Mathcad - MAEBODE5 26 02

Bode Plots 1

(C) Copyright 2002 Richard R. Johnston, Ph.D., P.E.

The Bode magnitude and phase plots are tools for analyzing the frequency response of
linear circuits or systems that can be defined in terms of transfer functions. To construct the
bode plots, the transfer function H(s), must be given in factored form:

H(s) = nz = zero
\/
∏K⋅ (s + ωzi)
i=1 ( ) ( ) ( )K⋅ s + ωz1 ⋅ s + ωz2 ⋅.. ⋅ s + ωznz 
np ( ) ( ) ( )s + ωp1 ⋅ s + ωp2 ⋅.. ⋅ s + ωpnp
∏ (s + ωpi)
i=1 /\
pole

where numerator terms are called zeros, denominator terms are called poles, ωpi and ωzi are
constants, nz is the number of zeros, and np is the number of poles. Bode's insight was to
realize that if we take the log of the magnitude of both sides of the above equation we have:

(( )) (( )) (( ))log( H(s) ) =log 
s + ωz1 ⋅ s + ωz2 ⋅.. ⋅ s + ωznz 
s + ωp1 ⋅ s + ωp2 ⋅.. ⋅ s + ωpnp

( ) ( ) ( )= log s + ωz1 + log s + ωz2 + .. + log s + ωznz

( ) ( ) ( )- log s + ωp1 − log s + ωp1 − .. − log s + ωpnp

This formulation allows us to deal with the right-hand side as a sum of terms, rather than as a
quotient of products of terms. The basic plan is to find the graph of each of the terms on the
right-hand side, and then to "add" the graphs together "by eye" (or by machine with Mathcad).

Decibels

Before we go into the details of this procedure, it is necessary to examine the term on the
left-hand side of the above equation. What Bode understood was that the quantity on the
left-hand side is, in fact, already in standard units. The units to which we refer, were
developed by telephone engineers to conveniently represent ratios of power as an aid to
analyzing the attenuation of telephone signals as they are propagated down the telephone
wire. The Bel (after Alexander Graham Bell) is defined as

2

Loss in Bels = log Pin 
 Pout 


Unfortunately the Bel (like the Farad) is too large a unit to be useful, so the deciBel, which is
one tenth of a Bel, is the usual unit.

Loss in DeciBels = 10 ⋅log Pin 
 Pout 


Now in most electronic and control work the transfer function represents not a ratio of powers,
but a ratio of voltages or currents. However, since we may write:

V2 = I2⋅R

P=
R

 Vin2 
 
10 ⋅log R   Vin 
we have: Loss in dB =   Vout2   = 20 ⋅log Vout  or
    


 R 

 Vout2 
 
Gain in dB = 10 ⋅log R  ⋅log Vout 
  Vin2   = 20  Vin  and
    

 R 

Loss in dB = 10 ⋅log Iin2 ⋅R  = 20  Iin  or
Iout2 ⋅R  ⋅log Iout 



⋅log Iout2 ⋅R   Iout 
Gain in dB = 10 Iin2 ⋅R  20 ⋅log 
=  Iin 

Point-by-Point Addition of Functions 3

To gain some insight into "adding graphs by eye" the reader is encouraged to examine the
examples given below. ( u(t) is the Heaviside unit step function.)
Recall that given a function f(t), then the graph of f(t-k) is simply the graph of f(t) shifted to
the right by k units.

t := −1 , −0.999 .. 5 u(t) := if (t > 0 , 1 , 0)

f1(t) := t⋅u(t) f2(t) := −1 ⋅(f1(t)) g1(t) := t g2(t) := −t

f3(t) := u(t − 2) − u(t − 3) f4(t) := f1(t − 1) − 2 ⋅f1(t − 2)

2 2

1 u(t−2) 1
u(t) 0

0 1
024
10 2 4 t
t

5 4
4
3 3
f1(t) 2
1 f1(t−1) 2
0 1
10 2 4
0
t
10 2 4
1
0 t
1
f2(t) 2 5
3 4
4 3
50 2 4 g1(t) 2
1
t 0
10 2 4

t

1 1 4
0
1 g1(t)+g2(t) 0
g2(t) 2
3 1
4 024
5 t

024

t

2 1

f1(t)+g2(t) f2(t)+g1(t) 0
0
1
024 024
t t

4 4

f1(t−1) 2 f1(t−2) 2
0 0
024 024
t t

2

f1(t−1) ... 4 f1(t−1) ... 1
+ f1(t−2) + − f1(t−2) 0
2
4 1 5
0 0
02
t t

2 1 5

1 0
f3(t) f4(t)

0 1

10 2 4 6 20 2 4
t t

Normalization

Before we proceed with the derivation of the magnitude plots of the individual terms of a

transfer function, we introduce an algebraic normalization of the transfer function that makes

the individual plots simpler to manipulate. The transfer function is usually given in factored

form as:

nz

∏ ( )K⋅ s + ωz ( ) ( ) ( )K⋅ s + ωz1 ⋅ s + ωz2 ⋅.. ⋅ s + ωznz 
i ( ) ( ) ( )s + ωp1 ⋅ s + ωp2 ⋅.. ⋅ s + ωpnp
i=1
H(s) = =
np
∏ (s + ωpi)
i=1

where the coefficients of s have been set to one. A more convenient normalization, for our
purposes, is to set the constant coefficients to one as shown below:

nz ∏ ∏nz nz s + 1 
 
∏ ( )K⋅ s + ωz K⋅ ωzi  ωzi 
i i=1 ⋅i = 1
H(s) = i=1 =
np ∏np
np s + 1
∏ ωpi i=1  
∏ (s + ωpi)  ωpi
i=1
i=1

( ) ( ) ( )= K⋅ s + ωz1 ⋅ s + ωz2 ⋅.. ⋅ s + ωznz  so
( ) ( ) ( )s + ωp1 ⋅ s + ωp2 ⋅.. ⋅ s + ωpnp

 s 1 ⋅ s 1 ⋅.. ⋅ s 1  6
  ωz2   ωznz  
H(s) =  K ⋅ωz1 ⋅ωz2 ⋅ .. ⋅ωznz    ωz1 + + 1 ⋅.. ⋅ + 1 
 ωp1 ⋅ωp2 ⋅.. wpnp  ⋅  + +   s +  
⋅  ωpnz 
s 1 ⋅ s 
   ωp2
  ωp1

For example: 10 ⋅2 ⋅3  (s + 1) ⋅ s + 1 ⋅ s + 1 
5 ⋅6 ⋅7  + 2 3 + 1 
H(s) = 10⋅(s + 1) ⋅(s + 2) ⋅(s + 3) = ⋅ 
(s + 5) ⋅(s + 6) ⋅(s + 7)   s 1 ⋅ s 1 ⋅ s 
 5 6 + 7 

2  (s + 1) ⋅ s + 1 ⋅ s + 1 
7  + 2 3 + 1 
= ⋅ 
  s 1 ⋅ s 1 ⋅ s 
 5 6 + 7 

We now wish to find the magnitude plot for each of the terms of H(s), which we shall add
together to form the magnitude plot for the transfer function as a whole. The magnitude plot is
a graph of the magnitude of the transfer function (in dB) vs. log(ω). The transfer functions

2

shown above contain three distinct kinds of terms: constant terms ( ), zeros (s + 1), and poles

7

s

( + 1).

