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Lecture 17. System Response II • Poles & Zeros • Second-Order Circuits • LCR Oscillator circuit: An example • Transient and Steady States 1 Pole-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 (s 3)( s 3.5)( s 2 4s 5) H ( s) (s 5)( s 2 4)( s 1.5) Im Re 2 Second-Order Circuits i(t) • KVL around the loop: vr(t) + vc(t) + vl(t) = vs(t) + vr(t) – R + – + vc(t) C – vl(t) + L t – 1 di (t ) R i (t ) i ( x)dx L vs (t ) C dt di(t ) 1 d 2i(t ) dvs (t ) R i(t ) L 2 dt C dt dt d 2i(t ) R di(t ) 1 1 dvs (t ) i(t ) 2 dt L dt LC L dt 3 Second-Order Circuits • In general, a second-order circuit is described by d 2 x(t ) dx(t ) 2 2 0 0 x(t ) f (t ) 2 dt dt • For zero-initial conditions, the transfer function would be X( s) s 2 20 s 02 F( s) X( s ) 1 H ( s) 2 F( s) s 20 s 02 4 Characteristic Equation & Poles X( s ) 1 H ( s) 2 F( s ) s 20 s 02 • The denominator of the transfer function is known as the characteristic equation • To find the poles, we solve : s 2 2 0 s 02 0 which has two roots: s1 and s2 20 (20 ) 2 402 s1 , s2 0 0 2 1 2 Real and Unequal Roots: Overdamped • If > 1, s1 and s2 are real and not equal xc (t ) K1e 2 1 t 0 0 K 2e 2 1 t 0 0 • The amplitude decreases exponentially over time. This solution is overdamped 1 i(t) 0.8 0.6 0.4 0.2 0 0.0E+00 5.0E-06 Time 1.0E-05 6 Complex Roots: Underdamped 1 • If < 1, s1 and s2 are complex • Define the following constants: d 0 1 2 s1 , s2 j d 0.4 i(t) 0 0.8 0.6 0.2 0 -1.00E-05 -0.2 1.00E-05 3.00E-05 -0.4 -0.6 -0.8 -1 Time xc (t ) e t A1 cos d t A2 sin d t • This solution is underdamped 7 Real and Equal Roots • If = 1, then s1 and s2 are real and equal xc (t ) K1e 0t K2 t e 0t • This solution is critically damped 8 An Example i(t) 10W vs(t) + – 769pF • This is one possible implementation of the filter portion of an intermediate frequency (IF) amplifier 159mH d 2i (t ) R di(t ) 1 1 dvs (t ) i ( t ) dt 2 L dt LC L dt d 2i (t ) di(t ) 2 2 0 0 i (t ) f (t ) 2 dt dt 9 An Example (cont’d.) 1 1 0 2.86 106 rad/sec LC (159 μH)(769 pF) R 10 W 20 0.011 L 159 μH 2 0 i(t) • Note that 0 = 2pf = 2p 455,000 Hz) • Is this system overdamped, underdamped, or critically damped? • What will the current look like? 1 0.8 0.6 0.4 0.2 0 -0.2 -1.00E-05 -0.4 -0.6 -0.8 -1 1.00E-05 3.00E-05 10 Time An Example (cont’d) i(t) 1kW vs(t) + – 769pF • Increase the resistor to 1kW • Exercise: what are and 0? 159mH • The natural (resonance) frequency does not change: 0 = 2p455,000 Hz) • But the damping ratio becomes = 2.2 • Is this system overdamped, underdamped, or critically damped? • What will the current look like? 11 A Summary Damping Ratio Poles (s1, s2) Damping ζ>1 Real and unequal Overdamped ζ=1 Real and equal Critically damped Complex conjugate pair set Underdamped Purely imaginary pair Undamped 0<ζ<1 ζ=0 12 Transient 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 circuit 1.2 1 5 5 2 t 10 3t f (t ) e e 6 2 3 0.8 0.6 0.4 0.2 Steady State Response Transient Response 0 0 1 Transient Response 2 3 4 5 Steady-State Response 13 Class Examples • P7-6, P7-7, P7-8 14