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#### Fixed Point IIR implementation

```Hi pals,

I would like to implement a IIR Biquad filter using the fixed point
arithmetics...
Hence to reduced the intermediate states I plan to use the following trick:
s(k) = x(k) -a1*s(k-1) -a2*s(k-2)
y(n) = b0*s(k) + b1*s(k-1) + b2*s(k-2)

The I can compute each y, saving only 2 states (s(k-1) a�nd s(k-2))...

BUT ... in order to scale my coefficients or input, I need to know what
are the boundaries of the s(k) serie... in order to find a proper Fixed
Point format for my coefficients..
So my questions are :

* for which conditions, s(k) is bounded
* If the s(k) is bounded, what are the boundaries?

Thanks in advance for your help (and sorry for my poor englsih)

Fred.
```
 0
fred_nach (6)
11/30/2005 11:35:55 AM
comp.dsp 20333 articles. 1 followers. allnor (8510) is leader.

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```Fred Nach wrote:
> Hi pals,
>
> I would like to implement a IIR Biquad filter using the fixed point
> arithmetics...
> Hence to reduced the intermediate states I plan to use the following trick:
> s(k) = x(k) -a1*s(k-1) -a2*s(k-2)
> y(n) = b0*s(k) + b1*s(k-1) + b2*s(k-2)
>
> The I can compute each y, saving only 2 states (s(k-1) a�nd s(k-2))...
>
> BUT ... in order to scale my coefficients or input, I need to know what
> are the boundaries of the s(k) serie... in order to find a proper Fixed
> Point format for my coefficients..
> So my questions are :
>
> * for which conditions, s(k) is bounded
> * If the s(k) is bounded, what are the boundaries?
>
> Thanks in advance for your help (and sorry for my poor englsih)

Your English is good enough, I think. The s[k] are data you supply. They
can't be larger or smaller than the representation allows, but they may
have smaller limits.

I see no feedback in the formulas you give. Is that correct?

Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
```
 0
jya (12871)
11/30/2005 4:17:06 PM
```Fred Nach wrote:
>
> I would like to implement a IIR Biquad filter using the fixed point
> arithmetics...
> Hence to reduced the intermediate states I plan to use the following tric=
k:
> s(k) =3D x(k) -a1*s(k-1) -a2*s(k-2)
> y(n) =3D b0*s(k) + b1*s(k-1) + b2*s(k-2)
>
> The I can compute each y, saving only 2 states (s(k-1) a=E9nd s(k-2))...

this is the Direct 2 Form (sometimes called the "Direct II Canonical
Form").  unrecommeded for floating-point (if the Q is quite high, you
end up subtracting numbers very close to each other and  losing
precision) and *highly* unrecommended for fixed-point (saturation

try the Direct 1 Form:

y[n] =3D b0*x[n] + b1*x[n-1] + b2*x[n-2] - a1*y[n-1] - a2*y[n-2]

every product on the right side of the =3D sign is a double precision
fixed point and all of those products should be added together in
double precision.  this is trivial in the Mot 56K and the ADI SHArC and
maybe in the TI fixed-pointers.

> BUT ... in order to scale my coefficients or input, I need to know what
> are the boundaries of the s(k) serie... in order to find a proper Fixed
> Point format for my coefficients..

your coefs are defined strictly in terms of the frequency response you
want.  the gross range of a1 is -2 to +2 and a2 is -1 to +1 for any
stable biquad filter.  the b0, b1, b2 coefs can be nearly anything but
you will have to choose a range and possibly do some scaling using the
arithmetic shift operation on the result before saving to y[n].

> So my questions are :
>
> * for which conditions, s(k) is bounded
> * If the s(k) is bounded, what are the boundaries?

that is a purely artificial construct.  are your fixed-point signal
values considered 16 bit signed integers?  then it's  -32768 <=3D x[n] <
+32768 .  are they normalized?  then it's -1 <=3D x[n]< +1.  their range
is whatever you want them to be.

