Mathcad - MAEBODE5 26 02

 100 90   10000 90   10000 90  51
     10 
 10000 90  B :=  10 
 50000 
 10000 0   10⋅10000 0  D :=  90 
 10
A :=   
 50000 0 
 10⋅10000 0 
 50000 −90   50000   10 ⋅50000 0 
 90 
  C :=  10
 106 −90  
 10⋅50000 0 

100

A〈1〉 50

B〈1〉
C〈1〉 0

D〈1〉 50

100 1 .103 1 .104 1 .105 1 .106
100 A〈0〉 , B〈0〉 , C〈0〉 , D〈0〉
Auxilliary Diagram

 100 90  ΦT (ω) := 90 + Φ1(ω) + Φ2(ω)
 10000 
 10 90 

 50000 
 10 58.546 
points := 

 10000⋅10 −58.546 

 50000⋅10 −90 
 106 −90 

100 52

Φ T(ω) 50
0
points〈1〉

50

100 1 .103 1 .104 1 .105 1 .106
100 ω , points〈0〉
^ ^ ^^
^ ^
#2 #4 #5 #6
#1 #3

For ω < 1000, the only contribution is from the zero at 0 so points #1 and #2.

For 1000 < ω < 5000, the slope is -45 deg per decade so point #3.

 50000  = 58.546 For 5000 < ω < 105 , the slope is
90 − 45 ⋅log 10 

 10000
10

-45 -45 = -90 deg per decade so point #4. 58.546 − 90 ⋅log 10 ⋅10000  = −58.547
 
50000
10

For 105 < ω < 5 ⋅105 , the slope is -45 deg per decade so point #5.

−58.547 − 45 ⋅log 5 ⋅105  = -90. For ω > 5 ⋅105 the slope is zero, so point #6.
105 

Second Order Magnitude Approximations

Second Order Magnitude Approximations 53

When we gave the general form of the transfer function as:

zero
nz |
K⋅ (s + ωzi)
∏ ( ) ( ) ( )H(s) \/
∏ ( ) ( ) ( )=
i=1 = K⋅ s + ωz1 ⋅ s + ωz2 ⋅.. ⋅ s + ωznz 

np s + ωp1 ⋅ s + ωp2 ⋅.. ⋅ s + ωpnp

(s + ωpi) /\

i=1 |
pole

We allowed for the possibility that some of the constants ωzi and ωpi might be complex
numbers as a result of second order terms that do not factor into real first order terms. To

make magnitude and phase approximations of such terms, the terms are not factored into

complex form but are left as second order terms of the form:

s2 + 2⋅ ζ ⋅s + 1
ωo2 ωo

where ζ < 1, since if ζ = 1, the quadratic is a perfect square and is handled as a pole of
muliplicity 2, and if ζ >1 the quadratic is factorable into two real factors and the term is handled
as two distinct first order poles. The second order response has been normalized so that the

constant coefficient is one. Note that in the following discussions ω is a variable and ωo is the
name of a constant .

Magnitude Response of Second Order Poles
consider a transfer function of the form:

H(s) = 1 H(j⋅ω) = 1

s2 + 2⋅ ζ ⋅s + 1 1 − ω2 + 2⋅ζ ⋅ω
2 ωo 2 ω0
ω o ω 0

The low frequency asymptote is found by taking the limit as ω approaches zero of H(jω).

54

H(j⋅ω) = lim 20 ⋅log 1  = 0 dB
dB 
ω→0   −ω 2 2⋅ ζ ⋅j⋅ω 1 
 + ωo +  
 2  
 ω o

The high frequency asymptote is found by taking the limit as ω approaches infinity of H(jω).

H(j⋅ω) = lim 20 ⋅log 1 
dB 
ω→∞   −ω 2 2⋅ ζ 1 
 + ωo ⋅j⋅ω +  
  2  
 ω o

lim 20 ⋅log 1  = lim 20 ⋅log ω 2 
 o 
 −ω 2   ω→∞ 
= ω→∞ ω2


 
  ω 2  
 o 

( ( ) ( ))= 2
lim 20 ⋅log ω o − 20 ⋅log ω 2

ω→∞

( )= −20 ⋅log ω 2 = −40 ⋅log(ω)

The approximation strategy is the same as with first order magnitude plots: The magnitude

function is approximated by a horizontal straight line of 0 dB for frequencies less than ωo and
by a straight line of slope -40 dB per decade passing through ω = ωo for all frequencies greater
than ωo. The main difference between the first order case and the second order case, is that
the error depends heavily on ζ and may not usually be ignored. Several second order

magnitude functions are plotted below for various values of ζ (ωo has been normalized to 1.0).

