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Discussion Groups | Comp.DSP | FIR Filter Help

There are 45 messages in this thread.

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FIR Filter Help - 2005-02-13 08:19:00

Hi, I would like to apply a FIR filter to my frequency sample array.
Specifically, I want to use a band-pass filter against my data, but I'm
pretty new to DSP and I'm not sure exactly how to integrate this.

First, I have a complex array of frequency data (real, imaginary).

Second, I've used a FIR design program to come up with the following
information:

-Rectangular window FIR filter
-Filter type: BP
-Passband: 300 - 3000 Hz
-Order: 20
-Transition band: 368 Hz
-Stopband attenuation: 21 dB

-Coefficients:
a[0] =	-0.026050229
a[1] =	-0.038693827
a[2] =	0.008864745
a[3] =	-0.06381684
a[4] =	0.006034263
a[5] =	-0.028610367
a[6] =	-0.06422336
a[7] =	0.06343614
a[8] =	-0.16215464
a[9] =	0.14424528
a[10] =	0.68565154
a[11] =	0.14424528
a[12] =	-0.16215464
a[13] =	0.06343614
a[14] =	-0.06422336
a[15] =	-0.028610367
a[16] =	0.006034263
a[17] =	-0.06381684
a[18] =	0.008864745
a[19] =	-0.038693827
a[20] =	-0.026050229

Now, could someone please describe to me how I can use the above info.
and apply it to my freq. data? Thank you!


Re: FIR Filter Help - David Kirkland - 2005-02-13 09:08:00

Hi, I'm, assuming you know how to do a convolution - if not then you'll 
have to look it up.

Now to filter your complex series you really do the same thing, except 
that the multiplies and adds are now complex - you can think of you 
filter as having a complex componentm but it is just set to zero.

Since you filter has just a real component, it gets even easier. In this 
case you take the real part of the input data and apply your filter and 
it becomes the real part of your output data. You take the imaginary 
part of your input and apply the filter again (as if it were the real 
part - drop the i notation) and you end up with the imaginary part of 
your output. So you can just apply your filter twice to the real and 
imaginary parts of the input data to produce the corresponding part of 
your output data.

Note: This is only possible because your filter has just a real 
component. If it had a complex component you'd have to do the full 
complex adds and multiplies.

Hope that helps.

Cheers,
David

y...@yahoo.com wrote:
> Hi, I would like to apply a FIR filter to my frequency sample array.
> Specifically, I want to use a band-pass filter against my data, but I'm
> pretty new to DSP and I'm not sure exactly how to integrate this.
> 
> First, I have a complex array of frequency data (real, imaginary).
> 
> Second, I've used a FIR design program to come up with the following
> information:
> 
> -Rectangular window FIR filter
> -Filter type: BP
> -Passband: 300 - 3000 Hz
> -Order: 20
> -Transition band: 368 Hz
> -Stopband attenuation: 21 dB
> 
> -Coefficients:
> a[0] =	-0.026050229
> a[1] =	-0.038693827
> a[2] =	0.008864745
> a[3] =	-0.06381684
> a[4] =	0.006034263
> a[5] =	-0.028610367
> a[6] =	-0.06422336
> a[7] =	0.06343614
> a[8] =	-0.16215464
> a[9] =	0.14424528
> a[10] =	0.68565154
> a[11] =	0.14424528
> a[12] =	-0.16215464
> a[13] =	0.06343614
> a[14] =	-0.06422336
> a[15] =	-0.028610367
> a[16] =	0.006034263
> a[17] =	-0.06381684
> a[18] =	0.008864745
> a[19] =	-0.038693827
> a[20] =	-0.026050229
> 
> Now, could someone please describe to me how I can use the above info.
> and apply it to my freq. data? Thank you!
>


Re: FIR Filter Help - Andor - 2005-02-13 15:10:00

If you have frequency domain samples, you don't need the coefficients
of the FIR. You have specified:

"
-Rectangular window FIR filter
-Filter type: BP
-Passband: 300 - 3000 Hz
-Order: 20
-Transition band: 368 Hz
-Stopband attenuation: 21 dB
"

So all you do is attenuate the frequency samples that lie between 300
and 3000 Hz by 21 dB, et voila! Perhaps you might want to convert back
your frequency data into time domain data. I don't know.

The coefficients from the filter design program are only needed if you
want a time domain implementation of the FIR (that is, if your samples
are in time domain - but they are not, as you describe in your post).

Regards,
Andor


Re: FIR Filter Help - 2005-02-13 16:38:00

> If you have frequency domain samples, you don't need the coefficients
> of the FIR. You have specified:

Yes, I can get all my time-domain samples to frequency samples using my
FFT.

> So all you do is attenuate the frequency samples that lie between 300
> and 3000 Hz by 21 dB, et voila! Perhaps you might want to convert
back
> your frequency data into time domain data. I don't know.

This is the essence of my question. I'm not exactly sure how to
attenuate... And yes I will be converting back to time-domain with my
inverse FFT.

> The coefficients from the filter design program are only needed if
you
> want a time domain implementation of the FIR (that is, if your
samples
> are in time domain - but they are not, as you describe in your post).

This is correct, thank you.


Re: FIR Filter Help - 2005-02-13 19:40:00

Can I just apply a FIR filter in the time domain without even doing a
FFT? Is it faster in the time domain?


Re: FIR Filter Help - Jerry Avins - 2005-02-13 22:52:00



Re: FIR Filter Help - 2005-02-14 06:56:00

> Sometimes. It's certainly simpler.

Hmm, too bad I had to get my FFT and iFFT working before I realized
that. :)


Re: FIR Filter Help - 2005-02-14 08:29:00

is there a formula for filtering white noise from a time-domain signal?


Re: FIR Filter Help - Jerry Avins - 2005-02-14 14:36:00



Re: FIR Filter Help - 2005-02-14 20:43:00



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