TR TE Tissue Contrast

5 TR, TE, độ tương phản mô Giới thiệu Ở chương trước, thảo luận vai trò thành phần từ hóa dọc T1, thành phần từ hóa ngang T2 xung RF Thực tế Obviously, by doing the procedures described in the previous chapters only once, we won't be able to create an image To get any sort of spatial information, the process must be repeated multiple times, as we shall see shortly This is where TR and TE come into play The parameters TR and TE are related intimately to the tissue parameters T1 and T2, respectively However, unlike T1 and T2, which are inherent properties of the tissue and therefore fixed, TR and TE can be controlled and adjusted by the operator In fact, as we shall see later, by appropriate setting of TR and TE, we can put more “weight” on T1 or T2, depending on the type of clinical application How we actually measure a signal? With the patient in a large magnetic field (Fig 5-1a), we apply a 90° RF pulse, and the magnetization vector flips into the x-y plane (Fig 5-1b) Then, we turn off the 90° RF pulse, and the magnetization vector begins to grow in the z direction and decay in the x-y plane (Fig 5-1c) By convention, we apply the RF pulse in the x direction, and for that reason, in a rotating frame of reference, the vector ends up along the y axis (Fig 5-1b) Figure 5-1 A: Longitudinal magnetization before the RF pulse B: Immediately after the RF pulse, the magnet After a 90° RF pulse, we have decaying transverse magnetization Mxy (which is the component of the magnetization vector in the x-y plane) and recovering longitudinal magnetization Mz (which is the component of the magnetization vector along the z axis) Remember that the received signal can be detected only along the x axis, i.e., along the direction of the RF transmitter/receiver coil The receiver coil only recognizes oscillating signals (like AC voltage) and not nonoscillating voltage changes (like DC voltage) Thus, rotation in the x-y plane induces a signal in the RF coil, whereas changes along the z axis not At time t = 0, the signal is at a maximum As time goes by, because of dephasing (see Chapter 4), the signal becomes weaker in a sinusoidal manner (Fig 5-2) The decay curve of the signal is given by the following term: e-tT2* The sinusoidal nature of the signal is given by the equation cos ω t Therefore, the decaying sinusoidal signal is given by the product (e-t/T2*) (cos ωt) When t = 0: cos ω t = cos ω (0) = cos = e-t/T2* = e0 = When t = 0, (e-t/T2*)(cos ωt) = Thus, at time = 0, the signal is maximum (i.e., 100%) As time increases, we are multiplying a sinusoidal function (cos ωt) and a decaying function (e-t/T2*), which eventually decays to zero (Fig 5-3) Figure 5-2 The received signal (the FID) has a decaying sinusoidal waveform Figure 5-3 The product of a sinusoidal signal and an exponentially decaying signal results in a decaying sinus When we put a patient in a magnet, he or she becomes temporarily magnetized as his or her protons align with the external magnetic field along the z axis We then transmit an RF pulse at the Larmor frequency and immediately get back a free induction decay (FID; Fig 5-4) This process gives one signal—one FID—from the entire patient It doesn't give us any information about the location of the signal The FID is received from the ensemble of all the different protons in the patient's body with no spatial discrimination To get spatial information, we have to specify somehow the x, y, and z coordinates of the signal Here, gradients come into play The purpose of the gradient coils is to spatially encode the signal Figure 5-4 Immediately after transmission of the RF pulse, an FID is formed To spatially encode the signal, we have to apply the RF pulse multiple times while varying the gradients and, in turn, get multiple FIDs or other signals (e.g., spin echoes) When we put all the information from the multiple FIDs together, we get the information necessary to create an image If we just apply the RF pulse once, we only get one signal (one FID), and we cannot make an image from one signal (An exception to this statement is echoplanar imaging [EPI], which is performed after one RF pulse—see Chapter 22.) TR (The Repetition Time) After we apply one 90° pulse (the symbol we'll use for a 90° RF pulse is in Fig 5-5), we'll apply another The time interval between applications is called TR (the Repetition Time) Figure 5-5 The time interval between two successive 90° RF pulses is denoted TR What happens to the T1 recovery curve during successive 90° pulses (Fig 5-6)? Figure 5-6 The recovery curves after successive RF pulses • Immediately before time t = 0, the magnetization vector points along the z axis Call this vector M0 with magnitude M0 • Immediately after t = 0, the magnetization vector Mxy lies in the x-y plane, without a component along the z axis Mxy has magnitude M0 at t = 0+ • As time goes by and we reach time t = TR, we gradually recover some magnetization