20 ⋅log 1

:=( )Ha(ω)    (ζ = 1.0)
1 − ω2 
+ 2 ⋅ω ⋅(1) ⋅i

( )Hb(ω) 20 ⋅log 1 55

:=   1 − ω2 + 2 ⋅ω ⋅ 1  ⋅i   ζ = 1 
2  2


( )Hc(ω):=20 ⋅log 1   ζ 6 
  10
  1 − ω2 + 2 ⋅ω ⋅ 6  ⋅i   =
10

( )He(ω):=20 ⋅log 1   ζ 1 
  2
  1 − ω2 + 2 ⋅ω ⋅ 1  ⋅i   =
2

( )Hd(ω):=20 ⋅log 1   ζ 1 
  5
  1 − ω2 + 2 ⋅ω ⋅ 1  ⋅i   =
5

(ω) := 20 ⋅log 1   ζ 1 
  10
 
( )Hf  2 ⋅ 1  ⋅i  =
1 − ω2 10
+ ⋅ω

( )Hg(ω):=20 ⋅log 1   ζ 1 
  100
  1 − ω2 + 2 ⋅ω ⋅ 1  ⋅i   =
100

ω := 0.1 , 0.15 .. 100

40 56

20

Ha ( ω )
Hb(ω) 0
Hc ( ω )
Hd(ω) 20
He ( ω )
Hf(ω) 40
Hg ( ω )

60

800.1 1 10
ω

The above plot has been expanded around ω = ωo to expose the three distinct
characteristics that such transfer functions exhibit, depending on the value of ζ, and is shown

on the next page.

10 57

Hb ( ω ) 5
Hc ( ω ) 0
Hd ( ω )
He ( ω )

5

10 1 10
0.1 ω

1 the response does not peak. (Ha and Hb)
We see that for 1 > ζ >

2

The case ζ = 1 is the Butterworth maximally flat response. (Hb)
2

For all ζ such that 1 > ζ > 1 the maximum of the magnitude response occurs at ω = 0.
2

For values of ζ such that 1 1 the magnitude function peaks, but returns to 0
>ζ>
22

dB at a frequency smaller than ωo . (Hc) For values of ζ such that 1 > ζ
2

the response peaks and returns to 0 dB at a frequency larger than ωo. (He, Hf,Hg)

1
For ζ = the response peaks and returns to 0 dB at ω = ω (Hd)

2o

Peaked Responses 58

1
For values of ζ such that ζ < 2 , we define ωo, ωp, ωc , ω2 , Ao, Ap, and A2

Hp (ω ) := 20 ⋅log 1 
 
 
as shown in the plot below.  2 ⋅ 3  ⋅i 
10
( )( ωo = 1 in this example)
1 − ω2 + ⋅ω

Ao is the magnitude of the response at ω = ωo, Ap is the magnitude at ωp ,where the response
peaks , the response crosses through 0 dB at ωc , ω2 is one half ωo , and A2 is the magnitude
of the response at ω2 .

Ap := 4.85 Ao := 4.4 A2 := 1.9 (We'll give formulae for these presently.)

10 1 10

5 ω
ω 2 ω pω o ω c
Hp(ω) 0

Ap
5

Ao
A2

10

15

200.1

59

Derivation of ωp, ωc , ω2 , Ao, Ap, and A2

H(s) = 1 ( H(jω) )2 = 1

 s2 + 2⋅ζ ⋅s + 1   ω 2 2  2⋅ζ ⋅ω  2 
 ωo2 ωo     
 1 − 2  +  ωo  
o   
 ω

Now |H(jω)| reaches a peak at ω = ωp. Therefore:

d  1  = 0
dω    2  
   ω2   2 ⋅ζ ⋅ω  2  
   ωo2  ωo   
 1 − +   

ω = ωp

 2 ⋅1 − ω2  ⋅ −2 ⋅ω  + 2 ⋅ 2⋅ζ ⋅ω  ⋅ 2 ⋅ζ  
     ωo   ωo  
 ω 2  ω 2    The value of ω
 o o  that satisfies
=0 this equation is
= 2
  2   ωp
   1 ω2  2  
    2 ⋅ζ ⋅ω   
  −  +  ωo   ω = ωp
2  
ω o

In order for this function to equal zero the numerator must equal zero (there is no combination
of positive ω, ωo, and ζ that could produce a zero denominator), so:

 − ωp 2  ⋅ −2 ⋅ωp   2 ⋅ζ ⋅ωp  ⋅ 2 ⋅ζ  4 ⋅ω p 4 ⋅ω 3 + 8⋅ζ 2⋅ωp
2 ⋅1 ωo 2   ωo2  + ⋅ ωo   ωo  −+ p
 2    =
  2 4 2
ω o ω o ω o

−1 + ω 2 + 2⋅ζ 2 =0
p

= 2
o
ω

Therefore, the response peaks at a frequency ωp such that : ωp = ωo ⋅ 1 − 2 ⋅ζ 2 60

The magnitude crosses back through 0 dB at ω = ωc, so

20 ⋅log( H(jω) ) = 0, and H( jω) ( ( ) )= 1 so 2
H jω = 1. As a result we have:

1= 1

 2  2 2
c 
 − ω  +  2 ⋅ζ ⋅ωc 
 ω 2   
1 o  ωo 



 ωc2  ω 4 4 ⋅ζ 2 ⋅ω 2
 ωo2  c c
1 −  2 ⋅  + + =1 So
  4 2
ω o ω o

 −2 ⋅ ωc2  + ω 4 + 4⋅ζ 2⋅ωc2 =0 and
 ωo2  c ωo2
 
  ω 4
o

( )ω2
c − 2 + 4⋅ζ 2 =0 Or ωc = ωo⋅ 2 ⋅ 1 − 2 ⋅ζ 2 = 2 ⋅ωp
Since ωp = ωo ⋅ 1 − 2 ⋅ζ 2
ω 2
o

