# Another fourier transform nmr question work shown need help!

1. Jan 4, 2010

### johnq2k7

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

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

2. Relevant equations

3. The attempt at a solution

2. Jan 4, 2010

### johnq2k7

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...

3. Jan 4, 2010

### phsopher

Are you sure it's not

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

4. Jan 4, 2010

### johnq2k7

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

5. Jan 4, 2010

### phsopher

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

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

Analogously

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

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
6. Jan 4, 2010

### johnq2k7

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

7. Jan 5, 2010

### phsopher

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

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

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.