> Thanks in advance for your help (and sorry for my poor englsih)

as Jerry said, your English is fine.

r b-j

```
 0
rbj (4086)
11/30/2005 5:57:01 PM
```> I see no feedback in the formulas you give. Is that correct?

The aX coeffecients should be the feedback terms and the bX
coefficients should be the feed forward terms (from drawing a direct
form II simulation diagram).  The s(k) terms are the inputs to the
delay buffers, which will be a function of the input, feed forward and
feedback terms.

I see the question he is asking, how to determine if these terms
overflow the fixed point number format in use, and how to determine a
proper scaling value to keep it from happening.  Unfortunately, I don't
know how to answer this question, but in the project I am presently
working on I have encountered the same problem, so I am hoping someone
has a good answer.  I got around the problem (? maybe ?) by using a
direct form I structure.

I have seen Matlab scripts that run an impulse response till it
stabilizes and then sum up the impulse responses and use the maximum
value as a scale factor, though I don't know if this is a correct
solution.

```
 0
no_spam_me2 (215)
11/30/2005 6:01:31 PM
```Noway2 wrote:
>>I see no feedback in the formulas you give. Is that correct?
>
>
> The aX coeffecients should be the feedback terms and the bX
> coefficients should be the feed forward terms (from drawing a direct
> form II simulation diagram).  The s(k) terms are the inputs to the
> delay buffers, which will be a function of the input, feed forward and
> feedback terms.
>
> I see the question he is asking, how to determine if these terms
> overflow the fixed point number format in use, and how to determine a
> proper scaling value to keep it from happening.  Unfortunately, I don't
> know how to answer this question, but in the project I am presently
> working on I have encountered the same problem, so I am hoping someone
> has a good answer.  I got around the problem (? maybe ?) by using a
> direct form I structure.
>
> I have seen Matlab scripts that run an impulse response till it
> stabilizes and then sum up the impulse responses and use the maximum
> value as a scale factor, though I don't know if this is a correct
> solution.

I expect that the b's are the denominator terms. Input samples are s[n],
output and feedback samples are y[n]. I see no coefficients operating on
any y[n].

I see what he's asking too. His problem lies with internal states, and I
don't know how to deal with that in the abstract. If there's a general
case, I haven't seen it.

Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
```
 0
jya (12871)
11/30/2005 6:36:47 PM
```Jerry Avins wrote:
>
> I expect that the b's are the denominator terms.

no, Jerry, the a's are in the denominator.  they reversed that
convention a while back.  in the 70's it was a_k on top and b_k on
bottom, but, i think because they think that poles are more important,
most texts have it switched around now.

r b-j

```
 0
rbj (4086)
11/30/2005 7:32:10 PM
```robert bristow-johnson wrote:
> Jerry Avins wrote:
>
>>I expect that the b's are the denominator terms.
>
>
> no, Jerry, the a's are in the denominator.  they reversed that
> convention a while back.  in the 70's it was a_k on top and b_k on
> bottom, but, i think because they think that poles are more important,
> most texts have it switched around now.
>
> r b-j

I knew that too, but old habits sometimes bubble up. The important point
was that Fred N. has no y terms to the right of an equal sign, so I
don't see any recursion. Somewhere along the line I wrote that. Twice, I
think.

Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
```
 0
jya (12871)
11/30/2005 9:21:45 PM
```Jerry Avins wrote:
> robert bristow-johnson wrote:
> > Jerry Avins wrote:
> >
> >>I expect that the b's are the denominator terms.
> >
> > no, Jerry, the a's are in the denominator.
>
> I knew that too, but old habits sometimes bubble up. The important point
> was that Fred N. has no y terms to the right of an equal sign, so I
> don't see any recursion.

there is recursion with the intermediate s[k] signal:

Fred Nach wrote:

> I would like to implement a IIR Biquad filter using the fixed point
> arithmetics...
> Hence to reduced the intermediate states I plan to use the following trick:
> s(k) = x(k) -a1*s(k-1) -a2*s(k-2)
> y(n) = b0*s(k) + b1*s(k-1) + b2*s(k-2)

this is the canonical Direct Form 2.  poles before zeros.  problem is,
if there is a high Q filter, those poles will amplify the signal so
much that it will saturate s[k] before the zeros get to beat the signal
back down.  DF2 ain't particularly good, especially for fixed-point
arithmetic.

r b-j

```
 0
rbj (4086)
11/30/2005 10:27:56 PM
```On Wed, 30 Nov 2005 11:01:31 -0800, Noway2 wrote:
> I have seen Matlab scripts that run an impulse response till it
> stabilizes and then sum up the impulse responses and use the maximum
> value as a scale factor, though I don't know if this is a correct
> solution.

It's pretty close.  This is the main reason why these "direct form 2"
implmentations are not preferred.  A direct form 1 implmentation will use
more memory but (a) memory is relatively cheap and (b) the values stored
will be as constrained as the input and output values.

--
Andrew

```
 0
12/1/2005 12:10:25 AM
```robert bristow-johnson wrote:
> Jerry Avins wrote:
>
>>robert bristow-johnson wrote:
>>
>>>Jerry Avins wrote:
>>>
>>>
>>>>I expect that the b's are the denominator terms.
>>>
>>>no, Jerry, the a's are in the denominator.
>>
>>I knew that too, but old habits sometimes bubble up. The important point
>>was that Fred N. has no y terms to the right of an equal sign, so I
>>don't see any recursion.
>
>
> there is recursion with the intermediate s[k] signal:

OK. Back to the books with my thinking cap on.

Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
```
 0
jya (12871)
12/1/2005 4:19:12 AM
```Jerry Avins wrote:
> robert bristow-johnson wrote:
> > Jerry Avins wrote:
....
> >>I knew that too, but old habits sometimes bubble up. The important point
> >>was that Fred N. has no y terms to the right of an equal sign, so I
> >>don't see any recursion.
> >
> >
> > there is recursion with the intermediate s[k] signal:
>
> OK. Back to the books with my thinking cap on.

no need.  here's a simple way to look at it:

s[k] =    x[k] - a1*s[k-1] - a2*s[k-2]
y[k] = b0*s[k] + b1*s[k-1] + b2*s[k-2]

think of this as two filters in cascade.  the first with input x[k] and
output s[k].  the second with input s[k] and output y[k].

the first is a all-pole filter (actually there are 2 zeros at the
origin, but don't worry about them) and the second is a all zero
filter, an FIR (like an FIR, there are 2 poles at the origin, but the
other 2 zeros kill 'em).  the transfer function of the first:

S(z)/X(z) = 1/(1 + a1*z^-1 + a2*z-2)

the transfer function of the second

Y(z)/S(z) = b0 + b1*z^-1 + b2*z^-2

and the transfer function of the whole sheee-bang is

H(z) = Y(z)/X(z) = ( Y(z)/S(z) ) * ( S(z)/X(z) )

and the reason it has only two states instead of four is that the
states for the first are identical to the states of the second (both
delays on s[k]) so they can be combined into a single delay line.

r b-j

```
 0
rbj (4086)
12/1/2005 7:09:00 AM
```Hi all !

Firstly thanks a lot for your fast and usefull help ;-)

It seems that my previous post was not clear enough so here are some
extra details :

* The filter structure is indeed the From II that robert B-J has
described very clearly.

* a1 and a2 are hence the DENOMINATORS coeff (I was born in the 80's !!)
and ... b1 and b2 are NUMERATOR (no b0 cause it is supposed scaled)

* The input signal is x(n) and it is a bounded signal : PCM 16 bits so
it varies from -32768 to 32767

* The s(k) are the internal states that I save between each sample
processing...

