[PPT] - Overview of Bode Plots Transfer function review Piece-wise linear PowerPoint Presentation

SLIDE 1

Overview of Bode Plots

Transfer function review
Piece-wise linear approximations
First-order terms
Second-order terms (complex poles & zeros)
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

1

SLIDE 2

Transfer Function Review H(s)

x(t) y(t)

Recall that if H(s) is known and x(t) = A cos(ωt + φ), then we can find the steady-state solution for y(t): yss(t) = A|H(jω)| cos (ωt + φ + ∠H(jω))

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

2

SLIDE 3

Bode Plots H(s)

x(t) y(t)

Bode plots are standard method of plotting the magnitude and

phase of H(s)

Both plots use a logarithmic scale for the x-axis
Frequency is in units of radians/second (rad/s)
The phase is plotted on a linear scale in degrees
Magnitude is plotted on a linear scale in decibels

HdB(jω) 20 log10 |H(jω)|

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

3

SLIDE 4

Decibel Scales It is important to become adept at translating between amplitude, |H(jω)|, and decibels, HdB(jω). Amplitude (|H(jω)|) Decibels (20 log10 |H(jω)|) 1 20 log10 1 = 10 20 log10 10 = 100 20 log10 100 = 1000 20 log10 1000 = 0.1 20 log10 0.1 = 0.01 20 log10 0.01 = 0.001 20 log10 0.001 =

1 2

20 log10

1 2

= -6.0206 2 20 log10 2 =

1

2

20 log10

1

2

=

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

4

SLIDE 5

Example 1: Bode Plots

vo(t)

+

vs(t) RL 20 nF

1 kΩ 10 kΩ

1. Find the transfer function of the circuit shown above.
2. Generate the bode plot.
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

5

SLIDE 6

Example 1: Workspace

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

6

SLIDE 7

Example 1: Bode Plot

10

1

10

2

10

3

10

4

10

5

10 20 |H(jω)| (dB) Active Lowpass RC Filter 10

1

10

2

10

3

10

4

10

5

100 120 140 160 180 ∠ H(jω) (degrees) Frequency (rad/s)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

7

SLIDE 8

Example 1: MATLAB Code

w = logspace(1,5,500); H = -50e3./(jw + 5e3); subplot(2,1,1); h = semilogx(w,20log10(abs(H))); set(h,’LineWidth’,1.4); ylabel(’|H(j\omega)| (dB)’); title(’Active Lowpass RC Filter’); set(gca,’Box’,’Off’); grid on; set(gca,’YLim’,[-5 25]); subplot(2,1,2); h = semilogx(w,angle(H)*180/pi); set(h,’LineWidth’,1.4); ylabel(’\angle H(j\omega) (degrees)’); set(gca,’Box’,’Off’); grid on; set(gca,’YLim’,[85 185]); xlabel(’Frequency (rad/s)’);

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

8

SLIDE 9

Example 2: Bode Plots

+

vs(t) RL

1 kΩ 1 kΩ 1 µF 2 µF

1. Find the transfer function of the circuit shown above.
2. Generate the bode plot.
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

9

SLIDE 10

Example 2: Workspace

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

10

SLIDE 11

Example 2: Bode Plot

10

1

10

2

10

3

10

4

10

5

−2 2 4 6 8 |H(jω)| (dB) Active Lead/Lag RC Filter 10

1

10

2

10

3

10

4

10

5

−180 −175 −170 −165 −160 ∠ H(jω) (degrees) Frequency (rad/s)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

11

SLIDE 12

Example 2: MATLAB Code

w = logspace(1,5,500); H = -2(jw+500)./(jw + 1000); subplot(2,1,1); h = semilogx(w,20log10(abs(H))); set(h,’LineWidth’,1.4); ylabel(’|H(j\omega)| (dB)’); title(’Active Lead/Lag RC Filter’); set(gca,’Box’,’Off’); grid on; set(gca,’YLim’,[-2 8]); subplot(2,1,2); h = semilogx(w,angle(H)*180/pi); set(h,’LineWidth’,1.4); ylabel(’\angle H(j\omega) (degrees)’); set(gca,’Box’,’Off’); grid on; set(gca,’YLim’,[-180 -160]); xlabel(’Frequency (rad/s)’);

