FIR Filter Help

>Design a single band-pass filter. Or rather, let a program design it
for you.

lol, well said.

Jerry Avins wrote:
> 
> I like the Lyons book. You'll find it and others cited on
> http://www.dspguru.com/. 

THat site is chock-full-o good DSP stuff.  In particular, it has a nice 
FAQ on FIRs, including working code.

http://www.dspguru.com/info/faqs/firfaq.htm


-- 
Jim Thomas            Principal Applications Engineer  Bittware, Inc
jthomas@bittware.com  http://www.bittware.com    (603) 226-0404 x536
Quidquid latine dictum sit, altum sonatur.
Whatever is said in Latin sounds profound.

samseed wrote:
> Coefficients:
> 
> a[0] =	-0.047739632
> a[1] =	-0.017229123
> a[2] =	1.0119492E-8
> a[3] =	-0.064445764
> a[4] =	-0.087204635
> a[5] =	-0.011658093
> a[6] =	-0.021648642
> a[7] =	-0.15716003
> a[8] =	-0.11321392
> a[9] =	0.221274
> a[10] =	0.43638006
> a[11] =	0.221274
> a[12] =	-0.11321392
> a[13] =	-0.15716003
> a[14] =	-0.021648642
> a[15] =	-0.011658093
> a[16] =	-0.087204635
> a[17] =	-0.064445764
> a[18] =	1.0119492E-8
> a[19] =	-0.017229123
> a[20] =	-0.047739632
> 
> 
> So, I have my time-domain samples patiently sitting in an array and I'm
> currently fumbling with the proper formulas! :O
> 

You have 21 coefficients, but the code snippet you posted uses only 
three of them.  You need to use them all.

-- 
Jim Thomas            Principal Applications Engineer  Bittware, Inc
jthomas@bittware.com  http://www.bittware.com    (603) 226-0404 x536
Quidquid latine dictum sit, altum sonatur.
Whatever is said in Latin sounds profound.

Is the following process correct?

//20 tap low-pass filter, Rectangular, coefficiants--->2000Hz, -10dB
double a0 =-0.028991204;
double a1 =0.008665845;
double a2 =0.033431053;
double a3 =-0.022146255;
double a4 =-0.037133045;
double a5 =0.043838896;
double a6 =0.039916355;
double a7 =-0.08839386;
double a8 =-0.041643664;
double a9 =0.28917018;
double a10 =0.50124127;
double a11 =0.28917018;
double a12 =-0.041643664;
double a13 =-0.08839386;
double a14 =0.039916355;
double a15 =0.043838896;
double a16 =-0.037133045;
double a17 =-0.022146255;
double a18 =0.033431053;
double a19 =0.008665845;
double a20 =-0.028991204;

for (int t=20; t < audioShortData.length; t++) {
audioShortData[t] = (short)((
a0*audioShortData[t]+
a1*audioShortData[t-1]+
a2*audioShortData[t-2]+
a3*audioShortData[t-3]+
a4*audioShortData[t-4]+
a5*audioShortData[t-5]+
a6*audioShortData[t-6]+
a7*audioShortData[t-7]+
a8*audioShortData[t-8]+
a9*audioShortData[t-9]+
a10*audioShortData[t-10]+
a11*audioShortData[t-11]+
a12*audioShortData[t-12]+
a13*audioShortData[t-13]+
a14*audioShortData[t-14]+
a15*audioShortData[t-15]+
a16*audioShortData[t-16]+
a17*audioShortData[t-17]+
a18*audioShortData[t-18]+
a19*audioShortData[t-19]+
a20*audioShortData[t-20]
}

Looks good to me!  I still would use a loop for the convolution step so it is
easier to change number of taps, but that's just a programming preference.
Based on your filter coefs and stated spec (2000Hz, -10dB), I'm guessing your
audio sample rate is something around 8kHz?

