“Bode Diagram”
Hendrik Wade Bode
(American ,1905-1982)
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode diagrams (de magnitude e de phase) are one way to characterize signals in
the frequency domain.
Transfer Function
s = 0 + jω = jωwe get the Fourier Transform from the Laplace Transform,
Signals are represented in the frequency domain by s functions:
X(s), Y(s), etc.
or by functions of jωX(jω), Y(jω), etc.
as we saw in chapter 8 (Fourier Transforms).
F { x(t) } = X(jω) e F { y(t) } = Y(jω)
In fact Laplace Transforms and the Fourier Transforms are representations that
are closely related to each other.
In many cases, if we replace ‘s’ with ‘jω’, that is, by making ‘s’ be a complex
number with zero real part and imaginary part ‘ω’,
as we saw in chapter 6 (Laplace Transforms)
L { x(t) } = X(s) and L { y(t) } = Y(s)
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
X(s) = X(0+jω) = X(jω), Y(s) = Y(0+jω) = Y(jω), etc.
If x(t) is the input of a system and y(t) is the output of the same system, in
certain applications it may be more interesting to represent on the block
diagrams these signals
X(s), X(jω), Y(s) and Y(jω)
in the frequency domain rather than the time domain.
Block diagrams with the input and output signals
represented in the frequency domain.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
On the block diagrams below G(s) and G(jω) are the impulse response of the
system as seen in the sections 5.10 (in chapter 5, Laplace Transform) and 8.5 (in
chapter 8, Fourier Transform) respectively.
Note that there the impulse response of the system was generally represented by
H(s) and H(jω) whereas here, in general, it will be G(s) and G(jω).
In chapter 4, on Systems and in chapter 8 on Fourier Transform, we have seen
some classic results about LTI (linear time invariant systems).
That is, the output y(t) is the convolution between the impulse response g(t) and
the input x(t). This implies that
.d)(g)t(x)t(g)t(x
d)(x)t(g)t(x)t(g)t(y
τ⋅τ⋅τ−=∗=
τ⋅τ⋅τ−=∗=
∞+
∞−
+∞
∞−
).j(G)j(X
)j(X)j(G)j(Y
ω⋅ω=ω⋅ω=ω
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Knowing the impulse response g(t) of a linear time invariant system (LTI) we can
find the output y(t) for any other input x(t).
For example, in the particular case of input x(t) = unit impulse,
x(t) = uo(t)
then the output y(t) = g(t) = the “impulse response of the system”.
where
X(jω) = F { x(t) } X(jω) = Fourier Transform of x(t),
Y(jω) = F { y(t) } Y(jω) = Fourier Transform of y(t), and
G(jω) = F { g(t) } G(jω) = Fourier Transform of g(t)
This result is due to:
the convolution transform is the product of the convolution.
a well-known property of the Convolution (chapter 8).
)j(X
)j(Y)j(G
ωω=ω
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Therefore the transfer function of a linear time invariant system (LTI) represented in
the frequency domain:
G(s) or G(jω),
q (s)G(s)
p (s)=
where q(s) and p(s) are polynomials in ‘s’ of the type
an sn + an-1 sn-1 + ... + a1 s + ao
or, alternatively,
)j(p
)j(q)j(G
ωω=ω
where p(jω) and q(jω) are polynomials in ‘s = jω’ of the type
an (jω)n + an-1 (jω)n-1 + ... + a1 (jω) + ao
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
that very commonly are rational fractions, that is, fractions which numerator and
the denominator are polynomials in ‘s’:
Poles and zeros of the Transfer Function
)s(p
)s(q)s(G =
Consider now the transfer function G(s) of a system after being reduced to the
rational fraction
and suppose that all the eventual commons roots of q(s) and p(s) have been
cancelled and therefore the above expression is a irreducible forma.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Characteristic Equation:
The polynomial p(s) is called characteristic polynomial of G(s), or the
characteristic polynomial of the system. The equation
p(s) = 0is called the “characteristic equation” of system.
Poles:
The roots of the characteristic polynomial p(s) = 0 are called the poles of G(s), or
the poles of the system. That is, the poles are the solutions of the characteristic
equation.
Zeros:
The roots of the numerator of G(s), q(s) are called zeros of G(s) or zeros of the
system. That is, the zeros are the solutions of the equation q(s) = 0.
