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Math Help - signals

  1. #1
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    signals

    Hi,
    I am taking a signal and system course. The book with all respect is not very helpful.

    I have this problem:


    show that

    \delta_{\epsilon}(t)=\frac{e(^\frac{-t}{\epsilon})}{\epsilon}* u(t)

    has the properties of a delta functionin the limit as \epsilon \rightarrow  0



    please can someone help me with this problem.
    I would like to know what to do because I have another problem of the same type.

    Thank you
    B
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  2. #2
    Senior Member
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    Quote Originally Posted by braddy View Post
    Hi,
    I am taking a signal and system course. The book with all respect is not very helpful.

    I have this problem:


    show that

    \delta_{\epsilon}(t)=\frac{e(^\frac{-t}{\epsilon})}{\epsilon}* u(t)

    has the properties of a delta functionin the limit as \epsilon \rightarrow  0




    please can someone help me with this problem.
    I would like to know what to do because I have another problem of the same type.

    Thank you
    B
    I gather that u(t) is the unit step function u(t) = 0 \text{ for } t < 0,\ u(t) = 1 \text{ for } t \ge 0.

    From this, you must show \lim_{\epsilon \to 0} \delta_{\epsilon} (t) = 0 \text{ for } t \ne 0 and \int_{-\infty}^{\infty} \delta_{\epsilon} (t) dt = 1.

    So given u(t) = 0 \text{ for } t < 0, you just have to show \lim_{\epsilon \to 0} \frac{e^{-t/\epsilon}}{\epsilon} (t) = 0 \text{ for } t > 0 and \int_{0}^{\infty}  \frac{e^{-t/\epsilon}}{\epsilon} (t)\ dt = 1.
    Last edited by JakeD; September 2nd 2007 at 11:25 AM. Reason: Added unit step function.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by braddy View Post
    Hi,
    I am taking a signal and system course. The book with all respect is not very helpful.

    I have this problem:


    show that

    \delta_{\epsilon}(t)=\frac{e(^\frac{-t}{\epsilon})}{\epsilon}* u(t)

    has the properties of a delta functionin the limit as \epsilon \rightarrow 0


    please can someone help me with this problem.
    I would like to know what to do because I have another problem of the same type.

    Thank you
    B
    To have the property of the \delta function as \epsilon \to 0 means that:

    <br />
\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \delta_{\epsilon}(t) f(t) dt = f(0)<br />

    for all functions f(t) from a suitable class of functions.

    RonL
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  4. #4
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    Quote Originally Posted by JakeD View Post
    I gather that u(t) is the unit step function u(t) = 0 \text{ for } t < 0,\ u(t) = 1 \text{ for } t \ge 0.

    From this, you must show \lim_{\epsilon \to 0} \delta_{\epsilon} (t) = 0 \text{ for } t \ne 0 and \int_{-\infty}^{\infty} \delta_{\epsilon} (t) dt = 1.

    So given u(t) = 0 \text{ for } t < 0, you just have to show \lim_{\epsilon \to 0} \frac{e^{-t/\epsilon}}{\epsilon} (t) = 0 \text{ for } t > 0 and \int_{0}^{\infty}  \frac{e^{-t/\epsilon}}{\epsilon} (t)\ dt = 1.
    Quote Originally Posted by CaptainBlack View Post
    To have the property of the \delta function as \epsilon \to 0 means that:

    <br />
\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \delta_{\epsilon}(t) f(t) dt = f(0)<br />

    for all functions f(t) from a suitable class of functions.

    RonL
    From the theory of weak convergence of probability distributions, it is a theorem that the conditions I gave

    \lim_{\epsilon \to 0} \delta_{\epsilon} (t) = 0 \text{ for } t \ne 0 and \int_{-\infty}^{\infty} \delta_{\epsilon} (t) dt = 1

    imply

    <br />
\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \delta_{\epsilon}(t) f(t) dt = f(0)<br />

    for all continuous functions f. So it may not be expected that this condition has to be shown for specific nascent delta functions in the exercises.
    Last edited by JakeD; September 2nd 2007 at 12:29 PM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by JakeD View Post
    From the theory of weak convergence of probability distributions, it is a theorem that the conditions I gave

    \lim_{\epsilon \to 0} \delta_{\epsilon} (t) = 0 \text{ for } t \ne 0 and \int_{-\infty}^{\infty} \delta_{\epsilon} (t) dt = 1

    imply

    <br />
\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \delta_{\epsilon}(t) f(t) dt = f(0)<br />

    for all continuous functions f. So it may not be expected that this condition has to be shown for specific nascent delta functions in the exercises.
    It is obvious that your condition implies mine (for suitably well behaved
    functions) and vice versa, but by the delta function property I understand:

    <br />
\int_{-\infty}^{\infty} \delta(t) f(t) dt = f(0)<br />

    so it is

    <br />
\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \delta_{\epsilon}(t) f(t) dt = f(0)<br />

    that has to be shown, which we may do by showing that it is sufficient to show that
    your version of the property holds.

    RonL
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  6. #6
    Member
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    Thank you Captain and Jake D.
    I will work on the suggestions and I 'll get back if I still dont get it.
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