Transient Analysis of First- and Second-Order Cont- 123docz.net

CHAPTER 6 Application to Control and Communications

6.3.2 Transient Analysis of First- and Second-Order Control Systems

Although the input to a control system is not known a-priori, there are many applications where the system is frequently subjected to a certain type of input and thus one can select a test signal. For instance, if a system is subjected to intense and sudden inputs, then an impulse signal might be the

appropriate test input for the system; if the input applied to a system is constant or continuously increasing, then a unit step or a ramp signal would be appropriate. Using test signals such as an impulse, a unit-step, a ramp, or a sinusoid, mathematical and experimental analyses of systems can be done.

When designing a control system its stability becomes its most important attribute, but there are other system characteristics that need to be considered. The transient behavior of the system, for instance, needs to be stressed in the design. Typically, as we drive the system to reach a desired response, the system’s response goes through a transient before reaching the desired response. Thus, how fast the system responds and what steady-state error it reaches need to be part of the design considerations.

First-Order Systems

As an example of a first-order system consider an RC serial circuit with a voltage sourcevi(t)=u(t)as input (Figure 6.9), and as the output the voltage across the capacitor,vc(t). By voltage division, the transfer function of the circuit is

H(s)= Vc(s)

Vi(s) = 1 1+RCs

Considering the RC circuit, a feedback system with input vi(t) and outputvc(t), the feedforward transfer functionG(s)in Figure 6.9 is 1/RCs. Indeed, from the feedback system we have

E(s)=Vi(s)−Vc(s) Vc(s)=E(s)G(s)

ReplacingE(s)in the second of the above equations, we have that Vc(s)

Vi(s) = G(s)

1+G(s)= 1 1+1/G(s)

so that the open-loop transfer function, when we compare the above equation toH(s), is G(s)= 1

RCs

The RC circuit can be seen as a feedback system: the voltage across the capacitor is constantly com- pared with the input voltage, and if found smaller, the capacitor continues charging until its voltage coincides with it. How fast depends on the RC value.

FIGURE 6.9

Feedback modeling of an RC circuit in series.

e(t) vi(t)

vi(t) vc(t)

vc(t) + −

G(s)= 1

RCs +

− +

− R

6.3 Application to Classic Control 373

−1 −0.8 −0.6 −0.4 −0.2 0

(a) (b)

−0.5 0 0.5

jΩ

RC=1

RC=10

0 10 20 30 40 50

0 0.2 0.4 0.6 0.8 1

t vc(t)

RC=10 RC=1

FIGURE 6.10

(a) Clustering of poles and (b) time responses of a first-order feedback system for1≤RC≤10.

Forvi(t)=u(t), so thatVi(s)=1/s, then the Laplace transform of the output is Vc(s)= 1

s(sRC+1) = 1/RC

s(s+1/RC)= 1

s − 1

s+1/RC so that

vc(t)=(1−e−t/RC)u(t)

The following MATLAB script plots the poles Vc(s)/Vi(s) and simulates the transients of vc(t) for 1≤RC≤10, shown in Figure 6.10. Thus, if we wish the system to respond fast to the unit-step input we locate the system pole far from the origin.

%%%%%%%%%%%%%%%%%%%%%

% Transient analysis

%%%%%%%%%%%%%%%%%%%%%

clf; clear all syms s t num = [0 1];

for RC = 1:2:10, den = [ RC 1];

figure(1)

splane(num, den) % plotting of poles and zeros hold on

vc = ilaplace(1/(RC∗s ˆ 2 + s)) % inverse Laplace figure(2)

ezplot(vc, [0, 50]); axis([0 50 0 1.2]); grid hold on

end hold off

Second-Order System

A series RLC circuit with the input a voltage source, vs(t), and the output the voltage across the capacitor,vc(t), has a transfer function

Vc(s)

Vs(s) = 1/Cs

R+Ls+1/Cs = 1/LC s2+(R/L)s+1/LC If we define

Natural frequency:n= 1

√CL (6.9)

Damping ratio:ψ=0.5R rC

L (6.10)

we can write

Vc(s)

Vs(s) = 2n

s2+2ψns+2n (6.11)

A feedback system with this transfer function is given in Figure 6.11 where the feedforward transfer function is

G(s)= 2n s(s+2ψn) Indeed, the transfer function of the feedback system is given by

H(s)= Vc(s)

Vs(s) = G(s) 1+G(s)

= 2n s2+2ψns+2n

The dynamics of a second-order system can be described in terms of the parametersnandψ, as these two parameters determine the location of the poles of the system and thus its response. We adapted the previously given script to plot the cluster of poles and the time response of the second-order system.