Example 9.1: Consider the transfer function G(s) given by
2)+2s+(s2)+(ss
)30s(2)s(G
2
+⋅=
It is easy to verify that G(s) has a zero
s = –30
and four poles:
s = 0, s = –2, e s = –1 ± j
where 2 are real and 2 are complex.
s4s6s4s
2)+s2+(s2)+(ss)s(p
234
2
+++=
==
The characteristic polynomial of this system is:
Since s = 0 is a pole of G(s), we say that this system has a “pole at the origin”.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Example 9.2: Consider now the transfer function G1(s) given by
Clearly G1(s) has a “zero at the origin”:
s = 0
)10+s10+(s10)+(s
s10)s(G
422
5
1 =
s 50 j 50 3= − ± ⋅
5323
4221
10s1011s110s
)10+s10+(s10)+(s)s(p
+×++=
==
and three poles:
and
where 1 is real and 2 are complex.
The characteristic polynomial of this system is:
10s −=
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Example 9.3:
G(s) has a double “zero at the origin” (s = 0) and four poles:
,c)-(s)b+(sa)+(s
s10)s(G
22
2
=
s = –a (double), s = –b2 and s = c.
Consider now the transfer function G(s) given by
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The BASIC FACTORS IN ‘s’ for the construction of a Bode diagram
s
1)s(G =
2s
1)s(G =
3s
1)s(G =
L
3. Derivative factors [zeros at the origem]: sn , n = 1, 2, ...
G(s) = s , G(s) = s2 , G(s) = s3,
2. Integrative factors [poles at the origin]: (1/s)n , n = 1, 2, ...
G(s) = KB
1. Bode gain ( KB )
...
...
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
4. 1st order factors of the type “real poles”: 1/(Ts + 1)n, n = 1, 2, ...
( )1Ts
1)s(G
+= ( )2
1Ts
1)s(G
+=
( )31Ts
1)s(G
+=
5. 1st order factors of the type “real zeros”: (Ts+ 1)n , n = 1, 2, ...
( )1Ts)s(G += ( )21Ts)s(G += ( )3
1Ts)s(G +=
...
...
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
6. 2nd order factors or quadratic, of the type “complex poles”:
1/[1+2ζ(s/ωn)+(s/ωn)2]n, n = 1, 2, ...
ω+
ωζ+
=
2
n
2
n
ss
21
1)s(G
2
2
n
2
n
ss
21
1)s(G
ω+
ωζ+
=
3
2
n
2
n
ss
21
1)s(G
ω+
ωζ+
=
...
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
7. 2nd order factors or quadratic, of the type “complex zeros”:
[1+2ζ(s/ωn)+(s/ωn)2]n, n = 1, 2, ...
2
n
2
n
ss
21)s(G
ω+
ωζ+=
2
2
n
2
n
ss
21)s(G
ω+
ωζ+=
3
2
n
2
n
ss
21)s(G
ω+
ωζ+=
...
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Dismembering the functions G(s) in basic factors
)2s2s()2s(s
)30s(2)s(G
2 ++++=
)2s2s()2s(s
130
s302
)s(G2 +++
+⋅=
++⋅
+⋅⋅⋅
+⋅=
1s2
s1
2
ss22
130
s302
)s(G2
Any transfer function G(s) can easily be rewritten with a combination of these basic
factors.
Example 9.4: Consider now the function G(s) seen in Example 9.1 which is given by
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Finally, grouping all the constants (from the numerator and from the denominator):
we obtain the expression bellow:
1522
302 =×⋅×
++⋅
+⋅
+⋅=
1s2
s1
2
ss
130
s15
)s(G2
which is entirely written in terms of basic factors in
the form:
( )( )
+
ωζ+
ω⋅+⋅
+⋅=1s
2s1Tss
1s'TK)s(G
n
2
n
2
B
KB = 15
T = ½
T’ = 1/30
2n =ω
707,02
2
2
1 ===ζ
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
( )
ω+
ω−⋅
ω⋅+⋅ω
ω⋅+⋅=
=
+ω+ω⋅
+ω⋅ω
+ω⋅=ω
j2
12
j1j
30j115
1j2
j1
2
jj
130
j15
)j(G
2
2
Doing this we get:
because this is the only difference between the two forms G(s) and G(jω).
s = jωthat is,
s = 0 + jω,
Example 9.5: To write the transfer function G(s) from the previous example in the
form of basic factors in jω and then get G(jω), we just have to substitute ‘s’ in the
result obtained for G(s) by
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Bode diagrams of the Basic Factors
The Bode diagrams are built for transfer functions G(jω) and are two:
Bode diagrams of magnitudeand
Bode diagrams of phase.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Knowing the Bode diagrams of the basic factors it is possible to use them to construct
the Bode diagram of any other transfer function G(jω) that we break apart in terms
of basic factors.
