Adding IIR filter together

Hi Experts,

I am newbie to DSP. If I have a filter which is sum of some other filters
of same signal source x(n), for example:

y1(n) = a1*y1(n-1)+b1*x(n);
y2(n) = a2*y2(n-1)+b2*x(n);
y3(n) = a1*y3(n-1)+b3*x(n);

y4(n) = y1(n)+y2(n)+y3(n);

What is overall structure of y4(n) then? can any one help me with this.

Many thanks.

Datu

On 24 Nov, 04:52, "datu" <daturas...@gmail.com> wrote:
> Hi Experts,
>
> I am newbie to DSP. If I have a filter which is sum of some other filters
> of same signal source x(n), for example:
>
> y1(n) = a1*y1(n-1)+b1*x(n);
> y2(n) = a2*y2(n-1)+b2*x(n);
> y3(n) = a1*y3(n-1)+b3*x(n);
>
> y4(n) = y1(n)+y2(n)+y3(n);
>
> What is overall structure of y4(n) then? can any one help me with this.

Try to draw a block diagram fro this filter. Maybe you can see the
overall structure in the drawing.

Rune

>>
>> y4(n) = y1(n)+y2(n)+y3(n);
>>
>> What is overall structure of y4(n) then? can any one help me with
this.
>
>Try to draw a block diagram fro this filter. Maybe you can see the
>overall structure in the drawing.
>
>Rune
Thanks Rune, I can draw the block diagram, But I want some overall format
only with y4(n) and x(n)?

Can I use z-transform to do
 
Y4(z) = Y1(Z)+Y2(Z)+Y3(Z);

Then transform this back to y4(n)?

On 24 Nov, 08:20, "datu" <daturas...@gmail.com> wrote:
> >> y4(n) = y1(n)+y2(n)+y3(n);
>
> >> What is overall structure of y4(n) then? can any one help me with
> this.
>
> >Try to draw a block diagram fro this filter. Maybe you can see the
> >overall structure in the drawing.
>
> >Rune
>
> Thanks Rune, I can draw the block diagram, But I want some overall format
> only with y4(n) and x(n)?
>
> Can I use z-transform to do
>
> Y4(z) = Y1(Z)+Y2(Z)+Y3(Z);
>
> Then transform this back to y4(n)?

Yes. The Z transform is what is known as a "linear transform".
The term "linear" means that

ZT{a(n)+b(n)} = ZT{a(n)} + ZT{b(n)}.

Rune

>> Y4(z) = Y1(Z)+Y2(Z)+Y3(Z);
>>
>> Then transform this back to y4(n)?
>
>Yes. The Z transform is what is known as a "linear transform".
>The term "linear" means that
>
>ZT{a(n)+b(n)} = ZT{a(n)} + ZT{b(n)}.
>
>Rune

thanks Rune, since we can do that I can get final y(n) format. Why some
implementations use the y1(n) y2(n)... and then y(n) = y1(n)+y2(n)...
format?
for example this pink filter: 
http://www.musicdsp.org/showone.php?id=76 
use "a weighted sum of first order filters". Is there any advantages to do
so or some design approaches lead to this?

Many thanks

On 24 Nov, 16:20, "datu" <daturas...@gmail.com> wrote:
> >> Y4(z) = Y1(Z)+Y2(Z)+Y3(Z);
>
> >> Then transform this back to y4(n)?
>
> >Yes. The Z transform is what is known as a "linear transform".
> >The term "linear" means that
>
> >ZT{a(n)+b(n)} = ZT{a(n)} + ZT{b(n)}.
>
> >Rune
>
> thanks Rune, since we can do that I can get final y(n) format. Why some
> implementations use the y1(n) y2(n)... and then y(n) = y1(n)+y2(n)...
> format?
> for example this pink filter:http://www.musicdsp.org/showone.php?id=76
> use "a weighted sum of first order filters". Is there any advantages to do
> so or some design approaches lead to this?

I don't know. I can guess that one advantage with parallel form
filters
in audio applications is that the gain of each filter can be easily
be scaled independently of the others. That way one can implement
some sort of bass and treble controls.

Rune

On 24 Nov., 04:52, "datu" <daturas...@gmail.com> wrote:
> Hi Experts,
>
> I am newbie to DSP. If I have a filter which is sum of some other filters
> of same signal source x(n), for example:
>
> y1(n) = a1*y1(n-1)+b1*x(n);
> y2(n) = a2*y2(n-1)+b2*x(n);
> y3(n) = a1*y3(n-1)+b3*x(n);
>
> y4(n) = y1(n)+y2(n)+y3(n);
>
> What is overall structure of y4(n) then? can any one help me with this.

