1. Dec 19, 2009

### 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 S(t) is given by:

S(t)= S(0)exp (i*w_0*t)exp(-t/T2*), t>=0
S(t)=0 , t<0

You showed the spectrum G(w) corresponding signal is given by:

G(w)= S(0) {((T2* + (w-w_0)(T2*)^2))/ (1+[(w-w_0)T2*]^2}

However, the exact form of the spectrum above is obtained assuming that the signal is acquired for an infinite period of time. In reality the signal is acquired over a finite time that will be denoted here as T. In that case the shape of the spectrum is obtained by assuming that S(t)=0, for t>T and that leads to the following spectrum:

G(w)= S(0){(T2* + i*(w-w_0)(T2*)^2 / (1 + [(w-w_0)T2*]^2} (1- X)

where X is a complex parameter that depends on T,T2*,w, and w_0

a.) Determine the value of the magnitude of X for the case where T=T2*

b.) if T=a*T2*, determin the min. value of the parameter 'a' such that |X|<0.01
(|X| rep. the magnitude of 'X')

Work shown:

For a.)

do ignore the imaginary part for the exact eqn. and sub in T for T2* and solve for 'X'?

If so,

I get X= 1- S(0){(T/ (1 + [(w-w_0)T]^2}

For b.)

if |X|<0.01 and T= aT2* then, T2*= T/a

then i get 1- S(0){((T/a)/ (1 + [(w-w_0)(T/a)]^2} < 0.01

how do i solve for a.

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

2. Relevant equations

3. The attempt at a solution

2. Dec 20, 2009

### phsopher

I don't think a Fourier transform of S(t) gives G(w), there appears to be a sign mistake somewhere (either in the expression of S(t) or in that of G(w). But in any case, in the first case the G(w) appears to be supposed to be given by

$$\int_0^{\infty} S(t) e^{-iwt} dt$$

The upper limit gives 0. In the second case G(w) is given by

$$\int_0^{T} S(t) e^{-iwt} dt$$

Now both limits give a contribution - the lower limit gives the same result as in previous case and the upper limit gives the same thing multiplied by something. That something is X. Carry out the integration to find out what it is. Then just plug in the desired values for T and find the things you are after.

3. Dec 21, 2009

### johnq2k7

therefore,

since, S(t) is given as S(0)exp(iw_0*t) exp (-t/T2*)

therefore would the integral of the 2nd one be:

integral from (0 to T) of S(t)exp(-i*w*t)
= integral from (0 to T) of S(0){(T/ (1 + [(w-w_0)T]^2} exp (-iw*t)

i'm confused how these expression lead to the value of X?

what do i do from here?

4. Dec 21, 2009

### phsopher

$$S(t)= S(0)e^{iw_0t}e^{-t/T_2}$$

so

$$\int_0^{T} S(t) e^{-iwt} dt = \int_0^{T} S(0)e^{iw_0t}e^{-t/T_2}e^{-iwt} dt$$

Do the integration and you'll see what the X should be.

5. Dec 21, 2009

### johnq2k7

Would you only integrate the exp (-t/T2*) expression, which is equivalent to exp(-t/T)
since T=T2* in that case

so do you just find the integral of exp(-t/T) treating t as a constant?

therefore the integral is (-t/1)(exp(-t/T)

therefore the expression becomes S(0){(T/ (1 + [(w-w_0)T]^2} exp (-iw*t)) (-t/1)(exp(-t/T)

what's X?

6. Dec 21, 2009

### phsopher

Why would you only integrate one term? I don't get it. Just integrate normally, it's not a difficult integral:

$$\int_0^{T} S(0)e^{iw_0t}e^{-t/T_2}e^{-iwt} dt = S(0) \int_0^{T} \exp\left[-(\frac{1}{T_2} + i(w-w_0))t\right]dt$$

Replace the stuff that doesn't depend on t with some constant if that makes it easier.

Last edited: Dec 21, 2009
7. Jan 4, 2010

### johnq2k7

for the that integral... the expression consists of an imaginary number i with t.. and additionally your integrating the expression from 0 to T for t, therefore u can't integrat the T2 expression right?.. how do u you go about integrating the expression...

i'm confused.. i'm sure it's easier than i'm thinking

8. Jan 4, 2010

### phsopher

Ok, denote

$$a = \frac{1}{T_2} + i(w-w_0)$$

This is just a constant, i.e. a number that has nothing to do with integration. So your integral becomes

$$\int_0^{T} S(0)e^{iw_0t}e^{-t/T_2}e^{-iwt} dt = S(0) \int_0^{T} \exp\left[-(\frac{1}{T_2} + i(w-w_0))t\right]dt = S(0) \int_0^{T} e^{-at}dt$$

Do you know how to integrate it now?