5

Magnitude Plots of Constant Terms

Since the magnitude of a transfer function H(s) = K is not a function of ω, such transfer
functions have magnitude plots as shown below:

2 ω := 0.1 , 0.2 .. 10 K2 := 15
K1 := 7

20 ⋅log(K1) = −10.881 20 ⋅log(K2) = 23.522

20⋅log(K1) 0 1 10 20 1 7
ω ω
200.1 20 ⋅ log ( K2) 10

0

0.1

Magnitude Plots of First order Poles

To begin our study of individual terms of the general transfer function H(s), let us use as an
example a transfer function that has only one term. Consider the frequency response of the
circuit shown below. (This experiment is performed in virtually every undergraduate circuits
laboratory. ) The transfer function of this circuit is:

1

H(s) = Vout (s) = s 1

Vin ( s)  1 + 1 =
s
s+1

Since we have agreed to deal with magnitudes in dB (in order to decouple the terms of the 8
transfer function from one another), we have plotted the magnitude of the response in dB
versus frequency by point-by-point calculation below.

ω := 0.01 , 0.02 .. 100 H(ω) := 20 ⋅log 1 
1 + ω ⋅i

10

0

10

H(ω)

20

30

40
20 40 60 80 100

ω

We see that for large frequencies (ω > 50) the plot is nearly linear, but for small frequencies
the plot is strongly nonlinear. In fact the plot appears nearly exponential. Since exponential
functions have linear graphs when plotted against a logarithmic frequency axis, we plot |H(s)|
versus log(ω) below. (We obtain the semi-log plot by using Format X Y Axes, and changing the
x axis to a log scale.)

10

0

10

H(ω)

20

30

400.01 0.1 1 10 100

ω

We see that this plot is nearly linear over most of the frequency range. Our basic strategy is to

We see that this plot is nearly linear over most of the frequency range. Our basic strategy is to
find equations for the two different straight lines that describe the magnitude response and to 9
extend the straight lines until they intersect as shown below.

( ) ( ( ) )ω0 := 1 (We will derive this equation
HSL ω := if ω > ω0 , −20 ⋅log ω , 0 for HSL presently.)

10

0

H(ω) 10
HSL(ω) 20

30

400.01 0.1 1 10 100

ω
Consider a more general transfer function of the form:

H(s) = 1 where ωo is a constant which we shall call the
break frequency for reasons that we shall
s + 1
  make clear presently. (ω without subscript is
 ω o the frequency variable over which we desire

the frequency response.)

The magnitude of this transfer function is: 20 ⋅log 1 
 j⋅ ω + 1 
H(jω) =  ωo 

dB

We seek a straight-line approximation to this magnitude for our magnitude plot. We proceed by

finding the limit of the magnitude function as ω approaches zero, and using that limit as an
approximation of the magnitude function for all "low" frequencies. We then find the limit of the

magnitude function as ω approaches positive infinity and use that limit as an approximation of
the magnitude function for all "high" frequencies. We shall define what we mean by "low"
frequencies and "high" frequencies presently.

10

lim 20 ⋅log j⋅ω1  = 20 ⋅log(1) − 20 ⋅log(1) = 0 dB
ωo
ω→0  +1 
 

Since the limit of the magnitude function as ω approaches zero is 0 dB, we approximate the
magnitude function as 0 dB for all "low" frequencies. That is we approximate the magnitude

function as a horizontal straight line at 0 dB for all "low" frequencies.

For "high" frequencies we take the limit of H(jω) as ω approaches positive infinity.

lim 20 ⋅log 1  = lim 20 ⋅log 1 
 j⋅ ω + 1  j⋅ ω 
ω→∞   ω→∞  
ωo ωo


 ωo 
20 ⋅log 
( ( ) )= = −20 ⋅log(ω)
lim = lim 20 ⋅log ωo − 20 ⋅log(ω)

ω→∞  j⋅ω  ω → ∞

This is the magnitude approximation at high frequencies.

Decades

To understand the significance of this last function, consider the function f(t) = 5t. This is the
equation of a straight line in t, with a slope of 5, passing through the point t = 0, f(t) = 0.

Similarly the function f(log(ω)) = -20 log(ω) is a straight line in log(ω) with a slope of -20 dB
passing through 0 dB at ω = 1.

(-20 log(1) = 0) The question is a slope of -20 dB per what? Not -20 dB per radian since we

are plotting against log(ω) not against ω. To answer our question of the units for the slope of

our straight line, we must calculate two frequencies ω1 and ω2 , such that log(ω2) - log(ω1) = 1.

Since the slope is the "rise" over the "run", and we desire a slope of −20 we need a "rise" of

1

-20 and a "run" of 1. So we have:

( ) ( )log ω2 − log ω1  ω2  11
log 
=  ω1  =1

If we exponentiate (base 10) both sides of the right hand equation, we obtain:

 ω2   ω 2  = 10
log   
10  ω1  = 101 or  ω1 

Two frequencies related in this way are said to be one decade apart. The slope of our "high"
frequency approximation is then -20 dB per decade. Of course simply specifying the slope
does not uniquely determine a straight line; there are infinitely many straight lines with slope
-20 dB per decade. Recalling our earlier analogy with straight lines f(t), we recognize that
f(t)=a t is not the most general equation of a straight line in t: it is the equation of the specific
straight line in t passing through (0,0). The most general equation of a straight line in t is
f(t) = a t + b. To uniquely specify our straight line we must specify one point on the line. The
question of which point to specify, and the question of what constitutes a "low" or "high"
frequency are considered together. Since the low frequency approximation holds when ω
approaches zero, i.e. when

ω

<< 1

ωo

We approximate the magnitude function by its low frequency approximation
for all ω such that

ω
ωo < 1(i.e. for all ω such that ω < ωo.)

That is, we use the low frequency approximation (asymptote) for all ω such that ω < ωo.
Similarly we use the high frequency approximation (asymptote) for all ω such that

ω
ωo > 1 (i.e. for all ω such that ω > ωo.)

The question of what point on the -20 dB per decade line to specify is settled by the above

approximation strategy: we select the point ω = ωo. The magnitude of a transfer function of
the form:

H(s) = 1

s + 1 
 
 ω o 

is approximated by a straight line of 0 dB for all ω < ω and a straight line of slope -20 dB per

is approximated by a straight line of 0 dB for all ω < ωo and a straight line of slope -20 dB per
decade passing through the point having magnitude 0 dB at ω = ωo, for all ω > ωo. That is, 12
|H(jω)| is approximated as 0 dB for

ω < ωo, and |H(jω)| is approximated by 20 log(ω) -20 log(ωo) for ω > ωo.

e.g. 1 1 1
H1(s) := s H2(s) := s H3(s) := s
+1
0.5 +1 +1
1 3

H4(s) := 1 ω := 0.1 , 0.2 .. 100
s
12 + 1

H1 (ω ) := if ω > 0.5 , −20 ⋅log ω  , 0  H2 (ω ) := if ω > 1 , −20 ⋅log ω  , 0
0.5 1

H3 (ω ) := if ω > 3 , −20 ⋅log ω  , 0 H4 (ω ) := if ω > 12 , −20 ⋅log ω  , 0
3 12

(We shall discuss these Mathcad functions for plotting straight-line magnitude
approximations later when we do the first "First Order Magnitude Example.")