along the z axis and lose some (or all) magnetization in the x-y plane Let's assume at time TR the transverse magnetization Mxy is very small What happens if we now apply another 90° RF pulse? We flip the existing longitudinal magnetization vector (Mz) back into the x-y plane However, what is the magnitude of the magnetization vector Mz at the time TR? Because Mz (t) = M0 (1 - e-t/T1) then at t = TR, Mz (TR) = M0 (1 - e-TR/T1) • As we see in the T1 recovery curve, the magnetization vector (Mz) at time TR is less than the original magnetization vector M0 because the second 90° RF pulse was applied before complete recovery of the magnetization vector Mz • After the magnetization vector is flipped back into the x-y plane, it will begin to grow again along the z axis (according to the T1 recovery curve) until the next TR, when it will again be flipped into the x-y plane We now have a series of exponential curves that never reach full magnetization Received Signal Let's now take a look at the signal we are receiving (S) Because we are only applying a series of 90° pulses, the signal will be a series of FIDs: • At time t = 0, the initial signal will be a strong FID similar to that shown in Figure 57a • At time t = TR, the signal will be slightly less in magnitude but will also be an FID (Fig 5-7b) • At time t = 2TR, the signal will be equal in magnitude to that in b (Fig 5-7c) Figure 5-7 The FIDs after successive RF pulses: A: at t = 0; B: at t = TR; C: at t = 2TR Because the T1 recovery curve is given by the formula - e-t/T1, if we could measure the signal immediately after the RF pulse is given with no delay, then each FID signal would be proportional to: - e-TR/T1 (This cannot really happen in practice.) Up to now, the signal S is given by the formula: S ∝ - e-TR/T1 Remember that the word “signal” is really a relative term The signal that we get is a number without dimension, i.e., it has no units If we are dealing with a tissue that has many mobile protons, then, regardless of what the TR and T1 of the tissue are, we'll get more signal with more mobile protons (see Chapter 2) Thus, when considering the signal, we must also consider the number of mobile protons N(H) S ∝ N(H) (1 - e-TR/T1) For a given tissue, the T1 and the proton density are constant, and the signal received will be according to the above formula If we measure the FID at time TR immediately after the application of the second 90° RF pulse, it will measure maximal and be equal to N(H) (1 - eTR/T1) Therefore, the FIDs, which are acquired at TR intervals (i.e., 1TR, 2TR) are maximal if they can be measured right after the 90° pulse, i.e., right at the beginning of the FID However, in reality, we have to wait a certain period until the system electronics allows us to make a measurement TE (Echo Delay Time or Time to Echo) TE stands for echo delay time (or time to echo) Instead of making the measurement immediately after the RF pulse (which we could not anyway), we wait a short period of time and then make the measurement This short time period is referred to as TE Let's go back to the T2* decay curve and see what happens In the x-y plane, the FID signal decays at a very rapid rate because of two factors: • External magnetic field inhomogeneities • Spin-spin interactions The signal decays at rate T2* according to the decay function: e-t/T2* From this we see that if we take the signal measurement right away, before there is any chance for signal decay, the signal will be equal to the original magnetization (M 0) flipped into the x-y plane (point in Fig 5-8) However, if we wait a short time period (TE) before we make a measurement, the signal will look like point in Figure 5-8: M0 = (e-TE/T2*) Figure 5-8 The value of the FID at time is M0, whereas at time TE it is M0·e-TE/T2* Now we have to put the two curves together because both T1 recovery and T2 decay processes are occurring simultaneously (Fig 5-9) Figure 5-9 The recovery and decay curves plotted on the same graph Let's go back to the T1 recovery curve After the 90° RF pulse, the spins are flipped into the x-y plane After a time interval TR, the amount of received longitudinal magnetization is M0 (1 - e-TR/T1) Superimposed on this T1 recovery curve, we'll draw another curve, which is the T2* decay curve, with two new axes The T2* decay curve starts out at the value of M0 (1 - e-TR/TE) on the T1 recovery curve and then decays very quickly The decay rate of the new curve is given by T2* according to the formula e-t/T2* After a period of time TE, we can measure the signal The value of the signal at TE will be a fraction of the maximum signal intensity on the T1 recovery curve In other words, it will be the product of Equations 5-3a and 5-3b: Signal = S ∝ M0 (1 - e-TR/T1) (e-TE/T2*) The confusing thing about the diagram is that there are two sets of axes (Fig 5-9): • The first set of axes is associated with TR • The second set of axes is associated with TE If we