While Bode diagrams of phase are graphs of
∠G(jω) in degrees
×ω (with logarithmic scale)
Bode diagrams of magnitude are graphs of
| G(jω) | in dB ( |G(jω) |dB )
×ω (with logarithmic scale)
1. The Bode gain ( KB )
Since G(jω) = KB is a constant (do not vary with ω), we have that |KB| in dB is
given by:
B10dBB Klog20K ⋅=
while ∠ KB is 0º or –180º, ∀ω, that is:
∠ KB = 0º if KB is a positive constant,
or
∠ KB = – 180º if KB is a negative constant.
Once you are familiar with the Bode diagrams graphs of the basic factors we
present here, building the Bode diagrams of the other transfer functions is
made easier, as we will see in the Examples.
So now let's show the Bode diagrams (magnitude and phase) for each of the
basic factors.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
<−
>=∠=ω∠
0Kse,º180
0Kse,º0
K)j(G
B
B
B
Of course the phase angle for negative KB, –180º is the same as +180º
which is π.
Thus, as stated above in the definition of Bode diagram of phase, it is
normal to represent the phase of KB (i.e., the angle ∠KB) in degrees
(rather than radians).
This is because, since G(jω) has a number of poles greater (or at most
equal) than the number of zeros, then ∠G(jω) will always tend towards
the negative part (downwards, below 0º).
However, for the purpose of Bode diagram there is a tendency to
adopt ∠KB = –180º in these situations.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode diagram
(magnitude and
phase) of
G(jω) = KB
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
If KB>1, then 0)j(GdB
>ω
If KB=1, then 0)j(GdB
=ω
If 0<KB<1, then 0)j(GdB
<ω
That is, by increasing the value of KB we make the whole Bode diagram of
magnitude “go up” while decreasing the value of KB makes the whole Bode
diagram of magnitude “go down”.
The effect that a gain variation KB in a Bode diagram with several
basic factors is that it shifts the magnitude curve up (if KB > 0) or
down (if KB < 0) and does not affect the phase angle curve.
On the other hand, the Bode diagram of phase is unchanged to variations
of KB if KB > 0 , or shifted downwards of 180º, in the case of KB < 0.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
2. Integrative factors (jω)-1, (jω)-2, … , (jω)-n
[ ]dBlog20
j
1log20)j(G
10
10dB
ω⋅−=
ω⋅=ω
which is actually the equation of a line with a slope of –20 dB/decade
since ω is represented in the logarithmic scale.
To see this, first note that
|G(jω)|dB intercepts 0 dB in ω = 1,
a detail that makes it easy for us to sketch it.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Firstly, for G(jω) = (jω)-1, we have that |G(jω)| in dB is given by:
for ω = 0,01 G(jω) = 40 dB
for ω = 0,1 G(jω) = 20 dB
for ω = 1 G(jω) = 0 dB
for ω = 10 G(jω) = – 20 dB
for ω = 100 G(jω) = – 40 dB
which allows us to see clearly that it is a straight line with a slope of
–20 dB/decade
In fact, we have to look at a few consecutive decades, that in the Bode
diagram of magnitude of G(jω) (i.e., |G(jω)|dB ):
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
for ω = 0,5 G(jω) = 6 dB
for ω = 1 G(jω) = 20 dB
for ω = 2 G(jω) = –6 dB
for ω = 4 G(jω) = – 12 dB
It is also customary to look at a few consecutive octaves (rather than
decades) of the Bode diagram of magnitude of G(jω) ( |G(jω)|dB ).
which is an alternative way of looking at this straight line since the
slope of –20 dB/decade is equivalent to – 6 dB/octave.
That is: a octave corresponds to: the double /or the half, depending
on the direction (to the right or to the left / increasing / or
decreasing).
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode diagram
(magnitude and
phase) of
G(jω) = (jω)-1
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
For the phase ∠G(jω), we have that:
Note that, since ω is represented in a logarithmic scale, then ω is
always positive (ω > 0) and therefore ∠ jω = 90º, and thus
–∠ jω = – 90º.
So, the Bode diagram of phase ∠G(jω), ∀ω, is a constant
equal to – 90º.
The effect of the basic factor G(jω) = 1/jω in a Bode diagram of phase
with several basic factors is that it shifts the curve of phase downwards
of 90º.
