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From: cfy30 on 6 Jul 2010 02:26
I have a question on how to apply the Fourier transform time shift equality
on a sinusoidal signal.

A cosine signal after a 1/4 period delay is a sine signal.

Consider the follow Fourier transform equality
x(t) <-> X(f)
x(t - a) <-> exp(-j*omega*a)*X(f)

x(t) = cos(omega1*t)
X(f) => {imp(omega + omega1) + imp(omega - omega1)}/2

x(t - 1/4T) => exp(-j*pi/2)*X(f)
= -j*X(f)

I expect to see {imp(omega + omega1) - imp(omega - omega1)}/(2*j) but using
the Fourier transform equality, I got -j*X(f). What is wrong?

From: Rune Allnor on 6 Jul 2010 02:38
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 equality
> on a sinusoidal signal.
>
> A cosine signal after a 1/4 period delay is a sine signal.  
>
> Consider the follow Fourier transform equality
> x(t) <-> X(f)
> x(t - a) <-> exp(-j*omega*a)*X(f)
>
> x(t) = cos(omega1*t)
> X(f) => {imp(omega + omega1) + imp(omega - omega1)}/2
>
> x(t - 1/4T) => exp(-j*pi/2)*X(f)
>             =  -j*X(f)
>
> I expect to see {imp(omega + omega1) - imp(omega - omega1)}/(2*j) but using
> the Fourier transform equality, I got -j*X(f). What is wrong?  

Nothing. You just need to massage the expression a bit:

-j = 1 * (-j) = (j/j)* (-j) = (j*(-j)) / j = -(-1)/j = 1/j.

I other words

-j == 1/j.

Rune
From: robert bristow-johnson on 6 Jul 2010 03:02
On Jul 6, 2:26 am, "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 equality
> on a sinusoidal signal.
>
> A cosine signal after a 1/4 period delay is a sine signal.  
>
> Consider the follow Fourier transform equality
> x(t) <-> X(f)
> x(t - a) <-> exp(-j*omega*a)*X(f)
>
> x(t) = cos(omega1*t)
> X(f) => {imp(omega + omega1) + imp(omega - omega1)}/2
>
> x(t - 1/4T) => exp(-j*pi/2)*X(f)

what is this function of t being equated to a function of f??

perfesser gonna slap the back of yer hand fer that. we'll watch it on
youtube.


>             =  -j*X(f)
>
> I expect to see {imp(omega + omega1) - imp(omega - omega1)}/(2*j) but using
> the Fourier transform equality, I got -j*X(f). What is wrong?  

delay of 1/4 T means multiplying, the DTFT X(omega) by exp(-j*omega*T/
4) the sign for the imaginary part will be different for omega = -
omega1 as for +omega1.

r b-j
From: cfy30 on 6 Jul 2010 10:50
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.


cfy30

>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
>
From: cfy30 on 6 Jul 2010 10:52
Hi r b-j,

I think I understand now. omega is a variable in frequency domain. The
multiplication factor is different between -omega and +omega. Thanks!


cfy30

>On Jul 6, 2:26=A0am, "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)
>
>what is this function of t being equated to a function of f??
>
>perfesser gonna slap the back of yer hand fer that. we'll watch it on
>youtube.
>
>
>> =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
>
>delay of 1/4 T means multiplying, the DTFT X(omega) by exp(-j*omega*T/
>4) the sign for the imaginary part will be different for omega =3D -
>omega1 as for +omega1.
>
>r b-j
>
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