analise sinais ppt09e (bode diagram)webx.ubi.pt/~felippe/texts2/analise_sinais_ppt09e.pdf · the...

“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______________________________________________________________________________________________________________________________________________________________________________________