Signals&Circuits
BODE PLOTS
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 with 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 of decibels. Curves with this type of display are known as Bode
plots and find wide applications in circuit analysis.
Consider simple network chain consisting of RC eleR
ments and having filter properties with aim to explain Bode
plots advantages.
Low-pass filter RC chain is shown in figure 1. Its voltage
transfer function
C
1
1
1
(1)
H ( jϖ ) =
=
=
1 + jϖRC 1 + jϖτ 1 + jϖ / ω b
Fig.1. Simplest low-pass filter
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
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
HPF: H ( jω ) = 1 /(1 − jω / ω b )
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 will be 1db away from the straight-line approximation.. At plus or minus 2
octaves, the deviation is only ¼ db.
1
Signals&Circuits
BODE PLOTS
So, to construct a Bode plot for the network consisting of RC elements it is necessary to do
the following.
H, dB
1. First, calculate time constant
τ=RC and locate the corner fre0
quency ωb=1/τ.
-20
2. Draw a straight-line segment
-40
along the zero dB ordinate up to
the corner frequency.
0.001 0.01
0.1
1
10
ω/ω
3.
From
the corner frequency point
0
(a)
ϕ rad
draw the line which falls at a rate
of 20 decibels per decade or 6
0
decibels per octave.
4. If greater accuracy is desired draw
a point 3 dB below the corner frequency, draw the points 1 dB be-2
low straight line segments at fre0.001 0.01
0.1
1
10
ω/ω
quencies plus minus 1 octave from
0
(b)
corner frequency and the points
Fig. 2. Bode plots of frequency responses of the low-pass
0.25 below at frequencies 2 ocfilter (continuous curves – results of precise calculation, dotted curves – approximation by straight-line segments)
taves above and below corner frequency.
5. Connect these points with a smooth curve.
C
High-pass filter chain consists of RC elements (figure 3). Its
transfer function
R
H ( jϖ ) =
1
1
1
=
=
.
1 + 1 / jϖRC 1 + 1 / jϖτ 1 − jϖ b / ω
(2)
Fig.3. Simple high-pass filter
The results of calculations done in the same manner as for lowpass filter are presented in the right side of the table 1. Plots of
the frequency responses are drawn in the
figure 4 assuming ratios of ω/ωb. AS for
0
low-pass filter magnitude response can
-20
be approximated with two straight line
-40
segments with intersect point at corner
-60
frequency equal to ωb. At this frequency
0.001 0.01
0.1
1
10
ω/ωb the output voltage will be 3 dB below the
(a)
input voltage (see continuous line)
ϕ rad
A band-pass filter can be build
2
on RC-coupled circuit (figure 5). Suppose we wished to sketch the response
curve in the form of a Bode plot. Since
C2 and R2 act as load upon C1 lets con0
sider circuit in which C2 is one - tenth of
ω/ω b C1. Therefore, we might assume a negli0.001
0.01
0.1
1
10
(b)
gible loading effect of C2 and R2 upon
C1 .
Fig.4. Bode plots of frequency responses of the high-pass
Let’s redraw figure 5a as figure
filter (continuous curves – results of precise calculation, 5b. In this figure dotted block containing
dotted curves – approximation by straight line segments)
R1 and C1 may be considered as a lowH, dB
2
Signals&Circuits
BODE PLOTS
pass filter having a gain or transfer function of
H1 =
V 12
V1
R1
C2
1
− jX C
1
.
=
R1 − jX C 1 + jω / ω 1
C
R1
R
C
=
2
V1
C1
V12
V21
(3)
2
R2
V2
(a)
(b)
Fig. 5. Simple coupled RC-circuit acting as pass-band filter
Here ω1 = 1 / R1C1 boundary (stop-band)
frequency of the lefthand part of the circuit showed in the
Fig 5.
The
output
voltage of the first
block V 12 is the input
to the middle block which represents an ideal, unity gain amplifier having infinite input and
zero output impedance. The isolation provided by this 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
H2 =
V2
V 12
=
R2
1
=
R 2 − jX C 2 1 − jω 2 / ω
(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
over-all 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 . On our graph paper we
may may make a mark 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).
3
Signals&Circuits
BODE PLOTS
H, dB
The overall phase response can
be determinated graphically adding
the individual phase response curves
0
(figure 6).
Now let’s try to sketch the response curve of the network shown as
-20
the figure 7. As each section barely
loads the preceding section, we can
-40
sketch the approximate response of
0.1
1
10
100
1000 ω/ω
each individual section. Since a loga(a)
ϕ rad
rithmic scale is used, the overall re2
sponse is fire sum of the individual
0
repose curves (Fig.7). Then break
points of each section are the same,
-2
sections 1 and 3 have identical repose
0.1
1
10
100
1000 ω/ω
(b)
curves as do sections 2 and 4.