0 0

H1(ω) 20 H2(ω) 20

400.1 1 10 100 400.1 1 10 100
ω ω

0 13

0 H4(ω) 20

H3(ω) 20

400.1 1 10 100 40 1 10 100
ω 0.1 ω

These plots are constructed on semi-log paper by hand as follows. First locate the break

frequency ωo, then draw a straight line of slope -20 dB per decade (downward and to the right)
starting at ωo. In some cases, for instance when ωo= 12 as above, the frequency that is one
decade above ωo is not on the graph. In these cases it is helpful if one knows the equivalent of
-20 dB per decade in dB per octave.

Octaves

Frequencies ω1 and ω2 are said to be one octave apart if ω2 = 2 ω1. (This terminology results
from musical theory where notes that are separated by eight diatonic pitches are said to be one
octave apart. For example middle C and the C one reaches by playing
C D E F G A B C, are an octave apart. Two notes that are an octave apart have a
frequency ratio of 2:1.) To find the slope in dB per octave that is equivalent to -20 dB per
decade we need to know how many decades there are in an octave. For ω1 and ω2 to be x
decades apart we require:

 ω2 
log  = x

 ω1 

We can check the above result for pairs of frequencies whose separation in decades is known,
for example:

ω1 ω2 Separation
in decades

10 100 1
10 1000 2
10 10000 3

Now if ω2 = 2 ω1, then

 ω2   2 ⋅ω 1 
log  log 
 ω1  =  ω1  = log(2) = 0.30103

So frequencies ω1 and ω2 that are one octave apart are 0.30103 decades apart, and 14

1 octave = 0.30103 decades, and 1 decade = 1 octaves.

0.30103

Then

dB dB decade dB
−20 ⋅ = −20 ⋅0.30103 ⋅ = −6.0206⋅
decade decade octave octave

or, -20 dB per decade is approximately -6 dB per octave. Referring again to the example

1

H4(s) = s
+1
12

while the decade related frequency ω = 120 is not on the graph, the octave related frequency
ω = 24 is on the graph. So in this case one would draw a straight line of slope -6 dB per
octave starting at ω = 12.

Magnitude Plots of First order Zeros

The situation for zeros is very similar. Consider a transfer function H(s) such that

H(s) = s + 1
 
 ω o

The low frequency approximation (ω < ω0) is:

H( jω) = lim 20 ⋅log j⋅ω +1  = lim 20 ⋅log(1) = 0 dB
ω0 
dB ω→0   ω→0

The high frequency approximation is:

H( jω) = lim 20 ⋅log j⋅ω + 1  = lim 20 ⋅log  j⋅ω  
ω→∞  ω0    ω0  
dB  ω→∞   

( ( ) )=
lim 20 ⋅log ω  = lim 20 ⋅log ω0 − 20 ⋅log(ω)
 ω0 
ω→∞  ω→∞

H(jω) = 20 ⋅log(ω) ⋅ dB 15

dB decade

First order zeros with break frequency "ω0", are handled in a manner similar to the manner in
which we handled first order poles with break frequency "ω0". The magnitude plot is a
horizontal straight line through 0 dB for ω < "ω0", and a straight line of slope + 20 dB per
decade (6 dB per octave) through "ω0", for all ω > "ω0".

e.g. H5(s) = s H5 (ω ) := if ω > 0.5 , 20 ⋅log ω  , 0
+1 0.5

0.5

H6(s) = s + 1 H6 (ω ) := if ω > 1 , 20 ⋅log ω  , 0
1

s H7 (ω ) := if ω > 3 , 20 ⋅log ω  , 0
H7(s) = 3 + 1 3

H8(s) = s H8 (ω ) := if ω > 12 , 20 ⋅log ω  , 0 
+1 12

12

40 40

H5(ω) 20 1 10 100 H6(ω) 20 1 10 100
ω ω
0 0
0.1 0.1

40 40

H7(ω) 20 1 10 100 H8(ω) 20 1 10 100
ω ω
0 0
0.1 0.1

Magnitude Plots of First Order Zeros and Poles at s = 0 16

First order magnitude plots for transfer functions that include zeros or poles at s = 0, that is for
terms where ω0 = 0, require special attention. Consider a transfer function having a zero at 0,
H(s) = s. The magnitude function H(ω) is H(ω) = |20 log(ω)| for all values of ω. The
magnitude plot is a straight line of slope 20 dB per decade (6 dB per octave), passing through
0 dB at the value of ω for which 20 log(ω) = 0. Since log(1) = 0, the magnitude function is a
straight line of slope 20 dB per decade (6 dB per octave) passing through ω = 1. While this
function is relatively easy to draw, it does complicate the process of adding the individual
graphs together since there is no frequency range where the slope of the function is zero: the
function contributes a slope of 20 dB per decade at ALL frequencies. Similarly transfer
functions containing poles at zero, i.e.,

1
H(s) =

s

have magnitude plots that are straight lines of slope -20 dB per decade (6 dB per octave)
passing through ω = 1. e.g.

H9(s) = s H9(ω) := 20 ⋅log(ω)

1 H10(ω) := −20 ⋅log(ω)
H10 (s) = s

40 40

H9(ω) 0 H10(ω) 0

40 1 10 100 40 1 10 100
0.1 ω 0.1 ω

Multiple poles and zeros

Multiple poles and zeros 17

s  n 1

Transfer functions having terms of the form  ωo + 1 or 1  n are handled exactly
 
 s +

 ω o

as above except that the slope of the magnitude plot of the term having multiplicity n is not 20

dB per decade (6 dB per octave), but 20 n dB per decade (6 n dB per octave). This is true for

both zeros and poles except of course that poles have negative slopes. The slope

multiplication is also correct when ω0 = 0, i.e. for poles and zeros at the origin (s = 0).

First Order Magnitude Examples

#1. As an example of the technique, consider the transfer function:

H(s) = 5 ⋅ s⋅(s + 3) 
 + 2) ⋅(s + 4)2 
(s 

First we normalize the transfer function as follows:

5 ⋅3  s ⋅ s + 1   s ⋅ s + 1 
⋅ 3  ⋅ 3 
H(s) = 15
=
2 ⋅42   s  ⋅ s 1 2  32  2 
 2 4   
+ 1 +  s + 1 ⋅ s + 1 
2 4

This example has several features that make it useful as an easy first example.

1. There are no break-points ω0 with magnitudes less than 1. (Control System Transfer
Functions commonly contain break-points smaller than ω0 = 1.)

2. None of the break points is more than 1 or 2 decades away from ω0 =1. (Electronic
Filter Transfer Functions commonly contain break-points 4, 5,6 or even 7 decades above
ω0 =1.)

15

The transfer function has 5 terms: a constant term of , a zero at zero, a zero at 3, a

32

pole at 2 and a pole of multiplicity 2 at 4. The individual term magnitude plots along with the
composite plot formed by the addition of the individual terms are given below.

Mathcad Functions for Magnitude Approximations 18

Consider the task of writing the equation of the linear function graphed below.

x := −10 , −9 × .9 .. 10 f (x) := 2 ⋅x

20 This is very easy, since the y intercept is
10
f (x) 0 zero we may write f(x) = m⋅x where m is
10
20 the easily calculated slope.

