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-]
Reply by Shytot●September 4, 20052005-09-04
"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
Reply by Gert Baars●September 4, 20052005-09-04
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.
>
Reply by Tim Wescott●September 4, 20052005-09-04
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
Reply by Gert Baars●September 4, 20052005-09-04
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?