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

12

SLIDE 13

Bode Plot Approximations

Until recently (late 1980’s) bode plots were drawn by hand
There were many rules-of-thumb, tables, and template plots to

help

Today engineers primarily use MATLAB, or the equivalent
Why discuss the old method of plotting by hand?

– It is still important to understand how the poles, zeros, and gain influence the Bode plot – These ideas are used for transfer function synthesis, analog circuit design, and control systems

We will discuss simplified methods of generating Bode plots
Based on asymptotic approximations
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

13

SLIDE 14

Alternate Transfer Function Expressions There are many equivalent expressions for transfer functions. H(s) = N(s) D(s) = bmsm + bm−1sm−1 + · · · + b1s + b0 ansn + an−1sn−1 + · · · + a1s + a0 = bm an s±ℓ (s − z1)(s − z2) . . . (s − zm) (s − p1)(s − p2) . . . (s − pn) = k s±ℓ

1 − s

z1

1 − s

z2

. . .
1 −

s zm

1 − s

p1

1 − s

p2

. . .
1 −

s pn

This last expression is called standard form
The first step in making bode plots is to convert H(s) to standard

form

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

14

SLIDE 15

Magnitude Components Consider the expression for the transfer function magnitude: |HdB(jω)| = 20 log10 |H(jω)| = 20 log10

k

s±ℓ(1 − s

z1 ) . . . (1 − s zm )

(1 − s

p1 ) . . . (1 − s pn )

s=jω

= 20 log10 |k| · |jω|±ℓ |1 − jω

z1 | . . . |1 − jω zm |

|1 − jω

p1 | . . . |1 − jω pn |

= 20 log10 |k| ± ℓ 20 log10 ω +20 log10

1 − jω

z1

+ · · · + 20 log10
1 − jω

zm

−20 log10
1 − jω

p1

− · · · − 20 log10
1 − jω

pn

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

15

SLIDE 16

Magnitude Components Comments |HdB(ω)| = 20 log10 |k| ± ℓ 20 log10 ω +20 log10

1 − jω

z1

+ · · · + 20 log10
1 − jω

zm

−20 log10
1 − jω

p1

− · · · − 20 log10
1 − jω

pn

Thus, |HdB(ω)| can be written as a sum of simple functions
This is similar like using basis functions {δ(t),u(t),& r(t)} to write

an expression for a piecewise linear signal

We will use this approach to generate our piecewise linear

approximations of the bode plot

Note that there are four types of components in this expression

– Constant – Linear term – Zeros – Poles

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

16

SLIDE 17

Magnitude Components: Constant

20

20
40

(dB) (rad/sec) 40

|H(jω)| ω

The constant term, 20 log10 |k|, is a straight line on the Bode plot.

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

17

SLIDE 18

Magnitude Components: Linear Term

20

20
40

(dB) (rad/sec) 40

|H(jω)| ω

The linear term, ±ℓ 20 log10 |ω|, is a line on the magnitude plot with a slope equal to ±ℓ 20 dB per decade. The x-axis intercept occurs at ω = 1 rad/s. Plot the bode magnitude plots for H(s) = s, 1

s,s2, 1 s2 .

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

18

SLIDE 19

Magnitude Components: Real Zeros Consider two limiting conditions for a term containing a zero, 20 log10

1 − jω

z

First condition: ω ≪ |z|

lim

ω z →0 20 log10

1 − jω

z

= 0

Thus, if

ω |z| ≪ 1, then 20 log10

1 − jω

z

≈ 0.