* The filter is indeed stable hence  |a1| <= 2 and |a2| <= 1

I think I understood that this implementation is not recommended because
it compute first the feedback terms and leads to big variations...
But is there a way to retrieve the boudary of s(k) mathematicaly ?

I think that as long as a1 and a2 are given for a stable filter s(k)
should be bounded...

For my program, this implementation is very interesting cause i code in
Assembly and It save a lot of MPIS to only read 2 states variable (s1
and s2) then 4 (x1 x2 y1 y2)...

Fred.

robert bristow-johnson a �crit :
> Fred Nach wrote:
>
>>I would like to implement a IIR Biquad filter using the fixed point
>>arithmetics...
>>Hence to reduced the intermediate states I plan to use the following trick:
>>s(k) = x(k) -a1*s(k-1) -a2*s(k-2)
>>y(n) = b0*s(k) + b1*s(k-1) + b2*s(k-2)
>>
>>The I can compute each y, saving only 2 states (s(k-1) a�nd s(k-2))...
>
>
> this is the Direct 2 Form (sometimes called the "Direct II Canonical
> Form").  unrecommeded for floating-point (if the Q is quite high, you
> end up subtracting numbers very close to each other and  losing
> precision) and *highly* unrecommended for fixed-point (saturation
> clipping will be your lot).
>
> try the Direct 1 Form:
>
>     y[n] = b0*x[n] + b1*x[n-1] + b2*x[n-2] - a1*y[n-1] - a2*y[n-2]
>
> every product on the right side of the = sign is a double precision
> fixed point and all of those products should be added together in
> double precision.  this is trivial in the Mot 56K and the ADI SHArC and
> maybe in the TI fixed-pointers.
>
>
>>BUT ... in order to scale my coefficients or input, I need to know what
>>are the boundaries of the s(k) serie... in order to find a proper Fixed
>>Point format for my coefficients..
>
>
> your coefs are defined strictly in terms of the frequency response you
> want.  the gross range of a1 is -2 to +2 and a2 is -1 to +1 for any
> stable biquad filter.  the b0, b1, b2 coefs can be nearly anything but
> you will have to choose a range and possibly do some scaling using the
> arithmetic shift operation on the result before saving to y[n].
>
>
>>So my questions are :
>>
>>* for which conditions, s(k) is bounded
>>* If the s(k) is bounded, what are the boundaries?
>
>
> that is a purely artificial construct.  are your fixed-point signal
> values considered 16 bit signed integers?  then it's  -32768 <= x[n] <
> +32768 .  are they normalized?  then it's -1 <= x[n]< +1.  their range
> is whatever you want them to be.
>
>
>>Thanks in advance for your help (and sorry for my poor englsih)
>
>
> as Jerry said, your English is fine.
>
> r b-j
>
```
 0
fred_nach (6)
12/1/2005 3:18:16 PM
```Me again ...

Ok, so i've just launch a Matlab plot of the filter derived from the
s(k) sequence :

s(k) = x(k) -a1*s(k-1) -a2*s(k-2)
hence considering x(k) as input and s(k) as output
H(Z) = 1/(1 + a1*z^-1 + a2*z^-2)

the frequency response give a curve that decrease gradualy, *BUT* start
witha gain of about 80 dB ... hence it means (as long as i
understand...) that for almost constant x(k) (=low freq) the gain is
very important so s(k) will take big values ... (gain of 80dB is like
multiplying by 10 000 !) hence it for a Fixed Point architecture it
requires ceil(log2(10000)) = 14 guard bits !!!

So yes... it's definitively not a good solution to implement a filter
like that ... and I will do it using the direct form 1.

Salut !

Fred
```
 0
fred_nach (6)
12/1/2005 4:06:39 PM
```Fred Nach wrote:
>
> It seems that my previous post was not clear enough

it was clear enough for me.

....

