero cancel
and if at the same
=
al=
-21' and=
the systemresponse. Lo Ihe input with transform = I(a unit pulse) is - rcos
=
transform. when compared with the transform E, given in Eq.
(4.101).is seentobe
and conclude that under these circumstances the system pulse response is a delayed version ofe,(kl.a typical second-order system pulse response.
For our first we consider the effect of zero location. We let - PI and explore the effect of the (remaining) zero location. on the step-response overshoot for three sets of values of(II and We select(II and so that the poles of the correspondtoa response with an equivalent s-plane damping ratio = 0.5and consider values of of18.45.and72degrees. In every case.
we will take the gain K to be such that the steady-state output value equals the step size. The situation in the is sketched in Fig.4.28 for = 18.The curve for = 0.5is also for reference. Tn addition to the two poles and one zero of we show the pole at = and the zero = O.which come from the transform of the input step. given by - I).
Figure4.28
Pole-zero pattern ofY(z) for the system of Fig.
4.27,withZ, = P" U(z)= z/(z - a, anda2selected for
= 18,and =0.5
-= of Eq. is many-to-one. There are many values ofsfor each value of In fact. if
Description of corresponding lines in s-plane and z-plane
=
+
j - N .- T
then , = r.The (great) significance of this fact will be explored in Chapter5.
Lines of constant damping in the s-plane are mapped into the ac-cording to Eq.(4.103)for several values of in Fig.4.24.We often refertothe damping of a pole in the in terms of this equivalent s-plane damping. or sometimes we simply refer to the damping of a pole. Likewise. lines of constant natural frequency. in the s-plane (semi-circles centered at the origin) are also mapped into the accordingtoEq.(4.103)for several values of in Fig.4.24. interesting to note that in the immediate vicinity = +L map of and looks exactly like the s-plane in the vicinity ofs= O. Because of the usefulness of this mapping. the Control System Toolbox has the function zgrid.mthat allows one to superimpose this mapping on various plots to help in the interpretation of the results. You will see its use in the figure files of discrete root loci in Chapter 7.
4.4.6 Step Response
Our eventual purpose. of course, is to design digital controls. and our interest in the relation between poles and zeros and time-domain response comes from our need to know how a proposed design will respond in a given dynamic situation.
The generic dynamic test for controls is the step response. and we will conclude this discussion of discrete system dynamic response with an examination of the relationships between the pole-zero patterns of elementary systems and the corresponding step responses for a discrete transfer function from IIto of a hypothetical plant. Our attention will be restricted to the step responses 'of the discrete shown in Fig. 4.27 for a selected set of values of the parameters.
Table 4.2
128 Chapter-+ Discrete