Second condition: ω ≫ |z| lim

ω z →∞ 20 log10

1 − jω

z

= 20 log10 | − jω

z | = 20 log10 |ω| − 20 log10 |z|

Thus, if

ω |z| ≫ 1, then this term is linear (on a log scale) with a slope

f 20 dB per decade and an x-axis intercept at ω = |z|.
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

19

SLIDE 20

Magnitude Components: Real Zeros Continued

20

20
40

(dB) (rad/sec) 40

|H(jω)| ω

Our piecewise approximation joins these two linear asymptotic approximations at ω = |z|. Plot the piecewise approximation of the term 20 log10

1 − jω

z

.
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

20

SLIDE 21

Magnitude Components: Real Zeros Continued 2

10

−2

10

−1

10 10

1

10

2

−10 10 20 30 40 50 Bode Magnitude Real Zero: z = ±1 rad/s Mag (dB) Frequency (rad/sec)

The approximation is least accurate at ω = |z|. The true magnitude is 3 dB higher than the approximation at this corner frequency.

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

21

SLIDE 22

Magnitude Components: Real Poles

20

20
40

(dB) (rad/sec) 40

|H(jω)| ω

Consider two limiting conditions for a term containing a pole,

−20 log10

1 − jω

p

This is just the negative of the expression for a zero
The piecewise approximation is the mirror image of that for a zero
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

22

SLIDE 23

Magnitude Components: Real Poles Continued

10

−2

10

−1

10 10

1

10

2

−50 −40 −30 −20 −10 10 Bode Magnitude Real Pole: p = −1 rad/s Mag (dB) Frequency (rad/sec)

The approximation is least accurate at ω = |p|. The true magnitude is 3 dB less than the approximation at this corner frequency.

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

23

SLIDE 24

Complex Poles & Zeros

Complex poles and zeros require special attention
Will discuss later
You will not be expected to plot approximations with complex

poles or zeros on exams

There are essentially 3 steps to generating piecewise linear

approximations of bode plots

1. Convert to standard form
2. Plot the components
3. Graphically add the components together
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

24

SLIDE 25

Example 3: Magnitude Components Draw the piecewise approximation of the bode magnitude plot for H(s) = (s + 10)(s + 100)2 10s2(s + 1000)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

25

SLIDE 26

Example 3: Workspace

20

20
40

(dB) (rad/sec) 40 60

60

101 102 103 100 104 105

|H(jω)| ω

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

26

SLIDE 27

Example 3: Solution

10

−1

10 10

1

10

2

10

3

10

4

10

5

−40 −20 20 40 60 Bode Magnitude Example 1 Mag (dB) Frequency (rad/sec)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

27

SLIDE 28

Example 4: Magnitude Components Draw the piecewise approximation of the bode magnitude plot for H(s) = 1011s(s + 100) (s + 10)(s + 1000)(s + 10, 000)2

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

28

SLIDE 29

Example 4: Workspace

20

20
40

(dB) (rad/sec) 40 60

60

101 102 103 100 104 105

|H(jω)| ω

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

29

SLIDE 30

Example 4: Solution

10 10

2

10

4

10

6

−30 −20 −10 10 20 30 40 50 60 70 80 Bode Magnitude Example 2 Mag (dB) Frequency (rad/sec)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

30

SLIDE 31

Phase Components H(jω) = k (jω)ℓ(1 − jω

z1 ) . . . (1 − jω zm )

(1 − jω

p1 ) . . . (1 − jω pn )

Each of these terms can be expressed in polar form: a + jb = Aejθ. Note that (jω)ℓ = ωℓjℓ = ωℓ(ej π

