Bode Plots - Engineering Electrical Circuits - Lecture Slides, Slides of Electrical Circuit Analysis

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Bode Plots

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Outline: Bode Plots

• A full Bode DIAGRAM consists of two Separate

Plots

1. |H(f)| (in dB) vs. f in Hz (or ω in rads/S)

2. ∠H(f) in Degrees vs f in Hz (or ω in rads/S)

• Recall: ω = 2π∙f
• “Poles” & “Zeros” in H(f) with one or the

Other

• A “Pole” Frequency, fBp, causes |H|→∞
• A “Zero” Frequency, fBz, causes |H| →

ProtoTypical H( f ) for Bode

• The 1-”Pole” and

1-”Zero” Xfer fcn:

• Then The “dB” Magnitude
• And the Phase Angle

( ) [ ]



















 

 

 



=

Bp

Bz

f

f jf j

f

f j

H f K

1

1

0

( ) 













 ∠ = 













 ∠ =∠ −∠ °−∠ ∠ =

Bz f Bp

f

f

f H f α 90 β α arctan β arctan

Bz Bp

dB f

f f j f

f H f = 20 log K 0 + 20 log 1 + j − 20 log − 20 log 1 +

Bode Plot Construction

• Consider a Transfer

Function with one pole, and

one zero

• K ≡ CONSTANT
• First, put the Transfer

Function into STANDARD

FORM

( )

( )

jf ( f jf )

K f jf

H f

Bp

Bz

=















 ⋅ +



 

 

 



=

Bp

Bp

Bz

Bz

f

f f jf j

f

f Kf j

H f

1

1



















 

 

 





 





 



Bp

Bz

Bp

Bz

f

f jf j

f

f j

f

Kf H f

1

1

( ) [ ]



















 

 

 



=

Bp

Bz

f

f jf j

f

f j

H f K

1

1

0

Bode Plot Construction

• Bode Magnitudes are typically plotted in

deciBels.

• So from the

last Slide

( ) [ ] ⇒

=

Bp

Bz

f

f f j

f

f j

H f K

1

1

0

( ) [ ]

Bz Bp

dB

Bp

Bz dB

f

f f j f

f H f K j

f

f f j

f

f j

H f K

= + + − − +





























=

20 log 20 log 1 20 log 20 log 1

1

1

20 log

0

0

Magnitude Equation

• Examine the dB Magnitude Equation
• 20logK 0

is a Constant

• Positive for K 0 > 1
• Zero for K 0 = 1
• Negative for K 0 < 1
• If K 0 = 2, then

Bz Bp

dB f

f f j f

f H f = 20 log K 0 + 20 log 1 + j − 20 log − 20 log 1 +

101 102 103 104

0

5

10

Mag

dB

f(Hz) f=0:10:10000; Mag = 20log10(2)ones(1,length(f)); semilogx(f,Mag, 'LineWidth',3), grid axis([10 10000 -20 10]) ylabel('Mag_{dB}') xlabel('f(Hz)')

H ( f ) 20 log K 0

dB

=

Bode Plot Example:

• Taking f Bz

= 100 Hz, approximate |H(f)|

• Note the

“Corner” at

2

Hz

fBz

f H f = 20 log 1 + j

for 100 ( 20dB/decadeSlope) 100

20 log 100

20 log 1

0 for 100 Hz 100

20 log 1

^ ≥ 

  



• ≈

• ≈ ≤

f

f f j

f

f j

100 101 102 103 104

0

10

20

30

40

50

Mag

dB

f (Hz)

f = logspace(0,4,500); Mag = 20log10(abs(1+jf/100));; semilogx(f,Mag, 'LineWidth',2), grid axis([1 10000 -10 50]) ylabel('Mag_{dB}') xlabel('f (Hz)')

Straight-Line

Approximations

Bode Plot Example:

• Converting to dB form

with f

B

= 300Hz

• Note again the

Corner

• This time at

300 Hz

• 1885 rads/sec

( ) j (^ f fBp )

H f

= 1

1

( ) [ ( )]















 = − +

= +

−

Bp

dB

Bp

f

f H f j

H f j f f

20 log 1

1

1

3000 ( 20dB/decadeSlope)

~ for 300

20 log 300

20 log 1

30 Hz

~ 0 for 300

20 log 1

 >^ − 

  

 − + ≈−

− + ≈ <

f

f f j

f

f j

100 101 102 103 104

0

5

10

Mag

dB

f (Hz)

