Hi All, I have a beginner-esqe question: all of my DSP literature shows the IIR Difference Equation as: y(n) = b(0)x(n) + b(1)x(n-1) + b(2)x(n-2) + ... + a(1)y(n-1) + a(2)y(n-2) + ... But then the C code implementations I see are something along the lines: FeedForwardVariable = (b[0] * x[n]) + (b[1] * x[n-1]) + (b[2] * x[n-2]); FeedBackwardVariable = (a[1] * y[n-1]) + (a[2] * y[n-2]) + (a[3] * y[n-3]); ResultVariable = FeedForwardVariable - FeedBackwardVariable; The implementation uses a subtraction operation on the feed back calculations that I don't see in the Difference Equation. Why is that? I realize I'm setting myself up for a "Duh!!", but I'm really confused why this is the case. Thanks much, Ben

# IIR Filter Difference Equation Implementation

"Benjamin M. Stocks" <stocksb@ieee.org> wrote in message news:132e56ad.0307101125.17d67c7e@posting.google.com...> Hi All, > I have a beginner-esqe question: all of my DSP literature shows the > IIR Difference Equation as: > y(n) = b(0)x(n) + b(1)x(n-1) + b(2)x(n-2) + ... + a(1)y(n-1) + > a(2)y(n-2) + ... > > But then the C code implementations I see are something along the > lines: > FeedForwardVariable = (b[0] * x[n]) + (b[1] * x[n-1]) + (b[2] * > x[n-2]); > FeedBackwardVariable = (a[1] * y[n-1]) + (a[2] * y[n-2]) + (a[3] * > y[n-3]); > ResultVariable = FeedForwardVariable - FeedBackwardVariable; > > The implementation uses a subtraction operation on the feed back > calculations that I don't see in the Difference Equation. Why is that? > > I realize I'm setting myself up for a "Duh!!", but I'm really confused > why this is the case. > > Thanks much, > > Beny(n) = b(0)x(n) + b(1)x(n-1) + b(2)x(n-2) + ... - [a(1)y(n-1) + a(2)y(n-2) + ...] n.b^ or y(n) = b(0)x(n) + b(1)x(n-1) + b(2)x(n-2) + ... -a(1)y(n-1) - a(2)y(n-2) - ... is the normal expression for the difference equation. Fred

On Thu, 10 Jul 2003 13:25:13 -0700, Benjamin M. Stocks wrote:> I have a beginner-esqe question: all of my DSP literature shows the > IIR Difference Equation as: > y(n) = b(0)x(n) + b(1)x(n-1) + b(2)x(n-2) + ... + a(1)y(n-1) + > a(2)y(n-2) + ... > > But then the C code implementations I see are something along the > lines: > FeedForwardVariable = (b[0] * x[n]) + (b[1] * x[n-1]) + (b[2] * > x[n-2]); > FeedBackwardVariable = (a[1] * y[n-1]) + (a[2] * y[n-2]) + (a[3] * > y[n-3]); > ResultVariable = FeedForwardVariable - FeedBackwardVariable; > > The implementation uses a subtraction operation on the feed back > calculations that I don't see in the Difference Equation. Why is that? > > I realize I'm setting myself up for a "Duh!!", but I'm really confused > why this is the case.That is not a dumb question at all; it causes a lot of confusion. It has to do with a few different conventions. Oppenheim and Schafer, and Lyons use the following: y[n] = sum(k=0..M)(b[k]*x[n-k]) + sum(k=1..N)(a[k]*y[n-k]) sum(k=0..M)(b[k]*z^-k) H(z) = -------------------------- 1 - sum(k=1..N)(a[k]*z^-k) Rabiner and Gold, Parkus and Burrus (*), and Proakis and Manolakis use the following: y[n] = sum(k=0..M)(b[k]*x[n-k]) - sum(k=1..N)(a[k]*y[n-k]) sum(k=0..M)(b[k]*z^-k) H(z) = -------------------------- 1 + sum(k=1..N)(a[k]*z^-k) In addition, some sources use a[k] for the numerator and b[k] for the denominator coefficients. Also be warned, that some texts (like (*) above), don't always normalize a[0] to 1. They are all equivalent, except for the sign of the a[k]. With positive a[k] you subtract, and with negative a[k] you add. The moral of the story is to be triple sure of what you have when designing and implementing IIR filters, especially if you are using someone else's code. -- Matthew Donadio (m.p.donadio@ieee.org)

Matthew Donadio <m.p.donadio@ieee.org> wrote in message news:<pan.2003.07.10.20.38.48.728260@ieee.org>...> On Thu, 10 Jul 2003 13:25:13 -0700, Benjamin M. Stocks wrote: > > I have a beginner-esqe question: all of my DSP literature shows the > > IIR Difference Equation as: > > y(n) = b(0)x(n) + b(1)x(n-1) + b(2)x(n-2) + ... + a(1)y(n-1) + > > a(2)y(n-2) + ... > > > > But then the C code implementations I see are something along the > > lines: > > FeedForwardVariable = (b[0] * x[n]) + (b[1] * x[n-1]) + (b[2] * > > x[n-2]); > > FeedBackwardVariable = (a[1] * y[n-1]) + (a[2] * y[n-2]) + (a[3] * > > y[n-3]); > > ResultVariable = FeedForwardVariable - FeedBackwardVariable; > > > > The implementation uses a subtraction operation on the feed back > > calculations that I don't see in the Difference Equation. Why is that? > > > > I realize I'm setting myself up for a "Duh!!", but I'm really confused > > why this is the case. > > That is not a dumb question at all; it causes a lot of confusion. > > It has to do with a few different conventions. > > Oppenheim and Schafer, and Lyons use the following: > > y[n] = sum(k=0..M)(b[k]*x[n-k]) + sum(k=1..N)(a[k]*y[n-k]) > > sum(k=0..M)(b[k]*z^-k) > H(z) = -------------------------- > 1 - sum(k=1..N)(a[k]*z^-k) > > Rabiner and Gold, Parkus and Burrus (*), and Proakis and Manolakis use the > following: > > y[n] = sum(k=0..M)(b[k]*x[n-k]) - sum(k=1..N)(a[k]*y[n-k]) > > sum(k=0..M)(b[k]*z^-k) > H(z) = -------------------------- > 1 + sum(k=1..N)(a[k]*z^-k) > > In addition, some sources use a[k] for the numerator and b[k] for the > denominator coefficients. Also be warned, that some texts (like (*) > above), don't always normalize a[0] to 1. > > They are all equivalent, except for the sign of the a[k]. With positive a[k] > you subtract, and with negative a[k] you add. The moral of the story is > to be triple sure of what you have when designing and implementing IIR > filters, especially if you are using someone else's code.Thank you, I understand it much better now. Regards, Ben