> * The filter is indeed stable hence  |a1| <= 2 and |a2| <= 1
>
> I think I understood that this implementation is not recommended because
> it compute first the feedback terms and leads to big variations...

it means that s[n] is too large to fit in you 16 bit words.  if you
represent them as 32 bit words, then you have to do double precision
multiplies when you multiply be b0, b1, b2.

> But is there a way to retrieve the boudary of s[n] mathematicaly ?
>
> I think that as long as a1 and a2 are given for a stable filter s[n]
> should be bounded...

it is.  the worst case is that if a1 and a2 are known, you can obtain
the impulse response for the all-pole filter

H_s(z) = 1/(1 + a1*z^-1 + a2*z^-2)

call that impulse response h_s[n].

the worst case for the maximum value for |s[n]| is what happens if x[n]
shares the same sign as s[n] (for each value of n) and is at it at its
maximum magnitude.  from the convolution summation

s[n] = SUM{ h_s[k] * x[n-k] }
n

max |s[n]| = max |x[n]| * SUM{ |h_s[n]| }
n            n            n           (max's and SUM over all n)

you might find that max gain, SUM{ |h_s[n]| }, to be a very large
value.  much larger than 1. if you were to reduce you input to such a
degree that you guarantee that s[n] never saturates, you will likely
reduce

> For my program, this implementation is very interesting cause i code in
> Assembly and It save a lot of MIPS to only read 2 states variable (s1
> and s2) then 4 (x1 x2 y1 y2)...

there is no reason for the DF1 to require more MIPS in the inner loop
than the DF2.  at least for cascaded biquads.  if you're doing only one
biquad, that is your overall filter order is 2, then it will cost a
little more (but the cost is worth it).

the DF1 requires 2 additional states but, if you cascade 2nd order
biquads, the two output states can be shared with the 2 input states of
the next section.  so if you have N sections(an order 2N filter), the
number of states you need for DF1 is 2N+2 only 2 more than DF2.  not a
very high price to pay.

the DF2 is useless in your case (assuming that your filter will have
some decent Q making the poles very close to the unit circle).  but, if
you like learning this yourself (we call this "learning the hard way"),
feel free to implement it.  but you will find that either you
pre-scaling of the input will drive your signal into the noise floor
or,if you do not pre-scale the input, your intermediate state signal,
s[n], will saturate.  only with double precision (that will cost more
MIPS than DF1) will you avoid that.

r b-j

```
 0
rbj (4086)
12/1/2005 4:28:55 PM
```Fred Nach wrote:
> Me again ...
>
> Ok, so i've just launch a Matlab plot of the filter derived from the
> s(k) sequence :
>
> s(k) = x(k) -a1*s(k-1) -a2*s(k-2)
> hence considering x(k) as input and s(k) as output
> H(Z) = 1/(1 + a1*z^-1 + a2*z^-2)
>
> the frequency response give a curve that decrease gradualy, *BUT* start
> witha gain of about 80 dB ... hence it means (as long as i
> understand...) that for almost constant x(k) (=low freq) the gain is
> very important so s(k) will take big values ... (gain of 80dB is like
> multiplying by 10 000 !) hence it for a Fixed Point architecture it
> requires ceil(log2(10000)) = 14 guard bits !!!
>
> So yes... it's definitively not a good solution to implement a filter
> like that ... and I will do it using the direct form 1.

our posts "crossed in the mail".  i'm glad that you "get it" now.  :-)

when using the DF1, make sure that all five terms (b0*x[n] b1*x[n-1],
etc) are added together in a double precision accumulator before
quantizing that back to a single precision output.  also, consider
"noise shaping" or "fraction saving" (Google Groups the latter term or
look up the DC Blocking filter in fixed-point at dspguru.com).  with
only 16 bits, you'll probably need it.