The transfer function has a magnitude Ao at ω = ωo, so

Ao = 20 ⋅log 1

  ω 2  2 ⋅ζ ⋅ω  
  o ⋅ ωo o  
 1 − + i   
2 
 ω o

20 ⋅log 1  61

=  2
 
  ωo2   2 ⋅ζ ⋅ω  2 
 1 ωo2   ωo  
 − +  o  


  ω 2 2  2 ⋅ζ ⋅ω  2 
20 ⋅log(1) − 20 ⋅log ω o    
 1 −  + o
  2 
= o  ωo  



( )= −20 ⋅log 0 + 4 ⋅ζ 2  = −20 ⋅log(2 ⋅ζ ) = Ao

At ω = ωp the magnitude is:

Ap = 20 ⋅log 1

  ω 2  2 ⋅ζ ⋅ωp  
  p ⋅ ωo  
 1 − + i   
2 
 ω o

= 20 ⋅log 1

  −  ω o ⋅ 1 − 2 ⋅ζ 22 +  2 ⋅ζ ⋅ ω o ⋅1 − 2 ⋅ζ 2  
 1   ⋅ ωo   
i   
 ωo2

   22  2  2 ⋅ωo⋅ 1 2  2
 1     
= 20 ⋅log(1) − 20 ⋅log − ω o ⋅ 1−2 ⋅ζ  + ⋅ζ ωo − 2 ⋅ζ
 ωo2
 


   2  62
 
 22   2 ⋅ωo⋅ 1 2  2 
    
= −10 ⋅log  1 − ωo ⋅ 1−2 ⋅ζ +  ⋅ζ ωo − 2 ⋅ζ
  ωo2



 2 ⋅ζ 2 2 + 
( ) ( )=⋅log 1 
−10 − 1 − 4⋅ζ 2⋅ 1 − 2⋅ζ 2

( ) ( )= −10 ⋅log 4 ⋅ζ 4 + 4 ⋅ζ 2 − 8 ⋅ζ 4 = −10 ⋅log4 ⋅ζ 2⋅ 1 − ζ 2 

( )So Ap = −10 ⋅log4 ⋅ζ 2⋅ 1 − ζ 2 

When ω = ωo the magnitude is:

2

A2 = 20 ⋅log 1

   ωo 2  ωo  
   2 ⋅ ⋅ζ ⋅ 
  2 
 2
1 − 2 + i ωo  
o 
 ω

  2  2 2 
  
−20 ⋅log   ω o   2 ⋅ζ ⋅ ωo 
   2
 2  
=  1 −  +  ωo  
2 
 ω o

( )= 9
−10 ⋅log 16 + ζ 2 = −10 ⋅log ζ 2 + 0.5625

1 ωp = ωo⋅ 1 − 2⋅ζ 2 63
2
In summary for ζ < we have: ωc = 2⋅ωp

( ) ( )Ao = −20 ⋅log(2 ⋅ζ ) Ap = −10 ⋅log4 ⋅ζ 2⋅ 1 − ζ 2  A2 = −10 ⋅log ζ 2 + 0.5625

For example when ζ = 0.3 and ωo = 1 (This is the value of ζ used in the illustration of the

definition of ωo, ω p, ωc , A, A, and A .), we obtain ωp = 0.9055, ωc = 1.281, Ao = 4.437 dB,
o p 2

Ap = 4.847 dB, and A2 = 1.854 dB.

Non-peaked Responses

1

For non-peaked responses, i.e. when ζ >

2

ωp, Ap, and ωc are not defined so we plot only Aoand A2. For example if ωo = 1 and ζ = 0.8,
we have:

( ) ( )Hnp(ω):= 20 ⋅log 1 
 
  1 − ω2 + 2 ⋅ω ⋅ 8  ⋅i   A0 := −20 ⋅log(2 ⋅0.8)
10 A2 := −10 ⋅log 0.82 + 0.563

10

5

Hnp(ω) 0

A0 5

A2
10

15

200.1 1 10
ω

64

Extraction of ζ and ωo from polynomial coefficients

Suppose we are given the polynomial s2 + a ⋅s + b and we wish to calculate ζ and ωo.
Since s2 + a ⋅s + b = s2 + 2 ⋅ζ ⋅ωo + ωo2 we have

ωo = b and 2 ⋅ζ ⋅ωo = a = 2 ⋅ζ ⋅ b a

so ζ =

2⋅ b

1

Example #1: suppose we are given the transfer function H(s) =

s2 + 2⋅s + 25

Find the Bode magnitude plot.

a := 2 b := 25 ω0 := b ω0 = 5 a ζ = 0.2
ζ :=

2⋅ b

11 (Notice that the normalization does not affect the
H(s) = ⋅
25  s2  values of ω0 and ζ, so ω0 and ζ can be calculated
 2 ⋅s + 1
+ before the normalization.)
 25 25 

After the normalization we have H(s) = s2 a s2 + 2 ⋅ζ ⋅s + 1 and we get
+ ⋅s + 1 =
bb 2 ω0
ω o

precisely the same formulae for ω0 and ζ as before, so the calculation of ω0 and ζ can be

carried out either before or after the normalization.

