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
Noise-equivalent bandwidth of digital filter
Started by ●November 22, 2004
Reply by ●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... > > JimI 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 ●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 ●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 ●November 23, 20042004-11-23
"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. > > > TomWell 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