Straight-Line Phase-Angle Plots

• The Phase Angle of constant K 0 is zero degrees
• The phase angle of a 1 st^ order ZERO Xfer fcn

has 3 straight lines (jf/fBz in numerator)

• 1. For values of f = f (^) Bz, 45° point (by α = atan[1/1])
• 1. For values of f>=10 f (^) Bz, it is a straight line at 90° (α = atan[10/1] ≈ 90°)
• 1. For values of f<=0.1fBz, it is a straight line at 0° (α = atan[0.1/1] ≈ 0°)
• 1. For values of 0.1fBz <= f <= 10fBz, it is a straight line with a slope of 45°/decade
• The phase angle of 1 st^ order POLE Xfer fcn

has 3 straight lines (jf/fBp in denominator)

• 1. For values of f = f (^) Bp, −45° point (by −β = −atan[1/1])
• 1. For values of f>=10 f (^) Bp, it is a straight line at −90° (−β = −atan[10/1] ≈ − 90°)
• 1. For values of f<=0.1fBp, it is a straight line 0° ° (−β = −atan[0.1/1] ≈ 0°)
• 1. For values 0.1fBp <= f <= 10fBp, it is a straight line with a slope of −45°/decade
• The approximate phase angle of −20 log(f/f (^) B) is a straight line having a

slope of −45°/decade that intersects the 0 degree axis at f=0.1fB, and intersects the −90° line at f=10fB



 





 

 ∠ = f Bz

f α arctan















 −∠ =− f Bp

f β arctan

Example φ

H

• ReCall from the dB development
• In this case
• Also Recall that for this transfer fcn

( )

( ) (^ )^  

  



 

  



=



 

 

 



 

  



=

100

1

10

1 2

1

1

0 f jf j

f j

f

f jf j

f

f j

H f K

Bp

Bz

∠ = °















 ∠ = 













 =∠ −∠ °−∠ ∠ =

Recalling 90

90 arctan arctan

j

f

f

f

f

Bz Bp

φ H α β α β

 

  

  ∠ = 

  

 ∠ = 100

arctan 10

arctan

f f α β

( )

Bz Bp

dB f

f f j f

f H f = 20 log K 0 + 20 log 1 + j − 20 log − 20 log 1 +

0 °^ α^90 °^ β

Angle, φ

H

, Exact:(^ ) ( ) (^)  

  



 

  



=

100

1

10

1

2 f jf j

f j

H f

10

• 10

0 10

1 10

2 10

3 10

-100 4

0

20

40

60

80

100

φ

H

(°)

f (Hz)

f = logspace(-1,4,500); a = atand(f/10); b = atand(f/100); angj = 90*ones(1,length(f)); q = a - angj -b; semilogx(f,a, f,-b, f,-angj, f,q, 'LineWidth',2), grid axis([0.1 10000 -100 100]) ylabel('\phi_{H} (°)') xlabel('f (Hz)')

Angle, φ

H

, Exact by:(^ ) ( ) (^)  

  



 

  



=

100

1

10

1

2 f jf j

f j

H f

10

• 10

0 10

1 10

2 10

3 10

4

φ

H

(°)

f (Hz)

f = logspace(-1,4,500); h = 2(1+jf/10)./((jf).(1+jf/100)) a=angle(h); deg=a180/pi; [degmax, Nmax] = max(deg); fmax = f(Nmax); semilogx(f,deg, fmax, degmax, '*', 'LineWidth',3), grid axis([0.1 10000 -90 -30]) ylabel('\phi_{H} (°)') xlabel('f (Hz)') fmax degmax

(31.26 Hz, -35.10°)

Second Order Transfer Function

• So we have:
• To find the poles/zeros, put H in Standard

form:

• One zero at DC frequency (f=0)  can’t

conduct DC due to capacitor

R j fC

j fL

R

V

V H f

s

O

= =

2

1 2

( f ) LC j ( f ) RC

j f CR

V

V H f

s

O

1 2 2

2

2 − +

= =

−

V O

Resonance

• For the Series Resonant

ckt at Right the Series

Impedance:

• At the “Resonant”

Frequency, f 0 ,The

imaginary impedances

CANCEL EXACTLY

• Thus the Frequency,

f 0 , that produces a

PURELY Resistive

Impedance where VO

= VS :

−

V O

( ) R

fC

Z (^) s f = j fL − j +

2

1 2

LC

f

f C

j f L j

π

π

π

2

1 Or

2

1 2

0

0

0

=

=