adding up coefficients

Hello,

Recently I designed my first IIR filter from a 1st orde Butterworth.
After implementation it works as it should.
My second filter a 2nd order Butterworth is ready on paper.
What occured to me is when I add up the coefficients from the 1st
order filter the sum = 1. Surprisingly adding up the coefficients
of the second order IIR filter the sum is also = 1.

Spare me the Math but is this the case for every filter so I can use
this as a check method?

Gert Baars wrote:

> Hello,
> 
> Recently I designed my first IIR filter from a 1st orde Butterworth.
> After implementation it works as it should.
> My second filter a 2nd order Butterworth is ready on paper.
> What occured to me is when I add up the coefficients from the 1st
> order filter the sum = 1. Surprisingly adding up the coefficients
> of the second order IIR filter the sum is also = 1.
> 
> Spare me the Math but is this the case for every filter so I can use
> this as a check method?
> 
> 
Do you mean that the ratio of the sums of the numerator coefficients and 
the denominator coefficients adds to 1?  Such as:

           0.1
H(z) = ---------  ?
         z - 0.9

If so then all that you're showing is that the filter has a DC gain of 1 
(z = 1 corresponds to DC).  This is the barest scratch on the surface of 
the frequency domain interpretation of the z transform -- take a look at 
http://www.wescottdesign.com/articles/zTransform/z-transforms.html for a 
slightly deeper scratch.

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

c(0).Z^0 + c(1).(Z^-1) + ... + c(m).(Z^-m)
-------------------------------------------
1 + (d(1).(Z^-1) + .... + d(n).(Z^-n)


IF Z=1 at W=0 which is DC

then H(Z=1) = C / (1 + D)       C is sum of all c's and D of all d's.

since d(n) is implemented negative then

If C-D = 1 then H(1) = (1+D)/(1+D) = 1

So indeed it shows the DC gain and that adding up all coefficients is a 
good method to check the Math.




Tim Wescott wrote:
> Gert Baars wrote:
> 
>> Hello,
>>
>> Recently I designed my first IIR filter from a 1st orde Butterworth.
>> After implementation it works as it should.
>> My second filter a 2nd order Butterworth is ready on paper.
>> What occured to me is when I add up the coefficients from the 1st
>> order filter the sum = 1. Surprisingly adding up the coefficients
>> of the second order IIR filter the sum is also = 1.
>>
>> Spare me the Math but is this the case for every filter so I can use
>> this as a check method?
>>
>>
> Do you mean that the ratio of the sums of the numerator coefficients and 
> the denominator coefficients adds to 1?  Such as:
> 
>           0.1
> H(z) = ---------  ?
>         z - 0.9
> 
> If so then all that you're showing is that the filter has a DC gain of 1 
> (z = 1 corresponds to DC).  This is the barest scratch on the surface of 
> the frequency domain interpretation of the z transform -- take a look at 
> http://www.wescottdesign.com/articles/zTransform/z-transforms.html for a 
> slightly deeper scratch.
> 

"Gert Baars" <g.baars13@chello.nl> wrote in message
news:21317$431b2a6d$3ec23590$23823@news.chello.nl...
>
>
> c(0).Z^0 + c(1).(Z^-1) + ... + c(m).(Z^-m)
> -------------------------------------------
> 1 + (d(1).(Z^-1) + .... + d(n).(Z^-n)
>
>
> IF Z=1 at W=0 which is DC
>
> then H(Z=1) = C / (1 + D)       C is sum of all c's and D of all d's.
>
> since d(n) is implemented negative then
>
> If C-D = 1 then H(1) = (1+D)/(1+D) = 1
>
> So indeed it shows the DC gain and that adding up all coefficients is a
> good method to check the Math.
>
>
>
Or check the Maths.

Shytot

On Sun, 04 Sep 2005 19:09:17 +0200, Gert Baars <g.baars13@chello.nl>
wrote:

>
>
>c(0).Z^0 + c(1).(Z^-1) + ... + c(m).(Z^-m)
>-------------------------------------------
>1 + (d(1).(Z^-1) + .... + d(n).(Z^-n)
>
>
>IF Z=1 at W=0 which is DC
>
>then H(Z=1) = C / (1 + D)       C is sum of all c's and D of all d's.
>
>since d(n) is implemented negative then
>
>If C-D = 1 then H(1) = (1+D)/(1+D) = 1
>
>So indeed it shows the DC gain and that adding up all coefficients is a 
>good method to check the Math.

Hi Gert,

   Your post was interesting to me.

I don't know if my memory has failed me, or maybe 
I never did know the simple rule (which you are  
developing here) that the DC gain of an IIR filter is
 

                Sum of feed forward coefficients
  DC gain = ----------------------------------------
             One  minus sum of feedback coefficients


Looking into this situation a little further, and being 
more general by analyzing the following IIR network,   

  x --->(+)---a0---->--------b0----->(+)---> y   
         ^            |               ^
         |          [z^-1]            |
         |            |               |
        (+)<--a1------+-------b1---->(+)
         ^            |               ^
         |          [z^-1]            |
         |            |               |
         +-<--a2------+-------b2---->-+


I think the rule should be that the DC gain of 
an an IIR filter is:

                Sum of feed forward coefficients
  DC gain = -----------------------------------------
             1/a0  minus sum of feedback coefficients

Of course this agrees with your "rule" so 
long as your a0 coefficient is unity.

Thanks for teachin' me some DSP Gert.

[-Rick-]