# Exponential Decay function in NMR

## Homework Statement

How many oscillations occur before Mxy decays to approximately 1/3 of its initial value, for a Larmor frequency of 100 MHz and T2 of 100ms?

## Homework Equations

According to what I learned, we can express transverse relaxation with cosine, sine, complex number and exponential decay function (Fd(t)=exp(-t/T2)).
Thus, Mxy(t)= lMxy(t)l Fd(t)exp(-i(wt)).

## The Attempt at a Solution

I'm not sure where to put Larmor frequency in the formula. Are we trying to determine Fd(t) or Mxy?

If I assume that we are determining Fd(t) which is the exponential decay function, then I would do the following:
Set Fd(t)=100 MHz and t=100ms. Then, calculate for T2. From what I heard in the lecture, the answer should be 10^7 which I'm not getting.

Any help would be appreciated!

ehild
Homework Helper

## Homework Statement

How many oscillations occur before Mxy decays to approximately 1/3 of its initial value, for a Larmor frequency of 100 MHz and T2 of 100ms?

## Homework Equations

According to what I learned, we can express transverse relaxation with cosine, sine, complex number and exponential decay function (Fd(t)=exp(-t/T2)).
Thus, Mxy(t)= lMxy(t)l Fd(t)exp(-i(wt)).

## The Attempt at a Solution

I'm not sure where to put Larmor frequency in the formula. Are we trying to determine Fd(t) or Mxy?

If I assume that we are determining Fd(t) which is the exponential decay function, then I would do the following:
Set Fd(t)=100 MHz and t=100ms. Then, calculate for T2. From what I heard in the lecture, the answer should be 10^7 which I'm not getting.

Any help would be appreciated!

You need the time elapsed when Fd(t) decreases to one third of its initial value, that is, Fd(t)/Fd(0)=1/3. If you know the time you can calculate the number of oscillation from the period.

ehild

You need the time elapsed when Fd(t) decreases to one third of its initial value, that is, Fd(t)/Fd(0)=1/3. If you know the time you can calculate the number of oscillation from the period.

ehild

I'm still lost..
The time elapsed was not given in the question, but how would you approach to calculate the number of oscillation from the period if the time was given? We are eventually calculating for Mxy(t)?

ehild
Homework Helper
Mxy(t) is a product of an exponentially decreasing function |Mxy| Fd(t) and a periodic function exp(-iwt). I quess |Mxy| is a constant, independent of t.

Fd(t)=exp(-t/T2). T2 was given, T2=100 ms. Fd(0)=1. What is t if Fd(t)=1/3?

The period T is the time of a complete oscillation. The number of periods in one second is the frequency f, so T=1/f. f was given as 100 MHz, that means 105 oscillations in one second.
You get the time T1/3 needed for the amplitude to decrease 1/3 of its initial value from exp(-T1/3/T2)=1/3. How many oscillations are performed during that time?

ehild

Mxy(t) is a product of an exponentially decreasing function |Mxy| Fd(t) and a periodic function exp(-iwt). I quess |Mxy| is a constant, independent of t.

Fd(t)=exp(-t/T2). T2 was given, T2=100 ms. Fd(0)=1. What is t if Fd(t)=1/3?

The period T is the time of a complete oscillation. The number of periods in one second is the frequency f, so T=1/f. f was given as 100 MHz, that means 105 oscillations in one second.
You get the time T1/3 needed for the amplitude to decrease 1/3 of its initial value from exp(-T1/3/T2)=1/3. How many oscillations are performed during that time?

ehild

Thanks for explaining it for me. Just to clear things up, I want to confirm my understanding:
105 oscillations/sec. How many oscillations per 109 sec? (I got 109 by calculating Fd(t)=1/3).

105 oscillations : 1 second = x oscillations : 109 seconds
x = 1.09 x 107 oscillations

Then this is why the prof said ~ 107 in the class. Correct?

ehild
Homework Helper
I made a mistake, 100 MHz = 108 oscillation per one second, or 105 oscillations per one millisecond (ms).
You calculated the time needed to decrease the amplitude in milliseconds: it was 109 ms.
The method and the result is correct at the end.

ehild

1 person
thank you so much for your help!