ω0 = 2.5 ωp := ω0⋅ 1 − 2⋅ζ 2 ωp = 4.796 ωc := 2⋅ωp ωc = 6.782
2

( )A0 := −20 ⋅log(2 ⋅ζ ) A0 = 7.959 Ap := −10 ⋅log4 ⋅ζ 2⋅ 1 − ζ 2  Ap = 8.136

( )A2 := −10 ⋅log ζ 2 + 0.5625 A2 = 2.2 The resulting plot is shown below.

ω := 0.1 , 0.15 .. 100

 ω0  20 ⋅log 1  65
 10  25
K0 K0 := K0 = −27.959

 ω0  L(ω) if ω ⋅log ω  
    ω0  
2 K0 + A2  := < ω0 , K0 , K0 − 40  

Corr :=  ωp 
 K0 + Ap 
  H(ω) 20 ⋅log 1 1 
  :=  25 ⋅ 
ω0 K0 + A0  1 ω 2  
 ⋅ ω  
 ωc K0    − 2  + j  2 ⋅ζ ⋅ ω0  
  0 
 ω

 10 ⋅ω0 K0 − 40  Note that Ap and A0 are too close to resolve. This is
usually the case for small values of ζ .

10

20

30

L(ω) 40
50
Corr〈1〉

H(ω)

60

70

800.1 1 10 100
ω , Corr〈0〉 , ω

1 ω0 := 4 ω0 = 2 0.36 66
ζ :=
Example #2 H(s) = ζ = 0.09
2⋅ 4
s2 + 0.36 + 4

ω0 1 − 2⋅ζ 2 ωp = 1.984 ωc := 2⋅ωp ωc = 2.805
2 = 1 ωp := ω0⋅

( )A0 := −20 ⋅log(2 ⋅ζ ) A0 = 14.895 Ap := −10 ⋅log4 ⋅ζ 2⋅ 1 − ζ 2  Ap = 14.93

 ω0 K0  ( )A2 := −10 ⋅log ζ 2 + 0.5625 A2 = 2.437
 10 
  K0 := 20 ⋅log 1  K0 = −12.041
 ω0 4

2 K0 + A2  if ω 40 ⋅log ω 
 
Corr :=   L(ω) := < ω0 , K0 , K0 − ω0  
 K0 + Ap   
ωp

 ω0 K0 + A0  H(ω) := 20 ⋅log 1 ⋅ 1 
   4 
  
 ωc K0   1 − ω 2  + j ⋅ 2 ⋅ζ ⋅ ω  
  ω0  
 10 ⋅ω0 K0 − 40   ω 2  
 0

0

L(ω) 20
40
Corr〈1〉

H(ω)

60

800.1 1 10 100
ω , Corr〈0〉 , ω

1

1 ω0 := 49 ω0 = 7 ζ := 8.4 67

Example #3 H(s) = 2⋅ 49 ζ = 0.6

s2 + 8.4⋅s + 49

ω0 ωp := ω0⋅ 1 − 2⋅ζ 2 ωp = 3.704 ωc := 2⋅ωp ωc = 5.238
= 3.5

2

( )A0 := −20 ⋅log(2 ⋅ζ ) A0 = −1.584 Ap := −10 ⋅log4 ⋅ζ 2⋅ 1 − ζ 2  Ap = 0.355

Note that since ωc < ω0 , their order in Corr is reversed.

 ω0 K0  ( )A2 := −10 ⋅log ζ 2 + 0.5625 A2 = 0.35
 10 
 
 ω0  K0 := 20 ⋅log 1  K0 = −33.804
K0 + A2  49
2
 
Corr :=  ωp K0 + Ap  L(ω) := if ω < ω0 , K0 , K0 − 40 ⋅log ω  
  ω0  
   
 ωc K0 
20 ⋅log 1 1 
  H(ω) :=  49 ⋅ 
 ω0 K0 + A0   1 ω 2  
⋅ ω  
 10 ⋅ω0 K0 − 40   − 2  + j  2 ⋅ζ ⋅ ω0  
0  
 ω

35

L(ω) 40
45
Corr〈1〉

H(ω)

50 1 10 100
0.1 ω , Corr〈0〉 , ω

1 12.8 68
Example #4 H(s) = s2 + 12.8 ⋅s + 64 ω0 := 64 ω0 = 8 ζ :=
2⋅ 64 ζ = 0.8

ω0 Since ζ > 1 , ωp , ωc , Ap are not defined
=4 2

2

( )A0 := −20 ⋅log(2 ⋅ζ ) A0 = −4.082 A2 := −10 ⋅log ζ 2 + 0.5625 A2 = −0.801

 ω0 K0  ⋅log 1 
 10  64
  K0 := 20 K0 = −36.124
ω0
Corr :=  2 K0 + A2  L(ω)  ⋅log ω  
    ω0  
 ω0  := if ω < ω0 , K0 , K0 − 40  

 K0 + A0 

 10 ⋅ω0 K0 − 40  H(ω) := 20 ⋅log 1 ⋅ 1 
64 
  ω 2  ⋅ ω  
  1 − + j  2 ⋅ζ ⋅ ω0  
 2   
 ω 0 

35

40

L(ω) 45
50
Corr〈1〉

H(ω)

55

600.1 1 10 100
ω , Corr〈0〉 , ω

1 4 ζ = 1.155 69
ζ :=
Example #5 H(s) =
2⋅ 3
s2 + 4⋅s + 3

Since ζ > 1, H(s) can be factored into two real factors.