Fig. 6. Bode plots of frequency responses of the pass-band Curve 1-3 is the sum of curves 1 and 3
filter (continuous curves – results of precise calculation, - and slopes 40 dB per decade. Curve 2dotted curves – approximation by straight-line segments)
4 is similarly obtained. If a better approximation is desired, the actual decibel levels for ω = ω1 , ω = 2ω 0 and ω = ω0 / 2 can be
sketched. Therefore, at
R
0.1C
ω = ω0 (ω / ω0 = 1) the
100R
0.001C
over-all response is
down 12 dB. At an ocC
0.01C
10R
1000R
tave above or below ω0 ,
the actual response will
be 2 dB below the over 1
2
3
4
all line segments.
Fig. 7. Ladder-type network of the complicated pass-band filter
A lag network.
Let’s consider circuit
shown in the figure 9.
H, dB
Transfer function for this circuit
1,3
2,4
0
-20
2,4
-40
2-4
H ( jω ) =
1,3
1-3
=
-60
0.01
0.1
ϕ rad
4
2
0
(a)
1
1-2-3-4
10
ω/ω
R2 − jX C 2
=
R1 + R2 − jX C 2
R 2 − j / ωC 2
=
R1 + R2 − j / ωC 2
1 + jω C 2 R 2
=
1 + jωC 2 (R1 + R2 )
1 + jω / ω1 H 1 ( jω)
.
(8)
=
=
1 + jω / ω 2 H 2 ( jω )
=
b
1,2,3,4
-2
We can see that nominator and denominator of this function has the
0.01
0.1
1
10
ω/ω
same form as function in low-pas filb
(b)
ter transfer function - so, the transfer
Fig.8. Bode plots of frequency responses of the ladder-type
pass-band filter (continuous ad dotted curves – results of function magnitude in decibels can be
precise calculation, - dashed curves – approximation by written as follows
-4
straight line segments)
4
Signals&Circuits
BODE PLOTS
H dB (ω) = H 2 dB (ω) + H 1dB (ω) = H 2 dB (ω) + [− H 1dB (ω)]
R
1
(9)
H, dB
R2
20
C
0
1
2
-20
1-2
-40
Fig. 9. A lag - network
2
Here H 2 dB (ω) - low pass filter response in
decibels with corner frequency ω 2 ,
100
1000
0.1
1
10
ω/ω
b
H 1dB (ω) - low-pass filter magnitude response with corner frequency ω 1 . So,
Fig. 10. Bode plots of magnitude responses of a lag function 1 + jω / ω1 in the numerator has network type filter (continuous and dotted curves – rea break at ω1 , is flat at zero dB below ω1 , sults of precise calculation, - dotted curves – approximation by straight line segments)
and rises at 20 dB per decade above ω1 .
A Bode plot of these components can be sketched in the following manner. Suppose
ω 2 = 10ω1 . Then magnitude frequency over-all response can be drawn as a sum of two simple responses (figure 10).
A lead - network. Such type of network is shown in the figure 10. Let’s develop the
transfer function for network:
-60
H ( jω ) =
R2
1
R2 + R1 //
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 / ω 1
= H 0 ( jω )
R2 + R1 1 + jωRe C1
1 + jω / ω 2
(11)
where
H 0 ( jω ) =
R1
C
R2
;
R1 + R2
Re =
R1 R2
R1 + R2
(12)
1
R2
Fig. 10. A lead-network
So Bode plot can be built on their plots - straight flat
line with slope [R1 / (R1 + R2 )] and two Bode plots of lowpass filters.. Because Re < R1 , ω 2 will be above ω1 . The
Bode plot is sketched in the Fig.11
An amplifier equivalent network. Let’s analyse network which magnitude response is shown in figure 12. This frequency response it is possible
to approximate using three straight line segments. According to the previous discussion about
5
Signals&Circuits
BODE PLOTS
Bode plots it is possible to say that magnitude
response of amplifier consist as three plots:
high-pass filter with corner frequency ω 1 and
1-3
3
time constant τ 1 , low-pass filter with corner
20
frequency ω 2 and time constant τ 2 , and ideal
amplifier with voltage gain K. Equivalent cir2
0
cuit can be build as shown in figure 13. In
1
this circuit RS represents input resistance, CS -20
total wiring, output and input capacitance, CC
- coupling capacitance and Ra - input resis-40
tance of the following stage. Voltage gain K
0.1
1
10
100
ω/ω represents amplifying coefficient in the mid1000
b
frequency range. Time constant τ1 = RS C S
Fig. 11. Bode plots of magnitude responses of a lead can be called as low -frequencies time con- network type filter (continuous and dotted curves – stant, τ 2 = Ra C a - high frequencies time conresults of precise calculation, - dotted curves – apstant.
proximation by straight line segments)
H, dB
6