UNIVERSITY OF ILLINOIS, URBANA

UNIVERSITY OF ILLINOIS, URBANA-CHAMPAIGN
Department of Electrical and Computer Engineering
ECE 480: Magnetic Resonance Imaging
Homework 6 Solution
Fall 2014
5.14 The pulse excitation sequence and the corresponding k-space trajectory are depicted in the corresponding
γ
γ
figure below. In the figure, ∆ky = 2π
∆Gτ, Gy,n = n∆G and kx,max = 2π
Gx τ . Note that the acquisition
interval, Tacq , is [τ, 3τ ].
5.16 Phase-encoding occurs in a direction at an angle of 45◦ to the kx −axis. Let ∆ky0 be the frequency
increment along the phase-encode axis. The corresponding phase encoding gradient follows from the
γ
relationship ∆ky0 = 2π
∆GTpe , and
Gx,n = n∆G cos 45◦
Gy,n = n∆G sin 45◦
Frequency encoding is at an angle of −45◦ to the kx −axis. The required frequency-encoding gradients
are:
√
2G
◦
Gx = G cos(−45) =
√2
2G
Gy = G sin(−45)◦ =
2
γ
0
= 2π
where kx,max
GTacq /2. The designed excitation sequence is shown in Figure 2. Correspondingly,
Figure 5 depicts the k-space trajectory for one cycle of the pulse sequence. Please note that only the
larger solid arrow (corresponding to the Tacq interval in the pulse sequence) is acquired in each excitation
cycle.
5.19 (a) Note that the strength of Gx and Gy is given in the figure,
Z
+∞
Sa (t) =
ρ(x, y)e
− Tt
2
e
−iγ(B0 + 12 Gx+
√
3
Gy)t
2
dxdy
−∞
The corresponding k-space trajectory is a radial line −60◦ about the kx axis.
(b) Using a similar approach as 5.19(a)
Z +∞
√
1
3
− t
Sa (t) =
ρ(x, y)e T2 e−iγ(B0 + 2 Gx− 2 Gy)t dxdy
−∞
The corresponding k-space trajectory is a radial line 60◦ about the kx axis.
5.20 (a) For S1 (t), signal expression given the frequency-encoding gradient Gx is,
Z +∞
− t
S1 (t) =
ρ(x, y)e T2 e−iγ(B0 +Gx x)t dx.
−∞
1
If we denote the strength of the new frequency-encoding gradient as G0x , given that the strength of G0x is
twice as Gx , i.e., G0x = 2Gx , we can express S2 (t) as
Z +∞
0
− t
S2 (t) =
ρ(x, y)e T2 e−iγ(B0 +Gx x)t dx.
−∞
(b) The largest sampling interval allowed for S2 (t) is 21 ∆t.
(c) If we denote ρˆ(x) =
R +∞
−∞
ρ(x, y)dy, the relation between S1 (t) and ρˆ(x) can be expressed as
Z
+∞
S1 (t) =
ρˆ(x)e−iγGx xt dx.
−∞
Apply inverse Fourier transform with respect to t, ρˆ(x) can be expressed as
Z
γGx +∞
ρˆ(x) =
S1 (t)eiγGx t dt.
2π −∞
Note that ρe1 (x) =
1
2π
R +∞
−∞
S1 (t)eixt dt. By comparing the expression of ρˆ(x) and ρe1 (x), it is obvious that
ρe1 (x) =
Similarly, we can express ρe2 (x) as
ρe1 (x) =
1
x
ρˆ(
).
γGx γGx
1
x
ρˆ(
).
γ2Gx γ2Gx
2
Figure 1: Problem 5.14 (a) Excitation sequence and (b) desired k-space trajectory.
Figure 2: Problem 5.16 The designed pulse sequence.
3
Figure 3: Problem 5.16 Trajectory for one repetition of the pulse sequence.
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