∠ G(jω) = ∠ (1/ jω) =
= – ∠ jω =
= – 90º , ∀ω
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Now, to the other integrative factors (jω)-2, (jω)-3, …, (jω)-n
For G(jω) = (jω)-n, we have a situation very similar to the factors
(jω)-1 which we have seen above: the slope of the magnitude
curve is multiplied by n, as well as the value of the phase curve.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The following graphic of the complete Bode diagram illustrates all this.
that is, the Bode diagram of magnitude |G(jω)|dB = |(jω)-n |dB is a line
with a slope of –20⋅n dB/decade or, equivalently, –6 dB/octave.
Besides, ∠G(jω) = ∠(jω)-n = –90º⋅n, ∀ω, that is, the Bode
diagram of phase ∠ G(jω) = ∠(jω)-n, ∀ω, is a constant with value
– 90º⋅n, its effect is to shift the curve of phase downwards of 90º⋅n.
Summarizing, for the integrative factors:
G(jω)|dB = –20 ⋅n ⋅ log10 ω [dB] and
|G(jω)|dB = |(jω)–n |dB intercepts 0 dB at ω = 1.
The Bode diagram
(magnitude and
phase) of
G(jω) = (jω)-n
,
n = 1, 2, ...
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
3. Derivative factors (jω), (jω)2, … , (jω)n
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The derivative factors G(jω) = (jω)n, n = 1, 2, ... they are very similar to
the integrative factors, only with the difference that the lines go up
rather than down, that is, the slope is positive (+20⋅n dB/decade) rather
than negative (–20⋅n dB/decade)
The following graphic of the complete Bode diagram illustrates all this.
Summarizing, for the derivative factors:
that is, the Bode diagram of phase |G(jω)|dB = |(jω)n |dB is a line with a
slope of +20⋅n dB/decade or, equivalently, +6 dB/octave.
Besides, ∠G(jω) = +90º⋅n, ∀ω, that is, the Bode diagram of phase
∠G(jω) = ∠(jω)n, ∀ω, is a constant with value +90º⋅n, its effect is
to shift the curve of phase upwards of 90º⋅n.
G(jω)|dB = +20 ⋅n ⋅ log10 ω [dB] and
|G(jω)|dB = |(jω)n |dB intercepts 0 dB at ω = 1.
The Bode diagram
(magnitude and
phase) of
G(jω) = (jω)n,
n = 1, 2, ...
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
4. First order pole factors (1 + jωT)-1
For G(jω) = 1/ (1 + jωT), we have that the magnitude |G(jω)| in dB is given
by:
( )
( )2
10
10dB
T1log20
Tj1
1log20)j(G
⋅ω+⋅⋅−=
ω+⋅=ω
Which we are going to divide into 2 intervals: ω << 1/T and ω >> 1/T,
that is, for low and high frequencies.
( ) ≅⋅ω+<<⋅ω 1T11T2
( )
( ) dB01log20
T1log20)j(G
10
2
10dB
=⋅⋅−≅
⋅ω+⋅⋅−=ω
In the interval, ω << 1/T (low frequency), we note that:
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
and therefore:
( )
>>ω⋅ω⋅−
<<ω=ω
T
1,Tlog20
T
1,0
)j(G
10
dB
Point where the 2 asymptotes lines intercept each other
0 dB for ω = 1/T
T
1c =ω T
1c =ω
The frequencyIs called “corner frequency”).
So, we have 2 approximations for the curve G(jω)|dB = 1/ (1 + jωT)|dB,
both lines, to which we call
“asymptotes lines”
for high and low frequencies.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Bode diagram of magnitude.
Simple pole factor G(jω) = 1/ (1 + jωT)
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The actual, not approximate, curve of G(jω)|dB coincides with the
asymptotes only when ω << ωc or when ω >> ωc, which in practice
corresponds to
( )T10
1
⋅<ω for low frequencies
T
10<ω for high frequencies
That is, the asymptotes are valid for a decade before the corner frequency
ωc = 1/T (in the case of asymptote for low frequencies) or one decade
after the corner frequency ωc = 1/T (in the case of asymptote for high
frequencies).
Indeed, it is easily shown that for ω = 1/10T (a decade below ωc), as well
as for ω = 10T (a decade above ωc), the errors shown by the magnitude
curve G(jω)|dB are negligible, practically null:
G(jω)|dB = – 0,04 db ≅ 0 dB
for ω = 1/(10T) or for ω = 10T.
and
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Near the corner frequency ωc the asymptotes only approximate the actual
curve of G(jω)|dB.
