Download BODE PLOTS

1
BODE PLOTS
A most useful means of displaying the amplitude and phase characteristics of network is to plot
the magnitude of the transfer function versus frequency on one curve and the phase
characteristics as a separate curve but the same frequency axis. The curves may be drawn on
semi log paper so that some decades or octaves of frequency can be included. As you know
decade is frequency range where frequency changes 10 times and octave – frequency range
where frequency changes twice. The amplitude characteristics
can be conveniently plotted in terms in decibels. Curves with
R
this type of display are known as Bode plots and find wide
applications in circuit analysis.
Lets consider simple network chain consisting of RC
elements and having filter properties with aim to explain Bode
C
plots advantages.
Low-pass filter RC chain is shown in figure 1. Its voltage
transfer function
1
1
(1)
H ( jϖ ) =
=
Fig.1. Simplest low-pass filter
1 + jϖτ 1 + jϖ / ω b
Here τ = RC - circuit time constant and ω b = 1 / τ - boundary (or cutoff) frequency. Let us
assume frequency as ratios of ϖ / ω b , calculate voltage gain H(ω) and then express the results in
decibels (table 1).
Table 1
HPF: H ( jω ) = 1 /(1 − jω / ω b )
LPF: H ( jω ) = 1 /(1 + jω / ω b )
ω /ωb
0.01
0.04
0.10
0.40
1.0
4.0
10.0
40.0
100.0
H(ω)
1.0
1.0
1.0
0.93
0.707
0.24
0.1
0.03
0.01
φ (ω)
0
-0.6
-2.30
-5.70
-220
-450
-760
-840
-88.50
-89.40
HdB
0
0
0
-0.7
-3.0
-12.3
-20.0
-32.0
-40.0
ωb /ω
100
25
10
2.5
1.0
0.25
0.1
0.025
0.01
H(ω)
φ (ω)
0.01
0.03
0.1
0.24
0.707
0.93
1.0
1.0
1.0
0
89.4
88.50
840
760
450
220
5.70
2.30
0.60
HdB
-40.0
-32.0
-20.0
-12.3
-3.0
-0.7
0
0
0
The results of table 1 are plotted in figure 2 as continuous line. It is possible to see that the
amplitude characteristics for this single time constant low-pass filter could be very well
constructed by drawing two straight lines, which intersect at the frequency ω = ω b . This
frequency is called the corner or break frequency because of the abrupt change in the slope of the
amplitude response. The slope above of the straight-line approximation for values of ω greater
then ωb is seen to be 20 decibels per decade or to be 6 decibels per octave. The greatest error
between the approximate and actual curve occurs at the corner frequency and is equal to 3 dB
away from the point of intersection. At an octave above or below the corner frequency the actual
response octaves. The deviation is only ¼ dB.
2
So, to construct a Bode plot for the network consisting on RC elements it is necessary to do
following.
1. First, calculate time constant τ=RC
H, dB
and locate the corner frequency
ωb=1/τ.
0
2. Draw a straight-line segment along
-20
the zero dB ordinate up to the
-40
corner frequency.
3. From the corner frequency point
0.001 0.01
0.1
1
10
ω/ω
draw the line which falls at a rate
0
ϕ rad
(a)
20 decibels per decade or 6
decibels per octave.
0
4. If grater accuracy is desired draw a
point 3 dB below the corner
frequency, draw the points 1 dB
-2
below straight line segments at
frequencies plus minus 1 octave
ω/ω
0.001 0.01
0.1
1
10
from corner frequency and the
0
(b)
points 0.25 below at frequencies 2
Fig. 2. Bode plots of frequency responses of the low-pass
octaves above and below corner
filter (continuous curves – results of precise calculation, dotted curves – approximation by straight-line segments)
frequency.
5. Connect these points with a smooth
curve.
C
High-pass filter chain consists on RC elements (figure 3). Its
transfer function
R
H ( jϖ ) =
1
1
=
.
1 + 1 / jϖτ 1 − jϖ b / ω
(2)
The results of calculations done in the same manner as for low-pass
filter are presented in the right side of the table 1. Plots of the
Fig.3. Simple high-pass filter
frequency responses are drawn in the figure 4 assuming ratios of
ω/ωb. So as for low-pass filter magnitude
response can be approximated with two
H, dB
straight line segments with intersect point at
0
corner frequency equal to ωb. At this
-20
frequency the output voltage will be 3 dB
-40
below the input voltage (see continuous line)
-60
A band-pass filter can be build on
RC-coupled circuit (figure 5). Suppose we
0.001 0.01
0.1
1
10
ω/ω
b
(a)
ϕ rad
wished to draw the response curve in the
form of a Bode plot. Because C2 and R2 act
2
as load upon C1 lets consider circuit in which
C2 is one - tenth of C1. Therefore, we might
assume a negligible loading effect of C2 and
R2 upon C1.