s1 := −4 + 16 − 12 s1 = −1
2 2

s2 := −4 − 16 − 12 s2 = −3
2 2

11

So =

s2 + 4⋅s + 3 (s + 1) ⋅(s + 3)

M1 (ω ) := if ω < 1 , 0 , −20 ⋅log ω  
1

M2 (ω ) := if ω < 3, 0, −20 ⋅log ω  
3

MT (ω) := M1 (ω) + M2 (ω)

10 1 10 100
ω
0

10

MT( ω )

20

30

40
0.1

Magnitude Response of Second Order Zeros 70

Transfer functions of the form H(s) = s2 + 2 ⋅ζ ⋅s + 1 (i.e. second order zeros)
ωo
ω 2
o

are handled in a manner similar to the approach we used on second order poles. The

values of ω0 , ωc , and ωp are as before, and

( )Ao := 20 ⋅log(2 ⋅ζ ) Ap := 10 ⋅log4 ⋅ζ 2⋅ 1 − ζ 2  A2 := 10 ⋅log(ζ + 0.5625)

Note the change in sign for all the magnitudes. 2
ζ :=
Example #1 H(s) = s2 + 2 ⋅s + 25 ω0 := 25
2⋅ 25

ω0 = 5 ζ = 0.2

ω0 = 2.5 ωp := ω0⋅ 1 − 2⋅ζ 2 ωp = 4.796 ωc := 2 ⋅ω p ωc = 6.782
2

( )A0 := 20 ⋅log(2 ⋅ζ ) A0 = −7.959 Ap := 10 ⋅log4 ⋅ζ 2⋅ 1 − ζ 2  Ap = −8.136

( )A2 := 10 ⋅log ζ 2 + 0.5625 A2 = −2.2  s 2 2 
⋅ + 1
H(s) = 25 + ⋅s
 25 25 

ω := 0.1 , 0.15 .. 100 K0 := 20 ⋅log(25) K0 = 27.959

 ω0 K0 
 10 
 
 ω0  L(ω) if ω ⋅log ω  
K0 + A2  :=  < ω0 , K0 , K0 + 40  ω0  
2  
 
Corr :=  ωp K0 + Ap  20 ⋅log 25 ⋅  ω2  
   
 ω0 K0 + A0  H(ω) := 1 − 2 + j ⋅2 ⋅ζ ⋅ ω 
  0  ω0 
 ω 
 
 ωc K0 

 10 ⋅ω0 K0 + 40 

71

80

L(ω) 60
40
Corr〈1〉

H(ω)

20 1 10 100
0.1 ω , Corr〈0〉 , ω

Example #2 H(s) = s2 + 12.8 ⋅s + 64 ω0 := 64 ω0 = 8 ζ := 12.8 ζ = 0.8
2⋅ 64

ω0 = 4 1
2 Since ζ > 2 , ωp , ωc , Ap are not defined

( )A0 := 20 ⋅log(2 ⋅ζ ) A0 = 4.082 A2 := 10 ⋅log ζ 2 + 0.5625

 ω0 K0 
 10 
  K0 := 20 ⋅log(64) K0 = 36.124
ω0
Corr :=  2 K0 + A2  L(ω) if ω ⋅log ω  
    ω0  
 ω0  := < ω0 , K0 , K0 + 40  

 K0 + A0

 10 ⋅ω0 K0 + 40  H(ω) := 20 ⋅log 64 ⋅  1 − ω2  + j ⋅2 ⋅ζ ⋅ ω  
   ω0  
 ω 2   
0

60 72

55

L(ω) 50
45
Corr〈1〉 40

H(ω)

35

30 1 10 100
0.1 ω , Corr〈0〉 , ω

Second Order Phase Approximations

The second order phase response is the difference in phase between the input and the output

sinusoids of a system characterized by a transfer function of the form:

H(s) = s2 + 2 ⋅ζ ⋅ωo⋅s + ω 2 The phase function is Φ(ω) = atan 2 ⋅ζ ⋅ωo ⋅ω 
o  2 2 
o
ω − ω

First Straight-Line Approximation (Inadequate)

We begin as with the first order case by finding the slope of the phase function at ω = ω0
and passing a straight line with that slope through the phase funvtion at ω = ω0. The slope

of the phase function is:

 ωo + 4⋅ζ ⋅ωo⋅ ω2 
2⋅ζ ⋅ ωo2 − ω2 
( ) ( )d 2 2 2
 ω o − ω 

Φ(ω) =   ω2 
( )dω  
 1 + 4 ⋅ζ 2 ⋅ω 2 ⋅ 2
o 

ω 2 − ω 2
o


2( )ω + ω 2 73
o
( )2⋅ζ ⋅ωo⋅
which simplifies to 4 2 ⋅ωo2 ⋅ω 2 ω4 2 2 2
o o
ω − + + 4 ⋅ζ ⋅ω ⋅ω

Evaluating this expression at ω = ωo gives the slope of Φ(ω) at ω = ωo as

1 radians

( )ζ ⋅ωo radian

To convert a slope of k radians per radian to an equivalent slope in radians per decade we

( ) ( )proceed as follows: the rise is the limit as ∆ -> 0 of , k⋅ ω + ∆ − ω − ∆  and the run is

 ωo + ∆
log .
the limit as ∆ -> 0 of  ωo − ∆  The slope in radians per decade is then:

( ) ( )lim k⋅ ωo + ∆ω − ωo − ∆ω  k⋅ωo
 ωo + ∆ω  = k⋅ωo⋅(ln(2) + ln(5)) = log(e)
∆ω → 0 log 

 ωo − ∆ω 

1

( )In our case the slope in radians per radian is k = ζ ⋅ωo which gives the slope at ω = ω0 as

1 radians 360 131.92841 degrees
ζ ⋅log(e) ⋅ decade or as 2 ⋅π ⋅ζ ⋅log(e) = ζ ⋅ decade

The slope of the phase function Φ(ω) at ω = ω0 is independent of ωo but depends on ζ .