( )10 10 cdB-3dB
1 1 1G(j ) 20 log 20 log , para
1 j 2 Tω = ⋅ = − ⋅ ⋅ = ω = ω =
+
For the phase angle ∠ G(jω), we have to:
∠ G(jω) = ∠ 1/ (1 + jωT) =
= – ∠ (1 + jωT) =
= – arctg (ωT)
Here you can also think of the ranges: ω << 1/T and ω >> 1/T, i.e.,
for low and high frequencies.
The maximum error is 3 dB and occurs exactly at the corner frequency
ωc = 1/T, the point where the two asymptotes meet, because for this
value of ω,
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
At low frequencies, ω << 1/T, we observe that:
( ) º01)j(G1T11T ≅∠=ω∠≅⋅ω+<<⋅ω
whereas at high frequencies, ω >> 1/T, we observe that:
( ) ( ) ( ) º90Tj)j(GTjTj11>>T −≅ω⋅∠−=ω∠ω⋅≅ω⋅+⋅ω
results that could also easily be obtained using eq. (9.9) with ωT ≅ 0 and
ωT ≅ ∞, respectively, since
>>ω−
<ω<ω−
<<ω
=ω∠
T
1,º90
T100100
T,)T(arctg
T
1,0
)j(G
and therefore:
arctg (0) = 0º and – arctg(∞) = – 90º
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Note that for ωc = 1/T, G(jωc) = – arctg(ωcT)= – arctg(1)= –45º, thus,
at the corner frequency ωc = 1/T we have that:
Bode diagram of phase.
Factors real poles (simple): G(jω) = 1/ (1 + jωT)
the graph of ∠ G(jω) passes in –45º at ω = 1/T,
that is, halfway between 0º and –90º; a detail to keep in mind when
sketching the Bode diagram of phase.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
That is, Bode diagram of phase ∠G(jω) tends asymptotically
ccfro m to 1 0
1 0
ω ⋅ ω
That is, since
a decade before the corner frequency ωc = 1/T(asymptotes for low frequencies)
to
a decade after the corner frequency ωc = 1/T(asymptotes for high frequencies).
to 0º (at the left) and to – 90º (at the right).
In practice we consider that ∠G(jω) varies from 0º to – 90º while the
frequency ω varies
Bode Diagram ______________________________________________________________________________________________________________________________________________________________________________________
Multiple pole factors (1 + jωT)-2, (1 + jωT)-3, ..., (1 + jωT)-n
Bode diagram of magnitude.
Multiple pole factor G(jω) = 1/ (1 + jωT)n, n = 2, 3, …
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Bode diagram of phase.
Multiple pole factor G(jω) = 1/ (1 + jωT)n, n = 2, 3, …
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
These factors are all analogous to the first order pole
factors, single and multiple, we have seen above.
The main differences are that the asymptotes in high
frequency magnitude curves rise with a slope of +20n dB/dec
instead of declining with a slope of –20n dB/dec and the phase
curves go from 0º to +90ºn instead of from 0º to –90ºn.
That is, the magnitude and phase curve for the first order zero factors
can be obtained by inverting the sign of the magnitude and phase
curves of the first order pole factors.
5. First order zero factors simple and multiple
(1 + jωT)1, (1 + jωT)2, ..., (1 + jωT)n
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Bode diagram de magnitude.
Simple and multiple zero factors G(jω) = (1 + jωT)n, n = 1, 2, 3, …
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Bode diagram de phase.
Simple and multiple zero factors G(jω) = (1 + jωT)n, n = 1, 2, 3, …
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Basic Factors with negative signs
The case of basic factors with negative signs such as
( )1Ts
1)s(G
−= ( )2
1Ts
1)s(G
−=
( )31Ts
1)s(G
−=
...
or
( )1Ts)s(G −= ( )21Ts)s(G −= ( )3
1Ts)s(G −= ...
however for the construction of the
Bode diagram of phase is required
a greater care in the analysis.
it easy to show that the
Bode diagram of magnitude is identic to the basic factor
corresponding to signal “+” ,
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
+
+=
++=ω
1100
s
)1s(100
1
)100s(
)1s()j(G
Example 9.6
11
(s 1) and s 1100
− + ⋅ +
)100/j1()j1()j(G ω+∠−ω+∠=ω
Besides, the phase of G(jω) is given by
Note that in this case KB = 1/100 = –40 dB and G(jω) has still other two
basic factors:
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode
diagram of
magnitude
and phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
+
−=
+−=ω
1100
s
)1s(100
1
)100s(
)1s()j(G
Example 9.7
1
1s100
1e)1s(
−
+⋅−
)100/j1()j1(º180
)100/j1()j1()j(G
ω+∠−ω−∠+=
=ω+∠−ω+−∠=ω
Besides, the phase of G(jω) is given by
Thus, the Bode diagram of magnitude is equal to the previous example (Example 9.6).
