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" 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" 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" 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" 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 >  |  Next  |  Last Pages: 1 2 Prev: SI System urged to adopt Next: soft decisions: what decision function to use?