Recent content by cathode-ray

Hi Lets consider first a small example with 3 students. Initially all the lockers are closed: locker 1 2 3 status 0 0 0 After the first student goes all the lockers are opened cause all are multiple of 1: locker 1 2 3 status 1 1 1 Then comes the second...

Im glad i could help :) I saw your table and you made a mistake in the calculation of the argument. Given a complex number the argument is given by: arg(a+jb)=arctan(b/a) and given two complex numbers z1,z2 you have the following properties: arg(z1*z2)=arg(z1)+arg(z2)...

The only mistake you made in 20*LOG10(SQRT(1+w/2127.659)) is that you not squared the imaginary part. I wrote 20*LOG10(SQRT(1+(w*47E3*10E-9)^2) which is equal to 20*LOG10(SQRT(1+(w/2127.659)^2). I used the multiplication because 2127.659 is an approximated value and 47E3*10E-9 is the exact...

I saw what you posted in the other thread and it is what i have.

Yup, thinking that way, when the real part is equal to the imaginary you have either a pole or a zero. I think you posted here the wrong attachment cause that one is the initial one...

Oh but is not 20*LOG10(SQRT(1+w/2127.659)) its 20*LOG10(SQRT(1+(w*47E3*10E-9)^2).

No I was asking if you are calculating 20log(|1+jwRC|) as 20log(sqrt(1+(wRC)^2))?

Nope I am getting the blue one, in the real measure. I am seeing your other post at the same time too and the last plot you posted its what is giving to me. How are you calculating |1+jwRC| ?

Yup the error in the static gain was the resistance R2. Now the amplitude plot is equal to the one you measured. However the calculations you made in that table for the numerator and the denominator of the transfer function, are still giving me different results. Did you used w=1,10,..[rad/sec]...

It gave me the same transfer function you got with the only difference that A=\frac{R_{2}}{R+R_{2}}. However as R=R_{2}=47[kΩ] this doesn't make any difference. So it seems to me that according to the data and circuit you gave that the transfer function is right Regarding to the plots you...

You can split your function in two new functions as you did. However you don't get any particular advantage by doing that, specially in this case where the transfer function is relatively simple. Essentially what you need to have in mind when doing the bode plot by hand is that: -Either...

Hi, Can you be more specific about your doubt? What got you stuck in the plotting of the Bode plot?

I finally found the answer and the mistake i was making :D. To design the filter i was considering initially a second order low pass Chebyshev filter, with the specified characteristics, and then i applied the frequency transformation to make it a band pass. However to make the transformation...

I made a search in the web before coming here and also found that document, but it didnt helped :S .

Homework Statement Get the transfer function of a second order Chebyshev passband filter, with central frequency f0 = 1 [kHz], lower cutoff frequency fc=670 [Hz], 3dB ripple in the pass-band and 30dB of gain in the central frequency. Homework Equations Maximum allowed variation in...