Hello Everyone, I've recently created a 2-Pole Chebyshev filter and have worked on a Bode plot for it. Initially I fed in sin waves of different frequencies by hand to the filter and measured the attenution to get data points. I know the independent axis of a Bode plot is log f/fc, but I was confused when I got different attenuation of frequencies for the same ratio of f/fc in my testing. For example, inputting a 100Hz wave into a filter with fc = 10Hz (f/fc = 10) the gain of the filter was 0.009128. However, inputting a 1000 Hz wave into a LP filter with fc = 100 Hz, the gain of the filter was 0.008163. I know this difference is not huge, but its giving me worry that my Bode plot may not be accurate for the filter at all cut off frequencies. Should the attenuation of the filter really improve with higher cut off frequencies? If so, what happens when fc gets near the nyquist frequency? Thanks for any help Andrew

# Bode Plots of Digital Filters

Started by ●August 3, 2010

Reply by ●August 3, 20102010-08-03

AndrewDSPdev <aritchie@n_o_s_p_a_m.vernier.com> wrote:> For example, inputting a 100Hz wave into a filter with fc = > 10Hz (f/fc = 10) the gain of the filter was 0.009128. However, > inputting a 1000 Hz wave into a LP filter with fc = 100 Hz, > the gain of the filter was 0.008163. I know this difference > is not huge, but its giving me worry that my Bode plot may > not be accurate for the filter at all cut off frequencies. > Should the attenuation of the filter really improve with higher > cut off frequencies?This is not immediately alarming and is probably just due to fc moving closer to the sample rate. As fc approaches the sample rate you get warping and your plot will no longer resemble a classical Bode plot. Steve

Reply by ●August 3, 20102010-08-03

On Aug 3, 8:54=A0pm, "AndrewDSPdev" <aritchie@n_o_s_p_a_m.vernier.com> wrote:> Hello Everyone, > > I've recently created a 2-Pole Chebyshev filter and have worked on a Bode > plot for it. Initially I fed in sin waves of different frequencies by han=d> to the filter and measured the attenution to get data points. I know the > independent axis of a Bode plot is log f/fc, but I was confused when I go=t> different attenuation of frequencies for the same ratio of f/fc in my > testing. > =A0 =A0 =A0For example, inputting a 100Hz wave into a filter with fc =3D =10Hz (f/fc> =3D 10) the gain of the filter was 0.009128. However, inputting a 1000 Hz > wave into a LP filter with fc =3D 100 Hz, the gain of the filter was > 0.008163. I know this difference is not huge, but its giving me worry tha=t> my Bode plot may not be accurate for the filter at all cut off frequencie=s.> Should the attenuation of the filter really improve with higher cut off > frequencies? If so, what happens when fc gets near the nyquist frequency? > > Thanks for any help > AndrewYou didn't say what method you used for transforming the Chebyshev filter to the digital domain, but I'll assume you used the bilinear transform (the most common method). If you did it by hand, you'll remember the "prewarping" step that you used to map the frequency response of the analog filter (measured along the jw-axis in the s- plane) to a digital approximation (measured around the unit circle in the z-plane). Since you're mapping an infinitely-long axis in the analog domain to a circle of finite circumference in the digital domain, the frequency responses are not exactly the same. More precisely, you'll find that as you approach the Nyquist frequency, the response of your digital Chebyshev filter will vary noticeably from the response of your prototype filter. So, while you're right, for the analog filter you care about f/fc only, for the digital implementation you also need to consider how close the frequency of interest f is to half of the sampling frequency. Jason