The Basics of MRI

Chapter 5



A detailed description of the Fourier transform ( FT ) has waited until now, when you have a better appreciation of why it is needed. A Fourier transform is an operation which converts functions from time to frequency domains. An inverse Fourier transform ( IFT ) converts from the frequency domain to the time domain.

The concept of a Fourier transform is not that difficult to understand. Recall from Chapter 2 that the Fourier transform is a mathematical technique for converting time domain data to frequency domain data, and vice versa. You may have never thought about this, but the human brain is capable of performing a Fourier transform. Consider the following sine wave and note.

A musician with perfect pitch will tell us that this is middle C (261.63 Hz) on the western music scale. This musician will be able to also tell us that this sine wave is the first G above middle C (392 Hz),

and that this sine wave this note is a C one octave above middle C (523.25 Hz).

Some can tell the notes when more than one are played simultaneously, but this process becomes more difficult when more notes are played simultaneously. Play all of the above notes simultaneously. Can you hear which frequencies are simultaneously being played? The Fourier transform can! Change the relative amplitudes of the notes. Can you determine their relative amplitudes with your ear? The Fourier transform can!

The Fourier transform ( FT ) process is like the musician hearing a tone (time domain signal) and determining what note (frequency) is being played. The inverse Fourier Transform ( IFT ) is like the musician seeing notes (frequencies) on a sheet of music and converting them to tones (time domain signals).

The + and - Frequency Problem

To begin our detailed description of the FT consider the following. A magnetization vector, starting at +x, is rotating about the Z axis in a clockwise direction. The plot of Mx as a function of time is a cosine wave. Fourier transforming this gives peaks at both +ν and -ν because the FT can not distinguish between a +ν and a -ν rotation of the vector from the data supplied.

A plot of My as a function of time is a -sine function. Fourier transforming this gives peaks at +ν and -ν because the FT can not distinguish between a positive vector rotating at +ν and a negative vector rotating at -ν from the data supplied.

The solution is to input both the Mx and My into the FT. The FT is designed to handle two orthogonal input functions called the real and imaginary components.

Detecting just the Mx or My component for input into the FT is called linear detection. This was the detection scheme on many older NMR spectrometers and some magnetic resonance imagers. It required the computer to discard half of the frequency domain data.

Detection of both Mx and My is called quadrature detection and is the method of detection on modern spectrometers and imagers. It is the method of choice since now the FT can distinguish between +ν and -&nu, and all of the frequency domain data be used.

The Fourier Transform

An FT is defined by the integral

Think of f(ω) as the overlap of f(t) with a wave of frequency ω.

This is easy to picture by looking at the real part of f(ω) only.

Consider the function of time, f( t ) = cos( 4t ) + cos( 9t ).

To understand the FT, examine the product of f(t) with cos(ωt) for &omega values between 1 and 10, and then the summation of the values of this product between 1 and 10 seconds. The summation will only be examined for time values between 0 and 10 seconds.


The inverse Fourier transform (IFT) is best depicted as an summation of the time domain spectra of frequencies in f(ω).

Phase Correction

The actual FT will make use of an input consisting of a REAL and an IMAGINARY part. You can think of Mx as the REAL input, and My as the IMAGINARY input. The resultant output of the FT will therefore have a REAL and an IMAGINARY component, too.

Consider the following function:

f(t) = e-at e-i2πνt

In FT NMR spectroscopy, the real output of the FT is taken as the frequency domain spectrum. To see an esthetically pleasing (absorption) frequency domain spectrum, we want to input a cosine function into the real part and a sine function into the imaginary parts of the FT. This is what happens if the cosine part is input as the imaginary and the sine as the real.

In an ideal NMR experiment all frequency components contained in the recorded FID have no phase shift. In practice, during a real NMR experiment a phase correction must be applied to either the time or frequency domain spectra to obtain an absorption spectrum as the real output of the FT. This process is equivalent to the coordinate transformation described in Chapter 2

If the above mentioned FID is recorded such that there is a 45° phase shift in the real and imaginary FIDs, the coordinate transformation matrix can be used with = - 45°>. The corrected FIDs look like a cosine function in the real and a sine in the imaginary.

Fourier transforming the phase corrected FIDs gives an absorption spectrum for the real output of the FT.

The phase shift also varies with frequency, so the NMR spectra require both constant and linear corrections to the phasing of the Fourier transformed signal.

= m ν + b

Constant phase corrections, b, arise from the inability of the spectrometer to detect the exact Mx and My. Linear phase corrections, m, arise from the inability of the spectrometer to detect transverse magnetization starting immediately after the RF pulse. The following drawing depicts the greater loss of phase in a high frequency FID when the initial yellow section is lost. From the practical point of view, the phase correction is applied in the frequency domain rather then in the time domain because we know that a real frequency domain spectrum should be composed of all positive peaks. We can therefore adjust b and m until all positive peaks are seen in the real output of the Fourier transform.