r b-j

```
 0
rbj (4086)
12/1/2005 4:35:02 PM
```On 1 Dec 2005 08:35:02 -0800, "robert bristow-johnson"
<rbj@audioimagination.com> wrote:
>
>Fred Nach wrote:
>> Me again ...
>>
>> Ok, so i've just launch a Matlab plot of the filter derived from the
>> s(k) sequence :
>>
>> s(k) = x(k) -a1*s(k-1) -a2*s(k-2)
>> hence considering x(k) as input and s(k) as output
>> H(Z) = 1/(1 + a1*z^-1 + a2*z^-2)
>>
>> the frequency response give a curve that decrease gradualy, *BUT* start
>> witha gain of about 80 dB ... hence it means (as long as i
>> understand...) that for almost constant x(k) (=low freq) the gain is
>> very important so s(k) will take big values ... (gain of 80dB is like
>> multiplying by 10 000 !) hence it for a Fixed Point architecture it
>> requires ceil(log2(10000)) = 14 guard bits !!!
>>
>> So yes... it's definitively not a good solution to implement a filter
>> like that ... and I will do it using the direct form 1.
>
>our posts "crossed in the mail".  i'm glad that you "get it" now.  :-)
>
>when using the DF1, make sure that all five terms (b0*x[n] b1*x[n-1],
>etc) are added together in a double precision accumulator before
>quantizing that back to a single precision output.  also, consider
>"noise shaping" or "fraction saving" (Google Groups the latter term or
>look up the DC Blocking filter in fixed-point at dspguru.com).  with
>only 16 bits, you'll probably need it.
>

Isn't this a case where a lattice filter implementaion helps?
```
 0
just5217 (63)
12/1/2005 4:46:28 PM
```Just Cocky wrote:

> Isn't this a case where a lattice filter implementaion helps?

problem.  it preserves the mean power in those internal states but does
not guarantee against saturation. in the DF1, you can scale down b0,
b1, b2 to such a point that the summer (and output y[n]) does not
saturate (and then scale it up later, if you can get away with it).

r b-j

```
 0
rbj (4086)
12/1/2005 5:07:45 PM
```Just Cocky wrote:

> Isn't this a case where a lattice filter implementaion helps?

problem.  it preserves the mean power in those internal states but does
not guarantee against saturation. in the DF1, you can scale down b0,
b1, b2 to such a point that the summer (and output y[n]) does not
saturate (and then scale it up later, if you can get away with it).

r b-j

```
 0
rbj (4086)
12/1/2005 6:02:21 PM
```robert bristow-johnson wrote:
> Jerry Avins wrote:
>
>>robert bristow-johnson wrote:
>>
>>>Jerry Avins wrote:

...

>>OK. Back to the books with my thinking cap on.
>
>
> no need.  here's a simple way to look at it:

...

Thanks much. Part of my problem was misreading the original equations,
the ones with () instead of [].

Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
```
 0
jya (12871)
12/1/2005 6:29:17 PM
```"Andrew Reilly" <andrew-newspost@areilly.bpc-users.org> wrote in message
news:pan.2005.12.01.00.10.23.587775@areilly.bpc-users.org...
> On Wed, 30 Nov 2005 11:01:31 -0800, Noway2 wrote:
>> I have seen Matlab scripts that run an impulse response till it
>> stabilizes and then sum up the impulse responses and use the maximum
>> value as a scale factor, though I don't know if this is a correct
>> solution.
>
> It's pretty close.  This is the main reason why these "direct form 2"
> implmentations are not preferred.  A direct form 1 implmentation will use
> more memory but (a) memory is relatively cheap and (b) the values stored
> will be as constrained as the input and output values.