It's a parallel connection of three first-order lowpass filters
(provided the a_k are positive). You can easily calculate the
resulting transfer function:

http://groups.google.com/group/comp.dsp/browse_frm/thread/cc743c73701025fe/f465eacd6979d25a?#f465eacd6979d25a

Regards,
Andor

On Nov 24, 4:25 pm, Andor <andor.bari...@gmail.com> wrote:
> On 24 Nov., 04:52, "datu" <daturas...@gmail.com> wrote:
>
> > If I have a filter which is sum of some other filters
> > of same signal source x(n), for example:
>
> > y1(n) = a1*y1(n-1)+b1*x(n);
> > y2(n) = a2*y2(n-1)+b2*x(n);
> > y3(n) = a1*y3(n-1)+b3*x(n);

i think you mean "a3" here, right datu?  also, just a form, so we know
what are continuous functions and what are discrete sequences, might i
suggest using brackets [] for discrete-time indices?  it's not a C
thing (even though C uses brackets for discrete indices), it's a
notational convention that i think is used reasonably often in DSP
texts now.

    y1[n] = a1*y1[n-1] + b1*x[n]
    y2[n] = a2*y2[n-1] + b2*x[n]
    y3[n] = a3*y3[n-1] + b3*x[n]


    y4[n] = y1[n] + y2[n] + y3[n]

    Y4(z) =   Y1(z) + Y2(z) + Y3(z)
          = ( H1(z) + H2(z) + H3(z) )*X(z)

> > What is overall structure of y4(n) then? can any one help me with this.

    H4(z) = H1(z) + H2(z) + H3(z)

where

    H_k(z) = b_k/( 1 - a_k*z^(-1) )  k=1,2,3

so stick it together.  if you want a convolution for y4[n] in the time
domain, the ensemble impulse response is the sum of the 3 impulse
response and those are exponential functions for n>=0.

r b-j

robert bristow-johnson wrote:

(Someone wrote)

>     y1[n] = a1*y1[n-1] + b1*x[n]
>     y2[n] = a2*y2[n-1] + b2*x[n]
>     y3[n] = a3*y3[n-1] + b3*x[n]

>     y4[n] = y1[n] + y2[n] + y3[n]

>     Y4(z) =   Y1(z) + Y2(z) + Y3(z)
>           = ( H1(z) + H2(z) + H3(z) )*X(z)

It seems to me, though, that the result won't be any easier
to compute, most likely harder.

I have somewhere a Motorola application note on building a digital
graphic equalizer with the 56001.  (I believe it is available
on the web.)  I am pretty sure it adds up the result of all
the filters.

-- glen

>On Nov 24, 4:25 pm, Andor <andor.bari...@gmail.com> wrote:
>> On 24 Nov., 04:52, "datu" <daturas...@gmail.com> wrote:
>>
>> > If I have a filter which is sum of some other filters
>> > of same signal source x(n), for example:
>>
>> > y1(n) = a1*y1(n-1)+b1*x(n);
>> > y2(n) = a2*y2(n-1)+b2*x(n);
>> > y3(n) = a1*y3(n-1)+b3*x(n);
>
>i think you mean "a3" here, right datu?  also, just a form, so we know
>what are continuous functions and what are discrete sequences, might i
>suggest using brackets [] for discrete-time indices?  it's not a C
>thing (even though C uses brackets for discrete indices), it's a
>notational convention that i think is used reasonably often in DSP
>texts now.
>
>    y1[n] = a1*y1[n-1] + b1*x[n]
>    y2[n] = a2*y2[n-1] + b2*x[n]
>    y3[n] = a3*y3[n-1] + b3*x[n]
>
>
>    y4[n] = y1[n] + y2[n] + y3[n]
>
>    Y4(z) =   Y1(z) + Y2(z) + Y3(z)
>          = ( H1(z) + H2(z) + H3(z) )*X(z)
>
>> > What is overall structure of y4(n) then? can any one help me with
this.
>
>    H4(z) = H1(z) + H2(z) + H3(z)
>
>where
>
>    H_k(z) = b_k/( 1 - a_k*z^(-1) )  k=1,2,3
>
>so stick it together.  if you want a convolution for y4[n] in the time
>domain, the ensemble impulse response is the sum of the 3 impulse
>response and those are exponential functions for n>=0.
>

Hi robert, thank you let me know this convention. I manually calculated
this the 
numerator of H4(Z)=
(-b1*a2*a3-b2*a1*a3-b3*a1*a2)*z^-2+(b1*a3+b1*a2+b2*a3+b2*a1+b3*a2+b3*a1)*z-1-b1-b2-b3
and the denominator of of H4(Z) =
-1+a1*a2*a3*z^-3+((-a2-a1)*a3-a1*a2)*z^-2+(a3+a2+a1)*z-1

So this is third order IIR filter! WHY people design three first-order
filters and parallel connect them, why not directly design this third
order filter? Rune replied earlier that this might because easy adjust
individual parameters. Is there any literature for this?