"Country_Chiel" <Chiel@bothy.nichts.co.uk> wrote in message news:<1101181914.985975@ftpsrv1>...
> > Jim
> I think your solution to Parsevals integral is right provided the impulse
> response is FIR. I assume you have divided it out and are applying it -
> summing to infinity -I wonder what the accuracy of that is? For the IIR case
> there is a set of tables for the solution of such discrete time integrals
> for various orders.
>
>
> Tom
Well at first I was trying with a first-order IIR then a moving
average FIR. For example, if I have a simple averaging IIR y(n) =
a*y(n-1) + (1-a)*x(n), then my h(n) = (1-a)*a^n, n=0,1,...,inf.
Using the aforementioned formula, summing h(n)^2 from 0 to inf gives
you a geometric series that equals (1-a)/(1+a). Summing h(n) from 0
to inf gives you 1.
So my NEBW for this IIR filter is
Wn = pi*(1-a)/(1+a).
This kind of matches up with the NRR (noise reduction ratio) outlined
in Orfanidis (without the pi multiplier), so I wanted to make sure I
was going down the correct path before marrying myself to the result.
Thanks,
Jim
Reply by Stephan M. Bernsee●November 23, 20042004-11-23
On 2004-11-23 10:25:38 +0100, dpenev@yahoo.com (Dimitar Penev) said:
> Hi Jim,
>
> Please read this very good article http://dsp-bg.info//viewtopic.php?t=23
> There is definition of ENBW of window. (The definition for the FIR is
> the same.)
>
> I hope this helps.
> penev
> -------------------
> DSP forum www.dsp-bg.info
>
Hi Penev,
are you sure you are allowed to post the scanned article on that page?
Not that I doubt your motives, but there may be IP issues with doing
so...
--
Stephan M. Bernsee
http://www.dspdimension.com
Reply by Dimitar Penev●November 23, 20042004-11-23
Hi Jim,
Please read this very good article
http://dsp-bg.info//viewtopic.php?t=23
There is definition of ENBW of window. (The definition for the FIR is the same.)
I hope this helps.
penev
-------------------
DSP forum www.dsp-bg.info
jshima@timing.com (JS) wrote in message news:<c15bb83f.0411221008.4e42c2e1@posting.google.com>...
> Hello all,
>
> Havent been around in some time, but I have a question that has been
> needling me. I want to determine the noise-equivalent bandwidth of an
> FIR or IIR filter. I started by attempting to convert the analog
> definition to a digital equivalent. Although my attempts dont appear
> to give me the correct answer. Hoping someone could lend a hand in
> this seemingly simple problem (not enough coffee today).
>
> I started with the standard NEBW definition:
>
> Wn = integral(0,inf) { |H(w)|^2 dw}
> -----------------------------
> |H(0)|^2
>
> I then converted this to a sampled digital equivalent limited to the
> sample band:
>
> Wdn = integral(0,pi) { |H(e^jw)|^2 dw }
> -------------------------------
> |H(0)|^2
>
> where Wdn is the digital frequency = Wn*T
>
> Next applied Parseval to simplify the numerator and denominator:
>
> sum(-inf, inf) {h(n)^2} =
>
> 1/2pi * integral(-pi, pi) { |H(e^jw)|^2 dw }
>
> giving:
>
> Wdn = pi * sum(inf,-inf) { h(n)^2}
> -----------------------
> [ sum(-inf, inf) { h(n) } ] ^2
>
>
> This doesnt seem to give me the proper answer. Can anyone shed any
> light on my error? Be gentle...
>
> Jim
Reply by Country_Chiel●November 22, 20042004-11-22
"JS" <jshima@timing.com> wrote in message
news:c15bb83f.0411221008.4e42c2e1@posting.google.com...
> Hello all,
>
> Havent been around in some time, but I have a question that has been
> needling me. I want to determine the noise-equivalent bandwidth of an
> FIR or IIR filter. I started by attempting to convert the analog
> definition to a digital equivalent. Although my attempts dont appear
> to give me the correct answer. Hoping someone could lend a hand in
> this seemingly simple problem (not enough coffee today).
>
> I started with the standard NEBW definition:
>
> Wn = integral(0,inf) { |H(w)|^2 dw}
> -----------------------------
> |H(0)|^2
>
> I then converted this to a sampled digital equivalent limited to the
> sample band:
>
> Wdn = integral(0,pi) { |H(e^jw)|^2 dw }
> -------------------------------
> |H(0)|^2
>
> where Wdn is the digital frequency = Wn*T
>
> Next applied Parseval to simplify the numerator and denominator:
>
> sum(-inf, inf) {h(n)^2} =
>
> 1/2pi * integral(-pi, pi) { |H(e^jw)|^2 dw }
>
> giving:
>
> Wdn = pi * sum(inf,-inf) { h(n)^2}
> -----------------------
> [ sum(-inf, inf) { h(n) } ] ^2
>
>
> This doesnt seem to give me the proper answer. Can anyone shed any
> light on my error? Be gentle...
>
> Jim
I think your solution to Parsevals integral is right provided the impulse
response is FIR. I assume you have divided it out and are applying it -
summing to infinity -I wonder what the accuracy of that is? For the IIR case
there is a set of tables for the solution of such discrete time integrals
for various orders.
Tom
Reply by JS●November 22, 20042004-11-22
Hello all,
Havent been around in some time, but I have a question that has been
needling me. I want to determine the noise-equivalent bandwidth of an
FIR or IIR filter. I started by attempting to convert the analog
definition to a digital equivalent. Although my attempts dont appear
to give me the correct answer. Hoping someone could lend a hand in
this seemingly simple problem (not enough coffee today).
I started with the standard NEBW definition:
Wn = integral(0,inf) { |H(w)|^2 dw}
-----------------------------
|H(0)|^2
I then converted this to a sampled digital equivalent limited to the
sample band:
Wdn = integral(0,pi) { |H(e^jw)|^2 dw }
-------------------------------
|H(0)|^2
where Wdn is the digital frequency = Wn*T
Next applied Parseval to simplify the numerator and denominator:
sum(-inf, inf) {h(n)^2} =
1/2pi * integral(-pi, pi) { |H(e^jw)|^2 dw }
giving:
Wdn = pi * sum(inf,-inf) { h(n)^2}
-----------------------
[ sum(-inf, inf) { h(n) } ] ^2
This doesnt seem to give me the proper answer. Can anyone shed any
light on my error? Be gentle...
Jim