Lecture 13: H(s) Poles-Zeros & BIBO Stability. (a) What are poles and zeros? Answer: HS math and calculus review.

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1 1. Introduction Lecture 13: H(s) Poles-Zeros & BIBO Stability (a) What are poles and zeros? Answer: HS math and calculus review. (b) What are nice inputs? Answer: nice inputs are bounded inputs; if you excite a circuit with a nice bounded input, like a step function, then one hopes to get a nice (bounded) output signal. Bounded input Bounded output (BIBO) stability. (c) Hey Prof Ray, I read this article about meta stable which is in some books. Answer: Yeah right! No such thing. A flawed concept made up by those who never studied stability. (d) Prove it Prof Ray you are always talking about proving things. Huh! Answer: the Tacoma Narrows bridge the one that did the twist and shout (girders screaming as they twisted) to the symphony of the Tacoma narrows wind that bridge like London bridge it tumbled down. So much for metastable; what do ya think? (e) OK so what are transfer functions good for? 2. General Structure of a Rational H(s). (a) Assumption: Generally, in 202, H(s) = n(s) d(s) meaning that deg[n(s)] deg[d(s)]. is assumed to be PROPER (b) Structure of the transfer function:

2 H(s) = K sm + b 1 s m b m s n + a 1 b n a 1 = K (s z 1 )(s z 2 )...(s z m ) (s p 1 )(s p 2 )...(s p n ) (a) z i is a finite zero of H(s) ; (b) p i is a finite pole of H(s) (c) if lim s H(s) = 0 then is an infinite zero of H(s) ; (d) n m is the number of infinite zeros of H(s) ; (e) dc gain is H(0) = K( 1) n m m z i i=1 n p i i=1. Example 1. Find H(s) from the pole-zero plot below assuming H(0) = 5.

3 Step 1. H(s) = K (s + jb)(s jb) (s + a + jb)(s + a jb) = K s 2 + b 2 (s + a) 2 + b 2 Step 2. H(0) = 5 = K b2 a 2 + b 2 K = 5 a 2 + b 2 b 2 3. BIBO stability: nice input leads ALWAYS to a nice output. (a) What is nice? Think about what you consider a nice person and what others may consider a nice person. This is difficult to pin down. In mathematics we often define nice by what is called a norm. Probably the most familiar norm to most students in the Euclidean norm of a vector: x 1 x = x 2 then x = x x 2 2, the length of the vector. For us, nice refers to a property of the input signal and output signal. For us, nice means bounded. A function f (t) is bounded if f (t) K < for all t. So if nice in always leads to nice out means bounded inputs lead to bounded outputs. so called BIBO stability. Yeah right Ray. So how much time do we have to check all those bounded inputs to see if they ALL lead to bounded outputs??? No way. I ain t got time for that. I ll spend the rest of my life catching my breath, letting it go. Good point. That is exactly the problem. The idea is NOT to check all those input-output pairs but to characterize the property of BIBO stability in terms of the transfer function H(s). BIBO stability is true if and only if ALL poles of H(s) lie in the open left half complex plane. It s all so simple now.

4 Example 2. H(s) = V out = 12 V in s 3. Suppose v in (t) = u(t) V. Then 12 V out = s(s 3) = 4 s 4 s 3 to infinity as t. implies v out (t) = 4u(t) 4e3t u(t) which blows up Conclusion: poles in the right half complex plane are very bad. Example 3. H(s) = V out V in = 12 s. Suppose v in (t) = u(t) V. Then V out = 12 t. s 2 implies v out (t) = 4r(t) = tu(t) which blows up to infinity as Conclusion: poles on the imaginary axis are bad. Example 4. H(s) = V out = 12 V in s + 3. Suppose v in (t) = u(t) V. Then 12 V out = s(s + 3) = 4 s 4 s + 3 implies v out (t) = 4u(t) + 4e 3t u(t) which exponentially goes to zero as t. Conclusion: poles in the left half complex plane are very good. Big Conclusion: A circuit is BIBO stable if and only if all poles of H(s) lie in the open left half complex plane.

5 Example 5. The bridge over the narrows problem: Example 6. Find the range of a for instability. Strategy: nodal analysis.

6 1. Output node: 1 V out + 1 ( s V out V 1) = 0. Solving for V 1 yields: (s +1)V out = V 1 2. The V 1 node. ( V 1 V in ) av out + 1 ( s V 1 V out ) = 0 From 1, upon substitution (s +1)V out av out +V out = V in 3. Conclusion: H(s) = V out V in = 1 s + 2 a which requires that a 2 < 0 a < 2 for BIBO stability.

7 Worksheet 1. (s 1) 2 (s + 2) 2 H(s) = 36 (s +1) 2 (s + 4)(s 3) 2 1. How many finite zeros? 2. Location of finite zeros? 3. Multiplicity of finite zeros? 4. How many infinite zeros? 5. How many finite poles? 6. Location of finite poles? 7. Multiplicity of finite poles? 8. True-False. The impulse response contains a term of the form Kδ (t). 9. True-False. The circuit/system described by the above transfer function is BIBO stable.

8 Worksheet 2. Given H(1) = 15 and the pole-zero plot below, find H(s). 1. H(s) = K 2. H(1) = 15 = K Therefore K = 3. True-False: The circuit/system described by the above transfer function is BIBO stable.

9 Appendix to Lecture 13. Example 1. Two Component Model of the Ingestion and Metabolism of a Drug. Compartment 1 Model dm 1 dt = K 1 m 1 + r Compartment 2 Model dm 2 dt = K 1 m 1 K 2 m 2 Assumptions 1. Output = m 2 (t) = mass of drug in bloodstream at t. 2. All IC s are zero at t = 0.

10 Objective: Construct H(s) and discuss. Step 1. sm 1 (s) = K 1 M 1 (s) + R(s) implies M 1 (s) = Step 2. sm 2 (s) = K 1 M 1 (s) K 2 M 2 (s) M 2 (s) = K 1 M 1 (s) s + K 2. 1 s + K 1 R(s) Step 3. H(s) = M 2 (s) R(s) = K 1 (s + K 1 )(s + K 2 ) Step 4. Impulse Response: h(t) = K 1 K 2 K 1 e K 1 t u(t) K 1 K 2 K 1 e K 2 t u(t) Step 5. Pole-zero plot of H(s). Suppose K 2 > K 1.