# R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

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1 R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

2 8.1. Review of Bode plots Decibels Table 8.1. Expressing magnitudes in decibels G db = 0 log 10 G Actual magnitude Magnitude in db Decibels of quantities having units (impedance example): normalize before taking log Z db = 0 log 10 Z R base 1/ 6dB 1 0 db 6 db 5 = 10/ 0 db 6 db = 14 db 10 0dB 1000 = dB = 60 db 5Ω is equivalent to 14dB with respect to a base impedance of R base = 1Ω, also known as 14dBΩ. 60dBµA is a current 60dB greater than a base current of 1µA, or 1mA. 9

3 Bode plot of f n Bode plots are effectively log-log plots, which cause functions which vary as f n to become linear plots. Given: G = f n Magnitude in db is G db = 0 log 10 f Slope is 0n db/decade n = 0n log 10 f Magnitude is 1, or 0dB, at frequency f = 60dB 40dB 0dB 0dB 0dB 40dB 60dB 0dB/decade 0 db/decade 40dB/decade 40dB/decade n = f n = 1 n = 1 n = f f f 1 f log scale 10

4 Single pole response Simple R-C example R v + 1 (s) C + v (s) Transfer function is G(s)= v (s) v 1 (s) = 1 sc 1 sc + R Express as rational fraction: G(s)= 1 1+sRC This coincides with the normalized form G(s)= 1 1+ s 0 with 0 = 1 RC 11

5 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

6 Asymptotic behavior: low frequency For small frequency, ω << ω 0 and f << : 0 << 1 G(j ) db G(j ) = Then G(jω) becomes Or, in db, G(j ) 1 1 =1 0dB 0dB 40dB 0dB 0dB/decade f 1 G(j ) db 0dB 60dB f This is the low-frequency asymptote of G(jω) 13

7 Asymptotic behavior: high frequency For high frequency, ω >> ω 0 and f >> : 0 >> 1 G(j ) db G(j ) = dB 0dB Then G(jω) becomes G(j ) 1 0 = f 1 0dB 40dB 60dB 0dB/decade f 1 f The high-frequency asymptote of G(jω) varies as f -1. Hence, n = -1, and a straight-line asymptote having a slope of -0dB/decade is obtained. The asymptote has a value of 1 at f =. 14

8 Deviation of exact curve near f = Evaluate exact magnitude: at f = : G(j 0 ) = = 1 G(j 0 ) db = 0 log db at f = 0.5 and : Similar arguments show that the exact curve lies 1dB below the asymptotes. 15

9 Summary: magnitude G(j ) db 0dB 10dB 0dB 1dB 3dB 0.5 1dB 0dB/decade 30dB f 16

10 Phase of G(jω) Im(G(j )) G(j ) G(j ) G(j ) Re(G(j )) G(j )= 1 1+j = 1 j G(j )= tan 1 0 G(j )=tan 1 Im G(j ) Re G(j ) 17

11 Phase of G(jω) G(j ) 0 0 asymptote G(j )= tan ω G(jω) asymptote ω f 18

12 Phase asymptotes Low frequency: 0 High frequency: 90 Low- and high-frequency asymptotes do not intersect Hence, need a midfrequency asymptote Try a midfrequency asymptote having slope identical to actual slope at the corner frequency. One can show that the asymptotes then intersect at the break frequencies f a = e / / 4.81 f b = e /

13 Phase asymptotes G(j ) 0-15 f a = / 4.81 f a = e / / 4.81 f b = e / f b = f 0

14 Phase asymptotes: a simpler choice G(j ) 0-15 f a = / 10 f a = / 10 f b = f b = f 1

15 Summary: Bode plot of real pole 0dB G(j ) db 1dB 3dB 0.5 1dB G(s)= 1 1+ s 0 0dB/decade G(j ) 0 / /decade

16 8.1.. Single zero response Normalized form: Magnitude: G(s)= 1+ s 0 G(j ) = 1+ 0 Use arguments similar to those used for the simple pole, to derive asymptotes: 0dB at low frequency, ω << ω 0 +0dB/decade slope at high frequency, ω >> ω 0 Phase: G(j )=tan 1 0 with the exception of a missing minus sign, same as simple pole 3

17 Summary: Bode plot, real zero G(s)= 1+ s 0 +0dB/decade 0.5 1dB 0dB 1dB 3dB G(j ) db /decade 45 G(j ) 0 /

18 Right half-plane zero Normalized form: Magnitude: G(s)= 1 s 0 G(j ) = 1+ 0 same as conventional (left half-plane) zero. Hence, magnitude asymptotes are identical to those of LHP zero. Phase: G(j )= tan 1 0 same as real pole. The RHP zero exhibits the magnitude asymptotes of the LHP zero, and the phase asymptotes of the pole 5