2 )ℓ = ωℓej π 2 ℓ

H(jω) = |k|ejηkπ (ωℓejℓ π

2 )N1ejθ1 . . . Nmejθm

D1ejφ1 . . . Dnejφn = |k||ω|ℓN1 . . . Nm D1 . . . Dn × exp

j(ηkπ + ℓ π

2 + θ1 + · · · + θn − φ1 − · · · − φn)

where

ηk =

k ≥ 0

1 k < 0

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

31

SLIDE 32

Phase Components: Comments ∠H(jω) = ηkπ + ℓ π

2 + θ1 + · · · + θn − φ1 − · · · − φn

= ηkπ + ℓ π

2 + ∠

1 − jω

z1

+ · · · + ∠
1 − jω

zm

−∠
1 − jω

p1

− · · · − ∠
1 − jω

pn

Thus the phase of H(jω) is also a linear sum of the phases due to

each component

We will consider each of the four components in turn

– Constant – Linear term – Zeros – Poles

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

32

SLIDE 33

Phase Components: Constant

(deg) (rad/sec)

∠H(jω) ω 180◦ 90◦ −90◦ −180◦

The complex angle of the constant term, k, is either 0◦ if k > 0 or 180◦ if k < 0.

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

33

SLIDE 34

Phase Components: Linear Term

(deg) (rad/sec)

∠H(jω) ω 180◦ 90◦ −90◦ −180◦

The linear term, ∠(jω)ℓ = ∠jℓ = ℓ × 90◦, is a constant multiple of 90◦ Plot the bode phase plots for H(s) = s, 1

s,s2, 1 s2 .

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

34

SLIDE 35

Phase Components: Real Zeros Consider three limiting conditions for a term containing a zero, ∠1 − jω

z

First condition: ω ≪ |z| lim

ω z →0 ∠

1 − jω

z

= 0◦

Thus, if

ω |z| ≪ 1, then ∠

1 − jω

z

≈ 0◦.

Second condition: ω = |z| ∠

1 − jω

z

ω=|z| = ∠ (1 − jηz) = −ηz45◦

where ηz = sign(z) Third condition: ω ≫ |z| lim

ω z →∞ ∠

1 − jω

z

= ∠
− jω

z

= −ηz90◦

Thus, if

ω |z| ≫ 1, then ∠

1 − jω

z

≈ −ηz90◦.
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

35

SLIDE 36

Phase Components: Real Zeros Continued

(deg) (rad/sec)

∠H(jω) ω 180◦ 90◦ −90◦ −180◦

Our piecewise approximation joins these three linear asymptotic approximations at ω = 10−1|z| and ω = 10|z|. Plot the piecewise approximation of the term ∠

1 − jω

z

. Assume

that z is in the left-hand plane (i.e. Re{z} < 0)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

36

SLIDE 37

Phase Components: Real Zeros in Left Plane

10

−2

10

−1

10 10

1

10

2

−10 10 20 30 40 50 60 70 80 90 100 Phase (deg) Frequency (rad/sec)

The approximation is least accurate at ω = 0.1|z| and ω = 10|z|.

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

37

SLIDE 38

Phase Components: Real Zeros Continued 2

(deg) (rad/sec)

∠H(jω) ω 180◦ 90◦ −90◦ −180◦

If the zero is in the right half plane (i.e. Re{z} > 0), then the phase approaches −90◦ asymptotically. Plot the piecewise approximation of the term ∠

1 − jω

z

. Assume

that z is in the right half plane.

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

38

SLIDE 39

Phase Components: Real Zeros in Right Plane

10

−2

10

−1

10 10

1

10

2

−100 −90 −80 −70 −60 −50 −40 −30 −20 −10 10 Phase (deg) Frequency (rad/sec)

The approximation is least accurate at ω = 0.1|z| and ω = 10|z|.

