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 =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 >> > 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 Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
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