University of Toronto Department of Electrical and Computer Engineering ECE356S System and Signal Analysis II Spring 1999 Problem Set #5

1. Sketch the root locus of the feedback system with the loop transfer function L(s) =

\frac{N(s)}{D(s)} with N(s) = 1, and D(s) = s(s + 3)(s^{2} + 2s + 2). Check your sketch using

matlab.

Note: While it is not important in most cases, you can get a nicer and more accurate sketch for branches leaving imaginary open-loop poles using the so-called angle of departure γ_{i} at the ith pole p_{i}. It is given by

γ_{i} = \frac{1}{n_{i}} (∑_{j} arg(p_{i} − z_{j}) − ∑_{j} arg(p_{i} − p_{j}) + (2k + 1)π) k = 0, 1, …, n_{i} − 1

where n_{i} is the multiplicity of the pole p_{i}, p_{j} is the jth open-loop pole, z_{j} is the jth open-loop zero, and arg refers to the angle of the complex number. For simple poles, the expression simplifies to

γ_{i} = ∑_{j} arg(p_{i} − z_{j}) − ∑_{j} arg(p_{i} − p_{j}) + π

2. Consider the following feedback system

1/s(s + 2) K(1 + 2/s)  F r  y - + - +

The feedback gain F and the gain K in the forward path are both adjustable. This problem asks you to study the closed loop system behaviour in terms of K and F.

(i) Express the feedback system in the form of

r  + -  K  P (s; F)  y

and identify P (s; F). P (s; F) is thus the equivalent open loop transfer function,

expressed as \frac{N(s)}{D(s)}, and is a function of the feedback gain F.

(ii) Let F = 1 and plot the root locus of the closed loop system (CLS) as a function of K.

(iii) Discuss the open loop pole pattern (i.e. poles of P (s; F)) for F > 1. Describe in this case the 2 possible scenarios for the root locus of CLS.

(iv) One of the scenarios involves a real axis break-in point. The break-in point may be determined from the roots of ND′ −DN′ = 0. Draw the root locus of ND′ −DN′ as a function of F. Show that this root locus has a breakaway point at s = −\frac{2}{3}. Show also that this occurs for F = \frac{32}{27}. From this, show that there is a real axis break-in point for the root locus of CLS for F < \frac{32}{27}. Explain what happens to the real axis break-in point for F > \frac{32}{27}. (Hint: Recall that if s is a point on the root locus of 1+KL(s), the corresponding value of K = \frac{1}{|L(s)|}).

(v) Sketch the root locus of the CLS as a function of K for F = \frac{32}{27} and F = 4. (Hint: From the previous parts, you should be able to conclude that at F = \frac{32}{27}, s = −\frac{2}{3} is a point on the root locus of CLS)

3. Let

P CF (s) = \frac{1 + 2s}{(1 + s)(1 + 4s)(1 + 5s)}

Sketch the Nyquist plot for this system and find the range of values of K for which the closed loop system is stable.

4. Let

P CF (s) = \frac{s + 3}{s(s − 1)}

Sketch the Nyquist plot for this system and find the range of values of K for which the closed loop system is stable.