ECE 45 Bode Plots

University of California, San Diego J. Connelly

ECE 45 Bode Plots

Bode Plots: Recall the transfer function (frequency response) H(ω) of an LTI system is such that

A cos(ωot + θ) H(ω) A H(ωo) cos(ωot + ∠H(ωo)+ θ). −→ −→ | | H(ω) represents the change in magnitude and the change in phase when the input to the system is sinusoidal with frequency ω. Bode plots are a ways of plotting and approximating the magnitude H(ω) and phase ∠H(ω) for a very large range of frequencies without losing precision. | |

We generally plot the magnitude in decibels: 20 log10( H(ω) ) and the phase in radians. In each plot the ω-axis is on a logarithmic scale, i.e. the “ticks” are powers| | of 10. Bode plots make use of two mathematical approximations that are fairly accurate when considering orders of magnitude changes in x for x,a > 0:

2 x log10(x) x a log10 1+ ≥ a ≈ 0 x < a r ! a 0 x< 10 −1 x 10x a tan π/4 log10 x< 10a a  a 10 ≤ ≈ π/2 x 10a  ≥ Both of these are approximately linear (in certain regions) on a log scale. i.e.:

2 x −1 x log10 1+ tan a ! ≈ a ≈ r 40 30 π 20 2 π 10 4 x x a a a a 100 10 a 10a 100a 100 10 a 10a 100a We can use these approximations to plot the magnitude and phase of a transfer function H(ω).

Please report any typos/errors to [email protected] Suppose we have a transfer function

2 m x0 + x1(jω)+ x2(jω) + + xm(jω) H(ω)= 2 ··· n y0 + y1(jω)+ y2(jω) + + yn(jω) ··· where x0,...,xm and y0,...,y1 are complex numbers. (Nearly all transfer functions we encounter are of this form). By factoring polynomials, we can write H(ω) in standard form:

jω jω 1+ a1 1+ ar H(ω)= Cωk ··· 1+ jω 1+ jω b1 ··· bs where a1,...,ar and b1,...,bs are real numbers, C is some constant complex number, and k is some positive or negative integer. Having a transfer function in standard form allows us to use our earlier approximations, since we can write each linear term in polar form as

2 jω ω tan−1 ω 1+ = 1+ ej ( a ) a a r

Approximating the magnitude of a transfer function in standard form: The magnitudeof a product of complex numbers is the product of the magnitudes (i.e. XY = X Y ), and the log of a product is the sum of the logs (i.e. log(AB) = log(A) + log(B)). Thus| | | || |

r 2 s 2 ω ω 20 log10 ( H(ω) )=20 log10( C )+ k log10(ω)+ log10 1+ log10 1+   a   b  | | | | n=1 s n − n=1 s n X X     

2 ω 0 when ω < an 20 log10 1+ an ! ≈ adds 20 dB/dec when ω an r ≥ 20 log10( C )is present at all frequencies. | | 20k log10(ω) is zero when ω =1 and adds 20k dB/dec.

Approximating the phase of a transfer function in standard form: The phase of a product of complex numbers is the sum of the phases (i.e. ∠XY = ∠X + ∠Y ). Thus

r s ω ω ∠H(ω)= ∠(C)+ ∠(ωk)+ tan−1 tan−1 a b n=1 n − n=1 n X X ∠(C) is present at all frequencies, and ωk is a real number, so ∠(ωk)=0 .

0 when ω < an/10 −1 ω tan adds π/4 rads/dec when an/10 ω < 10an an ≈  ≤ π/2 when ω 10an  ≥  Approximating Values from the Bode Plot: We can use the fact Bode plots are a logrithmic linear approximation of the magnitude and phase to ﬁnd the values of points:

∆y y2 y1 y2 y1 Slope = = − = − . ∆x log(x2) log(x1) log(x2/x1) − Example 1: Draw the Bode plot (amplitude and phase) and ﬁnd all critical points of the transfer function jω H(ω)= 1000 + jω

H(ω) is a simple high pass ﬁlter. We ﬁrst write H(ω) in standard form: jω H(ω)= jω 1000 1+ 1000 Then

ω 2 20 log10 ( H(ω) ) = 20 log10 (ω) 60 20 log10 1+ | | − − 1000 ! r π ω ∠H(ω)= tan−1 . 2 − 103

