(a) For \E < 2, we have
M 422 5
QSFECT(aOD) = f feXPcfw_ (/J MdyJr/ Midy'de
: f feXpcfw_ (mm/4 P 301m?) v4 ends:
: fexpwm/zl 7 e2)expcfw_m<5 , 4 7 0 [mammary
(a) The xray source detector apparatus rotates at a speed of 47r radians/s, so it takes 0.5 s to rotate a full Circle
(271'). During this period of time, the patient table moves 2 cm/ s X 0.5 s = 1 cm. So the pitch of the helix is
(a) From Table 10.1, we know that Z1 : 1.52 X 106 kgmZs1 and Z2 : 1.35 X 106 kgmzsl. Assume 20
is far away. Therefore the only return is from point (0, 0,20) on the interface. Therefore, 6? = 9,- : 6t 2 0;
cos 6,. = cos 6,- : cos 6,; = 1.
The basic relationship between the pulse sequence parameters and the Fourier frequencies is
u 2 nyt,
1) : cyAy.
There are four regions to consider.
(1) O S t S 1 ms: The 1.: component is given by
u : met
: 4, 258 Hz/G x 10 G/cmt
: 4,258 mm1
(a) From tables (internet or physics or chemistry textbooks),
mass of carbon-12 = 199264663 x 10726 kg.
From the information given in the problem statement, we calculate
mass of (6p + 6n + 6e) 2 20090759569 x 1026 kg.
The mass defect is there
The mass of an electron mE is 0000548 u. 50 IU = more. The equivalent energy of an electron is 511 keV.
So the equivalent energy of 1 u is 0.0010548 X 511 keV = 931 MeV.
(3) Using (7.8), the decay constant A is found as
The highest energy is determined by the peak x-ray tube voltage. For example, if the peak voltage is p kV,
then the peak x-ray energy will be p keV. The energy spectrum is determined by several factors. First, it will
be zero abov
SIGNALS AND THEIR PROPERTIES
(a) 65(33, 9) : Z:=_w Z:_Do 6U: m, y n) : E:=_OO 5(x m) -:=_OO 5(y n), therefore it is a
(d) 5(33, y) is a separable signal when ugvg : O. For example, if no : 0, scfw_x, y) : sincfw_27rv0y) is t
The magnetic eld B at 2 : 0 and z : 1 cm is
3(0) : 1 Tesla and B(1) : 1.5 Tesla.
The Larmor frequencies at these positions are
f(0) : 42.58 MHz and f(1) : 63.87 MHz.
The next time when the magnetization vectors on two planes have same phase
(3) Given hl (I) we rst nd the Fourier Transform H1(u) as follows:
f ex /5321ru1d$
co 2 .
: f e(I +3107rux)/5dI
00 2 - 2 2 2 2
: f e(I +3107ruI25'rr 11: 5851 H. d1?
2 2 - 2
: e57r 'u. f e(I+J5K'. [Sal-7:
Hence, the MTF is
(a) The system is separable because Mr, y) : eUIIHyI) : elmleiyl.
(b) The system is not isotropic since Mm, y) is not a function of?" : \/ $2 + yz.
Additional comments: An easy check is to plug in a: : 1, y : 1 and a: : O, y : x/E into h(55,
(3) Use Beers law for calculating the path length to in the septa to allow less than 60% incident photons to
pass through. If a is the linear attenuation coefcient for lead at 140 keV, then 67 S 0.60. This gives
u: 2 0.51 / it. From geometry,
By taking the derivatives of w1(z, t) with respect to z and t, we have
It is obvious that
: E(z at) + "(z + ct), : CZEH(Z at) + CZEH(Z + at) .
So, w2cfw_z, t) : cfw_z ct) +(ztct) is also a solution to the w