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

39

SLIDE 40

Phase Components: Real Poles Real poles in the left half plane have the same phase as real zeros in the right half plane. We will only discuss poles in the left half plane because only these systems are stable. First condition: ω ≪ |p| lim

ω p →0 −∠

1 − jω

p

= −0◦

Second condition: ω = |p| −∠

1 − jω

p

ω=|p| = −∠ (1 − sign(p) × j) = −∠ (1 + j) = −45◦

Third condition: ω ≫ |p| lim

ω p →∞ −∠

1 − jω

p

= −∠
− jω

p

= −∠j = −90◦
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

40

SLIDE 41

Phase Components: Real Poles

(deg) (rad/sec)

∠H(jω) ω 180◦ 90◦ −90◦ −180◦

Plot the piecewise approximation of the term −∠

1 − jω

p

. Assume

that p is in the left-hand plane (i.e. Re{p} < 0)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

41

SLIDE 42

Phase Components: Real Poles in Left Plane

10

−2

10

−1

10 10

1

10

2

−100 −90 −80 −70 −60 −50 −40 −30 −20 −10 10 Phase (deg) Frequency (rad/sec)

The approximation is least accurate at ω = 0.1|p| and ω = 10|p|.

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

42

SLIDE 43

Example 5: Phase Components

(deg) (rad/sec) 101 102 103 100 104 105 270

270

∠H(jω) 180◦ 90◦ −90◦ −180◦

Draw the piecewise approximation of the bode phase plot for H(s) = (s + 10)(s + 100)2 10s2(s + 1000)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

43

SLIDE 44

Example 5: Solution

10

−1

10 10

1

10

2

10

3

10

4

10

5

−150 −100 −50 50 Phase (deg) Frequency (rad/sec)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

44

SLIDE 45

Example 6: Phase Components

(deg) (rad/sec) 101 102 103 100 104 105 270

270

∠H(jω) 180◦ 90◦ −90◦ −180◦

Draw the piecewise approximation of the bode phase plot for H(s) = 1011s(s + 100) (s + 10)(s + 1000)(s + 10, 000)2

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

45

SLIDE 46

Example 6: Solution

10 10

2

10

4

10

6

−150 −100 −50 50 100 Phase (deg) Frequency (rad/sec)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

46

SLIDE 47

Example 7: Magnitude

20

20
40

(dB) (rad/sec) 40 60

60

101 102 103 100 104 105

|H(jω)| ω

Plot the magnitude of H(s) = 10 s + 10 s + 1000

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

47

SLIDE 48

Example 7: Phase

(deg) (rad/sec) 101 102 103 100 104 105 270

270

∠H(jω) 180◦ 90◦ −90◦ −180◦

Plot the phase of H(s) = 10 s + 10 s + 1000

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

48

SLIDE 49

Example 7: Solution

10

−1

10 10

1

10

2

10

3

10

4

10

5

−20 20 40 Bode Example 1 Mag (dB) 10

−1

10 10

1

10

2

10

3

10

4

10

5

50 100 Phase (deg) Frequency (rad/sec)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

49

SLIDE 50

Example 8: Magnitude

20

20
40

(dB) (rad/sec) 40 60

60

101 102 103 100 104 105

|H(jω)| ω

Plot the magnitude of H(s) = −10 s − 10 s + 1000

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

50

SLIDE 51

Example 8: Phase

(deg) (rad/sec) 101 102 103 100 104 105 270

270

∠H(jω) 180◦ 90◦ −90◦ −180◦

Plot the phase of H(s) = −10 s − 10 s + 1000

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

51

SLIDE 52

Example 8: Solution

10

−1

10 10

1

10

2

10

3

10

4

10

5

−20 20 40 Bode Example 2 Mag (dB) 10

−1

10 10

1

10

2

10

3

10

4

10

5

−150 −100 −50 Phase (deg) Frequency (rad/sec)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

52

SLIDE 53

Example 9: Circuit Example

vs(t) vo(t)

+

100 mH 10 mF

11 Ω

Draw the straight-line approximations of the transfer function for the circuit shown above. Hint: Recall from one of the Transfer Functions Examples H(s) =

