# signals

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• September 2nd 2007, 08:59 AM
braddy
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
• September 2nd 2007, 09:45 AM
JakeD
Quote:

Originally Posted by braddy
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 $\int_{0}^{\infty} \frac{e^{-t/\epsilon}}{\epsilon} (t)\ dt = 1.$
• September 2nd 2007, 10:32 AM
CaptainBlack
Quote:

Originally Posted by braddy
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:

$
\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \delta_{\epsilon}(t) f(t) dt = f(0)
$

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

RonL
• September 2nd 2007, 11:16 AM
JakeD
Quote:

Originally Posted by JakeD
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
To have the property of the $\delta$ function as $\epsilon \to 0$ means that:

$
\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \delta_{\epsilon}(t) f(t) dt = f(0)
$

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

$
\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \delta_{\epsilon}(t) f(t) dt = f(0)
$

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.
• September 2nd 2007, 12:40 PM
CaptainBlack
Quote:

Originally Posted by JakeD
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

$
\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \delta_{\epsilon}(t) f(t) dt = f(0)
$

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:

$
\int_{-\infty}^{\infty} \delta(t) f(t) dt = f(0)
$

so it is

$
\lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \delta_{\epsilon}(t) f(t) dt = f(0)
$

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
• September 2nd 2007, 02:50 PM
braddy
Thank you Captain and Jake D.
I will work on the suggestions and I 'll get back if I still dont get it.