# Discrete Convolution confusion

• December 23rd 2007, 08:36 PM
chopet
Discrete Convolution confusion
I am confused by the definition of convolution:

given 2 series {1,1} and {1,1},
what is their convolution?

Do I perform as below, taking the product of values only when Row1 and Row2 overlap:
Step 1:
......1 1....
...1 1.........
-------------
......1

Step 2:
... 1 1...
... 1 1...
-------------
... 1+1 = 2

Step 3:
...1 1...
..... 1 1
----------------
..... 1

Total = 1+2+1=4

Hence, the convolution of {1,1} and {1,1} is 4.

But I have a textbook which stated:

If $a_k = b_k = 1$ for all $k\geq0$,
then $c_k=a_k*b_k = k+1$

where { $c_k$} is defined as the convolution of sequences { $a_k$} and { ${b_k}$}.

If I use the value of the first example, k=2, then $c_k=3$

If you can clear the air, I would very much appreciate it.
• December 23rd 2007, 09:15 PM
CaptainBlack
Quote:

Originally Posted by chopet
I am confused by the definition of convolution:

given 2 series {1,1} and {1,1},
what is their convolution?

:
:

If you can clear the air, I would very much appreciate it.

Go back to your notes and look up the definition of discrete convolution you
have been given. In particular we need to know if you are doing circular
convolution, if not how you are treating values of the sequences for index
outside the range you have defined them for.

RonL
• December 23rd 2007, 11:06 PM
Isomorphism
Quote:

Originally Posted by chopet
I am confused by the definition of convolution:

given 2 series {1,1} and {1,1},
what is their convolution?

Do I perform as below, taking the product of values only when Row1 and Row2 overlap:
Step 1:
1 1
1 1
-------------
1

Step 2:
1 1
1 1
-------------
1+ 1 = 2

Step 3:
1 1
1 1
----------------
1

Total = 1+2+1=4

Why are you adding it? {1,1} * {1,1} = {1,2,1}
Quote:

But I have a textbook which stated:

If $a_k = b_k = 1$ for all $k\geq0$,
then $c_k=a_k*b_k = k+1$

where { $c_k$} is defined as the convolution of sequences { $a_k$} and { ${b_k}$}.

Hmm this notation confuses me and It is definitely not true from the context. Look at $c_3$ for example in the above example.
• December 25th 2007, 05:16 AM
chopet
Quote:

Originally Posted by CaptainBlack
Go back to your notes and look up the definition of discrete convolution you
have been given. In particular we need to know if you are doing circular
convolution, if not how you are treating values of the sequences for index
outside the range you have defined them for.

RonL

Hi. The convolution was defined in the context of random varialbes in the textbook as below:

Let X and Y be 2 random variables taking on integer values only.
P{X=j} = $a_j$
P{Y=j} = $b_j$

Then we have a new random variable S = X + Y.
Given a certain integer S = r, we have the following:

$c_r = P\{S=r\} = a_0 b_r + a_1 b_{r-1} + a_2 b_{r-2} + .. + a_r b_0$

Introducing the convolution sign, we can put it as:
$\{c_k\} = \{a_k\} * \{b_k\}$

This definition doesn't seem to be similar to the more commonly defined convolution:
$\sum f(n)g(m-n)$
• December 25th 2007, 05:36 AM
CaptainBlack
Quote:

Originally Posted by chopet
Hi. The convolution was defined in the context of random varialbes in the textbook as below:

Let X and Y be 2 random variables taking on integer values only.
P{X=j} = $a_j$
P{Y=j} = $b_j$

Then we have a new random variable S = X + Y.
Given a certain integer S = r, we have the following:

$c_r = P\{S=r\} = a_0 b_r + a_1 b_{r-1} + a_2 b_{r-2} + .. + a_r b_0$

Introducing the convolution sign, we can put it as:
$\{c_k\} = \{a_k\} * \{b_k\}$

This definition doesn't seem to be similar to the more commonly defined convolution:
$\sum f(n)g(m-n)$

These are the same definition as long as you take $a_k=0, b_k=0,\ k<0$.

RonL
• December 25th 2007, 09:22 AM
chopet
Oh. Suddenly I see it now.
Thanks for your help, and have a blessed Christmas.