Even though memory is cheap, sometimes I find the overhead of the extra memory
stores/fetches slows the filter down a bit (depending on the architecture).

```
 0
12/3/2005 6:16:47 AM

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Hi, I searched the internet to find out the difference between a fixed point & a floating point DSP. To my surprise, I did not find a well defined difference. Plz, can anyone differentiate between the two so that a layman can also understand ? Thanx in advance... Sandeep Chikkerur wrote: > > Hi, > > I searched the internet to find out the difference between a fixed > point & a floating point DSP. > To my surprise, I did not find a well defined difference. > > Plz, can anyone differentiate between the two so that a layman can > also understand ...

IIR fixed-point implementation
HI everyone, I was wondering if anyone can recommend me good books or papers on fixed-point IIR implementation strategies on FPGAs or CPUs. Especially regarding fc versus Fs and coefficient quantization. I have issues with an order-1 DC-removal filter with Fs = 100 MHz and Fcut = 250 kHz. Basically, I need lots of precision in the intermediate calculations which impact the number of bits for the multiplier and thus the maximum frequency of my FPGA design. Best regards On 11/14/2011 1:15 PM, Benjamin Couillard wrote: > HI everyone, > > I was wondering if anyone can recommend me good...

IIR Filter fixed point DSP
Hi I am trying to design a IIR filter using a 16-bit fixed point DSP. I nee to calculate the coeff in the form of second order sections. I would lik to do this with MATLAB. Does anyone have some MATLAB code that will d this. I found a routine on Texas Instrument site but as well as the coef it creates some scaling factors for the inputs. I dont want this as I hav seen other programs such as Filter Express that just give the Coeff. Thanks J This message was sent using the Comp.DSP web interface o www.DSPRelated.com OK. You have several kinds of IIR filter, for instance elliptic (the be...

IIR biquad implementation in fixed point / integer
Hello, I tried implementing a LPF biquad (that works perfectly in floating point) with: a = [1.00000, -1.90887, 0.91129], b = [6.1150e-04, 1.2230e-03, 6.1150e-04] in fixed point / integer, with 16 bit coefficients, 10 bit input signal and a 32 bit accumulator. I suspect the coefficient resolution is not enough. I tried: - the direct form 1 and direct form 2 transposed structures, but the output is zero if the input signal is not very high -- because of the small b coefficients, the signal is attenuated very much in the FIR portion of the filter and is lost; int biquad_ab(d, a0...

multirate implementation on fix point DSP and efficiency
Hye, I have a question to try to find a way to improve my processing. I use a multirate filter with decimation. There is a decimation by 40. One FIr is used then decimation, and the IIR with decimation, then a second IIR with decimation, then again one IIR and decimation and at last one FIr . I use a fix point DSP. The filter are designed in 16 bits inputs / output (32 bits for computation). It appears that the results is that i have in output a 12bits resoutio and 4 lsb bits are lost during the stages. How can i improve this? Should i change the IIR design? or thedecimation? thank yo...

Fixed point implementation of 4'th order IIR filters
Hi Does anyone have some guidelines on how to implement a 4'th order low-pass Butterworth IIR filter in fixed point. My cut-off frequency is relatively close to the DC frequency so high precision is needed for the coefficients. What about realization structure and so on! I have implemented the bit-flipping algorithm in http://www.cmsa.wmin.ac.uk/~artur/pdf/Paper16.pdf for quantization of coefficients and it indeed works, but does some other techniques allow for further reductions of number of bits used to represent the coefficients. Thomas I've found the normalized lattice-ladder...

couple of questions on practical IIR filter implementation on fixed point systems
Hello. I am a professional programmer, but I am doing DSP just as a hobby. Three years ago I designed a morse / PSK31 decoder on a 16bit no- multiplier controller. After rouhgly 3 years of absence in the DSP field I am back to design a HAM shortwave software defined radio using a beautiful new part from Texas Instruments - TLV320AIC3254 codec with miniDSP core. The part could do all the signal processing I need with just a 5x5mm board space including earphone driver. That's an engineer's dream come true. The goal is to downconvert a SSB voice channel down to audio with a quadrature m...