Phase functions for various values of ζ are shown below along with the approximation
obtained by extending a line of slope 132 / ζ degrees per decade through the point on the

phase function where ω = ω0 . (We have normalized ωo to 1.)

ωo := 1 ζ := 1 74

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

00.1 1 10 100
ω

ζ := 0.8

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

00.1 1 10 100
ω

ζ := 0.6

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

0 1 10 100
0.1 ω

ζ := 0.4 ω

75

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

0 1 10 100
0.1 ω

ζ := 0.2

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

00.1 1 10 100
ω

ζ := 0.1

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

0 1 10 100
0.1 ω

As can be seen from the above plots (and as we shall show by direct calculation), this 76
straight-line approximation suffers from the same shortcomings that made it useless for the
first order case. To demonstrate the large errors at ωlo and ωhi, we begin by calculating ωlo
the value of ω at which the straight line crosses 0o, and ωhi the value of ω at which the

straight line crosses 180o. ωlo is the frequency where

90 + ζ 1 e) ⋅ 360 ⋅log ω  =0 Then
⋅log( 2 ⋅π  ωo 


−90 ⋅2 ⋅π ⋅ζ ⋅log(e)  ω lo   −π   − π ⋅ζ 
360 log  2 log  e 2 
we have = = ⋅ζ ⋅log(e) =
 ωo 

 − π ⋅ζ  ωo⋅4.81− ζ
ωo⋅e 2 
so ω lo = =

 π ζ
 
Similarly for ωhi we have ω hi = ω o ⋅ e 2  = ωo⋅4.81ζ

The error at ωlo is given by substitution of ωlo into the expression for Φ(ω), which gives

  π − ζ    −1  
     2 
atan 2 ⋅ζ ⋅ω o ⋅ω o ⋅ e 2   −atan exp ⋅π ⋅ζ 
   
Errorlo =   2  = 2 ⋅ζ ⋅ (−1 + exp(−π ⋅ζ ))
  
  π − ζ
  2 
ωo2 − ω o ⋅ e 

The errors (in degrees) at ωlo are shown below for various values of ζ .

ζ := 0.1 , 0.2 .. 1.0

 exp  −1 ⋅π ⋅ζ  
 2 
Errorlo(ζ ) := −atan  2 ⋅ζ ⋅ (−π ⋅ζ))  ⋅ 360

(−1 + exp  2⋅π

Errorlo(ζ ) = 77

32.375
32.058
31.536

30.82
29.924
28.868
27.672
26.359
24.955
23.487

Similarly for the error at ωhi we have: Errorhi =

   π ζ    exp  1  
  ⋅    2 
π  π atan 2 ⋅ζ ⋅ω o ⋅ωo e 2   =  ⋅π ⋅ζ 
2  
− 2 +   2   atan 2 ⋅ζ ⋅ (−1 + exp(π ⋅ζ ))
   
  2   π ζ  
 o   2 
ω − ωo⋅e 

 exp  1 ⋅π ⋅ζ  
 2 
Errorhi(ζ ) := atan 2⋅ζ ⋅  ⋅ 360

(−1 + exp(π ⋅ζ ))  2⋅π

Errorhi(ζ ) =

32.375
32.058
31.536

30.82
29.924
28.868
27.672
26.359
24.955
23.487

To extend our results for the first order phase approximation to the second order case, let us 78

consider the second order transfer function with ζ = 1. Proceeding as before we set ωhi to

10 ω and ωlo to 10 o

o , giving a straight line with slope 90 per decade shown below along

ω0

with the 132 degrees per decade line.

ζ

ζ := 1

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

90+90 ⋅ log ( ω )

00.1 1 10 100
ω

This new approximation is a better fit to the phase function Φ(ω) for the case ζ = 1 than the

approximation whose slope is 132 degrees per decade. To extend the result to the case

ζ

where ζ does not equal 1, let us examine the behavior of the phase function as ζ increases.
From the plots of Φ(ω) for various values of ζ , we see that as ζ decreases, the slope of the

phase function evaluated at ω0 increases. To make our new approximation's slope

90

increase as ζ decreases, we take as the slope of our straight-line approximation

ζ

degrees per decade. The results are shown below (along with the 132 degree per decade

ζ

lines) for several values of ζ .

