> Oli Filth said the following on 28/08/2006 14:25:
>> Randy Yates said the following on 28/08/2006 14:24:
>>> According to both [proakiscomm] and [bracewell], the convolution integral
>>> is defined to be
>>>
>>> y(t)= \int_{-\infty}^{+\infty} x(\tau) g(t - \tau) d\tau.
>>>
>>> Notice that there is no dependence in the limits of integration on the
>>> independent variable "t."
>>>
>>> HOWEVER, in Proakis' second equation in deriving the matched filter,
>>> equation 5.1-15, which is describing the matched filter output, he
>>> writes:
>>>
>>> y_k(t) = \int_{0}^{t} r(\tau) h_k(t-tau) d\tau.
>>>
>>> Now since both r(t) and h_k(t) are non-zero only in the range
>>> 0 <= t < T, then the expression above is true. However, it seems
>>> to be non-standard, and since there is nothing subsequently
>>> that requires this form, unnecessarily complex.
>>> Thoughts? Why would Proakis write the equation in this form?
>>>
>> As it says in the line above that equation:
>> "h_k(t) = 0 outside of the interval 0 <= t <= T"
>>
>
> Oops, sorry, I missed that you already said that yourself. I think
> the point is that as we only need to integrate over T, we can
> implement it as an integrate-and-dump architecture.
Hi Oli,
The obvious alternate [non-t-dependent in the limits] representation
of the convolution given the domains of the functions in use would be
y_k(t) = \int_{0}^{T} r(\tau) h_k(t-tau) d\tau.
In that representation, the sum-and-dump becomes apparent when we let
t = T,
y_k(T) = \int_{0}^{T} r(\tau) h_k(T-tau) d\tau.
Again, I don't see the need to use the confusing form of the convolution.
--
% Randy Yates % "Maybe one day I'll feel her cold embrace,
%% Fuquay-Varina, NC % and kiss her interface,
%%% 919-577-9882 % til then, I'll leave her alone."
%%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr
Reply by Randy Yates●August 28, 20062006-08-28
Randy Yates <yates@ieee.org> writes:
> Hi Folks,
>
> According to both [proakiscomm] and [bracewell], the convolution integral
> is defined to be
>
> y(t)= \int_{-\infty}^{+\infty} x(\tau) g(t - \tau) d\tau.
Sorry, I meant to write
According to both [signalsandsystems] and [bracewell], the convolution integral
is defined to be
.
.
.
@BOOK{signalsandsystems,
title = "{Signals and Systems}",
author = "{Alan~V.~Oppenheim, Alan~S.~Willsky, with Ian~T.~Young}",
publisher = "Prentice Hall",
year = "1983"}
@BOOK{bracewell,
title = "{The Fourier Transform and Its Applications}",
author = "{Ronald~N.~Bracewell}",
publisher = "McGraw-Hill",
edition = "second",
year = "1986"}
--
% Randy Yates % "Remember the good old 1980's, when
%% Fuquay-Varina, NC % things were so uncomplicated?"
%%% 919-577-9882 % 'Ticket To The Moon'
%%%% <yates@ieee.org> % *Time*, Electric Light Orchestra
http://home.earthlink.net/~yatescr
Reply by Oli Filth●August 28, 20062006-08-28
Oli Filth said the following on 28/08/2006 14:25:
> Randy Yates said the following on 28/08/2006 14:24:
>> According to both [proakiscomm] and [bracewell], the convolution integral
>> is defined to be
>>
>> y(t)= \int_{-\infty}^{+\infty} x(\tau) g(t - \tau) d\tau.
>>
>> Notice that there is no dependence in the limits of integration on the
>> independent variable "t."
>>
>> HOWEVER, in Proakis' second equation in deriving the matched filter,
>> equation 5.1-15, which is describing the matched filter output, he
>> writes:
>>
>> y_k(t) = \int_{0}^{t} r(\tau) h_k(t-tau) d\tau.
>>
>> Now since both r(t) and h_k(t) are non-zero only in the range
>> 0 <= t < T, then the expression above is true. However, it seems
>> to be non-standard, and since there is nothing subsequently
>> that requires this form, unnecessarily complex.
>> Thoughts? Why would Proakis write the equation in this form?
>>
>
> As it says in the line above that equation:
>
> "h_k(t) = 0 outside of the interval 0 <= t <= T"
>
Oops, sorry, I missed that you already said that yourself. I think the
point is that as we only need to integrate over T, we can implement it
as an integrate-and-dump architecture.
--
Oli
Reply by Oli Filth●August 28, 20062006-08-28
Randy Yates said the following on 28/08/2006 14:24:
> According to both [proakiscomm] and [bracewell], the convolution integral
> is defined to be
>
> y(t)= \int_{-\infty}^{+\infty} x(\tau) g(t - \tau) d\tau.
>
> Notice that there is no dependence in the limits of integration on the
> independent variable "t."
>
> HOWEVER, in Proakis' second equation in deriving the matched filter,
> equation 5.1-15, which is describing the matched filter output, he
> writes:
>
> y_k(t) = \int_{0}^{t} r(\tau) h_k(t-tau) d\tau.
>
> Now since both r(t) and h_k(t) are non-zero only in the range
> 0 <= t < T, then the expression above is true. However, it seems
> to be non-standard, and since there is nothing subsequently
> that requires this form, unnecessarily complex.
>
> Thoughts? Why would Proakis write the equation in this form?
>
As it says in the line above that equation:
"h_k(t) = 0 outside of the interval 0 <= t <= T"
--
Oli
Reply by Randy Yates●August 28, 20062006-08-28
Hi Folks,
According to both [proakiscomm] and [bracewell], the convolution integral
is defined to be
y(t)= \int_{-\infty}^{+\infty} x(\tau) g(t - \tau) d\tau.
Notice that there is no dependence in the limits of integration on the
independent variable "t."
HOWEVER, in Proakis' second equation in deriving the matched filter,
equation 5.1-15, which is describing the matched filter output, he
writes:
y_k(t) = \int_{0}^{t} r(\tau) h_k(t-tau) d\tau.
Now since both r(t) and h_k(t) are non-zero only in the range
0 <= t < T, then the expression above is true. However, it seems
to be non-standard, and since there is nothing subsequently
that requires this form, unnecessarily complex.
Thoughts? Why would Proakis write the equation in this form?
--Randy
@BOOK{proakiscomm,
title = "{Digital Communications}",
author = "John~G.~Proakis",
publisher = "McGraw-Hill",
edition = "fourth",
year = "2001"}
@BOOK{bracewell,
title = "{The Fourier Transform and Its Applications}",
author = "{Ronald~N.~Bracewell}",
publisher = "McGraw-Hill",
edition = "second",
year = "1986"}
--
% Randy Yates % "Midnight, on the water...
%% Fuquay-Varina, NC % I saw... the ocean's daughter."
%%% 919-577-9882 % 'Can't Get It Out Of My Head'
%%%% <yates@ieee.org> % *El Dorado*, Electric Light Orchestra
http://home.earthlink.net/~yatescr