5. The curves of Bode
a. System theory
1. Basic knowledge
2. Harmonic analyse of a linear system
b. Filters with OPAMs
In telecommunications filters equiped with OPAMs are generally used. Different
types of filters exist. The characteristics of these filters can be shown
by the curves of Bode. That is what we will do here. But before drawing
the curves we need to calculate H(p). This is something you can't do with
Eureka. We will make one complete example, with calculation of H(p). In
the other examples is H(p) given.
An OPAM is an active electronic component. This means it needs a voltage
to function. An ideal OPAM has an infinite amplification, an infinite input
impedance R_{i} and an output impedance R_{o} of 0. The
amplification is independent of the frequency on the input.
The symbol of an OPAM 

Each time you must try to write H(p) like this (note that it isn't always
possible) :
A filter with OPAM has this form :
We calculate H(p) like this :
Z_{1}(p) and Z_{2}(p) are the Fourier transforms of
impedances of the passive components that are situated at the place 1 or
2. The calculation of the Fourier transform will not be shown here. This
would take us to far away from our objective.
This filter has an isomorphic transmitance H(p) as shown in formula 4.2.1.
But how do we make an electronic circuit that gives this H(p) ? The circuit
is shown here below :
Let us control if it is right what we say. We find for :
When we use formula 4.2.2, we get this result:
Identify this formula with formula 4.2.1 and you'll find :
C_{1}R_{1}=K, T_{1}
C_{1}R_{2}=T_{2}
When we now introduce values for K, T_{1} and T_{2},
we will be able to draw the curves of Bode. Don't forget to make the change
of p = jw = (in Eureka) i*10^x w represents
little omega here The curves must be introduced like this in Eureka.

First we have the curve of the gain A.

This curve was defined as : A = 20 log (abs(H(jw)))
in function of log(w).
Remark :
w is here what our variable x is in Eureka and because the curves
of Bode are given in function of log(w), we have to use 10^x (see definition
base 10 logarithm in part 6).
Like always we have to define our limits of the plane. We use cartesian
analytic coordinates.
Pay attention we you enter values for limits of the abscissa. Don't
forget this axis is a logarithmic axis. So don't use values higher than
20.
20 = log (1e20), this means a 1 with 20 zeros.
Kilo = 1e3 = 1.000 
Mega = 1e6 = 1.000.000 
Giga = 1e9 = 1.000.000.000 
Tera = 1e12 = 1.000.000.000.000 
So if you want to view the curves between 1 Hz and 6.500 Hz, you must
define your abscissa limits from 0 (= log 1) to 4 (= log 10.000). The limits
of the ordinate vary usually from 60 till +60. (Can even be more or less.
There is a way to estimate it.) It depends how strong the amplification
is.
Then it is time to introduce the function with graph>describe.
Introduce this :
20*log(abs((K*i*10^x)/((T_{1}*i*10^x+1)
(T_{2}*i*10^x+1))))
Draw graph [A].
Change the Xaxis to an axis with a logarithmic scale.

Second we have the curve of the phasing.

This curve was defined as : phi = arg(H(jw)) in function of log(w).
The limits for the abscissa stay the same.
The ordinate axis gives us values in degrees. These values can vary
from 180^{o} till +180^{o} and they can only go to 5 asymptotic
values : 180^{o}, 90^{o}, 0^{o}, 90^{o},
180^{o}. The curve will never have this value, except on infinity,
but will always tend to one of these values. (Again there is a way to estimate
the values.)
When the limits are set, introduce the function :
arg((K*i*10^x)/((T_{1}*i*10^x+1)
(T_{2}*i*10^x+1)))*180/pi
Draw graph [B].
Change the Xaxis to an axis with a logarithmic scale.
Of course it is also possible to get both curves on one graph. Then
you should set the limits from the plane from the first time to those for
the phasing graph, because this graph usually has the greatest. See the
recording "bode.rec" made for us by François. Why do we call this a bandpass
filter? When you draw the gain curve, you'll see that only a specific band
of the frequency spectrum will be amplified (relatively seen (<= 0),
higher amplification than the other frequencies). The other frequencies
will also be amplified, but less (mostly a negative amplification) than
our specific band. (see 5.c.2)
c. More examples
(all formulas also available as *.fmu file in directory /bode)
The circuit drawn in fig. 4.3.1 gives this for H(p) :
with : T1 = R1C1
T2 = R2C2
R1 = R2 = 1 Mega Ohm
C1 = 1 micro F
C2 = 0.063 micro F
This gives us a "low pass filter" with an amplification of 24 dB. Try
to draw this curves yourself. Both curves are supplied in the /bode directory.
It are /bode/bode_a1.img and /bode/bode_f1.img.

2. Filter from a telephone central
This filter has next H(p) :
The files are /bode/bode_a2.img and /bode/bode_f2.img.
Here we have a H(p) that can not be transformed to the form like formula
4.2.1.
The general form is this one :
In all the above cases we had dzeta = 1. The curve of the amplitude
(/bode/bode_3a.img) we get here, looks similar to the curves from a spring
(see mechanics). So does the formula.