79

ζ := .8

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

90+ 90 ⋅log(ω)

ζ

00.1 1 10 100
ω

ζ := .6

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

90+ 90 ⋅log(ω)

ζ

00.1 1 10 100
ω

ζ := .4 80

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

90+ 90 ⋅log(ω)

ζ

00.1 1 10 100
ω

ζ := .2

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

90+ 90 ⋅log(ω)

ζ

00.1 1 10 100
ω

ζ := .1 81

180

( )arg2⋅ζ  ⋅ 360
⋅ωo⋅ω ⋅ i+ ωo2−ω 2 2⋅π

90+ 132 ⋅log(ω) 90

ζ

90+ 90 ⋅log(ω)

ζ

00.1 1 10 100

ω

In order to calculate the error at ωlo and ωhi, we must produce formulae for ωlo and ωhi in

terms of ζ . Since

90 + 90  ω hi  ω hi ω o ⋅10 ζ
⋅log 
180 = ζ  ωo  we have =

90 + 90  ω lo  we have ωlo = ωo ⋅10− ζ
⋅log 
Similarly, since 0 = ζ  ωo 

Substituting these values into the phase function gives the error at ωlo and ωhi.

Error1lo = atan  2 ⋅ζ ⋅ω 2 ⋅ 10(− ζ ) 
 o 
  2 2 ⋅10(− ) 2  
  o o ζ  
ω − ω

= −atan  2 ⋅ζ ⋅  10(− ζ ) ζ )  
 + 100(− 
−1

ζ := 0.1 , 0.2 .. 1.0

 −atan 10(− ζ )   82
 + 100(−  
Error1lo(ζ ) := 2 ⋅ζ ⋅  −1 ζ )  ⋅ 360
2 ⋅π

Error1lo(ζ ) = .

23.291
22.749

21.88
20.73
19.36
17.834
16.221
14.58
12.967
11.421

π −  π +  2 ⋅ζ ⋅ω 2 ⋅ 10ζ 
( )Error1hi = 2  2 atan o 
  2 2 2 
   ω o − ω o ⋅ 10ζ  


atan 2 ⋅ζ ⋅ 10ζ 
−1 + 100ζ 
( )=

atan 2 ⋅ζ ⋅ 10ζ  360
−1 + 100ζ  2 ⋅π
( )Error1hi(ζ ) := ⋅

Error1hi(ζ ) =

23.291
22.749

21.88
20.73
19.36
17.834
16.221
14.58
12.967
11.421

Summary of Second Order Phase Approximations 83

90
We see that the error for the ζ degree per decade line is considerably smaller at both ωlo

132
and ωhi than the error for the ζ degree per decade line, demonstrating the superiority of

90

the fit to the degree per decade line. We have shown a straight-line approximation to the

ζ

second order phase function that is easy to draw and has easily remembered intercepts. The

line passes through ω0 at 90o and has a slope of 90 degrees per decade. The intercepts
ζ

are: ωhi = ωo ⋅10ζ and ωlo = ωo ⋅10− ζ

Example #1: H(s) = s2 + 2 ⋅s + 25 = 25  s2 + 2 
⋅ ⋅s + 1
 25 25 

The constant term is greater than zero and so contributes nothing to the phase plot.

90o o

ω0 = 5 ζ = 0.2 ωhi= 7.924 ωlo = 3.155 and the slope is ζ = 450 per decade.

ω := 0.01 , 0.02 .. 1000

phase (ω ) := if ω > 3.155 , if ω < 7.924 , 90 + 450 ⋅log ω  , 180 , 0
5

150 0.1 1 10 100 1 .103
ω
phase(ω) 100

50

0
0.01

The situation for poles is analagous as shown below. 84

1 11
Example #2: H(s) = =⋅
( )s2 + 2⋅s + 25 25  2 
 s 2 ⋅s + 1
+
 25 25 

The constant term is greater than zero and so contributes nothing to the phase plot.

90o o

ω0 = 5 ζ = 0.2 ωhi= 7.924 ωlo = 3.155 and the slope is ζ = 450 per decade.

ω := 0.01 , 0.02 .. 1000

phase (ω ) := if ω > 3.155 , if ω < 7.924 , −90 − 450 ⋅log ω  , −180 , 0
5

0

50

phase(ω) 100

150 0.1 1 10 100 1 .103
0.01 ω

Final Example Magnitude Response 85

( )H(s) =s + 1  ⋅ s2 1  s + 1  s2 + 4.8 + 
5 5   ⋅ 9 9 1
+ 4.8 ⋅s + 9 ⋅9 1
s 5 
80
( )(s + 80) ⋅ s2 + 8 ⋅s + 400 = 80 ⋅400 ⋅ +  s2 
⋅ 400 1
 1 + 8 +
400 ⋅s 

ω := 0.01 , 0.02 .. 1000 K0 :=  1  K0 = −84.998
20 ⋅log ⋅9 

5
80 ⋅400

M1 (ω) := if ω < 1 , 0 , 20 ⋅log ω  
 5   
1
5

M2 (ω ) := if ω < 80 , 0 , −20 ⋅log ω  
80

M3 (ω ) := if ω < 3 , 0 , 40 ⋅log ω   ω30 := 3 ζ3 := 4.8 ζ3 = 0.8
3 2⋅ 9

M4 (ω ) := if ω < 20 ,0, −40 ⋅log ω   ω40 := 20 ζ4 := 8 ζ4 = 0.2
20 2⋅ 400

 0.01 K0 
 0.2 K0 
 
3 ⋅log 3  
 K0 + 20 0.2 

points := 

 20 K0 + 20 ⋅log 3  + 60 ⋅log 20  
0.2 3

 80 0
 1000 0 

86

MT (ω) := K0 + M1 (ω) + M2 (ω) + M3 (ω) + M4 (ω)