draw them to scale, the T2 decay curve (time scale TE) will be decaying much faster than the T1 curve is recovering (time scale TR) However, the graph does give us a visual concept of what the final signal intensity is going to be Because the initial longitudinal magnetization M0 is proportional to the number of mobile protons, i.e., M0 ∝ N(H) then, in general, the signal intensity we measure is given by: Signal Intensity = SI ∝ N(H)(e-TE/T2*) (1 - e-TR/T1) (The difference between T2 and T2* is the correction for external magnetic field inhomogeneity achieved with spin-echo techniques.) Tissue Contrast (T1 and T2 Weighting) Let's see what happens when we deal with two different tissues So far we have been dealing with a single tissue, but now we'll consider two tissues: tissue A and tissue B Figure 5-10 A: Two tissues A and B with different T1s Which tissue has the longer T1? B: Consider two diffe If we draw just a tangent along each curve at the origin, tissue A has a longer T1 than tissue B However, just by looking at the curves, it takes tissue A longer to reach equilibrium than tissue B Let's say we have two different TRs: • Short TR = TR1 • Long TR = TR2 TR1 gives the better contrast Let's go back to Equation 5-4 and see if this answer makes sense: SI = N(H)(e-TE/T2*) (1 - e-TR/T1) where SI stands for signal intensity If TR goes to infinity, then - e-TR/T1 becomes If TR → ∞, then - e-TR/T1 → and SI → N(H)(e-TE/T2*) If TR is very long, we can get rid of the T1 component in the equation What this means in practice is that we eliminate (or, more realistically, reduce) the T1 effect by having a very large TR Long TR reduces the T1 effect We can't really achieve a long enough TR in practice to eliminate totally the T1 effect 100%, but we can certainly minimize the T1 effect with a TR of 2000 to 3000 msec (in general, if TR is to times T1, then the T1 effect becomes negligible) Let's go back to Figure 5-10b and see what happens at TR = TR1 At this point, the TR is not long enough to eliminate the T1 term in the equation (1 - eTR/T1) So we have: signal intensity (tissue A)/signal intensity (tissue B) = (1 - e-TR1/T1(tissue A))/(1 - e-TR1/T1(tissue B)) Because the T1s of tissue A and tissue B are different, the short TR brings out the difference in contrast between tissue A and tissue B Thus, for short TR, the two tissues can be differentiated on the basis of different T1s In other words, we get T1 tissue contrast with short TR Short TR enhances the T1 contrast We don't want TR to be very long when we're evaluating T1 because, as we've already learned, when TR → ∞, then (1 - e-TR/T1) approaches 1, thus eliminating the T1 effect However, we also don't want TR to be too short If TR is close to zero, then e -0/T1 = e0 = and - e-TR/T1 = - = In this situation, with very short TR, we end up with no signal Ideally, we would like to have a TR that is not much different from the T1 of the tissue under study T2* Tissue Contrast Let's consider T2* contrast between two tissues Again, if we graphically draw a tangent at t = for each curve, we see that tissue A has a longer T2* Put differently, it takes the signal from tissue A longer to decay than that from tissue B Figure 5-11 A: Two tissues A and B with different T2*s Which tissue has the longer T2? B: Consider two dif Let's pick two different TEs (Fig 5-11b) Here we have two TEs: • Short TE = TE1 • Long TE = TE2 E2 gives us more contrast Let's again look at the formula for signal intensity (Eqn 5-4): SI = N(H) (e-TE/T2*) (1 - e-TR/T1) If TE is very short (close to zero), then e-TE/T2* approaches TE → ⇒ e-TE/T2* → e0 = Then signal intensity = N(H)(1)(1 - e-TR/T1) = N(H) (1 - e-TR/T1) This means that, with a very short TE, we get rid of the T2* effect in the equation Therefore, we eliminate (or, again, in reality, reduce) the T2* effect by having a very short TE Short TE reduces the T2* effect We can see this graphically from the graph (Fig 5-11b) and mathematically from the equation (Eqn 5-3) When we have a long TE, we enhance T2* contrast between tissues Even though the signal to noise ratio is low (because there is greater signal decay for a longer TE), the tissue contrast is high Key Points • Long TR: reduces T1 effect • Short TR: enhances T1 effect • Short TE: reduces T2* (T2) effect • Long TE: enhances T2* (T2) effect ... have two different TRs: • Short TR = TR1 • Long TR = TR2 TR1 gives the better contrast Let's go back to Equation 5-4 and see if this answer makes sense: SI = N(H)(e -TE/ T2*) (1 - e -TR/ T1) where SI... different TEs (Fig 5-11b) Here we have two TEs: • Short TE = TE1 • Long TE = TE2 E2 gives us more contrast Let's again look at the formula for signal intensity (Eqn 5-4): SI = N(H) (e -TE/ T2*) (1 - e -TR/ T1)... e -TR/ T1) If TE is very short (close to zero), then e -TE/ T2* approaches TE → ⇒ e -TE/ T2* → e0 = Then signal intensity = N(H)(1)(1 - e -TR/ T1) = N(H) (1 - e -TR/ T1) This means that, with a very short TE,