10 5 0 5 10 Contrast the previous equation with the
x equation of the linear function graphed
below

g(x) := 2 ⋅x − 5

. g(x) 20 This is only slightly more difficult since
10 we must find the y intercept b as well as
the slope m, and the function is
0
10 g(x) = m⋅x + b. However, notice that
20 10 5 0 5 10
when we plot | H(jω) | vs log(ω), we are
x frustrated since there is no place on the
plot where log(ω) = 0. (However many
decades smaller than log(ω) = 1.0 we
go.)

The key to solving our problem is to note that the second fuction g(x) is merely the first
function f(x) shifted to the right by 2.5 units. By substitution of x -> (x - 2.5) in f(x) we have

g(x) = 2 ⋅(x − 2.5) = 2 ⋅x − 5. We are then able to write the equation for g(x) without

knowing the y intercept.

1 50 19
First consider | H(jω) | vs log(ω), where H(s) = s
H10(ω) 0
(Note that in Mathcad we Format X Y Axes to set
the x axis to a log axis)

50 1 10 100
0.1 ω

Now consider what happens when we replace 40
[log(ω)] in H10(ω) by [log(ω) -log(3)].
H10 ω  20
( ( ))Since H10(ω) = −20 ⋅ log ω , when we replace 3 0

[log(ω)] by [log(ω) -log(3)] , we get 20

−20 ⋅(log(ω ) − log(3))= −20 ⋅log ω 
3
400.1
1 1 10 100
which is | H(jω) | vs log(ω), where H(s) = s ω

shifted to the right by 2. (3 - 1 = 2)

Finally consider | H(jω) | vs log(ω), where 0

1 H3(ω) 20

H(s) = 400.1

s
3 +1

Note that H (ω) is 0 for ω < 3 and H10  ω for ω >3. 1 10 100
3 ω
3

The if statement in Mathcad is perfectly suited to generating piece-wise linear functions. 20

The syntax is if(condition,true,false), where condition is a logical (boolean) expression, true

is the expression to be returned if the condition is true, and false is the expression to be

returned if the condition is false. For example, the magnitude approximation of

H(s) = s 1 is H(ω) = if ω > 3 , −20 ⋅log ω  , 0. In general the magnitude
+ 3
1
3

approximation in Mathcad for a term of the form H(s) = 1 is

s +1

ω0

H(ω) = if  ω > ω 0 , −20 ⋅log ω  , 0.
  ω0  
 

s
The magnitude approximation for a term of the form H(s) = + 1 is

ω0

if ω > ω 0 , 20 ⋅log  ω  , 0 
  ω0  .
  

For our example then, we have:

M1 (ω ) := 20 ⋅log 15  M2 (ω ) := if ω > 3 , 20 ⋅log ω  , 0
32 3

6.57 20

M1 (ω ) 6.58 M2 ( ω )

6.590.1 1 10 100 0
ω
0.1 1 10 100
ω

M3(ω) := 20 ⋅log(ω) (See plot of H9 above.)

M4 (ω ) := if ω > 2 , −20 ⋅log ω  , 0 M5 (ω ) := if ω > 4 , −40 ⋅log ω  , 0
2 4

0 21

0 M5(ω) 20

M4 ( ω )

20

0.1 1 10 100 40
0.1 1 10 100

ωω
In Mathcad all we have to do to add these functions point-by-point is to

define MT (ω) as

MT (ω) := M1(ω) + M2(ω) + M3(ω) + M4(ω) + M5(ω) and plot MT (ω)

5

0

5

MT(ω) 10

15

20

25

30 1 10 100
0.1
ω
This is our composite Bode magnitude plot!

To proceed by hand with semi-log paper, we must first discuss some peculiarities of semi-log
paper.

Semi-log paper 22

Several remarks are in order concerning the use of semi-log paper. First, since the vertical

scale is linear, all magnitudes must be converted into dB. For example, the magnitude of the

constant term 15 must be converted to dB as 20 ⋅log 15  = -6.58 dB. Second, since the
32 32

frequency scale is a log scale it is never necessary to take the log of any frequency: by

locating frequencies according to their values as described by the log scale, the curve is

automatically scaled horizontally according to log(ω). Finally the frequency scale must be

oriented so that the distance between successive numbers decreases as one proceeds

from left to right; this usually means placing the log scale markings on top. The log scale

runs from 1 to 1 three times on 3 cycle semi-log paper. The scale markings must be

identified as to their magnitudes as shown above in our composite Bode magnitude plot .

Using the scale in our example #1 abovee.g., the first cycle is 0.1, 0.2, 0.3, ... 0.9, 1.0, the

second cycle is 1.0, 2.0, 3.0, ... 9.0, 10.0, the third cycle is 10.0, 20.0, 30.0, 90.0, 100.0.

The frequency scale can be set to span any three decade frequency range that is desired.

We could have for example, 0.01 to 10.0, or 1 to 1000.0, or 1000 to 1000000, or any range

that is necessary to represent the magnitude plot of the transfer function at hand. To follow

the steps we now give, refer to the figure that follows the steps.

The composite drawing is produced on semi-log paper as follows:

1) Add all the contributions to the magnitude at ω = 1. In the case of our present example
this would be:
-6.58 + 0 + 0 + 0 + 0 = -6.58 dB. Enter point #1.

2) Determine the behavior of the composite function for frequencies smaller that ω = 1.
In our example the only contribution to the magnitude function for ω < 1 is the
contribution from the zero at zero ( the s term in the numerator). Therefore at ω = 0.1
the magnitude is 20 dB smaller than it is at ω = 1 since the slope of the magnitude
function is +20 dB per decade. So the Magnitude at ω = 0.1 = -6.58 - 20 = -26.58.
Enter point #2

3) Determine the behavior of the composite function for ω > 1. Again in our example, for

1 < ω < 2, the only contribution to the magnitude function is from the zero at zero, so

extend the +20 dB per decade line to ω = 2 and enter point #3.

(-6.58 dB +  log  2  ⋅decades⋅ 20 ⋅dB  = -0.56 dB)
1 decade

4) For 2 < ω < 3, we have two contributions to the composite magnitude function: the 23
contribution of the zero at zero, and the contribution of the pole at 2. Since the zero at
zero contributes a slope of +20 dB per decade and the pole at 2 contributes a slope of
-20 dB per decade, the contributions cancel each other out and the composite

magnitude function does not change between ω = 1 and ω = 2, so draw a horizontal
line from point #3 to ω = 3. Enter point #4.