20 log10 H(ω) 40 20 | | B dB ω 20 10−1 1 10 102 103 104 105 106 20 −1 2 3 4 5 6 − 10 1 10 10 10 10 10 10 40 20 − − A 40 60 − − 60 80 − −

∠ π H(ω) C 2 π 2 π 4 π 4 D −1 2 3 4 5 6 rads ω 10 1 10 10 10 10 10 10 −1 2 3 4 5 6 π 10 1 10 10 10 10 10 10 4 π − 4 π − − 2 To ﬁnd the height of point A, we note 20 log10(1) = 0, so 60 is the only “active” term at ω =1. − To ﬁnd the height of point B, we use the fact the slope is 20 dB per decade. So the height of B is

3 60 + 20(log10(10 ) log10(1)) = 0. − − We also have C = (100,π/2) and D = (104, 0).

Example 2: Draw the Bode plot (amplitude and phase) and ﬁnd all critical points of the transfer function 500+5jω H(ω)= jω (1 + jω) 10 + 100 We ﬁrst write H(ω) in standard form:

jω 50 1+ 100 H(ω)= jω (1 + jω) 1+ 1000 Then

ω 2 20 log10 ( H(ω) ) = 20 + 20 log10(5) + 20 log10 1+ | | 100 ! r 2 2 ω 20 log10 √1+ ω 20 log10 1+ − − 1000 ! r ω ω ∠H(ω) = tan−1 tan−1 (ω) tan−1 . 100 − − 1000 The height at point A is 20(1 + log10(5)).

The height at point B is 20+20 log10(5), since the slope is 20 dB/decade and the frequency changes by a factor of 100. − −

20 log10( H(ω0) )=0 when | |

0 = 20(1 + log10(5)) 20 (log10(ω0) log10(1)) − − 1 + log10(5) = log10(ω0) ω0 = 50 → → 20 log10 H(ω) 40 40 | | A 20 20 20 log(5) B dB ω −2 −1 2 3 4 5 10−210−1 1 10 102 103 104 105 10 10 1 10 10 10 10 10 20 20 − − 40 40 − − 60 60 − −

π 2 π ∠ 4 H(ω) π 4 rads ω −2 −1 2 3 4 5 π 10 10 1 10 10 10 10 10 −2 −1 2 3 4 5 4 π 10 10 1 10 10 10 10 10 − 4 π − 2 π − − 2 Example 3: Draw the Bode plot (amplitude and phase) and ﬁnd all critical points of the transfer function 3 (400 ω2 + 40jω) H(ω)= − 2(jω + 6000)(jω 10−6 + 1)

We ﬁrst write H(ω) in standard form:

jω 2 ( 20 + 1) H(ω)= jω jω 10 6000 +1 106 +1

Then ω 2 ω 2 ω 2 20 log10 ( H(ω) ) = 40 log10 1+ 20 20 log10 1+ 20 log10 1+ 6 | | 20 ! − − 6000 ! − 10 ! r r r ω ω ω ∠H(ω)= 2tan−1 tan−1 tan−1 . 20 − 6000 − 106 The change in log(ω) from point A to B is log(6000) log(20), and the slope (dB per decade) in this region is 40, so the height at B is − 20 + 40(log(6000) log(20)) = 60 + 40 log(3). − − The height at C is (60 + 40log(3)) + 20(log(106) log(6000)) = 120 + 20log(3) 20 log(2) − − 20 log10( H(ω0) )=0 when | |

0= 20 + 40(log(ω0) log(20)) ω0 = 20 √10 63.2 − − −→ ≈ We also have D = (2, 0), E = (200, π), F = (600, π), G = (60000,π/2), H = (105,π/2), I = (107, 0)

120 C 20 log10 H(ω) 100 | | 80 B 40 60 20 40

1 10 102 103 104 105 106 107 20 20 − dB 2 3 4 5 6 7 ω 40 1 10 10 10 10 10 10 10 − 20 A − 40 −

E F π π ∠H(ω) 3π 3π 4 4 π π H 2 2 G π π 4 4 D I rads ω 1 10 102 103 104 105 106 107 1 10 102 103 104 105 106 107 π π − 4 − 4 π π − 2 − 2