Why Strange Form of Convolution Integral in Proakis' Matched Filter Discussion?

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

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

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

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

Oli Filth <catch@olifilth.co.uk> writes:

> 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