5) For ω > 3 we must add the contribution of the zero at 3. This adds another slope of

+20 dB per decade so we would like to draw a line with slope +20 dB per decade

between ω = 3 and ω = 30. Unfortunately the magnitude 19.5 dB is beyond the range

of our graph, so we draw a line with slope +6 dB per octave between ω = 3 and ω = 6,

giving us point #5. ( - 0.56 + log 6  ⋅20 = 5.46dB)
3

6) At ω = 4, the composite function changes character again as we must add the

contribution of the double pole (pole of multiplicity 2) at 4. Enter point #6

(Point #6 is where the line we drew in step 5 intersects ω = 4)

(-0.56 + log 4  ⋅20 = 1.94dB)
3

7) For ω > 4 the double pole at ω = 4 adds a slope of -40 dB per decade so the total
slope for ω > 4 is +20 - 20 + 20 - 40 = -20 dB per decade, so enter point #7.
(1.94 - 20 = -18.06)

This completes the straight-line approximation to the magnitude of the transfer function

H(s) = 5 ⋅ s⋅(s + 3)   0.1 −26.58 
 + 2) ⋅(s + 4)   1.0 −6.58 
(s 2   
 
2.0 −0.56

points :=  3.0 −0.56 
 
 4.0 1.94 

L(ω) is the construction line in step 5.  6.0 5.46 

L(ω) := −0.56 + 20 ⋅log ω   40 −18.06 
3

10 24

MT( ω ) 0
10
(points)〈1〉 20

L(ω)

30

400.1 1 10 100
ω , points〈0〉 , ω
Points ^ ^
^ ^ ^^ ^
#2 #7
# 1 #3 #4 #6 #5

Example #2

 s + 1 ⋅ s + 1
  5
 s + 1  ⋅(s + 5) 1 ⋅5   1 
4 4 4 
H(s) = = ⋅
s ⋅s 1  1 ⋅ s 1
+ 2 2 s   1  + 
 2 

  s  + 1 ⋅ s + 1
 1  5
 4
= 5 ⋅ 

2 s ⋅ s + 1

1  
2

ω := 0.1 , 0.15 .. 10

5 5

M1 (ω ) := 20 ⋅log 5  20 ⋅log 5  = 7.959 25
2 2

M2(ω) := if ω > 1 , 20 ⋅log   ω  , 0 M3 (ω ) := if ω > 5 , 20 ⋅log ω  , 0
 4 1    5
 4 

M4(ω) := −20 ⋅log(ω) M5 (ω ) := if ω > 1 , −20 ⋅log  ω   , 0
 2  1  
  2  


7.97 1 10 20 1 10
ω ω
7.965 M2 ( ω )

M1(ω) 7.96 0

7.955 0.1

7.95
0.1

10 1 10 20 1 10
ω ω
M3 ( ω ) M4(ω) 0

0 20
0.1
0.1

0 1 10
ω
M5 ( ω )

20

0.1

MT (ω) := M1(ω) + M2(ω) + M3(ω) + M4(ω) + M5(ω) 26

 0.1 27.96 
 0.25 20 
 
points :=  0.5 20 

 1.0 13.98 
 5.0 0 

At log(ω) = 1 the contribution from the pole at ω = 0 is 20 ⋅log   1   = 12.0412 dB and
4  1 
 4 

the contribution from the constant is 20 ⋅log 5  = 7.959 dB. So the total contribution at
2

1
log(ω) = 4 is 20 dB => point #1. Since the magnitude response is -20 dB per decade for

ω< 1 = 0.1 and magnitude 20 + 20   1   = 27.96. For
4 , we may plot point #2 as ω ⋅log 4 

 0.1 

11 1
4 <ω < the zero at cancels the pole at zero, and we draw a horizontal line to point #3
4
2

11

at ω = . For < ω < 5, we have slope -20 dB per decade so we plot point #4 at ω = 1.0

22

and magnitude 20 - 20 ⋅log  1   = 13.98 dB. Finally for 5 < ω , we have
 1 
 2 

-20 + 20 - 20 + 20 = 0 dB, hence point #5 at ω = 5 and magnitude

13.98 - 20 ⋅log 5  = 0dB.
1

30 27
25
20 1 10
ω , points〈0〉
MT(ω) 15 ^
^ ^^
points〈1〉 10 #5
5 #1 #3 #4
0
50.1

Points ^

#2

Example #3

H(s) = 100000 ⋅s = 100000 ⋅ s
(s + 10000) ⋅(s + 50000)
10000 ⋅50000  s + 1 ⋅ s + 1
10000 50000

ω := 100 , 200 .. 1000000

M1 (ω ) := 20 ⋅log 100000  M2(ω) := 20 ⋅log(ω)
 5 ⋅108 


M3 (ω ) := if ω > 10000 , −20 ⋅log ω  , 0 
10000

M4 (ω ) := if ω > 50000 , −20 ⋅log ω  , 0 
50000

28

M1(ω) 74

74.1 1 .103 1 .104 1 .105 1 .106
100 ω

200

M2(ω) 100

0 1 .103 1 .104 1 .105 1 .106
100 ω

0 1 .103 1 .104 1 .105 1 .106
ω
M3 ( ω )

20

100

0

M4 ( ω )

20

100 1 .103 1 .104 1 .105 1 .106
ω

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

 100 −33.979  29

 104 
 6.021 
points :=  
 5 ⋅104 6.021 

 5 ⋅105 
 −13.979 

At log(ω) = 100 we have -73.979 from the constant term and 0dB +

(2 ⋅decades) ⋅ 20 ⋅ dB  = 40 dB so the total magnitude at log(ω) = 100 is
decade

40 - 73.979 = -33.979 => pont #1. For 100 < ω < 104 we have

+20 dB per dacade so point #2 at ω = 104 and magnitude -33.979 +  104 
20 ⋅log 100 


=6.021. For ω > 104 , the pole at 104 cancels the zero at 0 and we have a horizontal

line to point #3. For 5 ⋅104 < ω we have slope - 20 dB per decade, so point #4 at

ω = 5 ⋅105 and magnitude 6.021 - 20 dB = -13.979

10

0

MT(ω) 10

points〈1〉
20

30

40 1 .103 1 .104 1 .105 1 .106
100 ω , points〈0〉
Points ^ ^
^ ^
#3 #4
#1 #2

First Order Magnitude Errors 30

It should be noted that there is an error associated with the straight-line approximation to the

first order magnitude function, and that the error is greatest at the break frequency "ω0" where
the assumptions about "high" and "low" frequencies are the weakest. The highest "low"

frequency is ω = ω0, and the lowest "high" frequency is also ω = "ω0". The magnitude of the
error can be found by evaluating H(jω) at ω = ω0. For pole terms, the error is given by the
difference between 0 dB and

20 ⋅log 1  1
 j⋅ ω + 1  ( )evaluated at ω = ω0, which is 20 ⋅log 
 ω0   
12 + 12

= −20 ⋅log( 2) = -3.103 dB

So for a pole the magnitude is actually at -3 dB rather than at 0 dB as predicted by the
straight-line approximation.

For zero terms the error is given by the difference between 0 dB and

20 ⋅log j⋅ ω + 1  evaluated at ω = ω0, which is ( )20 ⋅log 12 + 12 
 ω0 
 = 20 ⋅log( 2) = 3.103 dB

So for a zero the magnitude is actually at +3 dB rather that at 0 dB as predicted by the
straight-line approximation.

First Order Phase Approximations

The first order phase response is the difference in phase angle between the input sinusoid
and the output sinusoid as a function of frequency for a system characterized by a transfer
function of the form:

H(s) = s + ωo (First order zero at ωo)

where ωo is referred to as the break frequency for reasons that are apparent from our 31
investigation of the magnitude response, and which plays an important role in the phase
response as well.

The phase response as a function of frequency is:

Φ (ω) ω
= atan 
 ω  A plot of Φ vs. log(ω) for ωo= 1 is shown below:
o
ω := 0.01 , 0.02 .. 100

90

80

70

60

atan ω  ⋅ 360 50
1 2⋅π 40

30

20

10

00.01 0.1 1 10 100

ω

Our first approach to a straight-line approximation might be to calculate the slope of the phase

function at ω = ωo and draw a line with that slope through the point on the phase function
where ω = ωo. This is the basis for the approach that Nilsson [1] takes to second order phase
approximations. The slope of the phase function at ω = ωo is found by evaluating the
derivative of the phase function at ω = ωo.