In magnetic resonance imaging, the Mx or My signals are rarely displayed. Instead a magnitude signal is used. The magnitude signal is equal to the square root of the sum of the squares of Mx and My.

Fourier Pairs

To better understand how FT NMR functions, you need to know some common Fourier pairs. A Fourier pair is two functions, the frequency domain form and the corresponding time domain form. Here are a few Fourier pairs which are useful in MRI. The amplitude of the Fourier pairs has been neglected since it is not relevant in MRI.

Constant value at all time

Real: cos(2πνt), Imaginary: -sin(2πνt)

Comb Function (A series of delta functions separated by T.)

Exponential Decay: e-at for t > 0.

A square pulse starting at 0 that is T seconds long.

Gaussian: exp(-at2)

Convolution Theorem

To the magnetic resonance scientist, the most important theorem concerning Fourier transforms is the convolution theorem. The convolution theorem says that the FT of a convolution of two functions is proportional to the products of the individual Fourier transforms, and vice versa.

If f(ω) = FT( f(t) ) and g(ω) = FT( g(t) )

then f(ω) g(ω) = FT( g(t) ⊗ f(t) ) and f(ω) ⊗ g(ω) = FT( g(t) f(t) )

It will be easier to see this with pictures. In the animation window we are trying to find the FT of a sine wave which is turned on and off. The convolution theorem tells us that this is a sinc function at the frequency of the sine wave.

Another application of the convolution theorem is in noise reduction. With the convolution theorem it can be seen that the convolution of an NMR spectrum with a Lorentzian function is the same as the Fourier transform of multiplying the time domain signal by an exponentially decaying function.

The Digital FT

In a nuclear magnetic resonance spectrometer, the computer does not see a continuous FID, but rather an FID which is sampled at a constant interval. Each data point making up the FID will have discrete amplitude and time values. Therefore, the computer needs to take the FT of a series of delta functions which vary in intensity.

What is the FT of a signal represented by this series of delta functions? The answer will be addressed in the next heading, but first some information on relationships between the sampled time domain data and the resultant frequency domain spectrum. An n point time domain spectrum is sampled at δt and takes a time t to record. The corresponding complex frequency domain spectrum that the discrete FT produces has n points, a width f, and resolution δf. The relationships between the quantities are as follows.

f = (1/δt)

δf = (1/t)

Sampling Error

The wrap around problem or artifact in a magnetic resonance image is the appearance of one side of the imaged object on the opposite side. In terms of a one dimensional frequency domain spectrum, wrap around is the occurrence of a low frequency peak on the wrong side of the spectrum.

The convolution theorem can explain why this problem results from sampling the transverse magnetization at too slow a rate. First, observe what the FT of a correctly sampled FID looks like. With quadrature detection, the image width is equal to the inverse of the sampling frequency, or the width of the green box in the animation window.

When the sampling frequency is less than the spectral width or bandwidth, wrap around occurs.

The Two-Dimensional FT

The two-dimensional Fourier transform (2-DFT) is an FT performed on a two dimensional array of data.

Consider the two-dimensional array of data depicted in the animation window. This data has a t' and a t" dimension. A FT is first performed on the data in one dimension and then in the second. The first set of Fourier transforms are performed in the t' dimension to yield an ν' by t" set of data. The second set of Fourier transforms is performed in the t" dimension to yield an ν' by ν" set of data.

The 2-DFT is required to perform state-of-the-art MRI. In MRI, data is collected in the equivalent of the t' and t" dimensions, called k-space. This raw data is Fourier transformed to yield the image which is the equivalent of the ν' by ν" data described above.


  1. What is the Fourier transform of a 63 MHz sine wave?

  2. What is the Fourier transform of a 50 µs long square pulse?

  3. What is the Fourier transform of a 63 MHz sine wave which is turned on for only 50 µs?

  4. What is the width of the function in the answer to question 3 in Hz and in ppm when this signal is greater than 90% of its maximum value? Why is this result significant in NMR and MRI?

  5. What quadrature sampling rate is necessary to accurately record the FID from a 2000 Hz wide NMR spectrum?

  6. What is the Fourier transform of the quadrature signal from a 63 MHz sine wave? (Assume there are equal real 63 MHz cosine and imaginary 63 MHz sine components.)

  7. What is the Fourier transform of a 33 µs long square pulse?

  8. What is the Fourier transform of the 63 MHz sine wave of Question 1 which is turned on for only 33 µs?

  9. What is the width of this function in Hz and in ppm when this signal is greater than 90% of its maximum value? Why is this result significant in NMR and MRI?

  10. What quadrature sampling rate is necessary to accurately record the FID from a 500 Hz wide NMR spectrum?

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