Assumen=1 rad/sec and let 0≤ψ≤1 (so that the poles ofH(s)are complex conjugate for 0≤ ψ <1 and double real forψ=1). Let the input be a unit-step signal so thatVs(s)=1/s. We then have:

FIGURE 6.11

Second-order feedback system.

vs(t) e(t) G(s)

vc(t) + −

6.3 Application to Classic Control 375

−1 −0.5 0

−1

−0.8

−0.6

−0.4

−0.2 0 0.2 0.4 0.6 0.8 1

jΩ

(a)

0 10 20 30 40 50

0 0.5 1 1.5 2

0 10 20 30 40 50

0 0.2 0.4 0.6 0.8 1

t vc(t)vc(t)

(b)

(a) Clustering of poles and time responsesvc(t)of second-order feedback system for (b)√

2/2≤ψ ≤1and (c)0≤ψ≤√

2/2.

(a) If we plot the poles ofH(s)asψchanges from 0 (poles injaxis) to 1 (double real poles) the responsey(t)in the steady state changes from a sinusoid shifted up by 1 to a damped signal. The locus of the poles is a semicircle of radiusn=1. Figure 6.12 shows this behavior of the poles and the responses.

(b) As in the first-order system, the location of the poles determines the response of the system. The system is useless if the poles are on the jaxis, as the response is completely oscillatory and the input will never be followed. On the other extreme, the response of the system is slow when the poles become real. The designer would have to choose a value in between these two forψ.

(c) For values of ψ between√

2/2 to 1 the oscillation is minimum and the response is relatively fast (see Figure 6.12(b)). For values ofψfrom 0 to√

2/2 the response oscillates more and more, giving a large steady-state error (see Figure 6.12(c)).

nExample 6.4

In this example we find the response of an LTI system to different inputs by using functions in the control toolbox of MATLAB. You can learn more about the capabilities of this toolbox, or set of specialized functions for control, by running the demorespdemoand then usinghelpto learn more about the functionstf, impulse, step, andpzmap, which we will use here.

We want to create a MATLAB function that has as inputs the coefficients of the numeratorN(s) and of the denominatorD(s)of the system’s transfer functionH(s)=N(s)/D(s)(the coefficients are ordered from the highest order to the lowest order or constant term). The other input of the function is the type of responset wheret=1 corresponds to the impulse response,t=2 to the unit-step response, andt=3 to the response to a ramp. The output of the function is the desired response. The function should show the transfer function, the poles, and zeros, and plot the cor- responding response. We need to figure out how to compute the ramp response using thestep function.

Consider the following transfer functions:

(a) H1(s)= s+1 s2+s+1 (b) H2(s)= s

s3+s2+s+1 Determine the stability of these systems.

Solution

The following script is used to look at the desired responses of the two systems and the location of their poles and zeros. We consider the second system; you can run the script for the first system by putting % at the numerator and the denominator afterH2(s)and getting rid of % afterH1(s) in the script. The functionresponsecomputes the desired responses (in this case the impulse, step, and ramp responses).

%%%%%%%%%%%%%%%%%%%

% Example 6.4 -- Control toolbox

%%%%%%%%%%%%%%%%%%%

clear all; clf

% % H 1(s)

% nu = [1 1]; de = [1 1 1];

%% H 2(s)

nu = [1 0]; de = [1 1 1 1]; % unstable h = response(nu, de, 1);

s = response(nu, de, 2);

r = response(nu, de, 3);

function y = response(N, D, t) sys = tf(N, D)

poles = roots(D) zeros = roots(N) figure(1) pzmap(sys);grid if t == 3,

D1 = [D 0]; % for ramp response end

Transient Analysis of First- and Second-Order Control Systems

Frequency Response from Poles and Zeros

What Have We Accomplished? What Is Next?