19 Summary: Bode plot, RHP zero G(s)= 1 s 0 +0dB/decade 0dB 1dB 3dB G(j ) db 0.5 1dB G(j ) 0 / /decade

20 Frequency inversion Reversal of frequency axis. A useful form when describing mid- or high-frequency flat asymptotes. Normalized form, inverted pole: G(s)= 1 An algebraically equivalent form: G(s)= 1+ 0 s s 0 1+ s 0 The inverted-pole format emphasizes the high-frequency gain. 7

21 Asymptotes, inverted pole G(s)= s 3dB 1dB 1dB 0dB 0.5 G(j ) db +0dB/decade G(j ) +90 / /decade

22 Inverted zero Normalized form, inverted zero: G(s)= 1+ 0 s An algebraically equivalent form: G(s)= 1+ s 0 s 0 Again, the inverted-zero format emphasizes the high-frequency gain. 9

23 Asymptotes, inverted zero G(j ) db 0dB/decade G(s)= 1+ 0 s 0.5 1dB 3dB 1dB 0dB /decade 45 G(j ) 90 /

24 Combinations Suppose that we have constructed the Bode diagrams of two complex-values functions of frequency, G 1 (ω) and G (ω). It is desired to construct the Bode diagram of the product, G 3 (ω) = G 1 (ω) G (ω). Express the complex-valued functions in polar form: G 1 ( )=R 1 ( ) e j 1 ( ) G ( )=R ( ) e j ( ) G 3 ( )=R 3 ( ) e j 3 ( ) The product G 3 (ω) can then be written G 3 ( )=G 1 ( ) G ( )=R 1 ( ) e j 1 ( ) R ( ) e j ( ) G 3 ( )= R 1 ( ) R ( ) e j( 1 ( )+ ( )) 31

25 Combinations G 3 ( )= R 1 ( ) R ( ) e j( 1 ( )+ ( )) The composite phase is 3 ( )= 1 ( )+ ( ) The composite magnitude is R 3 ( )=R 1 ( ) R ( ) R 3 ( ) db = R 1 ( ) db + R ( ) db Composite phase is sum of individual phases. Composite magnitude, when expressed in db, is sum of individual magnitudes. 3

26 Example 1: G(s)= G 0 1+ s 1 1+ s with G 0 = 40 3 db, f 1 = ω 1 /π = 100 Hz, f = ω /π = khz G 40 db 0 db 0 db G 0 = 40 3 db G 0 db f Hz 0 db/decade G 0 db 40 db G 0 f 1 /10 10 Hz f /10 00 Hz 45 /decade f khz 40 db/decade db /decade 10f 10f 1 1 khz 0 khz 45 /decade Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f 33

27 Example Determine the transfer function A(s) corresponding to the following asymptotes: A f A db f 1 A 0 db +0 db/dec 10f 1 f /10 A +45 /dec /dec 0 0 f 1 /10 10f 34

28 Example, continued One solution: A(s)=A 0 1+ s 1 1+ s Analytical expressions for asymptotes: For f < f 1 1+ s 1 A 0 1+ s For f 1 < f < f s = j = A = A s 1 A 0 1+ s s = j = A 0 s 1 s = j 1 = A 0 1 = A 0 f f 1 35

29 Example, continued For f > f A s s s = j = A 0 s 1 s = j s s = j = A 0 1 = A 0 f f1 So the high-frequency asymptote is A = A 0 f f1 Another way to express A(s): use inverted poles and zeroes, and express A(s) directly in terms of A 1+ 1 s A(s)=A 1+ s 36

30 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 G(s)= 1 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

31 Approach 1: factor denominator G(s)= 1 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: G(s)= 1 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

32 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

33 Approach : Define a standard normalized form for the quadratic case G(s)= 1 1+ s + s or G(s)= 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 >

34 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

35 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 G(s)= 1 1+ s Q 0 + s 0 Result: = 0 = 1 LC Q = R C L 41

36 Magnitude asymptotes, quadratic form In the form G(s)= 1 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

37 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

38 Phase asymptotes! G(jω) = tan 1 1 Q ω ω0 1 ω ω 0 0 Increasing Q G 90 Low frequency asymptote o " High frequency asymptote of 180 " Change from 0 to 180 becomes more sharp as Q is increased" f /! 45!!

39 Mid-frequency phase asymptote! Match slope at f = :" or" Choose same approximation as in real pole case: " 0 0 f a 0 0 f a 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!!

40 Two-pole response: exact curves 0 Q = Q = 10dB Q = 5 Q = Q = 10 Q =5 Q = Q = 1 Q = Q = 1 Q = 0.7 Q = 0.5 0dB Q = 0. Q = 0.1 G db Q = 0.5 G dB Q = Q = 0.1-0dB f / f / 44

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