Filters: Differences between fvtool and bode

Hi,

I am willing to make a complex filter by multiplying the transfer function of several elementary filters (Butterworth for example).

While doing this, I found several questions about what is the differences in the plot of the functions bode and the one performed in fvtool .

Maybe, I'm not understanding the way fvtool express the filter. For example, the transfer function of a Butterworth 2nd order is 1/(1 + 1.41s + s^2), and bode plots its response correctly.

In the other hand, fvtool calculates a different expression for a Butterworth 2nd order, and the figures drawn by it has nothing to do from those done by bode - Even when I intgroduce the same transfer funcion!

Can anyone tell me what am I missing?

Thank you in advance

Answer 1

[Honglei Chen]

I'm not sure how you are using bode, but from your transfer function, you are working with analog filter while fvtool is for digital filters, so you may not be comparing apple to apple.

[Miguel]

It was that actually :)

In order to have the same answer with bode , I had to trace the discrete transfer function

bode( tf(Num,Den,Ts) )

[zhaohui luo]

Really works!Thanks