0
Let’s redraw figure 5a as figure 5b. In
0.001
ω/ω b
0.01
0.1
1
10
this
figure
left dotted block containing R1
(b)
and C1 may be considered as a low-pass filter
having a gain or transfer function of
Fig.4. Bode plots of frequency responses of the high-pass
filter (continuous curves – results of precise calculation, dotted curves – approximation by straight line segments)
3
o
H1 =
V12
o
=
− jX e
1
.
=
R1 − jX C 1 + jω / ω1
(3)
V1
Here ω1 = 1 / R1C1 - boundary (stopband) frequency of the left-hand part of the circuit showed in
the Fig 5.
The output voltage
o
of the first block V 12 is the
input to the middle block
which represents an ideal,
R
V1
V2
V21
C
C1
2
1
unity gain amplifier having
R2
V12
infinite input and zero
output impedance. The
(a)
(b)
isolation provided by this
Fig. 5. Simple coupled RC-circuit acting as pass-band filter
amplifier illustrates the fact
that C2 and R2 are assumed
as a negligible load upon C1. Using this preposition amplifier input and output voltages are equal
and its transfer function (gain) is equal to 1.
The last block in the right-hand side in the given network is high-pass filter and its
transfer function
R1
C2
R1
C
=
R2
1
=
R2 − jX e 2 1 + jω 2 / ω
o
H2 =
V2
o
V12
2
(4)
Here ω 2 = 1 / R2 C 2 - boundary frequency also.
The over-all transfer function is then the product of the individual gains
H ( jω ) = H 1 ( jω ) ⋅ H 12 ( jω ) ⋅ H 2 ( jω ) =
1
1
(1 + jω / ω 1 ) (1 − jω 2 / ω )
.
(5)
In polar form
H ( jω) = H 1 (ω)∠ϕ1 (ω) ⋅ H 2 (ω)∠ϕ 2 (ω) = H 1 (ω) ⋅ H 2 (ω)∠ϕ1 (ω) ⋅ ϕ 2 (ω) .
(6)
We see that the over-all magnitude response is the product of individual gains and overall phase angle - the algebraic sum of individual phase angles. This tells us how to plot the phase
response. The amplitude response in decibels
H dB (ω ) = 20 log[(H 1 (ω )H 2 (ω ))] = 20[log H 1 (ω ) + log H 2 (ω )] = H 1dB (ω ) + H 2 dB (ω ) .
(7)
Here we see that the over-all response in dB is the algebraic sum of the individual decibel
response curves.
Suppose we investigate a specific case in which ω 2 = 100ω1 . So we on our graph paper
we may make a mash at, say, ω1 = 1 and then locate ω 2 two decades above, at point ω 2 = 100 ,
as shown in figure 6. The approximations of magnitudes response can be done using there
straight line segments: with 0-dB line between frequencies and, straight line segment falling
down at a rate of 20 dB per decade to the right side from frequency and falling down to the left
side from frequency ω1 . Actual curve of magnitude response can be plotted using the points 3
dB below the corner points and 1 dB below the points 1 octave below and above corner
frequencies (figure 6).
4
The overall phase response can be
determinates by graphically adding the
individual phase response curves (figure
6).
0
Now lets try to sketch the response
curve of the network shown as the figure
7. At each section barely loads the
-20
preceding section, we can sketch the
approximate response of the individual
-40
section. Since a logarithmic scale is used,
0.1
1
10
100
1000 ω/ω
the overall response is fire sum of the
(a)
ϕ rad
individual repose curves (Fig.7). Then
2
breaks points of each section are the same,
0
sections 1 and 3 have the identical repose
curves as sections 2 and 4.
-2
Sum of the curves 1 and 3 so curve 1-3 is
0.1
1
10
100
1000 ω/ω
(b)
the sum of curves 1 and 3 and slopes 40
Fig. 6. Bode plots of frequency responses of the pass-band
dB per decade. Curve 2-4 is similarly
filter (continuous curves – results of precise calculation, obtained. If a better approximation is
dotted curves – approximation by straight-line segments)
desired, the actual decibel levels
for ω = ω1 , ω = 2ω0 and ω = ω0 / 2 can
be sketched. Therefore, at
R
0.1C
ω = ω0 (ω / ω0 = 1) the over100R
0.001C
all response is down 12 dB.
