Another fourier transform nmr question work shown

You know that Se(t) is even, so Se(-t)=Se(t). Also you know that Fe(t) is even. SoS(t) = Se(0) \exp(i2\pi f_0t)Fe(t)S(-t) = Se(0) \exp(-i2\pi f_0t)Fe(t)Now just add these two together and use the fact that Se(-t) = Se(t) and getS(t) + S(-t) = 2Se(0) \cos(2\pi f_0t)Fe(t)Thus the spectrum of the signal is given byG(f) = \int_{-\infty}^{\in
  • #1
johnq2k7
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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!

 
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  • #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!
 
  • #3
Are you sure it's not

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

Analogously

[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.
 
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  • #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!
 
  • #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.
 

FAQ: Another fourier transform nmr question work shown

1. What is a Fourier Transform in NMR?

A Fourier Transform in NMR (Nuclear Magnetic Resonance) is a mathematical technique used to convert a signal from its original form (time domain) to a different representation (frequency domain). In NMR, this is used to analyze the different frequencies of a sample's atomic nuclei and determine their chemical and structural properties.

2. How is a Fourier Transform performed in NMR?

In NMR, a sample is placed in a strong magnetic field and then subjected to a radiofrequency pulse. This pulse causes the atomic nuclei to absorb and emit energy, which is detected by a receiver coil. The resulting signal is then transformed using a Fourier Transform algorithm to produce a spectrum of frequencies.

3. What information can be obtained from a Fourier Transform NMR spectrum?

A Fourier Transform NMR spectrum provides information about the chemical and structural properties of a sample. The position and intensity of peaks in the spectrum correspond to the different types of atomic nuclei present in the sample and their relative abundance. This can be used to identify the chemical compounds present and determine their structure.

4. What are some applications of Fourier Transform NMR?

Fourier Transform NMR is widely used in chemistry, biochemistry, and material science for various applications. It is commonly used for structural analysis of organic molecules, identification of impurities in pharmaceuticals, and studying protein structures. It is also used in the quality control of food and beverages, as well as in environmental and forensic analysis.

5. What are the advantages of using Fourier Transform NMR over other techniques?

Fourier Transform NMR is a non-destructive and non-invasive technique, meaning the sample does not need to be altered or destroyed for analysis. It is also highly sensitive and can provide detailed information about the chemical and structural properties of a sample. Additionally, it is a relatively fast and cost-effective technique compared to other methods such as X-ray crystallography or mass spectrometry.

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