© François Le Coat
©Karl Samyn

Last modified :
         July 6 2012

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.
  • 1. OPAM

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 Ri and an output impedance Ro of 0. The amplification is independent of the frequency on the input.
The symbol of an OPAM [OPAM]
  • 2. Determination of H(p)

Each time you must try to write H(p) like this (note that it isn't always possible) :

[formula 4.2.1]

A filter with OPAM has this form :

[OPAM General case]

We calculate H(p) like this :

[formula 4.2.2]

Z1(p) and Z2(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.

  • 3. Band passage filter

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 :

[OPAM case 1]

Let us control if it is right what we say. We find for :

[formula 4.2.3]

[formula 4.2.4]

When we use formula 4.2.2, we get this result:

[formula 4.2.5]

Identify this formula with formula 4.2.1 and you'll find :

     C1R1=K, T1

When we now introduce values for K, T1 and T2, 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 = Tera = 1e12 =

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)/((T1*i*10^x+1) (T2*i*10^x+1))))
Draw graph [A].

Change the X-axis 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 -180o till +180o and they can only go to 5 asymptotic values : -180o, -90o, 0o, 90o, 180o. 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)/((T1*i*10^x+1) (T2*i*10^x+1)))*180/pi
Draw graph [B].

Change the X-axis 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 band-pass 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)
  • 1. Filter using 2 OPAM's

The circuit drawn in fig. 4.3.1 gives this for H(p) :

[figure 4.3.1]

[formula example 1]

     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) :

[formula example 2]

The files are /bode/bode_a2.img and /bode/bode_f2.img.

  • 3. Special case

Here we have a H(p) that can not be transformed to the form like formula 4.2.1.

[formula example 3]

The general form is this one :

[formula 4.3.1]

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.