EE225E/BIO265, Spring 2011Principles of MRI Miki Lustig

HW1 Solution

1. 78 RPM means 78 rotations per min which is 78 rotations per 60 secondsspeed at radius r: v = 78×2×π×r

60= 8.1681r per seconds

vouter = 8.1681× 4.75 insec

= 38.799 insec

vinner = 8.1681× 1.875 insec

= 15.315 insec

(a) vouter = fk

=⇒ k = fvouter

= 2038.799

kHzinsec

= 51.5 in−1

(b) vinner = fk

=⇒ k = fvinner

= 2015.315

kHzinsec

= 130.6 in−1

2. (a) The triangle waveform is best represented with a limited number of coefficients. Thisis because the triangle looks the most periodic and like a sign wave. Also, the othershapes such as the rectangular and bipolar pulse need high frequency components (i.e.more coefficients) to model the sharp edges. This can also be seen from the magnitudespectrum plots which show a rapid drop off for triangle 1/x2 and a lower dropoff for theother shapes (1/x for rectangle)

(b) The Gibbs ringing always occurs regardless of the number of coefficients used. The Gibbsringing phenomenon reflects the difficulty in modeling a discontinuous function with afinite number of harmonics. As the number of coefficients increases, the ringing centersat the discontinuities.

(c) As was discussed in lecture, we can thing of the k-space representation as:

F (k)rect(kx/W )

where rect(Kx/W ) is a windowing function of width W . When we inverse FT back tothe space domain we get:

F−1{F (k)rect(Kx/W )} = f(x) = f(x) ∗W sinc(Wx)

where f(x) is an approximated version of the “perfect” f(x). As W → ∞ the sincapproaches an impulse and so our reconstruction is almost perfect. Thus the width ofthe Gibbs ringing is reduced. However, the height of W sinc(Wx) increases as W → ∞wich is why the amplitude of the ringing persists.

3. (Thanks, Maryam!)

1

(a) Scaling Theorem

F[f(x

a,y

b)]

= |ab|F (akx, bky)

=

∫ ∞−∞

∫ ∞−∞

f(x

a,y

b)e(−i2πkxx)e(−i2πkyy)dxdy

=

∫ ∞−∞

∫ ∞−∞|ab|f(

x

a,y

b)e(−i2πakx

xa)e(−i2πkyb

yb)d(

x

a)d(

y

b)

let u =x

a

let v =y

b

=

∫ ∞−∞

∫ ∞−∞|ab|f(u, v)e(−i2πakxu)e(−i2πbkyv)dudv

= |ab|F [f(u, v)]

= |ab|F (akx, bky)

(b) Shift Theorem

F [f(x− a, y − b)] =

∫ ∞−∞

∫ ∞−∞

f(x− a, y − b)e−i2πkxxe−i2πkyydxdy

=

∫ ∞−∞

∫ ∞−∞

f(x− a, x− b)e−i2πkx(x−a)e−i2πky(y−b)e−i2πkxae−i2πkybd(x− a)d(y − b)

let u = x− alet v = y − b

= e−i2πkxae−i2πkyb∫ ∞−∞

∫ ∞−∞

f(u, v)e(−i2πkxu)e(−i2πkyv)dudv

= e−i2πkxae−i2πkybF (kx, ky)

2

(c) Convolution Theorem

F [f(x, y) ∗ g(x, y)] =

∫ ∞−∞

∫ ∞−∞

f(x, y) ∗ g(x, y)e−i2πkxxe−i2πkyydxdy

=

∫ ∞−∞

∫ ∞−∞

[∫ ∞−∞

∫ ∞−∞

f(u, v)g(x− u, y − v)dudv

]e−i2πkxxe−i2πkyydxdy

=

∫ ∞−∞

∫ ∞−∞

[∫ ∞−∞

∫ ∞−∞

f(u, v)g(x− u, y − v)e−i2πkxxe−i2πkyydxdy

]dudv

=

∫ ∞−∞

∫ ∞−∞

f(u, v)[G(kx, ky)e

(−i2πkxu)e(−i2πkyv)]dudv

= G(kx, ky)

∫ ∞−∞

f(u, v)e(−i2πkxu)e(−i2πkyv)dudv

= F (kx, ky)G(kx, ky)

(d) Derivate Theorem

F[

´f(x)]

= F

[d

dxf(x)

]= F

[d

dx

∫ ∞−∞

F (k)e(i2πkx)dk

]= F

[∫ ∞−∞

(i2πk)F (k)e(i2πkx)dk

]= (i2πk)F (k)

4. This is an important problem since the reconstruction resembles a method in MRI calledparallel imaging in which the sensitivity of local coils is used to accelerate scans by reducingthe phase encodes.

(a) The Nyquist rate of f(x) is 1[1/Day], so f(x) is bandlimited with maximum frequency1/2[1/Day].

The Fourier transform F{f(x)} = F (k). From the derivative property we get thatF{f(x)} = 2πikF (k). Which means that f(x) is also bandlimited with the same band-width as f(x)

The student samples both f(x) and f(x) every two days. So we get:

f(x) = f(x)1

2X(

x

2)

ˆf(x) = f(x)1

2X(

x

2)

3

-1/2 1/2 -1/2 1/2

-1/2 1/2-1/2 1/2

From the figure we see that we need only concentrate on the interval −12< k < 1

2

For the interval k > 0 we have,

F{f(x)}(k) =1

2

(F (k) + F (k − 1

2)

)F{ ˆf(x)}(k) = iπ

(kF (k) + (k − 1

2)F (k − 1

2)

)which we can write as,[

12

12

iπk iπ(k − 12)

] [F (k)

F (k − 12)

]=

[F{f(x)}(k)

F{ ˆf(x)}(k)

]so, [

F (k)F (k − 1

2)

]=i4

π

[iπ(k − 1

2) −1

2

−iπk 12

][ F{f(x)}(k)

F{ ˆf(x)}(k)

]. Similarly for the interval k < 0 we get[

F (k)F (k + 1

2)

]=−i4π

[iπ(k + 1

2) 1

2

−iπk 12

][ F{f(x)}(k)

F{ ˆf(x)}(k)

]. So, for the entire interval we get:

F (k) = 24(2k)F{f(x)}(k)− 2i

πsign(k)F{ ˆf(x)}(k)

where 4(k) is a triangle function. From F (k) we can get f(x).