R Ls

s2 + R

Ls + 1 LC

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

53

SLIDE 54

Example 9: Magnitude

20

20
40

(dB) (rad/sec) 40 60

60

101 102 103 100 104 105

|H(jω)| ω

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

54

SLIDE 55

Example 9: Phase

(deg) (rad/sec) 101 102 103 100 104 105 270

270

∠H(jω) 180◦ 90◦ −90◦ −180◦

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

55

SLIDE 56

Example 9: Solution

10

−1

10 10

1

10

2

10

3

10

4

−40 −20 20 Bode Example 3 Mag (dB) 10

−1

10 10

1

10

2

10

3

10

4

−100 −50 50 100 Phase (deg) Frequency (rad/sec)

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

56

SLIDE 57

Complex Poles Complex poles can be expressed in the following form: C(s) = ω2

n

s2 + 2ζωns + ω2

n

= 1 1 + 2ζ s

ωn +

s

ωn

2

ωn is called the undamped natural frequency
ζ (zeta) is called the damping ratio
The poles are p1,2 = (−ζ ±
ζ2 − 1) ωn
If ζ ≥ 1, the poles are real
If 0 < ζ < 1, the poles are complex
If ζ = 0, the poles are imaginary: p1,2 = ±jωn
If ζ < 0, the poles are in the right half plane (Re{p} > 0) and the

system is unstable

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

57

SLIDE 58

Complex Poles Continued The transfer function C(s) can also be expressed in the following form C(s) = 1 1 + 2ζ s

ωn +

s

ωn

2 = 1 1 +

s Qωn +

s

ωn

2 where Q 1 2ζ The meaning of Q, the Quality factor, will become clear in the following slides. C(jω) = 1 1 −

ω

ωn

2 +

jω Qωn

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

58

SLIDE 59

Complex Poles Magnitude 20 log10 |C(jω)| = 20 log10

1 −

ω ωn 2 + jω Qωn

−1

= −20 log10

1 − ω2

ω2

n

2 + ω Qωn 2 For ω ≪ ωn, 20 log10 |C(jω)| ≈ −20 log10 |1| = 0 dB For ω ≫ ωn, 20 log10 |C(jω)| ≈ −20 log10 ω2 ω2

n

= −40 log10 ω ωn dB At these extremes, the behavior is identical to two real poles.

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Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 60

Complex Poles Magnitude Continued 20 log10 |C(jω)| = −20 log10

1 − ω2

ω2

n

2 + ω Qωn 2 For ω = ωn, 20 log10 |C(jωn)| = −20 log10 1 Q = 20 log10 Q = QdB

J. McNames

Portland State University ECE 222 Bode Plots

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SLIDE 61

Complex Poles Magnitude

10

−1

10 10

1

−60 −50 −40 −30 −20 −10 10 20 30 40 Frequency (rad/sec) Mag (dB) Complex Poles Q = 0.1 Q = 0.5 Q = 0.707 Q = 1 Q = 2 Q = 10 Q = 100

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 62

Complex Poles Phase ∠C(jω) = ∠ 1 1 −

ω

ωn

2 +

jω Qωn

For ω ≪ ωn, ∠C(jω) ≈ ∠1 = 0◦ For ω ≫ ωn, ∠C(jω) ≈ ∠ 1 −

ω Qωn

= ∠ − Qωn ω = ∠ − 1 = −180◦ For ω = ωn, ∠C(jω) ≈ ∠ 1 Qj = ∠1 j = ∠ − j = −90◦ At the extremes, the behavior is identical to two real poles. At other values of ω near ωn, the behavior is more complicated.

