Overview of Bode Plots Transfer Function Review Transfer function - - PowerPoint PPT Presentation

SLIDE 1

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

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

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

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 2

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

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

Example 1: MATLAB Code

w = logspace(1,5,500); H = -50e3./(j*w + 5e3); subplot(2,1,1); h = semilogx(w,20*log10(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

Example 1: Workspace

• J. McNames

Portland State University ECE 222 Bode Plots

• Ver. 1.19

6

SLIDE 3

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

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

Example 2: MATLAB Code

w = logspace(1,5,500); H = -2*(j*w+500)./(j*w + 1000); subplot(2,1,1); h = semilogx(w,20*log10(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

Example 2: Workspace

• J. McNames

Portland State University ECE 222 Bode Plots

• Ver. 1.19

10

SLIDE 4

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

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

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

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 5

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

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

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

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 6

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

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

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

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 7

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

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

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

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 8

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

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

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

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 9

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

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

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

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 10

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

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

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

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 11

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

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

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

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 12

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

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

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

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 13

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

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

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

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 14

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

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

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

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 15

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.

• J. McNames

Portland State University ECE 222 Bode Plots

• Ver. 1.19

59

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

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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

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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

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SLIDE 16

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

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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

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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
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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.

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SLIDE 17

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
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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

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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

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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

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SLIDE 18

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

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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

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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

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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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SLIDE 19

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

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