The derivative of Φ(ω), is d atan  ω  = 1
dω  ωo 
  ω o ⋅ 1 + ω 2 

ω 2 
o 

Which evaluated at ω = ωo gives: slope = 1

(2 ⋅ω o)

Unfortunately this is the slope in radians per radian, not radians per decade. To calculate the
slope in radians per decade we must calculate the "rise" over the "run" where the "rise" is in 32
radians and the "run" is in decades. Since we are interested in the slope at the point where
ω = ωo , we take as the rise, the limit as ∆ approaches 0 of:

( ) ( ) ωo + ∆ − ωo − ∆  ⋅ 1 (The difference in ω is 2 ∆ )
2 ⋅ω o

For the run we use the limit as ∆ approaches 0, of 2∆ expressed in decades. Since one

decade is defined as the distance between two frequencies ω1 and ω2, where ω1 and ω2 are

related as ω2 = 10 ω1, we have:

( ) ( )∆ω (in decades) = log ω2  ω2 
− log ω1 = log 
 ω1 

which gives  ω2  = 1 decade if ω2 = 10ω1 as required. The run is given then, by the
log 
 ω1 

limit as ∆ approaches 0 of:

 ωo + ∆ 
log 
 ωo − ∆ 

The slope in radians per decade is given then as:

( ) ( )lim  ωo + ∆ − ωo − ∆  ⋅ 1 
 2 ⋅ω o 
 
    1
∆ω → 0  log ωo + ∆   which simplifies to 2 ⋅ln(10)

  ωo − ∆  (We must use L'Hopital's rule.)

An interesting aspect of this result is that the slope of the phase function (in radians per

decade) evaluated at the break frequency ωo is independent of the value of ωo. This is, of
course, one of the principle reasons for plotting the phase function against log (ω) rather than
against ω. Phase functions for various values of ωo are plotted below to demonstrate the
plausibility of the independence of the slope at ω = ωo from the value of ωo.

1 radians 1 360 degrees
⋅ln(10) ⋅ ⋅
⋅ln(10) ⋅ = 2 2 ⋅π decade
2 decade

Φ1(ω) := atan ω  ⋅ 360 Φ2(ω) := atan ω  ⋅ 360 33
0.1 2 ⋅π 0.5 2 ⋅π

Φ1(ω) 90 Φ2(ω) 90
80 80
70 1 10 100 70 1 10 100
60 60
50 50
40 40
30 30
20 20
10 10

00.01 0.1 00.01 0.1

ω ω

Φ3(ω) := atan ω  ⋅ 360 Φ4(ω) := atan ω  ⋅ 360
5 2 ⋅π 10 2 ⋅π

Φ3(ω) 90 Φ4(ω) 90
80 80
70 1 10 100 70 1 10 100
60 60
50 50
40 40
30 30
20 20
10 10

00.01 0.1 00.01 0.1

ω ω

First Straight-Line Approximation (Inadequate)

The slope of the phase function evaluated at the break frequency is

360 ⋅ 1 ⋅ln(10) = 65.964 Deg
2 ⋅π 2 Dec

If we plot a straight line with this slope through the phase function at the break frequency we
obtain the following result. (We have normalized the break frequency ωo to 1)

L (ω) := 45 + 65.964 ⋅log(ω) degrees per decade

L1(ω) := 45 + 65.964 ⋅log(ω) degrees per decade 34

90

80

atan ω  ⋅ 360 70
1 2⋅π 60
50
L1 ( ω ) 40
30

20

10

00.01 0.1 1 10 100

ω
We propose as our first straight-line approximation to the first order phase
function:

Φ approx (ω ) _ if ω < ωlo
|0 if ωlo < ω < ωhi
= | 45+65.964 log(w) if ωhi< ω
|_90

where ωlo is the frequency where L1(ω) intersects 0 degrees, and ωhi is the frequency where
L1(ω) intersects 90 degrees. Since 45 + 64.964 log(ωhi )= 90 we have:

ωhi =  10 45  ⋅ω ωhi= 4.928 ⋅ωo
 64.964 
o
or

Since 45 + 64.964 log( w) = 0 we have:

ω lo =  − 45  or ωlo = 0.2029 ⋅ω o
10 64.964  ⋅ωo

This approximation suffers from two principal shortcomings. The first disadvantage is that the

line L1(ω) underestimates the range of frequencies over which the phase function changes
with ω. In other words the error, i. e., the difference between Φ(ω) and Φapprox (ω) at ωlo and
at ωhi is too large. The second problem is the complexity of the expressions for ωlo and ωhi.
Let us devise a new straight-line approximation by applying one of the oldest, most

fundamental engineering rules of thumb to the selection of ωlo and ωhi. The rule of thumb in

1

question is the notion that if quantity a is less that of quantity b, then compared to

10

quantity b, quantity a may be ignored to a first order approximation.

Standard Phase Approximation (Best)

Standard Phase Approximation (Best)

35

For small frequencies, i.e. for frequencies smaller than ωlo, Φ(ω) -> 0, so let us choose for ωlo

ω0
not .2029 ω , but . Similarly for large frequencies, i.e. for frequencies larger than ω ,
o 10 hi

Φ(ω) −> 90, so let us choose not ωhi = 4.928 ωo, but ωhi = 10 ωo. The straight line resulting

from this selection changes value from 0 to 90 over a frequency range of ω0 to 10 ⋅ω0.
10

ω0 ω0
Since the distance between and ω is one decade and the distance between and
10 o 10

10 ⋅ω0 is one decade, the slope of the line is 90 degrees per 2 decades or 45 degrees per

decade. (45 deg per decade = log(2) ⋅45 deg per octave = 13.56 deg per octave.) The new

straight-line approximation as well as the original straight-line approximation, are plotted below.

L2(ω) := 45 + 45 ⋅log(ω)

90 L1 L2

80 10

atan ω  ⋅ 360 70
1 2⋅π 60
50
L1 ( ω ) 40
30
L2 ( ω ) 20

10

0 0.1 1 100
0.01

ω

In general for a zero at ω0, we have Φ(ω) =

 < ω0 , 0 , if  ω < 10⋅ω0 , 45 + 45 ⋅log ω  , 90
if ω 10   ω0  
 


The line L (ω) is the familiar 45 degree per decade straight-line approximation to which all

The line L2(ω) is the familiar 45 degree per decade straight-line approximation to which all
undergraduate engineering students are introduced. The result is too often simply stated with36

little or no explanation of the underlying derivation. In fact, well over 95% of the over 100

ω
electrical engineering students informally surveyed thought that the slope of atan 
 ω0 

evaluated at ω = ωo has a value of 45 degrees per decade. Before we demonstrate how this
same line of reasoning can be applied to the second order phase response, let us calculate the

errors of L1(ω) and L2(ω) at their respective ωlo and ωhi.