9. Jan 4, 2010

### johnq2k7

then you would get the expression, S(0){-(e^(a*t))/a} from 0 to T

so you would get: S(0)*-{((e^aT)/a)-(1/a))

how do you determine the expression for X then,

if a= 1/T2 + i(w-w_0)

therefore you get, S(0){-((e^(1/T2 + i(w-w_0)t)/a)) -1/(1/T2+ i(w-w_0)}

what is X in this case?

and if T=a*T2*, how do you determine the min. value of the parameter 'a' such that |X|<0.01
(|X| rep. the magnitude of 'X')?

10. Jan 4, 2010

### phsopher

$$G_{old}(w) = \int_0^{\infty} S(0)e^{iw_0t}e^{-t/T_2}e^{-iwt} dt = S(0) \int_0^{\infty} \exp\left[-(\frac{1}{T_2} + i(w-w_0))t\right]dt = S(0) \int_0^{\infty} e^{-at}dt = \frac{S(0)}{a} = S(0)\frac{T_2-i(w-w_0) T_2^2}{1+(w-w_0)^2 T_2^2}$$

Note the different sign from your original expression - like I said earlier I think one of the original expressions has a sign mistake. If there isn't one then the expression for G(w) is derived in a way unknown to me and you'll have to show how you did it.

Anyway, your new expression for G is

$$G_{new}(w) = \int_0^{T} S(0)e^{iw_0t}e^{-t/T_2}e^{-iwt} dt = \frac{S(0)}{a}\left(1-e^{-aT}\right)$$

According to your original post $$G_{new}(w) = G_{old}(w)(1-X)$$

Do you see what X is now?

As to the second question, set |X| < 0.01 and solve for a.

EDIT: of course this a is not the same a that I used earlier to denote the expression on the exponent. Poor choice of notation on my part.

11. Jan 4, 2010

### johnq2k7

so if it's gnew(w)= g_old(w)(1+X)

isn't X=e^-aT

or based on there expression

X= e^{(-(T2-i(w-wo)T2^2)/(1+(w-wo)^2*T2^2)*T}

for part b.) if T=a*T2

therefore how do u solve, do u simply sub in a*T2 for T into the expression

X= |e^{(-(T2-i(w-wo)T2^2)/(1+(w-wo)^2*T2^2)*(a*T2)}| <0.01

how do u find the min. value in this case?

sorry for disturbing, there is still some slight confusion

12. Jan 4, 2010

### johnq2k7

since, it asks for the magnitude of X, this is the 'real part of the solution so the imaginary part can be ignored right from my previous expression..

therefore i get

X= e^{(-((T2-i(w-wo)T2^2)/(1+(w-wo)^2*T2^2))*T}
=e^{(-((T2/(1+(w-wo)^2*T2^2))*T}

therefore for part b, it's

X= e^{(-((T2/(1+(w-wo)^2*T2^2))*(a*T2)}<0.01

i'm confused.. because apparently it's suppose to be X= e^-1 i think.. and then for b.

e^-1<0.01

13. Jan 4, 2010

### phsopher

Yes, X = e^-aT.

Remember that

$$a = \frac{1}{T_2} + i(w-w_0)$$

so

$$X = \exp\left[-\frac{T}{T_2} - i(w-w_0)T\right]$$

From which it follows that

$$|X| = \exp(-\frac{T}{T_2})$$

If T=aT2 then

$$|X| = \exp(-a)$$

What is the smallest a for which this is smaller than 0.01?

14. Jan 4, 2010

### johnq2k7

since

e^(-a)<0.01
-a<(ln 0.01)

therefore is the smallest value of a less than 0.01... ln 0.01?

15. Jan 4, 2010

### phsopher

-a < ln 0.01
a > -ln 0.01 = -ln (10^-2) = 2 ln 10

So the smallest value of a is 2 ln 10 (or rather with this a |X| is exactly 0.01)

16. Jan 4, 2010

### phsopher

In light of new information that the signal is supposed to be

$$S(t)= S(0)\exp (-iw_0t)exp(-t/T_2), t\geq0; S(t)=0 , t<0$$

The correct Fourier transform is

$$\int_0^{\infty} S(t) e^{iwt} dt$$

Otherwise the whole thing goes in similar way.

17. Jan 4, 2010

### johnq2k7

in that case.. is the solution for X and part b correct?

if that's the integral... it is an imaginary integral so how do u solve that part.. i'm confused?

18. Jan 5, 2010

### phsopher

You solve the integral in the exact same way that we've done previously, the only difference is that some signs have changed. Now instead of

$$\int_0^{T} S(0)e^{iw_0t}e^{-t/T_2}e^{-iwt} dt$$

we have

$$\int_0^{T} S(0)e^{-iw_0t}e^{-t/T_2}e^{iwt} dt$$

The integral is almost identical to the one that we've done except for a couple of minus signs. I've already explained how to calculate such an integral so reread the thread. Combine the exponentials, denote the part that doesn't depend on t with some letter if that makes it easier and find X.

The b part should be the same since it doesn't depend on the imaginary part.