"samseed" <agent3492003@yahoo.com> wrote in message
news:1108564946.710984.197470@l41g2000cwc.googlegroups.com...
> Is the following process correct?
>
> //20 tap low-pass filter, Rectangular, coefficiants--->2000Hz, -10dB
> double a0 =-0.028991204;
> double a1 =0.008665845;
> double a2 =0.033431053;
> double a3 =-0.022146255;
> double a4 =-0.037133045;
> double a5 =0.043838896;
> double a6 =0.039916355;
> double a7 =-0.08839386;
> double a8 =-0.041643664;
> double a9 =0.28917018;
> double a10 =0.50124127;
> double a11 =0.28917018;
> double a12 =-0.041643664;
> double a13 =-0.08839386;
> double a14 =0.039916355;
> double a15 =0.043838896;
> double a16 =-0.037133045;
> double a17 =-0.022146255;
> double a18 =0.033431053;
> double a19 =0.008665845;
> double a20 =-0.028991204;
>
> for (int t=20; t < audioShortData.length; t++) {
> audioShortData[t] = (short)((
> a0*audioShortData[t]+
> a1*audioShortData[t-1]+
> a2*audioShortData[t-2]+
> a3*audioShortData[t-3]+
> a4*audioShortData[t-4]+
> a5*audioShortData[t-5]+
> a6*audioShortData[t-6]+
> a7*audioShortData[t-7]+
> a8*audioShortData[t-8]+
> a9*audioShortData[t-9]+
> a10*audioShortData[t-10]+
> a11*audioShortData[t-11]+
> a12*audioShortData[t-12]+
> a13*audioShortData[t-13]+
> a14*audioShortData[t-14]+
> a15*audioShortData[t-15]+
> a16*audioShortData[t-16]+
> a17*audioShortData[t-17]+
> a18*audioShortData[t-18]+
> a19*audioShortData[t-19]+
> a20*audioShortData[t-20]
> }
>

yes.

what is the correct formula for a high-pass?

"samseed" <agent3492003@yahoo.com> wrote in message
news:1108581210.574553.72530@z14g2000cwz.googlegroups.com...
> yes.

Just a tip, include some of the message you are replying too so we can see the
context--what are you saying "yes" to?

> what is the correct formula for a high-pass?

Again, use your filter design program---there is no easy formula! I think there
are some ways to transform a LP into a HP, but I would still use the program.

my bad.

>I'm guessing your audio sample rate is something around 8kHz

correct. However, 44kHz is the only rate that has valid filter output,
but that may just be a bug in my app.

>Again, use your filter design program---there is no easy formula! I
think there
are some ways to transform a LP into a HP, but I would still use the
program.

yes, I'm using the design program. What I meant to ask was, is the
high-pass a multiply/add like a low-pass, or do you multiply/ subtract
like so:

audioShortData[t] = (short)(( a0*audioShortData[t] -
a1*audioShortData[t-1] - a2*audioShortData[t-2] - ...)

samseed wrote:
> my bad.
> 
> 
>>I'm guessing your audio sample rate is something around 8kHz
> 
> 
> correct. However, 44kHz is the only rate that has valid filter output,
> but that may just be a bug in my app.
> 
> 
>>Again, use your filter design program---there is no easy formula! I
> 
> think there
> are some ways to transform a LP into a HP, but I would still use the
> program.
> 
> yes, I'm using the design program. What I meant to ask was, is the
> high-pass a multiply/add like a low-pass, or do you multiply/ subtract
> like so:
> 
> audioShortData[t] = (short)(( a0*audioShortData[t] -
> a1*audioShortData[t-1] - a2*audioShortData[t-2] - ...)

Multiply and add is the usual. but id you always subtract, that just 
inverts the output signal. The operation is called MAC, for "Multiply 
And Accumulate".

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

"samseed" <agent3492003@yahoo.com> wrote in message
news:1108595391.559407.170670@z14g2000cwz.googlegroups.com...
> my bad.
>
> >I'm guessing your audio sample rate is something around 8kHz
>
> correct. However, 44kHz is the only rate that has valid filter output,
> but that may just be a bug in my app.

Probably.

> >Again, use your filter design program---there is no easy formula! I
> think there
> are some ways to transform a LP into a HP, but I would still use the
> program.
>
> yes, I'm using the design program. What I meant to ask was, is the
> high-pass a multiply/add like a low-pass, or do you multiply/ subtract
> like so:
>
> audioShortData[t] = (short)(( a0*audioShortData[t] -
> a1*audioShortData[t-1] - a2*audioShortData[t-2] - ...)

OK, I understand you.  The convolution formula is always the same regardless of
what filter type you are using--low-pass, high-pass, band-pass, something
totally arbitrary, etc..  So your filtering function can always be the same, but
depending on what filter coefficients you use, it may create a LP, HP, etc..