At ωlo we have:

atan(0.2029) ⋅ 360 = 11.47 So the error at ωlo is 11.47 degrees for L1(ω)
2 ⋅π

360 So the error at ωlo is 5.711 degrees for L2(ω)
and atan(0.1) ⋅ = 5.711

2 ⋅π

At ωhi we have:

atan(4.928) ⋅ 360 = 78.529 So the error at ωhi is 90 - 78.529 = 11.47 degrees
2 ⋅π for L1(ω)

360
and atan(10) ⋅ = 84.289
2 ⋅π So the error at ωhi is 90 - 84.289 = 5.711 degrees

for L2(ω)

A similar analysis for the related transfer function 1
gives the following result: H(s) =

s + ωo

L3(ω) := −45 − 65.964 ⋅log(ω)
L4(ω) := −45 − 45 ⋅log(ω)

L4 L3 37

0

10

 0−atan  ω   ⋅ 360 20
1 2⋅π 30
40
L3 ( ω ) 50
60
L4 ( ω ) 70

80

900.01 0.1 1 10 100

ω

In general for a pole at ω0 we have Φ(ω) =

 < ω0 , 0 , if  ω < 10 ⋅ω 0 , −45 − 45 ⋅log ω  , −90
if ω 10   ω0  
 


For both transfer functions we see that the 45 degree per decade approximation better
represents the phase function because it distributes the error on both sides of the line,
whereas the 66 degree per decade approximation (the one based on the slope of the phase
function at the break frequency) puts all of the error on one side of the line. In other words the
45 degree per decade line is a better linear curve fit to the phase function. This results in the
smaller errors at ωlo and ωhi.

Phase Plots of First Order Poles and Zeros at s = 0

The phase function Φ(ω) for a tranfer function H(s) = s is found by taking the angle of H(jω)
as follows:

( )H jω = jω = 0 + jω The angle of such a complex number is

Φ (ω) = atan Re (H( jω))  = atan ω  = 90 Deg for all ω
 Im(H( jω))  0

θ := 0, π .. 2 ⋅π f1 (θ) := if θ = π , 1 , 0
10 2

100 120 90 60 38

90 50 f1(θ) 150 0.6 30
180 0.2 0

00.01 0.1 1 10 100 210 330
ω
240 270 300
θ

Plot of 0 + j 1 = 1 at angle 90 deg
in the complex plane

1

The phase function Φ(ω) for a tranfer function H(s) =

s

is found by taking the angle of H(jw) as follows:

H(jω) = 1 = 0 − j The angle of such a complex number is

jω ω

Φ (ω) = atan Re (H( jω))  = atan   − 1  
 Im(H( jω))   ω

0 = -90 Deg for all ω

f2 (θ) := if θ = 3 ⋅π , 1 , 0
2

0 120 90 60

− 90 50 f2(θ) 150 0.6 30
180 0.2 0

1000.01 0.1 1 10 100 210 330
ω
240 270 300
θ

Plot of 0 - j 1 = 1 at angle -90 deg in the complex plane

Phase Plots of Constant Terms 39

Constant terms K produce no phase shift at all if K > 0 since the angle of K + j0 = 0.

f3(θ) := if (θ = 0 , 1 , 0)

1 120 90 60
00
f3(θ) 150 0.6 30
180 0.2 0

10.01 0.1 1 10 100 210 330
ω
240 270 300
θ

Plot of 1 + j 0 = 1 at angle 0 deg in the
complex plane

Constant terms K such that K < 0 produce a phase shift of 180o for all ω, since the angle of

-K + j0 is 180o.

f4(θ) := if (θ = π , 1 , 0)

400 120 90 60
180 200
f4(θ) 150 0.6 30
180 0.2 0

00.01 0.1 1 10 100 210 330
ω
240 300
270
θ

Plot of -1 + j 0 = 1 at angle180 deg in
the complex plane

First Order Phase Approximation Examples

First Order Phase Approximation Examples 40

s + 1 −5 ⋅  s + 1
⋅(s + 12) 1
#1. H(s) = −5 ⋅
s H(s) = 12 ⋅ s + 1
12
s

ω := 0.01 , 0.02 .. 1000

−5

Since the angle of + j0 = 180 degrees , the constant term contributes a phase of

12

180degrees for all ω. (A positive constant has angle = 0 and so would not produce any
phase contribution an all.) The pole at s = 0 contributes a phase of -90 degrees for all ω,
so the constant phase is 180 - 90 = 90 degrees. The zero at s = 1 contributes a phase
shift that varies from 0 degrees for ω < 0.1 to 90 degrees for ω > 10, varying linearly at
+ 45 degrees per decade for 0.1 < ω < 10. The pole at s = 12 contributes a phase shift
that varies from 0 degrees for ω < 1.2 to -90 degrees for ω > 120, varying linearly at -45
degrees per decade (-13.56 degrees per octave). We begin by calculating the phase shift
for all ω < 0.1 as 180 - 90 = 90 degrees, fixing point 1. The next step is to make an
auxiliary drawing like the one below, which assumes that all of the phase change
contributed by each pole and zero occurs suddenly at the break point ωo (the solid red line
with vertical segments). Next draw in the 45 degrees per decade lines at the proper
locations on the same drawing. (These are the blue dashed lines.) We now can see all
the contributing slopes at all frequencies. (The black line at 135 degrees help us
associate a blue line with a red vertical trasition as follows. The red vertical transition, the
black line at 135 degrees and the blue line associated with that red vertical trasition all
intersect at the same point.) The construction of the phase plot itself begins with a straight
line of +45 degrees per decade between ω = 0.1 and ω = 1.2 (point 2), fixing point 3.
Notice that for 1.2 < ω < 10 we have a slope of +45 -45 = 0 degrees, so construct a
horizontal line from point 3 to point 4. Finally we have a slope of -13.6 degrees per octave
(point 5) between ω = 12 and ω = 120, so construct a straight line of slope -45 degrees

per decade ( -13.6 degrees per octave) to point 6.

 0.01 90  1 90   1 90 
 1 90     10 
B :=  10 

 1 180   10 180   12 

A :=   D :=  10 180 
 
 12 180   12 180   10 180 
 12 90   
C :=  10   120 90 

 1000 90   120 90 

41

180

A〈1〉 160
B〈1〉 140
C〈1〉
D〈1〉 120
135 100

800.01 0.1 1 10 100 1 .103

A〈0〉 , B〈0〉 , C〈0〉 , D〈0〉 , ω

Auxilliary drawing

Φ1(ω) := if ω < 1 , 0 , if  ω < 10 , 45 + 45 ⋅log ω  , 90  
10 1

Φ2(ω) := −90 Φ3(ω) := 180

Φ4(ω) := if ω < 12 , 0 , if  ω < 10 ⋅12 , −45 − 45 ⋅log ω  , −90  
10 12

 0.01 90 
 
 0.1 90 

 1.2 138.563 
points :=  
 10 138.563 

 120 90 
 
 103 90 

ΦT (ω) := 0 + Φ1(ω) + Φ2(ω) + Φ3(ω) + Φ4(ω)

42

100

Φ 1(ω) 50

0 0.1 1 10 100 1 .103
0.01 ω

0

Φ 4(ω) 50

1000.01 0.1 1 10 100 1 .103
ω

140

Φ T(ω) 120

points〈1〉
100

Points 800.01 0.1 1 10 100 1 .103

^ ^ ω , points〈0〉 ^^
^^
#1 #2 #5 #6
#3 #4

43

At ω = 0.1 the contributions to the phase function are the constant -90 degrees from the
pole at ω = 0 and the 180 degress from the (-5) constant term so point #1.

1

For ω < , the slope is 0 degrees per decade so point #2.

10

1
For < ω < 1.2 we have slope +45 degrees per decade so point #3.

10

( 90 + 45 ⋅log 1.2  =138.563 )
0.1

For 1.2 < ω < 10 the slope is + 45 + -45 = 0 degrees per decade so point #4.

