Why does the cascade of a lowpass and a highpass Butterworth filter not produce the same

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Zayden - 2021-03-13T13:56:35+00:00
Question: Why does the cascade of a lowpass and a highpass Butterworth filter not produce the same

Why does the cascade of a lowpass and a highpass Butterworth filter not produce the same results as a bandpass Butterworth filter in Signal Processing Toolbox 6.10 (R2008b) ? I created a cascade of lowpass and highpass Butterworth filters and compared the result to a bandpass Butterworth filter - all created using either the Signal Processing Toolbox or the Filter Design Toolbox.   The following is an example of this process: %%8th order bandpass filter [b,a]=butter(4,[0.4 0.6],'bandpass'); Hbp=dfilt.df1(b,a); %%Cascade of 4th order lowpass and highass filters [b1,a1]=butter(4,0.6); % Lowpass [b2,a2]=butter(4,0.4,'high'); % Highpass H1=dfilt.df1(b1,a1); H2=dfilt.df1(b2,a2); Hcas=dfilt.cascade(H1,H2); I expected "Hcas" and "Hbp" to be equivalent filters. However, executing the following command shows that this is not true:     fvtool(Hbp,Hcas)

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    • Expert Answer

    Profile picture of Neeta Dsouza Neeta Dsouza answered . 2025-11-20

    This is actually the expected result and is due to the fact that the Signal Processing Toolbox does not cascade a lowpass and a highpass filter to create a bandpass IIR filter. Instead, it first creates a lowpass prototype and then uses a "Transformation Function" to convert it to a bandpass filter as described by the following help  documentation in Signal Processing Toolbox 6.10 (R2008b):
     
     
    web([docroot '/toolbox/signal/f4-9099.html#f4-9199'])
    For the theoretical background information, please refer to the references mentioned in the following help documentation:
     
     
    web([docroot '/toolbox/signal/f4-19497.html'])

web([docroot '/toolbox/filterdesign/ug/f2-4793.html'])


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