Note that in this case KB = 1/100 = –40 dB again and G(jω) still have two other
basic factors:
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode
diagram of
magnitude
and phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
−
+=
−+=ω
1100
s
)1s(100
1
)100s(
)1s()j(G
Example 9.8
11
(s 1) and s 1100
− + ⋅ −
)100/j1(º180)j1(
)100/j1()j1()j(G
ω−∠−+ω+∠=
=ω+−∠−ω+∠=ω
Besides, the phase of G(jω) is given by
So, the Bode diagram of magnitude is equal to the 2 previous Examples
(Examples 9.6 and 9.7).
Note that in this case KB = 1/100 = –40 dB again and G(jω) still have two other
basic factors:
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode
diagram of
magnitude
and phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
−
−=
−−=ω
1100
s
)1s(100
1
)100s(
)1s()j(G
Example 9.9
Note that in this case KB = 1/100 = –40 dB again and G(jω) still have two other
basic factors: 1
1(s 1) and s 1
100
− − ⋅ −
So, the Bode diagram of magnitude is equal to the 3 previous examples
(Examples 9.6, 9.7 and 9.8).
)100/j1()j1(
)100/j1(º180)j1(º180
)100/j1()j1()j(G
ω−∠−ω−∠=
ω−∠−−ω−∠+=
=ω+−∠−ω+−∠=ωBesides, the phase of G(jω) is given by
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode
diagram of
magnitude
and phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Example 9.10
( )
++
=++
=ω1
100
s1s
1
)100s()1s(
100)j(G
Note that in this case KB = 1 = 0 dB and G(jω) still have two other
basic factors:
1
1 1( s 1) s 1
100
−− + ⋅ +
and
)100/j1()j1()j(G ω+∠−ω+∠−=ωBesides, the phase of G(jω) is given by
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode
diagram of
magnitude
and phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
( )
−+
=−+
=ω1
100
s1s
1
)100s()1s(
100)j(G
Example 9.11
1
1 1(s 1) and s 1
100
−− + ⋅ −
So, the Bode diagram of magnitude is equal to the previous example (Example 9.10).
)100/j1(º180)j1(
)100/j1()j1()j(G
ω−∠−+ω+∠−=
=ω+−∠−ω+∠−=ω
Note that in this case KB = 1 = 0 dB again and G(jω) still have two other basic
factors:
Besides, the phase of G(jω) is given by
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode
diagram of
magnitude
and phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
( )
+−
=+−
=ω1
100
s1s
1
)100s()1s(
100)j(G
Example 9.12
1
1 1(s 1) and s 1
100
−− − ⋅ +
)100/j1()j1(º180
)100/j1()j1()j(G
ω+∠−ω−∠−=
=ω+∠−ω+−∠−=ω
Besides, the phase of G(jω) is given by
So, the Bode diagram of magnitude is equal to the 2 previous examples
(Examples 9.10 and 9.11).
Note that in this case KB = 1 = 0 dB again and G(jω) still have two other basic
factors:
Bode Diagram ______________________________________________________________________________________________________________________________________________________________________________________
The Bode
diagram of
magnitude
and phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
( )
−−
=−−
=ω1
100
s1s
1
)100s()1s(
100)j(G
Example 9.13
Note that in this case KB = 1 = 0 dB again and G(jω) still have two other basic
factors: 1
1 1s100
1e)1s(
−−
−⋅−
So, the Bode diagram of magnitude is equal to the 3 previous examples
(Examples 9.10, 9.11 and 9.12).