For 10 < ω < 120 the slope is -45 degrees per decade so point #5.

(138.563 - 45 ⋅log 120  = 90 degrees.) For ω > 120 we have slope = 0 so point #6.
10

#2 15  s ⋅ s + 1 
32 ⋅ 3 
H(s) =   ω := 0.1 , 0.2 .. 100
 1  2 
 s + 1 ⋅ s +
2 4

Φ1(ω) := if (ω) < 3 , 0 , if  ω < 3 ⋅10 , 45 + 45 ⋅log ω  , 90
10 3

Φ2(ω) := if ω < 2 , 0 , if  ω < 2 ⋅10 , −45 − 45 ⋅log ω  , −90  
10 2

100 0

Φ 1(ω) 50 Φ 2(ω) 50

0 1000.1 1 10 100
0.1 1 10 100

ωω

Φ3(ω) := if (ω) < 4 , 0 , if  ω < 4 ⋅10 , −45 − 45 ⋅log ω  , −90
10 4

0  0.1 90   2 90  44
 2 90   10 
Φ 3(ω) 50
 2 0 3 
1000.1  0
  10
 3 0 
A :=  3 90  E :=  4 
1 10 100 
ω   10 90

 4 90   20 0 
 4 0  
 30 90 
  40 0 
 100 0 

2 90  3 0  4 90 
     
B :=  10  C :=  10  D :=  10 

 20 0   30 90   40 0 

100 1 10 100
A〈0〉 , B〈0〉 , C〈0〉 , D〈0〉 , E〈0〉 , ω
A〈1〉 80
B〈1〉 Auxilliary Diagram

60
C〈1〉
D〈1〉 40
E〈1〉 20
45

0

0.1

ΦT (ω) := 90 + Φ1(ω) + Φ2(ω) + Φ3(ω)  0.1 90  45
 0.2 90 
100  0.3 82.076 

 0.4 82.076 
points :=  20 5.622 


 30 5.622 
 40 0 

 100 0 

Φ T(ω) 50

points〈1〉

0

Points 0.1 1 10 100
ω , points〈0〉
^ ^ ^^ ^ ^^ ^

#1 #2 #3 #4 #5 #6 #7 #8

( )2

For all ω < 10 , the slope of ΦT ω is zero, and the only contribution is from the zero at

23
ω = zero, so point #1 and point #2. For 10 < ω < 10 , the slope is -45 degrees per

decade so point #3. ( 90 - 45 ⋅log 0.3  = 82.076 degrees.) 34
0.2 For < ω < , the

10 10

4
slope is -45 + 45 = 0 so point #4. For < ω < 2 ⋅10 , the slope is -45 + 45 - 45 = -45

10

degrees per decade so point #5. (82.076 - 45 ⋅log 20  = 5.622)
0.4

For 20 < ω < 30 , the slope is +45 - 45 = 0 degrees per decade so point #6. 46

For 30 < ω < 40 , the slope is -45 degrees per decade so point #7. (5.622 - 45 ⋅log 40 
30

= 0) For all ω > 40 , the slope is 0 so point #8.

#3.  s + 1 ⋅ s + 1
  3
H(s) =  s + 1  ⋅(s + 3) 1 ⋅3   1 
4 4 4 
= ⋅
s ⋅s 1  1 ⋅ s 1
+ 2 s  + 
2   1  
2

⋅  s + 1 ⋅ s + 1
 3
1 
3 4 

= 2 ⋅ s 1
 1 
s   2  + 

ω := 0.01 , 0.015 .. 100

  1  
 4 
Φ1(ω) := if  (ω ) < , 0 , if ω < 10 ⋅ 1 , 45 + 45 ⋅log ω  ,  
10  4   1   90 
   4  
  


100

Φ 1(ω) 50 1 10 100
ω
0
0.01 0.1

  1   47
2
Φ2(ω) := if  (ω ) < , 0 , if ω < 10 ⋅ 1 , −45 − 45 ⋅log ω  , −90 
 10  2   1    
   2   
  

Φ3(ω) := if ω < 3 , 0 , if  ω < 10 ⋅3 , 45 + 45 ⋅log ω  , 90  
10 3

100 0

Φ 3(ω) 50 Φ 2(ω) 50

0 1 10 100 100 1 10 100
0.01 0.1 ω 0.01 0.1 ω

  1     1  
 4 −90   4 −90 

  10 
 
 0.01 −90  B :=  10   1  
1   2 0
1  4 ⋅10 0 
4 −90   10 

  E :=  3 
 1 0  10 −90 
  1  
4   2 0  
 
1  C :=  10   1 ⋅10 0
A :=  0 4
 2   
  1 −90 
1  2 ⋅10  1 ⋅10 −90 
2   2 
−90
 30 0 

 3 −90  3 −90 
 3 0 
 D :=  10
 100 0   
 30 0 

ΦT (ω) := −90 + Φ1(ω) + Φ2(ω) + Φ3(ω) 48

0 100
A〈1〉
B〈1〉 20

C〈1〉 40
D〈1〉
E〈1〉 60

− 90+45 80

1000.01 0.1 1 10
A〈0〉 , B〈0〉 , C〈0〉 , D〈0〉 , E〈0〉 , ω

 0.01 −90  Auxilliary Diagram

 1
 −90 
 40 
 
 1 −76.45 
 20

3 
 −76.45 
points :=  10 

 10 −35.02 
4 
 
 10 −35.02 

2 
 
 30 0 

 100 0 

49

0

20

Φ T(ω) 40
60
points〈1〉

80

1000.01 0.1 1 10 100
ω , points〈0〉
^
^ ^^ ^ ^^ ^
#8
#1 #2 #3 #4 #5 #6 #7

For ω < 0.01, the only contribution is from the pole at zero so point #1 and point #2.
11

For 4 <ω< 2 degrees so point #3.
the slope is + 45
10 10 decade

 1  1
−90 + 45 ⋅log 20 
1  = −76.454 For 2 <ω< 3
 40
the slope is +45 -45 = 0 so
10 10

31
point #4. For < ω < 10 ⋅ the slope is +45 -45 +45 = 45 so point #5.

10 4

 10  = −35.017 11
−76.454 + 45 ⋅log 4  For 10 ⋅ < ω < 10 ⋅ the slope is
3 
 10 42

-45 +45 = 0 so point #6.

1
For 10 ⋅ 2 < ω < 30 the slope is +45 so point #7.

−35.017 + 45 ⋅log 30  = 0. For ω > 30 the slope is 0 so point #8.
 1 

10 ⋅
 2

Example #4 50

H(s) = 100000 ⋅s = 100000 ⋅ s
(s + 10000) ⋅(s + 50000)
10000 ⋅50000  s + 1 ⋅ s + 1
10000 50000

ω := 100 , 200 .. 1000000

Φ1(ω) := if ω < 10000 , 0 , if  ω < 10 ⋅10000 , −45 − 45 ⋅log ω  , −90  
10 10000

0 1 .103 1 .104 1 .105 1 .106
ω
Φ1(ω) 50

100
100

Φ2(ω) := if ω < 50000 , 0 , if  ω < 10 ⋅50000 , −45 − 45 ⋅log ω  , −90  
10 50000

0

Φ2(ω) 50

100 1 .103 1 .104 1 .105 1 .106
100 ω