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Second-order Transfer Function-Signals and Systems-Lecture 32 Slides-Electrical and Computer Engineering, Slides of Signals and Systems Theory

Second-order Transfer Function, Step Response for Two Real Poles, Step Response for Two Complex Poles, Analysis of the Step Response, Implications in the S-plane, RC Circuit, Step Response for Two Complex Poles, Joseph Picone, Signals and Systems, Electrical and Computer Engineering, Mississippi State University, United States of America.

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2011/2012

Uploaded on 02/17/2012

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Download Second-order Transfer Function-Signals and Systems-Lecture 32 Slides-Electrical and Computer Engineering and more Slides Signals and Systems Theory in PDF only on Docsity! ECE 8443 – Pattern RecognitionECE 3163 – Signals and Systems • Objectives: Second-Order Transfer Function Real Poles Complex Poles Effect of Damping Circuit Example • Resources: MIT 6.003: Lecture 18 GW: Underdamped 2nd-Order Systems CRB: System Response RVJ: RLC Circuits LECTURE 32: SECOND-ORDER SYSTEMS Audio:URL: ECE 3163: Lecture 32, Slide 1 Second-Order Transfer Function • Recall our expression for a simple, 2nd-order differential equation: • Write this in terms of two parameters,  and n, related to the poles: • From the quadratic equation: • There are three types of interesting behavior of this system: )zeta""is( 2 )( 22 2    nn n ss sH   01 2 0 001 2 )()()( asas b sHtxbtya dt dy a dt yd   1, 221   nnpp d)(overdampe axis) real (negative poles two:1 damped)y (criticall at pole double:1 ed)(underdamp polescomplex:10       ns Impulse Response Step Response ECE 3163: Lecture 32, Slide 4 Step Response For Two Complex Poles • When 0 <  < 1, we have two complex conjugate poles: • The transfer function can be rewritten as:       22 2 222222 2 22 2 2 2 )( dn n nnnn n nn n s ss ss sH             • The step response, after some simplification, can be written as: • Hence, the response of this system eventually settles to a steady-state value of 1. However, the response can overshoot the steady-state value and will oscillate around it, eventually settling in to its final value.             n d d t n n wherettety n       1tan0,sin1)( dn nd nn jpp pp       21 2 2 21 , 1 1, ECE 3163: Lecture 32, Slide 5 Analysis of the Step Response For Two Complex Poles •  > 1: the overdamped system experiences an exponential rise and decay. Its asymptotic behavior is a decaying exponential. •  = 1: the critically damped system has a fast rise time, and converges to the steady-state value in an exponetial fashion. • 0 <  > 1: the underdamped system oscillates about the steady-state behavior at a frequency of d. • Note that you cannot control the rise time and the oscillation behavior independently! • What can we conclude about the frequency response of this system? Impulse Response Step Response ECE 3163: Lecture 32, Slide 6 Implications in the s-Plane Several important observations: • The pole locations are: • Since the frequency response is computed along the j-axis, we can see that the pole is located at ±d. • The bandwidth of the pole is proportional to the distance from the j-axis, and is given by n. • For a fixed n, the range 0 <  < 1 describes a circle. We will make use of this concept in the next chapter when we discuss control systems. • What happens if  is negative? dn nn j pp     1, 221
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