Hi,

I am going through the problems from "Signals and Systems" by Alan V. 
Oppenheim et. al. and I found this problem:

   To determine if x(t) is periodic and to find it's period if it is, 
where x(t) is:

   x(t) =  SIGMA (e^(-2t-n)) where n varies from minus infinity to plus 
infinity.

Now, as far as I can tell 'e' raised to anything that isn't a complex 
quantity is not going to be periodic.  Therefore, I concluded that x(t) 
as defined above is not periodic.

But then, I just can't tell what type of a signal this is.  Too many 
things that confuse me:

   1. For a given value of t, does the summation converge to a neat 
value or does it diverge?

   2. This mix of a continuous ('t') and discrete ('n') variable!  Never 
seen something like this.

   3. I wonder if this is a twisted representation of some 
familiar/popular/classic/obvious function.

What do you think?

Regards,
Anand

Am 28.06.2012 21:07, schrieb Anand P. Paralkar:
> Hi,
>
> I am going through the problems from "Signals and Systems" by Alan V.
> Oppenheim et. al. and I found this problem:
>
> To determine if x(t) is periodic and to find it's period if it is, where
> x(t) is:
>
> x(t) = SIGMA (e^(-2t-n)) where n varies from minus infinity to plus
> infinity.

Do you mean

\sum_{n=-\infty}^\infty e^{-2t - n}

and are you sure that you haven't forgotten a square somewhere in the 
exponent, as in

\sum_{n=-\infty}^\infty e^{-(t - n)^2} ?

For the former, use the functional relation of the exponential to write 
it as

.. =  e^{-2t} \sum_{n=-\infty}^\infty e^{-n}

where the latter sum clearly does not converge because -n->\infty as 
n->\infty and thus the sum explodes.

So this function is not defined. The latter sum, however, with the ^2 in 
the exponent, does converge for every t, it is actually absolutely 
convergent - it is bounded by the geometric series, and converges 
independent of t, so the limit is also continuous (actually, even 
smooth). It is obviously periodic with period one (substitute n->n+1). 
In fact, with a little substitution, you see that this is up to a phase 
just a special case of the theta function:

http://en.wikipedia.org/wiki/Theta_function

which is a complex function which is "as periodic" as a holomorphic 
function could ever be (periodic in one direction, periodic up to a 
phase in the second).

HTHH,
	Thomas

Hi,

probably it's a typo and the idea is that a sum 
c_k exp(i omega k t) 
over k = any set of integers
is periodic in 1/omega.

If so, each term is periodic on its own, so there is IMO not much left to
prove.

But I'm guessing on the nature of the question, so I may be completely
wrong.

What's the problem number? 

Here's the solution manual to the 2nd edition: 

http://www.scribd.com/doc/13377530/Signals-and-Systems-2E-Oppenheim-Solutions

Hi Thomas,

Thanks for your reply.  I checked the book again, I don't think I have 
forgotten a square.  I must however point-out that the first expression 
given in your reply should actually be:

   SIGMA(e^(-(2t-n)))

not:

   SIGMA(e^(-2t-n))


In either case, I am unable to make out anything of what's happening to 
the shape of the signal.

The reason why I posted this question is that this book has a learning 
at every possible nook and corner.  I am surprised to see a function 
that "leads to nowhere".

Thanks so much for looking into this,
Anand

On 29-06-2012 01:31, Thomas Richter wrote:
> Am 28.06.2012 21:07, schrieb Anand P. Paralkar:
>> Hi,
>>
>> I am going through the problems from "Signals and Systems" by Alan V.
>> Oppenheim et. al. and I found this problem:
>>
>> To determine if x(t) is periodic and to find it's period if it is, where
>> x(t) is:
>>
>> x(t) = SIGMA (e^(-2t-n)) where n varies from minus infinity to plus
>> infinity.
>
> Do you mean
>
> \sum_{n=-\infty}^\infty e^{-2t - n}
>
> and are you sure that you haven't forgotten a square somewhere in the
> exponent, as in
>
> \sum_{n=-\infty}^\infty e^{-(t - n)^2} ?
>
> For the former, use the functional relation of the exponential to write
> it as
>
> .. = e^{-2t} \sum_{n=-\infty}^\infty e^{-n}
>
> where the latter sum clearly does not converge because -n->\infty as
> n->\infty and thus the sum explodes.
>
> So this function is not defined. The latter sum, however, with the ^2 in
> the exponent, does converge for every t, it is actually absolutely
> convergent - it is bounded by the geometric series, and converges
> independent of t, so the limit is also continuous (actually, even
> smooth). It is obviously periodic with period one (substitute n->n+1).
> In fact, with a little substitution, you see that this is up to a phase
> just a special case of the theta function:
>
> http://en.wikipedia.org/wiki/Theta_function
>
> which is a complex function which is "as periodic" as a holomorphic
> function could ever be (periodic in one direction, periodic up to a
> phase in the second).
>
> HTHH,
> Thomas

:)

The problem number is 1.25 (f).

I tried the solution manual.  It has a terse "Not periodic".  There is 
an answer but no answer there! :)

Regards,
Anand


On 29-06-2012 19:46, David Drumm wrote:
> What's the problem number?
>
> Here's the solution manual to the 2nd edition:
>
> http://www.scribd.com/doc/13377530/Signals-and-Systems-2E-Oppenheim-Solutions