FIR filter bandpass filter from lowpass filter prototype

I want to make bandpass filter from lowpass filter prototype.

first i made lowpass filter and frequency translated using 
cosine function and i saw the freq. response of translated coefficient.
but it differs from lowpass filter response.

b  = firceqrip(368, 4.3/(122.88/2), [0.1, 0.0018], 'slope', 0);
fvtool(b);

n=0:length(b)-1;
cos_b = b.*(cos(((3*pi)/8)*n));
fvtool(cos_b)

> lowpass filter : attenuation is 55dB
> bandpass fitler : attenuation is about 50dB

and filter shape is different, especially filter edge part.
am i wrong in implement bandpass filter from lowpass protytype ?

if you frequency translate by a complex frequency instead of cos then you
get same response shifted. Otherwise there will be loss of 6 dB due to
mirroring of response.

Kaz

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Posted through http://www.DSPRelated.com

You are not wrong. What you see is the interaction of the negative
frequency components with the positive frequency components. That is
easier to realize if you split your cos multiplication into two complex
exponential multiplications instead (Euler's identity).

cos(w)=exp(jw)/2+exp(-jw)/2, w=2*pi*f

Now you can see that you are going to get a positive (w) and a negative
(-w) frequency component when multiplying your filter coefficients with
your cos. Both components have half the amplitude of the original filter
coefficients due to the "/2" in the equation above. That's why you see the
6dB loss.

Let's visualize it in MATLAB.

neg_b=b.*exp(-1j*3*pi*n/8)/2;
pos_b=b.*exp(1j*3*pi*n/8)/2;
p=fvtool(neg_b)
set(p, 'FrequencyRange', '[-Fs/2, Fs/2)')
p=fvtool(pos_b)
set(p, 'FrequencyRange', '[-Fs/2, Fs/2)')
p=fvtool(neg_b, 1, pos_b, 1)
set(p, 'FrequencyRange', '[-Fs/2, Fs/2)')
p=fvtool(neg_b+pos_b, 1, cos_b, 1)
set(p, 'FrequencyRange', '[-Fs/2, Fs/2)')

You will get four tabs in fvtool. The first tab show the negative
frequency component and the second one show the positive frequency
component. The third tab is the really interesting one. There you can see
a plot of both the negative and positive frequency components together.
You should clearly be able to see the interference patterns when the two
frequency spectrums are plotted together. In the fourth tab the negative
and positive components are added together to form your original frequency
response. Overlayed is your cos_b to verify that they are exactly the
same.

Hope that helps!

/ Tobias
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On Wed, 19 Aug 2015 18:41:07 -0700 (PDT), hyuckin kwon
<hikwon39@gmail.com> wrote:

>I want to make bandpass filter from lowpass filter prototype.
>
>first i made lowpass filter and frequency translated using 
>cosine function and i saw the freq. response of translated coefficient.
>but it differs from lowpass filter response.
>
>b  = firceqrip(368, 4.3/(122.88/2), [0.1, 0.0018], 'slope', 0);
>fvtool(b);
>
>n=0:length(b)-1;
>cos_b = b.*(cos(((3*pi)/8)*n));
>fvtool(cos_b)
>
>> lowpass filter : attenuation is 55dB
>> bandpass fitler : attenuation is about 50dB
>
>and filter shape is different, especially filter edge part.
>am i wrong in implement bandpass filter from lowpass protytype ?

Hi,
  The two freq magnitude plots sensible to me.  They 
both have VERY similar stopband attenuation.
You posted the words:

    "> bandpass fitler : attenuation is about 50dB"

Where did that 50dB value come from?

Keep this in mind: When you multiply your lowpass 
filter's coeffs by a cosine sequence the frequency 
response of your resultant bandpass filter will be 
the convolution of the lowpass filter's spectrum and 
the spectrum of a finite-duration cosine sequence.  

And the spectrum 
of a finite-duration cosine sequence is *NOT* a pair 
of impulses in the frequency domain.  The 
positive-frequency spectrum of a finite-duration 
cosine sequence has a sin(x)/x spectrum.
This means the bandpass filter's freq response will 
have a slightly different shape than the lowpass 
filter's frequency response shape.  

Try your code with 18 taps, instead of 368 taps, 
to see what I'm referring to here.

[-Rick-]
 

>Hi,
>  The two freq magnitude plots sensible to me.  They 
>both have VERY similar stopband attenuation.
>You posted the words:
>
>    "> bandpass fitler : attenuation is about 50dB"
>

Rick,

what tools are you using and why use 18 taps. Your tool is definitely
misleading you.

Kaz
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