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 bristowjohnson 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 bj
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 =A0j*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 bj, 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 =A0j*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 bj >

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