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000 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 047 048 049 050 051 052 053 ELE 301, Fall 2010 Laboratory No. 7 Stability and Root Locus Plots 1 Background 1.1 Transfer Function of a Circuit Model resistors, capacitors and inductors as linear time-invariant components. To obtain the transfer function of a circuit made up of these components, we determine each individual component’s current to voltage transfer function (impedance). The current-voltage relationship for a capacitance C is I (t)= C(dV (t)/dt). So when V (t)= Ve st , I (t)= CsV e st = Ie st . Hence V = Z C I , where Z C =1/(sC). The current-voltage relationship for inductance is V (t)= L(dI (t)/dt). As above consider V (t)= Ve st . Then I (t)= LsV e st = Ie st . So V = Z L I , where Z L = sL. You can apply the same analysis to a resistance R to obtain V = IR. So for inputs of the form e st , circuit components have voltage-current relationships that look like Ohm’s law, except that the impedances are now complex: Z R = R Z C =1/(sC) Z L = sL All the rules of resistor combination in series and in parallel apply to impedances, so sinu- soidal circuit analysis becomes very easy. As an example, we analyze the RC circuits in Figure 1. We use basic circuit laws to calculate Figure 1: Two RC circuits. v out = H(s)v in . Each circuit is a voltage divider. For the circuit on the left: v out = v in 1/sC R +1/sC = v in 1 1+ sRC Hence H 1 (s)= 1 1+ sCR For the circuit on the right: v out = v in R R +1/sC = v in sCR 1+ sRC Hence H 2 (s)= sCR 1+ sCR 1

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ELE 301, Fall 2010

Laboratory No. 7

Stability and Root Locus Plots

1 Background

1.1 Transfer Function of a Circuit

Model resistors, capacitors and inductors as linear time-invariant components. To obtainthe transfer function of a circuit made up of these components, we determine each individualcomponent’s current to voltage transfer function (impedance).

The current-voltage relationship for a capacitance C is I(t) = C(dV (t)/dt). So when V (t) =V est, I(t) = CsV est = Iest. Hence V = ZCI, where ZC = 1/(sC).

The current-voltage relationship for inductance is V (t) = L(dI(t)/dt). As above considerV (t) = V est. Then I(t) = LsV est = Iest. So V = ZLI, where ZL = sL. You can apply thesame analysis to a resistance R to obtain V = IR.

So for inputs of the form est, circuit components have voltage-current relationships thatlook like Ohm’s law, except that the impedances are now complex:

ZR = R ZC = 1/(sC) ZL = sL

All the rules of resistor combination in series and in parallel apply to impedances, so sinu-soidal circuit analysis becomes very easy.

As an example, we analyze the RC circuits in Figure 1. We use basic circuit laws to calculate

Figure 1: Two RC circuits.

vout = H(s)vin. Each circuit is a voltage divider. For the circuit on the left:

vout = vin1/sC

R+ 1/sC= vin

11 + sRC

Hence

H1(s) =1

1 + sCR

For the circuit on the right:

vout = vinR

R+ 1/sC= vin

sCR

1 + sRC

Hence

H2(s) =sCR

1 + sCR

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y(t)

(1- R

R

V 2

v(t)

0

C2

R0

R

C1

V 1

+

-

+

-

+-

Figure 2: Active low pass filter: Sallen-Key circuit.

1.2 A Second Order System

The circuit shown in Figure 2 is a variation on the Sallen-Key circuit. The circuit containstwo capacitors and gives rise to a second order differential equation. For this reason, it iscalled a second order system. Assuming that R0 is very large and that C1 = C2, the circuittransfer function from x to y is:

H(s) =τ2/β

s2 + (3− 1/β)τs+ τ2(1)

where τ = 1/(RC) and 0 < β ≤ 1.

We can parameterize the transfer function of a second order system as follows:

H(s) = Gω2n

s2 + 2ζωns+ ω2n

(2)

The two parameters ζ (zeta) and ωn are called the damping factor and the natural frequencyof the system, respectively, and G is a gain factor.

1.3 Questions

1. Find the gain G, natural frequency ωn, and damping factor ζ of the transfer functionH(s) in (2) as functions of β and τ .

2. For what value, if any, of β is the circuit critically damped?

3. Is the circuit underdamped for any the realizable values of beta? If so for whatvalues?

4. For what range of β ∈ (0, 1] is this circuit stable and why?

5. Derive the transfer function (2) of the Sallen-Key circuit in Figure 2. You can attachthis.

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2 Lab Procedure

2.1 Plotting Root Locus

A root locus of a transfer function is a plot of the poles of the transfer function (roots ofthe denominator polynomial) as a scalar parameter is varied. The roots of a polynomial arecontinuous functions of its coefficients and the coefficients of the denominator polynomialof H(s) are continuous functions of β. Hence as β is varied the roots trace out continuouscurves in the complex plane.

Create a root locus plot of the poles of the transfer function H(s) of the Sallen-Key circuitas β increases from 0.25 to 1. Do this by creating a loop that cycles through the desiredvalues of β in the range [0.25, 1] and for each value:

1. creates a vector of the coefficients of the denominator polynomial den of H(s)(highest to lowest powers of s).

2. calls the built-in MATLAB m-file roots to compute the roots of the polynomial.3. plots the position of the roots on the complex plane (use hold on and hold off).

To plot a 2-D points with (vectors of) coordinates (x, y) you can use:

plot(x,y,’r.’,’MarkerSize’,14);

Use the grid command:

sgrid

after the end of the loop to add a grid that indicates circles of constant ωn and radial linesof constant ζ . You can add text to the plot with the command:

text(xp,yp,’Root locus for β=1,...,0.25’);

Where (xp, yp) is the desired start position for the text. Include such annotation on yourplot. Make sure you also label all axis.

What does the root locus plot confirm about the stability of this circuit?

2.2 A Simple Feedback Loop

Now that we have a program to plot a root locus we will use it to explore the stability ofthe simple negative feedback system shown in Figure 3.

h(t)+

-

x(t) y(t)k

e(t)H(s)

+

-

X(s) Y(s)k

E(s)

Figure 3: Proportional error feedback regulation. (l) time domain, (r) Laplace domain.

H(s) is the transfer function of the open loop system. It is assumed to be a rational function.The closed loop transfer function from x to y is:

F (s) =kH(s)

1 + kH(s)=p(s)q(s)

where p(s) and q(s) are polynomials in s. The stability of the system requires that the rootsof q(s) be in the open left half of the complex plane (see Laplace transform notes).

For each of the transfer functions H(s) in the list below:

• Compute the closed loop transfer function F (s).

3

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• Plot a root locus of the poles of the closed loop system as k varies.• Interpret the root locus (explain what is happening).• Use the root locus plot to find the largest k (if it exists) such that the poles are

stable and either real of have a damping factor of approximately 0.75.

1. H(s) = 1s−1 (this system is unstable).

2. H(s) = ω2n

s2+2ζωns+ω2n

with ζ = 0.85 and ωn = 2.

3. H(s) = ω2n

s(s2+2ζωns+ω2n) with ζ = 0.85 and ωn = 2.

4. H(s) as in 3. except with ζ = 0.95 (almost critically damped).5. H(s) as in 3. except with ζ = 0.65 (lightly damped).

Attach all your plots and code to the lab handout prior to handing it in.

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