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Another fourier transform nmr question work shown need help!

  1. Jan 4, 2010 #1
    The "Free Induction Decay signal" (FID) is a particular type of NMR signal observed in both MRI and MRS. An idealized representation of the signal Sf(t) is given by

    Sf(t)= Sf(0) exp (-i2pi(f_0)(t))*exp(-t/T2*) t>=0
    Sf(t)= 0

    it was proven that Gf(f) corresponding to this signal is given by:

    Gf(f)= Sf(0) { [(T2*)/ (1+(2pi(f-f_f0)T2*)^2)] + [i2pi(f-f0)(T2*)^2/(1+ (2pi(f-f0)T2*)^2)]}

    a.) Show that the spectrum of the echo is given by

    Ge(f)= Se(0) { 2T2*/ (1+ (2pi(f-f0)T2*)^2}

    b.)using properties of even and odd func. and shift theorem, show that img. part of spect. must equal zero for any signal of form:

    S(t)= S(0) exp (i(2pi)(f0)t)Fe(t) , where Fe(t) is an even func. of t, Fe(t) does not need to be exp.

    Work shown:

    for part a.) if you ignore the img. part

    then Gf(f) is S(0) {{ T2*/ (1+ (2pi(f-f0)T2*)^2) + { T2*/ (1+ (2pi(f-f0)T2*)^2)}

    therefore, Ge(f)= Se(0) { 2T2*/ (1+ (2pi(f-f0)T2*)^2}

    however, i'm not sure my proof here is correct

    for part b.)

    i'm not sure how to use the even and odd func. to prove the signal equals zero. especial for img section

    Please help!!!

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 4, 2010 #2
    i forgot to include the echo signal info Se(t) info, which can be rep. by

    Se(t)= Se(0) exp (i(2pi(f0)(t))* exp (-|t|/T2*) from negative inf. to inf

    therefore, since the |t| is the magn. of t,

    therefore i assumed the img. part can be ignored which was helpful in my assumption. in my work shown for my solultion to part a. which i dont' believe is correct...

    please help!
  4. Jan 4, 2010 #3
    Are you sure it's not

    Se(t)= Se(0) exp (-i(2pi(f0)(t))* exp (-|t|/T2*)?
  5. Jan 4, 2010 #4
    your right!.. i made a mistake it's actually

    Se(t)= Se(0) exp (-i(2pi(f0)(t))* exp (-|t|/T2*)?

    can you please help me with part a and b please
  6. Jan 4, 2010 #5
    Then just do the integral. You can't just ignore things, you must keep the whole expression. Based on the info in the first post I presume Gf(f) comes from

    [tex]G_f(f) = \int_{-\infty}^{\infty} S_f(t) e^{i 2 \pi ft} dt = \int_{0}^{\infty} S_f(0) \exp \left(-i2\pi f_0t - t/T_2\right) e^{i 2 \pi ft} dt[/tex]


    [tex]G_e(f) = \int_{-\infty}^{\infty} S_e(t) e^{i 2 \pi ft} dt = \int_{-\infty}^{\infty} S_e(0) \exp \left(-i2\pi f_0t - |t|/T_2\right) e^{i 2 \pi ft} dt[/tex]

    Divide the integral into two parts - from -infinity to 0 and from 0 to infinity. That way you don't have to worry about |t| (think about the signs). You'll notice that the second integral is the same as in Gf(f) and that the first is similar only with a different sign.
    Last edited: Jan 4, 2010
  7. Jan 4, 2010 #6
    since t>0 and |t|=t and for t< 0 |t|= -t

    and since the inverse fourier transform integral for t>=0 is given .. .u can add the the results for the inverse transform for t<=0 to the result to get Ge(f)

    however, how do u integrate the expression for

    neg. infinity to zero for integral of Se(0)exp(-i*2pi*f0*t -t/T2)exp (i(2pi)ft) dt

    is it simply integral of exp (-t/T2) instead of integral of exp (-|t|/T2) dt from neg. inf. to zero

    therefore, it's exp (-t/T2) since the other parts are imaginary

    i'm confused please help!
  8. Jan 5, 2010 #7
    Why do you keep thinking that you can ignore the imaginary parts? YOU CAN'T!!! Just calculate the integral, it's almost identical to the ones we've done previously. When t is between -infinity and 0 it is negative so |t|=-t like you said. Thus

    [tex]\int_{-\infty}^{0}S_e(0) \exp \left(-i2\pi f_0t - |t|/T_2\right) e^{i 2 \pi ft} dt = \int_{-\infty}^{0}S_e(0) \exp \left(-i2\pi f_0t + t/T_2\right) e^{i 2 \pi ft} dt = \int_{-\infty}^{0}S_e(0) \exp \left[t/T_2 + i2\pi (f - f_0)t \right] dt[/tex]

    Now just calculate this like we did previously.

    As to the b part, I'm not familiar with the shift theorem, but I don't think it's necessary. Use the fact that eix = cosx +isinx.
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