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Transcript

PRESENTATION

CLIL:BODE PLOT

Hendrick Wade Bode, an American engineer that invented the "Bode Plot", a method used to represent F(Jω).

INTRODUCTION

This is the general representation of F(Jω) = Frequency response function
F(Jω)= [K * (Jω)^h* (1+ Jωτ1 )^n * (1- ω^2/ ω0^2+J2δ0 ω/ ω0)] / [ (Jω)^d * (1+ Jωτ2 )^m * (1- ω^2/ ω1^2+J2δ1 ω/ ω1)]
It describes the behaviour of the magnitude and phase of the transfer function of a system as ω varies.

WHAT IS THE BODE PLOT?

They are drawn on a semi-logarithmic graphic. • On the x-axis there are the values of ω in powers of 10.• On the Y axis we have:1) The module expressed in deciBel2) The phase expressed in degrees or radiants.

WHERE ARE THEY DRAWN?

1) II dB II = 20log IKI = 20log I30I = 29,54dB ---> LK=0°2) II dB II = 20log IKI = 20log I0,2I = 13,97dB ---> LK=0°3) II dB II = 20log IKI = 20log I-15I = 23,52dB ---> LK=-180°

Now we have to calculate the modules and the phases:

1) K=30 2) K=0,23) K=-15

Let's consider some random values of K:

some exercises

• When we solve the Module, we have to consider the absolute value of K, because the argument of a logarithm can't be zero or negative.• If the value of K >1, the result of the module will be always positive, while if the value of K is 0<K<1 the result of the module will be always negative.• For the phase, if K > 0, the phase is 0°. If K < 0, the phase is -180°

SOME TIPS TO SOLVE THIS EXERCISES

Math lab and LabVIEW allow us to draw the bode plot and the true F(J ω) graph too

IT'S IMPORTANT TO CONSIDER THAT

Links of useful videos to comprehend the K function:• https://www.youtube.com/watch?v=LQH7oJJ Okmw• https://youtu.be/CYaRer4vsdo?si=Y_6vu7A- mRRzGUzv• https://youtu.be/CSAp9ooQRT0?si=zpvU_ajc AI5ubgln
Group: Bajraku, Cambi and Cecconi

THANK YOU FOR WHATCHING!