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From: Tim Wescott on 6 Jul 2010 11:56
On 07/06/2010 07:50 AM, cfy30 wrote:
(top posting fixed)
>> On 6 Jul, 08:26, "cfy30"<cfy30(a)n_o_s_p_a_m.yahoo.com> wrote:
>>> I have a question on how to apply the Fourier transform time shift
> equali=
>> ty
>>> on a sinusoidal signal.
>>>
>>> A cosine signal after a 1/4 period delay is a sine signal. =A0
>>>
>>> Consider the follow Fourier transform equality
>>> x(t)<-> X(f)
>>> x(t - a)<-> exp(-j*omega*a)*X(f)
>>>
>>> x(t) =3D cos(omega1*t)
>>> X(f) =3D> {imp(omega + omega1) + imp(omega - omega1)}/2
>>>
>>> x(t - 1/4T) =3D> exp(-j*pi/2)*X(f)
>>> =A0 =A0 =A0 =A0 =A0 =A0 =3D =A0-j*X(f)
>>>
>>> I expect to see {imp(omega + omega1) - imp(omega - omega1)}/(2*j) but
> usi=
>> ng
>>> the Fourier transform equality, I got -j*X(f). What is wrong? =A0
>>
>> Nothing. You just need to massage the expression a bit:
>>
>> -j =3D 1 * (-j) =3D (j/j)* (-j) =3D (j*(-j)) / j =3D -(-1)/j =3D 1/j.
>>
>> I other words
>>
>> -j =3D=3D 1/j.
>>
>> Rune
>>
> Hi Rune,
>
> cosine(omega1) = {imp(omega + omega1) + imp(omega - omega1)}/2
>
> With 1/4T delay and by using the equality, I got {imp(omega + omega1)
> + imp(omega - omega1)}/(2j) which is not sine(omega1). Sine is
> {imp(omega + omega1) - imp(omega - omega1)}/(2j). The sign of
> imp(omega - omega1) is different.

You have not correctly interpreted the line

>>> x(t - a)<-> exp(-j*omega*a)*X(f)

Evaluate this at the frequency where each impulse occurs
(i.e. e^(j*b) != e^(-j*b)).

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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