8.1.6 Quadratic pole response: resonance

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1 8.1.6 Quadratic pole response: resonance Example G(s)= v (s) v 1 (s) = 1 1+s L R + s LC L + Second-order denominator, of the form 1+a 1 s + a s v 1 (s) + C R Two-pole low-pass filter example v (s) with a 1 = L/R and a = LC How should we construct the Bode diagram? 37

2 Approach 1: factor denominator 1+a 1 s + a s We might factor the denominator using the quadratic formula, then construct Bode diagram as the combination of two real poles: 1 s s 1 s with s 1 = a a a a 1 s 1 s = a 1 a a a 1 If 4a a 1, then the roots s 1 and s are real. We can construct Bode diagram as the combination of two real poles. If 4a > a 1, then the roots are complex. In Section 8.1.1, the assumption was made that ω 0 is real; hence, the results of that section cannot be applied and we need to do some additional work. 38

3 G(jω) and G(jω) Let s = jω: G(j )= 1 1+j = 1 j Magnitude is G(j ) = Re (G(j )) + Im (G(j )) = Im(G(j )) G(j ) G(j ) G(j ) Re(G(j )) Magnitude in db: G(j ) db = 0 log db 1

4 Approach : Define a standard normalized form for the quadratic case 1+ s + s or 0 1+ s + 0 Q s 0 0 When the coefficients of s are real and positive, then the parameters ζ, ω 0, and Q are also real and positive The parameters ζ, ω 0, and Q are found by equating the coefficients of s The parameter ω 0 is the angular corner frequency, and we can define = ω 0 /π The parameter ζ is called the damping factor. ζ controls the shape of the exact curve in the vicinity of f =. The roots are complex when ζ < 1. In the alternative form, the parameter Q is called the quality factor. Q also controls the shape of the exact curve in the vicinity of f =. The roots are complex when Q >

5 The Q-factor In a second-order system, ζ and Q are related according to Q = 1 Q is a measure of the dissipation in the system. A more general definition of Q, for sinusoidal excitation of a passive element or system is (peak stored energy) Q = (energy dissipated per cycle) For a second-order passive system, the two equations above are equivalent. We will see that Q has a simple interpretation in the Bode diagrams of second-order transfer functions. 40

6 Analytical expressions for and Q Two-pole low-pass filter example: we found that G(s)= v (s) v 1 (s) = 1 1+s L R + s LC Equate coefficients of like powers of s with the standard form 1+ s Q 0 + s 0 Result: = 0 = 1 LC Q = R C L 41

7 Magnitude asymptotes, quadratic form In the form 1+ s Q 0 + s 0 let s = jω and find magnitude: G(j ) = Q 0 Asymptotes are G 1 for << 0 G f for >> 0 G(j ) db 0 db 0 db 40 db 60 db 0 db f 40 db/decade f 4

8 Deviation of exact curve from magnitude asymptotes G(j ) = 1 At ω = ω 0, the exact magnitude is Q 0 G(j 0 ) = Q or, in db: G(j 0 ) db = Q db The exact curve has magnitude Q at f =. The deviation of the exact curve from the asymptotes is Q db G 0 db Q db 40 db/decade 43

9 Phase asymptotes! G(jω) = tan 1 1 Q ω ω0 1 ω ω 0 Increasing Q G 9 Low frequency asymptote o " High frequency asymptote of 180 " Change from 0 to 180 becomes more sharp as Q is increased" f /! 45!!

10 Mid-frequency phase asymptote! Match slope at f = :" or" Choose same approximation as in real pole case: " f a f a Slope: 180Q /decade G G f b f b f / f / 1/Q f a = e π/ Q 1 f a = 10 f b = e π/ 1 f Q b = 10 1/Q! 46!!

11 Two-pole response: exact curves 10dB Q = Q = 5 Q = Q = 1 Q = Q = Q = 10 Q =5 Q = Q = 1 Q = 0.7 Q = 0.5 G db 0dB Q = 0.5 G -9 Q = 0. Q = dB Q = Q = 0.1-0dB f / f /

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