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ASU EEE 202 - System Response II

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Slide 1Pole-Zero PlotSecond-Order CircuitsSlide 4Characteristic Equation & PolesReal and Unequal Roots: OverdampedComplex Roots: UnderdampedReal and Equal RootsAn ExampleAn Example (cont’d.)An Example (cont’d)A SummaryTransient and Steady-State ResponsesClass Examples• Poles & Zeros• Second-Order Circuits• LCR Oscillator circuit: An example• Transient and Steady StatesLecture 17. System Response II12Pole-Zero Plot•For a pole-zero plot place "X" for poles and "0" for zeros using real-imaginary axes•Poles directly indicate the system transient response features•Poles in the right half plane signify an unstable system•Consider the following transfer function)5.1)(4)(5()54)(5.3)(3()(22ssssssssHReIm3Second-Order CircuitsRC+–vc(t)+ –vr(t)L+–vl(t)i(t)+–dttdvdttidLtiCdttdiRs)()()(1)(22•KVL around the loop:vr(t) + vc(t) + vl(t) = vs(t))()()(1)( tvdttdiLdxxiCtiRstdttdvLtiLCdttdiLRdttids)(1)(1)()(224Second-Order Circuits•For zero-initial conditions, the transfer function would be 2002200221)()()()(2)(sssssHssssFXFX)()()(2)(20022tftxdttdxdttxd• In general, a second-order circuit is described byCharacteristic Equation & Poles•The denominator of the transfer function is known as the characteristic equation•To find the poles, we solve :which has two roots: s1 and s2022002ss124)2(2,2002020021ss200221)()()(sssssHFX6Real and Unequal Roots: Overdamped•If  > 1, s1 and s2 are real and not equal•The amplitude decreases exponentially over time. This solution is overdampedttceKeKtx1211200200)(7Complex Roots: Underdamped•If  < 1, s1 and s2 are complex•Define the following constants:•This solution is underdamped tAtAetxjssddtcddsincos)(,121212008Real and Equal Roots•If  = 1, then s1 and s2 are real and equal•This solution is critically dampedttcetKeKtx0021)(9An Example• This is one possible implementation of the filter portion of an intermediate frequency (IF) amplifier10769pFvs(t)i(t)159H+–)()()(2)()(1)(1)()(2002222tftidttdidttiddttdvLtiLCdttdiLRdttids10An Example (cont’d.)•Note that 0 = 2f = 2455,000 Hz)•Is this system overdamped, underdamped, or critically damped?•What will the current look like?011.0μH159102rad/sec1086.2)pF769)(μH159(1106020LRLC11An Example (cont’d)•Increase the resistor to 1k•Exercise: what are  and 0?1k769pFvs(t)i(t)159H+–•The natural (resonance) frequency does not change: 0 = 2455,000 Hz)•But the damping ratio becomes  = 2.2•Is this system overdamped, underdamped, or critically damped?•What will the current look like?12A SummaryDamping RatioPoles (s1, s2)Dampingζ > 1 Real and unequal Overdampedζ = 1 Real and equal Critically damped0 < ζ < 1 Complex conjugate pair set Underdampedζ = 0 Purely imaginary pair Undamped13Transient and Steady-State Responses•The steady-state response of a circuit is the waveform after a long time has passed, and depends on the source(s) in the circuit13tteetf323102565)(SteadyStateResponseTransientResponseTransientResponseSteady-StateResponse14Class Examples•P7-6, P7-7,


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ASU EEE 202 - System Response II

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