A30 := 20 ⋅log(2 ⋅ζ3) A32 := −10 ⋅log(0.ζ3)2 + 0.563

 0.3 K0 + 20 ⋅log 0.3  
 0.2 

 1.5 K0 + 20 ⋅log 1.5  + A32 
 0.2 
Corr3 :=  
3 20 ⋅log 3  
 K0 + 0.2 + A30 

 K0 + 20 ⋅log 3  + 60 ⋅log 10   ω40 = 10
 10 0.2 3  2
 

ω4p := ω40⋅ 1 − 2⋅ζ42 ω4p = 19.183 ω4c := 2⋅ω4p ω4c = 27.129

( )A40 := −20 ⋅log(2 ⋅ζ4) A40 = 7.959 A4p := −10 ⋅log4 ⋅ζ42⋅ 1 − ζ42 

( )A4p = 8.136 A42 := −10 ⋅log ζ42 + 0.5625 A42 = 2.2

 ω40 K0 + 20 ⋅log 3  + 60 ⋅log 10  + A42 
2 0.2 3 

Corr4 :=  ω40 K0 + 20 ⋅log 3  + 60 ⋅log 20  + A40 
 0.2 3 
 
 ω4c 
K0 + 20 ⋅log 3  + 60 ⋅log 20  + 20 ⋅log ω4c  
0.2 3  20 


20 87

MT( ω ) 0
20
points〈1〉 40
Corr3〈1〉 60
Corr4〈1〉

80

1000.01 0.1 1 10 100 1 .103

ω , points〈0〉 , Corr3〈0〉 , Corr4〈0〉

Final Example Phase Response

 0.01 0  A :=  0.02 0  ω4lo := ω40 ⋅10− ζ 4
 0.2 0   2 90 
 0.2 90  ω4hi := ω40⋅10ζ 4
ω3lo := ω30 ⋅10− ζ 3
 ω3hi := ω30⋅10ζ 3 C :=  ω4lo 270 
 3 90   ω4hi 90 
 3 270 
points :=  
 20 270 
 20 90 

 80 90  B :=  ω3lo 90  D :=  8 90 
 80 0   ω3hi 270   800 0 

 1000 0  90 = 112.5 90 = 450
ζ3 ζ4

88

300 0.1 1 10 100 1 .103

250
points〈1〉

200
A〈1〉
B〈1〉 150
C〈1〉

100
D〈1〉

50

0
0.01

points〈0〉 , A〈0〉 , B〈0〉 , C〈0〉 , D〈0〉

Auxilliary Drawing

For ω < 0.02 slope = 0, for 0.02 < ω < ω3lo slope = +45, for ω3lo < ω < 2 slope =
+45 +112.5 = 157.5, for 2 < ω < 8 slope = +112.5, for 8 < ω < ω4lo slope =
+112.5 -45 = 67.5, for ω4lo < ω < ω3hi slope = +112.5 -450 - 45 = -382.5, for
ω3hi < ω < ω4hi slope = -450 -45 = -495, for ω4hi < ω < 800 slope = -45

Point #1 = (0.01,0), point #2 = (0.02,0). 0 + 45 ⋅log ω3lo  = 61.924 so point #3,

 0.02 

61.925 + 157.5 ⋅log 2 = 160.191 so point #4.
 
ω3lo 

112.5 ⋅log 8  89
2
160.191 + = 227.923 so point #5,

227.923 +  ω4lo  = 241.284 so pint #6.
67.5 ⋅log
8

241.284 − 382.5  ω3hi  = 173.929 so point #7,
⋅log 
 ω4lo 

173.929 −  ω4hi  = 63.094 so point #8
495 ⋅log 
 ω3hi 

63.094 − 45 ⋅log 800  = 0 so point #9 = (800,0) and point #10 = (1000,0).
 ω4hi 


 0.01 0 
 
 0.02 0 

 ω3lo 61.924 
 
 2 160.191 
 
 8 227.933 
 ω4lo
points := 241.284 

 
 ω3hi 173.929 
 
 ω4hi 63.094 

 800 0 
 
 1000 0 

if ω 0 , if ω 45 ⋅log  90   90

Φ1 (ω ) := < 0.02 , < 2 , 45 + ω ,
0.2

Φ2 (ω ) := if ω < 8, 0, if ω < 800 , −45 − 45 ⋅log ω  , −90
80

Φ3 (ω ) := if  ω <  < ω3hi , 90 + 112.5 ⋅log ω  , 180  
 ω3lo , 0 , if ω  ω30   
   


Φ4 (ω ) := if  ω < ω4lo , 0 , if  ω < ω4hi , −90 − 450 ⋅log ω  , −180 
   ω40   
   

ΦT (ω) := Φ1(ω) + Φ2(ω) + Φ3(ω) + Φ4(ω)

250

200

Φ T(ω) 150

points〈1〉 100

50

0 0.1 1 10 100 1 .103
0.01
ω , points〈0〉 ^^
^^
^ ^ ^ ^^ ^ #9 #10
#1 #2
#3 #4 #5 #6 #7 #8

91

References

[1] Nilsson, James W. Electric Circuits . 4th ed. Reading, Mass.: Addison Wesley
Publishing Company, 1993