Time scaling property of Fourier transform

Hi,

I have a question about the time scaling property of the Fourier
transform.
One can find two different formulas of the time scaling property in the 
literature. These formulas are:
1. x(at) <-> 1/|a| X(f/a)
2. x(at) <->  1/a X(f/a)
The first formula uses the absolute value for 1/a, the second one does 
not use the absolute value for 1/a.

In fact, when I try to prove this property using the substitution rule in

the integral I obtain the second formula. Could somebody explain me 
please where the absolute value for 1/a comes from in the first formula.

Anatol

Check hyopthesis..

Anatol ha scritto:
> Hi,
> 
> I have a question about the time scaling property of the Fourier
> transform.
> One can find two different formulas of the time scaling property in the 
> literature. These formulas are:
> 1. x(at) <-> 1/|a| X(f/a)
> 2. x(at) <->  1/a X(f/a)
> The first formula uses the absolute value for 1/a, the second one does 
> not use the absolute value for 1/a.
> 
> In fact, when I try to prove this property using the substitution rule in
> 
> the integral I obtain the second formula. Could somebody explain me 
> please where the absolute value for 1/a comes from in the first formula.
> 
> Anatol
> 
> 
> 

Hi Anatol,
The first equation can't be correct as written, since it would mean: 

x(at) <-> 1/|a| X(f/a)   =    1/|-a| X(f/a)   <->  x(-at)

That is, x(at)is not equal to x(-at). 

My guess is that the author of this equation was referring to the
magnitude of the Fourier Transform, and was ignoring the phase.  You can
make the equation correct by including the statement that the sign of the
phase must be changed if a < 0. That is,

If x(t) <->  Mag X(f) and  Phase  X(f)
then x(-t) <->  Mag X(f) and -Phase X(f)

Hope this helps.
Regards,
Steve Smith  




     

On Nov 11, 9:52 pm, "Anatol" <uana...@yahoo.com> wrote:
>
> I have a question about the time scaling property of the Fourier
> transform.
> One can find two different formulas of the time scaling property in the
> literature. These formulas are:
> 1. x(at) <-> 1/|a| X(f/a)
> 2. x(at) <->  1/a X(f/a)
> The first formula uses the absolute value for 1/a, the second one does
> not use the absolute value for 1/a.
>
> In fact, when I try to prove this property using the substitution rule in
> the integral I obtain the second formula. Could somebody explain me
> please where the absolute value for 1/a comes from in the first formula.
>

On Nov 12, 8:08 pm, "SteveSmith" <Steve.Smi...@SpectrumSDI.com> wrote:
> Hi Anatol,
> The first equation can't be correct as written, since it would mean:
>
> x(at) <-> 1/|a| X(f/a)   =    1/|-a| X(f/a)   <->  x(-at)
>

no, the first one *is* correct (and appears as so in every
communications text i have in my possession).  when "a" is negative
when you substitute the variable of integration, the limits of the
Fourier Transform integral gets swapped (it becomes an integral from
+inf to -inf) and you need to swap them back to fit the definition of
the continuous Fourier Transform.

r b-j

robert bristow-johnson wrote:

> On Nov 11, 9:52 pm, "Anatol" <uana...@yahoo.com> wrote:

(snip)

>>x(at) <-> 1/|a| X(f/a)   =    1/|-a| X(f/a)   <->  x(-at)

> no, the first one *is* correct (and appears as so in every
> communications text i have in my possession).  when "a" is negative
> when you substitute the variable of integration, the limits of the
> Fourier Transform integral gets swapped (it becomes an integral from
> +inf to -inf) and you need to swap them back to fit the definition of
> the continuous Fourier Transform.

Somehow this reminds me that the Dirac delta function is even.

delta(at) = delta(t)/|a| (for constant a).

-- glen

>On Nov 11, 9:52 pm, "Anatol" <uana...@yahoo.com> wrote:
>>
>> I have a question about the time scaling property of the Fourier
>> transform.
>> One can find two different formulas of the time scaling property in
the
>> literature. These formulas are:
>> 1. x(at) <-> 1/|a| X(f/a)
>> 2. x(at) <->  1/a X(f/a)
>> The first formula uses the absolute value for 1/a, the second one does
>> not use the absolute value for 1/a.
>>
>> In fact, when I try to prove this property using the substitution rule
in
>> the integral I obtain the second formula. Could somebody explain me
>> please where the absolute value for 1/a comes from in the first
formula.
>>
>
>On Nov 12, 8:08 pm, "SteveSmith" <Steve.Smi...@SpectrumSDI.com> wrote:
>> Hi Anatol,
>> The first equation can't be correct as written, since it would mean:
>>
>> x(at) <-> 1/|a| X(f/a)   =    1/|-a| X(f/a)   <->  x(-at)
>>
>
>no, the first one *is* correct (and appears as so in every
>communications text i have in my possession).  when "a" is negative
>when you substitute the variable of integration, the limits of the
>Fourier Transform integral gets swapped (it becomes an integral from
>+inf to -inf) and you need to swap them back to fit the definition of
>the continuous Fourier Transform.
>
>r b-j
>