)100/j1()j1(
)100/j1(º180)j1(º180
)100/j1()j1()j(G
ω−∠−ω−∠−=
ω−∠−−ω−∠−=
=ω+−∠−ω+−∠−=ωBesides, the phase of G(jω) is given by
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode
diagram of
magnitude
and phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Quadratic basic factors
(poles and zeros quadratic factors)
7. Quadratic zeros factors
6. Quadratic polos factors
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
6. Quadratic pole factors [1 + 2ζ(jω/ωn) + (jω/ωn)2]-n, n = 1, 2, …,
Note that the transfer function G(jω)
ωω−
ωω⋅ζ+
=
ωω+
ωω⋅ζ+=ω
−
2
nn
12
nn j21
1jj21)j(G
has a couple of poles which will be:
a). complex poles if 0 ≤ ζ < 1
b). double poles (equal) if ζ =1
c). real poles (different) if ζ > 1
0 ≤ ζ ≤ 1
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The quadratic poles factors we deal with in this section are part of cases
(a) and (b) above, that is: 0 ≤ ζ ≤ 1, since case (c), real poles (different)
(ζ > 1), are already covered in the previous basic factors.
which corresponds to double poles (equal) jω/ω, a case that is also
already covered in the previous basic factors.
2
n
2
nn
j1
1
j21
1)j(G
ωω+
=
ωω−
ωω⋅+
=ω
In fact, even in case (b), when we have the limit situation, then
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Therefore the techniques that will be presented in this section for 0 ≤ ζ ≤ 1, will
coincide with others already presented in the particular case of ζ = 1.
( )
ω>>ω⋅ω⋅⋅−
ω⋅<ω<ω⋅
ωωζ+
ωω−⋅⋅⋅−
ω<<ω
=ω
n10
nn
2
n
2
n
10
n
dB
,Tlogn40
101,0,j
21logn20
,0
)j(G
For G(jω) = [1 + 2ζ(jω/ωn) + (jω/ωn)2]-n, n = 1,2, …,
we have that the magnitude |G(jω)| in dB is given by
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
10
nω
ζ = 1, n = 1G(jω) = [1 + 2ζ(jω/ωn) + (jω/ωn)
2]-n, n = 1,2, …
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Note that just as in the previous sections there was ωc, here too there is
a frequency ωn which is called
ωn = natural frequency of the system,
that separates the “high” and “low” frequencies and
the asymptotic line for high frequencies
a detail to keep in mind when sketching the Bode Diagram of magnitude.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
As the value of ζ decreases, ζ ≤ 1, the curves get higher and create peaks in
|G(jω)|dB , becoming higher and higher (from ζ <�
�= 0,707) and becoming
even higher as ζ → 0.
These peaks occur at frequencies ωr called
ωr = resonance frequency
that takes values
2
20,21 2 ≤≤−⋅= ζζωω fornr
In the vicinity of the natural frequency ωn the asymptotes only approximate
the real curve of G(jω)|dB showing a maximum error of 6×n dB that occurs
exactly at the corner frequency ωn , the point where the two asymptotes
meet.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
G(jω) = [1 + 2ζ(jω/ωn) + (jω/ωn)2]-1
The Bode
diagram of
magnitude
of G(jω)
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
As ζ increases, the resonance frequency ωr decreases slightly until,
when
707,02
2 ==ζ
then the resonance frequency ωr = ωn/2.
Note that for ζ = 0, ωr = ωn.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
On the other hand, these peaks reach values Mr
Mr = resonance peak
Which has values
r2
1 2M , for 0
22 1= ≤ ζ ≤
ζ⋅ −ζ
As ζ decreases, the resonance peak Mr increases.
For example,
Note that for 0,707 < ζ < 1, there is no resonance peak.
If ζ = 0,5 Mr = 1,155 ≅ 1,25 dB
If ζ = 0,25 Mr = 2,133 ≅ 6,6 dB
If ζ = 0,1 Mr = 5,025 ≅ 14 dB
If ζ = 0,05 Mr = 10,01 ≅ 20 dB
In particular, if ζ = 0,707, then
Mr = 1 = 0 dB (also there is no resonance peak).
Diagrama de Bode______________________________________________________________________________________________________________________________________________________________________________________
For the phase angle∠ G(jω), we have that:
ωω−
ωω⋅ζ
⋅−=
ωω+
ωω⋅ζ−∠=ω∠
−
2
n
n
n2
nn1
2
arctgnjj
21)j(G
Therefore:
∞→ω⋅−
ω=ω⋅−
→ω
=ω∠
,nº180
,nº90
0,º0
)j(G n
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
In practice we consider that ∠G(jω) varies from 0º to 180º × n
while the frequency ω varies
to
one decade after the natural frequency ωn(asymptote for high frequencies).
that is,
since
one decade before the natural frequency ωn(asymptote for low frequencies)
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
from ��
��to 10 · ω�
The Bode diagram of phase of G(jω) becomes steeper (with steeper
slope) as ζ → 0.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
7. Quadratic zero factors [1 + 2ζ(jω/ωn) + (jω/ωn)2]n, n = 1, 2, …,
they are at all analogous to quadratic pole factors we saw
above.