4

(b) Since we had to invert an non orthogonal 2× 2 matrix, noise is amplified. The amplifi-cation is different for different k, which results in colored noise.

5. Interpolation in k-space: In this problem we will see the effect of linear interpolation ink-space. This is a more involved version of the question that was given in the midterm anda simplified example of the “rotation in k-space challenge” I presented in class on severaloccasions.

The following image was resampled in the vertical direction in k-space by linear interpolationat 3

4the original rate.

before resampling after resampling

cross sectioncross section

There are two distinct artifacts that appear in the interpolated image and not in the original.The interpolated image is apodized (no longer has flat magnitude). In addition, there are somealiasing artifacts that appear as replication of the image in the interpolated dimension. Theseartifacts are the result of using a linear interpolation as opposed to a better interpolationscheme. In the following questions we will learn the source of these artifacts through a simple1D example.

You are given the following 1D signal that you want to sample in k-space. In this question,f(x) is the space signal and F (k) is its k-space.

16-16 0

x, [cm]

-32 32-48 48

f(x) 25.6

a) You decide to collect 256 k-space samples of F (k) with ∆k = 8256

cycles/cm. Expressf(x), the result of sampling F (k) in k-space, as a function of f(x). Draw the signal f(x).

5

Emphasize any artifacts in the drawing and point out the differences with the originalf(x).Solution: There are going to be two effects. Blurr due to the finite extent in k-spaceand replication due to the discrete sampling. Since ∆k = 8

256and the number of samples

is 256 the total extent in k-space is Wkz = 256 8256

= 8 cycles/cm. The finite extentcan therefore be written as a multiplication with the rect function u(k/8). The discretesampling at rate of ∆k = 8

256cycles/cm can be written as a multiplication with the Shah

function 2568X(256

8k). So, we can write F (k) as,

F (k) = F (k) · 256

8X(

256

8k) · u(

k

8).

Taking the inverse Fourier transform yields,

f(x) = f(x) ∗ 8sinc(8x)︸ ︷︷ ︸blurr

∗X(8

256x)

︸ ︷︷ ︸replication

=8

256

∞∑n=−∞

(f(x) ∗ 8sinc(8x)δ(x− 32n)

16-16 0 x, [cm]-32 32-48 48

1/32

ringingblurr^f(x)

b) You prepare to show the results from part (a) to your adviser, however, you suddenlyremember that he asked you to sample k-space at a finer interval of ∆k = 6

256cycles/cm.

Not wanting to repeat the experiment, you decide to interpolate F (k) and resampleit accordingly. Write an expression for F (k), the resampled signal, assuming linearinterpolation. Express F (k) as a function of F (k),HINT: linear interpolation is a convolution with an appropriately scaled triangle function

Λ(ak) =

{1− |ak| |ak| < 1

0 otherwise

Solution:Linear interpolation is a convolution with a triangle function. Our samples are spaced at8

256cycles/cm interval, so the convolution is with Λ(256

8k). The result Λ(256

8k) ∗ F (k) is a

continuous function. In order to resample, we need to multiply the continuous functionwith an impulse train at 6

256cycles/cm interval. Therefore we get that,

F (k) =

(Λ(

256

8k) ∗ F (k)

)256

6X(

256

6k)

6

c) Write an expression for f(x), the image domain representation of F (k) from part (b).First express f(x) as a function of f(x) and then as a function of f(x). Draw f(x),pointing out any artifacts due to the interpolation and resampling.(HINT: First try to draw the effect of the interpolation, then draw the effect of theresampling.)Solution:For this question we need to know that the Fourier transform of a triangle function is,F−1{Λ(k)} = sinc2(x). So, F−1{Λ(ak)} = 1

|a|sinc2(x/a). This can also be derived fromthe relation Λ(k) = u(k) ∗ u(k).So, the convolution and resampling yields,

f(x) =

(f(x) · 8

256sinc2(

8

256x)

)∗X(

6

256x).

and,

f(x) =

((f(x) ∗ 8sinc(8x) ∗X(

8

256x)

)· 8

256sinc2(

8

256x)

)∗X(

6

256x).

Drawinf f(x) · 8256

sinc2( 8256x) we get,

16-16 0 x, [cm]-32 32-48 48

~f(x)

in which the replicas have not been fully suppressed and the signal heavily apodized.

7

Convolving the result with X( 6256x) we get,

16-16 0 x, [cm]-32 32-48 48

16-16 0 x, [cm]-32 32-48 48

~f(x)

aliasing

d) Linear interpolation can be considered as interpolating with a (not so good) TBW=2filter. How would the result change if we use a TBW=8 windowed sinc filter for theinterpolation? Draw the result pointing out the differences.Solutions:A TBW=8 windowed sinc is a better filter and suppresses the replicas much better asillustrated below:

16-16 0 x, [cm]-32 32-48 48

~f(x)

slight apodization

8

The replication of the signal will have negligible aliasing within the field of view!

16-16 0 x, [cm]-32 32-48 48

~f(x)

almost no aliasing

9