C
0.01C
At an octave above or below
10R
1000R
ω 0 , the actual response will
be 2 dB below the over - all
1
2
3
4
line segments.
A lag network. Let’s
Fig. 7. Ladder-type network of the complicated pass-band filter
consider circuit shown in the
figure 9. Transfer function
for this circuit
H, dB
H, dB
1,3
H ( jω ) =
2,4
0
-20
2,4
-40
2-4
1,3
=
1-3
-60
R 2 − j / ωC 2
=
R1 + R2 − j / ωC 2
1 + jωC 2 R 2
=
1 + jωC 2 (R1 + R2 )
1 + jω / ω1 H 1 ( jω)
=
=
.
(8)
1 + jω / ω 2 H 2 ( jω)
=
0.01
0.1
ϕ rad
4
2
0
R2 − jX C 2
=
R1 + R2 − jX C 2
(a)
1
10
ω/ω
b
1-2-3-4
1,2,3,4
-2
-4
0.01
0.1
1
10
ω/ω
b
(b)
Fig.8. Bode plots of frequency responses of the ladder-type
pass-band filter (continuous ad dotted curves – results of
precise calculation, - dashed curves – approximation by
straight line segments)
We can see that nominator and
denominator of this function has the same
form as function in low-pas filter transfer
function - so, the transfer function
magnitude in decibels can be write as
follows
5
H dB (ω) = H 2 dB (ω) + H 1dB (ω) = H 2 dB (ω) + [− H 1dB (ω)]
Here H 2 dB (ω) - low pass filter
R
(9)
H, dB
20
1
1
0
R2
-20
C
-40
2
1-2
-60
2
Fig. 9. A lag - network
100 1000
0.1
1
10
ω/ω
b
response in decibels with corner
frequency ω 2 , H 1dB (ω) - lowFig. 10. Bode plots of magnitude responses of a lag pass filter magnitude response
network type filter (continuous and dotted curves –
with corner frequency ω1 . So,
results of precise calculation, - dotted curves –
approximation by straight line segments)
function 1 + jω / ω1 in the
numerator has a break at ω1 , is
flat at zero dB below ω1 , and rises at 20 dB par decade above ω1 .
A Bode plot of these components can be sketched in following manner. Let ω 2 = 10ω1 .
Then magnitude frequency over-all response can be drawn as sum of two simple responses
(figure 10).
A lead - network. Such type of network is shower in the figure 11. Let’s develop the
transfer function for network:
R2
H ( jω) =
R2 + R1 //
1
jωC1
=
R2
R1
R2 +
jωC1 (R1 + 1 / jωC1 )
.
(10)
After transforms such transfer function can be rewritten in following manner:
H ( jω ) =
R2 1 + jωR1C1 1 + jω / ω1
=
R2 + R1 1 + jωRe C1 1 + jω / ω 2
(11)
where
C1
H ( jω ) =
R1
C
2
Fig. 10. A lead-network
R1 R2
.
R1 + R2
(12)
So Bode plot can be build on their plots - straight
flat lines with slope [R1 / (R1 + R2 )] and bode plots of lowpass filter as for lag network. Because Re < R1 , ω 2 will be
above ω1 (in a lag network ω 2 below ω1 ).
6
H, dB
An amplifier equivalent network. Let’s
1-3
analyse network which magnitude response is
shown in figure 12. This frequency response it is
3
20
possible to approximate using three straight line
segments. According to the previous discussion
2
0
about Bode plots it is possible to say that
magnitude response of amplifier consist as three
1
plots: low-pass filter with corner frequency ω 2
-20
and time constant τ 2 and ideal amplifier with
-40
voltage gain K. Equivalent circuit can be build as
0.1
1
10
100
ω/ω
1000
shown
in figure 13. In this circuit RS represents
b
input resistance, CS - total wiring, output and
input capacitance, CC - coupling capacitance and
Fig. 11. Bode plots of magnitude responses of a lead
Ra - input resistance of the following stage.
- network type filter (continuous and dotted curves –
Voltage
gain K represents amplifying coefficient
results of precise calculation, - dotted curves –
in the mid-frequency range. Time constant
approximation by straight line segments)
τ1 = RS C S can be called as low -frequencies time
constant, τ 2 = Ra C a - high frequencies time constant.