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)

Answer 1

[MathWorks Support Team]

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'])