some suggestions on my octave-band spectrogram analysis in fixed point DSP implementation
i am designing one 1/3 octave band spectrogram analysis I use multirate filter bank to realize that spectrogram. It goes this way 1) first, from the biggest frequency value, I use three IIR bandpass filter, then calculate the std value. 2) Then decimate the input by 2 through(one 30 order FIR filter, then resampling the input), then use the same three coffeicents to calculate the successive three band, 3) then repeat the step 2) til the lowest frequency. there is 45 channels (one channel one ceter freuquency) I got the right result throught MATLAB But i got some different result ...

Floating-point to Fixed-point in (Sharc) DSP
Hello, I am sure this is a pretty simple question, but I am having some ambiguity. I need to confirm what is happening exactly. Any information is really appreciated. So this is a Sharc DSP I am working with. Question is pretty simple. Rn = FIX Fx; FIX description from the manual is as follows: "Converts the floating-point operand in Fx to a twos-complement 32-bit fixed-point integer result. If the MODE1 register TRUNC bit=1, the FIX operation truncates the mantissa towards –Infinity. If the TRUNC bit=0, the FIX operation rounds the mantissa towards the nearest integer....

floating point to fixed point conversion-in implementation of navigation system algorithm
Hi, I have a floating point c code to implement a navigation system algorithm. In order to implement it in the 64x+ fixed point DSP processor, i need to have the code in fixed point. I learnt that there are few processor specific IQmath libraries aiding this purpose. Is it practical to write the code using these libraries or should i manually write codes for each operation by scaling the inputs for each line of the code, in which case i have to create look up table for the trigonometric functions etc which is a harder process to begin with.. Thanks, Divya On 2/13/12 11:38 AM, divya_di...

Fixed point DSP
hello all, I am new in dsp programing and i an using fixed point DSP, as read FIxed point arithmatic it seems to be confusing so plz suggest an link or document having good practical explaination of the topic, or pl give some examples for understanding, Thanks in Advance, Rahul. jadhav_rahul wrote: > hello all, > I am new in dsp programing and i an using fixed point DSP, as i > read FIxed point arithmatic it seems to be confusing so plz suggest any > link or document having good practical explaination of the topic, or plz > give some examples for understan...

fixed point implementation
Hi, i want to implement the equation p3 = p3*0.5/cos(3*PI/8); into 16bit processor, where p3 is a 16 bit value; any pointers as to how to go about it -vijay n_vijay_anand wrote: > Hi, > > i want to implement the equation > > p3 = p3*0.5/cos(3*PI/8); > > into 16bit processor, where p3 is a 16 bit value; > > any pointers as to how to go about it It looks like '0.5/cos(3*PI/8)' is a constant, so simplify it. cos(3*PI/8) = 0.99978861 0.5/0.99978861 = 0.50010571 Depending on the algorithm you might get away with simply dividing by 2: p3 >...

fixed point IIR coeffs
Hi i am using built-in Matlab functions to design IIR fixed point filter. For example: [b a] = cheby1(2,6,0.25) hd = dfilt.df1(b,a) set(hd,'arithmetic','fixed') Is there a simple method to check the coefficient values for fixed arithmetic? Precisely the ones which will appear in HDL code ( generatehdl(hd) ). Because i want to make a parametrized filter with a coefficient lookup table for different cutoff frequencies. ...

LMS fixed point implementation
Hi everyone! I need a fixed point implementation for the LMS algorithm;I have to implement it on a TI dsp board (c6711 or a c5000); can anyone suggest me where I can find this code? thank you guys! f Use this: It is a single step for fixed point LMS. c_ptr points to coefficient, r_ptr points to history. muErr = e * mu; mu = 2048(trial), using fixed mu; accA, accB are at least 32 bit accumulators. /* Macro: Performs single step of filter and adapt */ #define ADAPT_AND_FILTER(accA, accB, muErr, c_ptr, r_ptr) \ accB = 16384; \ accB += ((i3...

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