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Math Help - Discrete Convolution confusion

  1. #1
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    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.
    Last edited by chopet; December 25th 2007 at 05:26 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by chopet View Post
    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
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  3. #3
    Lord of certain Rings
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    Quote Originally Posted by chopet View Post
    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}
    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.
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    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)
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by chopet View Post
    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
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  6. #6
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    Oh. Suddenly I see it now.
    Thanks for your help, and have a blessed Christmas.
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