FPGA Central

Hi all,

If h is the impulse response of an FIR filter then Sum[h(n) Exp(j w n)]
describes its frequency response. What is the range of n for an odd length
filter? Is it n = -N : N, or n = 0 : 2N? How about even-length filter; will
it be n = 0 : 2N-1 or n = -N/2 : N/2?

Regards,
Ishtiaq.

"I. R. Khan" <[email protected]> writes:

> Hi all,
>
> If h is the impulse response of an FIR filter then Sum[h(n) Exp(j w n)]
> describes its frequency response. What is the range of n for an odd length
> filter? Is it n = -N : N, or n = 0 : 2N? How about even-length filter; will
> it be n = 0 : 2N-1 or n = -N/2 : N/2?

It could be any of the above or something else yet.

In general, the n in the sum you listed runs from negative infinity to
positive infinity. The limiting factor is the non-zero extent of the
impulse response h[n], and that simply depends on the system being
represented.

For example, let's say you are given a filter with response from n =
-N to +N (an odd filter of length 2*N+1). If a delay of 2*N samples is
added to that system, the new system still has length 2*N+1 but runs
instead from N to 3*N, which is neither of your odd-length forms.
--
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%% Fuquay-Varina, NC % things were so uncomplicated?"
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http://home.earthlink.net/~yatescr

> "I. R. Khan" <[email protected]> writes:
>
> > Hi all,
> >
> > If h is the impulse response of an FIR filter then Sum[h(n) Exp(j w n)]
> > describes its frequency response. What is the range of n for an odd
length
> > filter? Is it n = -N : N, or n = 0 : 2N? How about even-length filter;
will
> > it be n = 0 : 2N-1 or n = -N/2 : N/2?

Ishtiaq,

It would help if you'd very carefully write the equation out completely.
Then you'd have the answer:
N
H(w)=sum [h(n)*e^-jwn] where N is the number of individual coefficients in
the filter polynomial expression.
n=1

N is the number of coefficients in the filter. So, you see?, this answers
the question.

If you choose to change how this is implemented for efficiency, then you
could change the number of multiplies and adds. Maybe that's what you were
wondering. And then, it's not "odd length" or "even length" so much as it's
"symmetrical" or "anti-symmetric" or not either one that makes the
difference.

Fred

Thanks Fred,

In fact the thing which confused me was representation of H in sin/cos forms
for (anti) symmetric filters.

Let h = {h1, h0, h1} be the impulse response, then

H1 = h1 e^jw + h0 + h1 e^(-jw )= h0 + 2h1 cos (w)

Or

H2 = h1 + h0 e^(-jw) + h1 e^(-2jw) = e^(-jw)[h0 + 2h1 cos (w)] = e^(-jw) H1

What is the importance of this e^(-jw)? How does it affect the magnitude and
phase responses? Is this some kind of measure of delay?

Regards,
Ishtiaq.

"I. R. Khan" <[email protected]> wrote in message news:c0ucjd$1asiv4$[email protected]..
> Thanks Fred,
>
> In fact the thing which confused me was representation of H in sin/cos forms
> for (anti) symmetric filters.
>
> Let h = {h1, h0, h1} be the impulse response, then
>
> H1 = h1 e^jw + h0 + h1 e^(-jw )= h0 + 2h1 cos (w)
>
> Or
>
> H2 = h1 + h0 e^(-jw) + h1 e^(-2jw) = e^(-jw)[h0 + 2h1 cos (w)] = e^(-jw) H1
>
> What is the importance of this e^(-jw)? How does it affect the magnitude and
> phase responses? Is this some kind of measure of delay?

The shifting property of the Fourier Transform where <-> refers to a time<->frequency dual
for:
f(t) <-> F[w]
then:
f(t-t0) <-> F[w]e^-jwt0

So, yes, to multiply by e^-jwt0 in frequency represents a shift in time > is a delay (or advance) ... exactly.

Similarly:
f(t)e^jw0t <-> F(w-w0) for a shift in frequency (which is less often useful)

Fred