The main differences are that the resonance peaks are
downwards instead of upwards and the phase curves goes
from 0º to 180ºn instead of 0º to –180ºn.
0 ≤ ζ < 1
That is, the magnitude and phase quadratic zero factors can be
obtained by inverting the sign of the magnitude and phase curves of
the quadratic pole factors.
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
+⋅+
+⋅
+⋅=
=+++
+=ω
1s400
5
400
s1s
100
1s
1s4
11,0
)400s5s()100s(s
)4s(1000)j(G
2
2
Example 9.14
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
BK 0,1 20 dB= = − n 20ω = 0,125ζ=2
r n 1 2 19, 68ω = ω ⋅ − ζ = 1
1
14 for zero of G( j )
Tω = = ω
2
2
1100 for single pole of G( j )
Tω = = ω
Note that:
dB1,1203,412
1M
2r ==
ζ−⋅ζ=
The Bode
diagram of
magnitude
and phase
of G(jω)
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
+⋅−
+⋅
+⋅=
=+−+
+=ω
1s400
5
400
s1s
100
1s
1s4
11,0
)400s5s()100s(s
)4s(1000)j(G
2
2
Example 9.15
The Bode diagram of magnitude is the same as the previous example
(Example 9.14).
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode diagram of phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Example 9.16
+⋅+
−⋅
+⋅=
=++−
+=ω
1s400
5
400
s1s
100
1s
1s4
11,0
)400s5s()100s(s
)4s(1000)j(G
2
2
The Bode diagram of magnitude is the same as the 2 previous example
(Examples 9.14 and 9.15).
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode diagram of phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Example 9.17
+⋅+
+⋅
−⋅=
=+++
−=ω
1s400
5
400
s1s
100
1s
1s4
11,0
)400s5s()100s(s
)4s(1000)j(G
2
2
The Bode diagram of magnitude is the same as the 3 previous example
(Examples 9.14, 9.15 and 9.16).
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode diagram of phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Example 9.18
The Bode diagram of magnitude is the same as the 4 previous example
(Examples 9.14, 9.15, 9.16 and 9.17).
+⋅+
−⋅
−⋅=
=++−
−=ω
1s400
5
400
s1s
100
1s
1s4
11,0
)400s5s()100s(s
)4s(1000)j(G
2
2
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
The Bode diagram of phase
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
( )
+⋅+
+⋅
+⋅=
=+++
+=ω
1s10
1
10
s1s
10
1s
1s10
)10s10s()10s(s
)1,0s(10)j(G
24
2
422
6
Example 9.19
dB01KB == 100n =ω 5,0=ζ
71,7021 2
nr =ζ−⋅ω=ω 1
1
10,1 for zero of G( j )
Tω = = ω
2
2
110 for single pole of G( j )
Tω = = ω
Note that:
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
r2
1M 1,155 1,25 dB
2 1= = =
ζ⋅ −ζ
The Bode
diagram of
magnitude
and phase
of G(jω)
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Example 9.20
( )( ) ( )1ss1s1,0s
1s101,0
)1ss()10s(s
)1,0s(10)j(G
2
2
+++⋅+⋅=
=+++
+=ω
dB201,0KB −== 1n =ω 5,0=ζ
707,021 2
nr =ζ−⋅ω=ω 1
1
10,1 for zero of G( j )
Tω = = ω
2
2
110 forsinglepole G( j )
Tofω = = ω
Note that:
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
dB25,1155,112
1M
2r ==
ζ−⋅ζ=
The Bode
diagram of
magnitude
and phase
of G(jω)
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
Example 9.21
Note that
dB01KB == 414,12n ==ω 354,0=ζ
224,121 2
nr =ζ−⋅ω=ω 1
1
1 1for zero G( j )
T 2ofω = = ω
2
2
120 for singlepole G( j )
Tofω = = ω
( )
++
+
+⋅=
=+++
+=ω
12
s
2
s1
20
ss
1s2
)2ss()20s(s
2
1s80
)j(G
2
2
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________
dB58,351,112
1M
2r ==
ζ−⋅ζ=
The Bode
diagram of
magnitude
and phase
of G(jω)
Bode Diagram______________________________________________________________________________________________________________________________________________________________________________________