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 63

Complex Poles Phase

10

−2

10

−1

10 10

1

10

2

−180 −160 −140 −120 −100 −80 −60 −40 −20 Frequency (rad/sec) Phase (deg) Complex Poles Q = 0.1 Q = 0.5 Q = 0.707 Q = 1 Q = 2 Q = 10 Q = 100

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 64

Complex Zeros

Left half plane:

– Inverted magnitude of complex poles – Inverted phase of complex poles

Right half plane:

– Inverted magnitude of complex poles – Same phase of complex poles

This is the same relationship real zeros had to real poles
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 65

Complex Zeros Magnitude

10

−1

10 10

1

−40 −20 20 40 60 80 Frequency (rad/sec) Mag (dB) Complex Zeros Q = 0.1 Q = 0.5 Q = 0.707 Q = 1 Q = 2 Q = 10 Q = 100

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 66

Complex Zeros Phase

10

−2

10

−1

10 10

1

10

2

20 40 60 80 100 120 140 160 180 Frequency (rad/sec) Phase (deg) Complex Zeros Q = 0.1 Q = 0.5 Q = 0.707 Q = 1 Q = 2 Q = 10 Q = 100

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 67

Complex Poles Maximum What is the frequency at which |C(jω)| is maximized? C(jω) = 1 1 +

jω

Qωn

+
jω

ωn

2 |C(jω)| = 1

1 − ω2

ω2

n

2 +

ω

Qωn

2

For high values of Q, the maximum of |C(jω)| > 1
This is called peaking
The largest Q before the onset of peaking is Q =

1 √ 2 ≈ 0.707

This curve is said to be maximally flat
This is also called a Butterworth response
In this case, |C(jωn)| = −3 dB and ωn is the cutoff frequency
J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 68

Complex Poles Maximum Continued If Q > 0.707, the maximum magnitude and frequency are as follows: ωr = ωn

1 −

1 2Q2 |C(jωr)| = Q 1 −

1 4Q2

ωr is called the resonant frequency or the damped natural

frequency

As Q → ∞, ωr → ωn
For sufficiently large Q (say Q > 5)

– ωr ≈ ωn – |C(jωr)| ≈ Q

Peaked responses are useful in the synthesis of high-order filters
Complex zeros (in the left half plane) have the inverted magnitude

and phase of complex poles

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 69

Complex Poles Example

vs(t) vo(t)

+

50 mH R 200 nF

Generate the bode plot for the circuit shown above. H(s) =

1 LC

s2 + R

Ls + 1 LC

= ω2

n

s2 + 2ζωns + ω2

n

= ω2

n

s2 + ωn

Q s + ω2 n

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 70

Complex Poles Example Continued 1 ωn =

1

LC = 10 k rad/s ζ = R 2L √ LC = R 2 √ CL = R × 0.001 Q = 1 √ LC L R =

L

C 1 R = 500 R R = 5 Ω ζ = 0.005 Q = 100 Very Light Damping R = 50 Ω ζ = 0.05 Q = 10 Light Damping R = 707 Ω ζ = 1.41 Q = 0.707 Strong Damping R = 1 kΩ ζ = 1 Q = 0.5 Critical Damping R = 5 kΩ ζ = 5 Q = 0.1 Over Damping

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Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 71

Complex Poles Example Continued 2

10

2

10

3

10

4

10

5

10

6

−120 −100 −80 −60 −40 −20 20 40 Frequency (rad/sec) Mag (dB) Resonance Example R = 5 Ω R = 50 Ω R = 707 Ω R = 1 kΩ R = 5 kΩ R = 50 kΩ

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

71

SLIDE 72

Complex Poles Example Continued 3

10

2

10

3

10

4

10

5

10

6

−180 −160 −140 −120 −100 −80 −60 −40 −20 Frequency (rad/sec) Phase (deg) Resonance Example R = 5 Ω R = 50 Ω R = 707 Ω R = 1 kΩ R = 5 kΩ R = 50 kΩ

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19

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SLIDE 73

Complex Poles Example Continued 4

0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Time (sec) Output (V) Resonance Example R = 5 Ω R = 50 Ω R = 707 Ω R = 1 kΩ R = 5 kΩ R = 50 kΩ

J. McNames

Portland State University ECE 222 Bode Plots

Ver. 1.19