Thanks,
The limits swapping explains the equation x(at) <-> 1/|a| X(f/a) 
for me now.
However, one can get very easely confused with the proof the authors 
provide in their textbooks. The variable substitution rule they use 
is based on substituting differentials by expressions on
differentials. For example, they substitute (dt) by (du/a). 
The situation when (du/a) is negative creates the impression that 
the limits swapping is already included in the equation.
 
In my opinion, the proof of this property is much more easy 
to follow if one uses this variable substitution rule, 

\int g(t) du/dt dt = \int g(u) du.

For this property, we have to apply the substitution rule given 
above rule for (u = at) and  (du/dt = a).
For that, we mupltiply first the Fourier transform by 
the neutral expression ((1/a)(a)), or equivalently by (1/a du/dt). 
In this way, we introduce (du/dt) in the Fourier integral and 
we can apply the variable substitution rule. Now, since we substitute 
(dt) by (du) and not by the expression (du/a), one is no more confused 
by the sign of (du/a). That is, he has always to consider the integral 
limits modifications because of the variable substition.

Anatol

Hi Anatol,
Sorry for the misinformation-- I missed a sign in my previous response,
and it is not correct. I agree that the first equation (with the absolute
value)is correct.  You can derive it from two other properties:

if x(t) <=> X(f),

then, x(at) <=> 1/a X(f/a), where a > 0

and also,  x(-t) <=> X(-f)

combining,

x(-at) <=> 1/a X(-f/a), where a > 0

which means,

x(at) <=> 1/|a| X(f/a), where a > 0 or a < 0

Regards,
Steve Smith

>Hi Anatol,
>Sorry for the misinformation-- I missed a sign in my previous response,
>and it is not correct. I agree that the first equation (with the
absolute
>value)is correct.  You can derive it from two other properties:
>
>if x(t) <=> X(f),
>
>then, x(at) <=> 1/a X(f/a), where a > 0
>
>and also,  x(-t) <=> X(-f)
>
>combining,
>
>x(-at) <=> 1/a X(-f/a), where a > 0
>
>which means,
>
>x(at) <=> 1/|a| X(f/a), where a > 0 or a < 0
>
>Regards,
>Steve Smith
>

Hi Steve,

Thank you for the posting. Your proof is nice. 
The only problem I see here is the fact that to prove 
x(-t) <=> X(-f)
we have to use the variable substitution rule 
that caused to me initially a lot of confusion 
because of isolated differentials used out of their 
natural context. 

Regards,
Anatol

On Nov 12, 7:44 pm, robert bristow-johnson <r...@audioimagination.com>
wrote:
> no, the first one *is* correct (and appears as so in every
> communications text i have in my possession).  when "a" is negative
> when you substitute the variable of integration, the limits of the
> Fourier Transform integral gets swapped (it becomes an integral from
> +inf to -inf) and you need to swap them back to fit the definition of
> the continuous Fourier Transform.

Now that we have that worked out, here is another puzzler.  Let

I = integral from -1 to +1 of 2/(1+x^2) dx and J = integral from -1 to
+1 of -2/(1+y^2) dy

so that J = -I.  Now, since the integrand is the derivative of
2.arctan(x), it is easy to
work out that I = pi and therefore J = -pi.   But, the substitution y
= 1/x transforms
I into J and so J = pi.  We conclude that pi = -pi so it must be that
pi = 0, which
simplifies many of the equations used in this group!

On Nov 13, 4:34 pm, "sarw...@YouEyeYouSee.edu" <dvsarw...@gmail.com>
wrote:
>
> Now that we have that worked out, here is another puzzler.  Let
>
> I = integral from -1 to +1 of 2/(1+x^2) dx and J = integral from -1 to
> +1 of -2/(1+y^2) dy
>
> so that J = -I.  Now, since the integrand is the derivative of
> 2.arctan(x), it is easy to
> work out that I = pi and therefore J = -pi.   But, the substitution y
> = 1/x

because division by zero is a problem, you have to bust the J integral
into two integrals (from -1 to 0 and from 0 to +1) when you make that
particular substitution and transform the limits of the integral in
accordance to the substitution.

r b-j