Notes 2-2-99

Question: Contrast-to-Noise = (Contrast between 2 tissues)/Noise

Suppose contrast is 1. Then, if SNR is 50:1, the noise is 0.2

Suppose image, look at SI as fxn of position.

Outside image, have noise, inside have signal plus noise.

We measure the difference between A and B, and subtract out the noise at the left. As can be easily imagined, if the noise increases, the distinction between A and B will increase.

Note: Multi-echo sequence.

90 180 180

In phase Out of Phase In Out In

Te1 = time between 90 and signal refocused by first 180

Te2 = time between 90 and signal refocused by second 180

Typical ranges from 10 - 150 msec.

New stuff for today: Spatial encoding

FT - > function of s , has both real and imaginary components

Square wave => sinc fxn

When we acquire thing in scanner, put spins in transverse plane, get signal.

Use two antenna systems, get 2 signals, both 90 out of phase. Label cosine one real, label sine one imaginary.

Our object will produce 2 signals, a real and imaginary component.

Somehow, we get twice as much data! Where is the redundancy?

Let's look at odd and even functions. Even functions => real, odd => all imaginary

f(x) = E(f(x)) + O(f(x))

E(x) = E(-x)

O(x)=-O(-x)

Any function => unique even and odd parts.

E(x) = (f(x)+f(-x))/2

O(x) = (f(x)-f(-x))/2

Let f(x) = E(x) + O(x)

If f(x) even, F(s) all real

If f(x) odd, F(s) all imaginary

If f(x) combo, F(s) complex

For a real fxn, the real part of FT is even.

The imaginary part of FT is odd.

The redundancy turns out to be that k-space is Hermitian for real images, that is, F(-s) = complex conjugate of F(s).

Hence, we need only listen to signal for 1/2 time to make a complete image.

Do a gradient echo sequence, put only positive x-gradient (assume ideal gradients), get right 1/2 of k-space.

Why do this? Less time.

Note te is time to zero-point in k-space.

Recall k-space: Time between successive lines, tr. Total imaging time tr*# lines.

Can cut imaging time in half!

Can also get very low te's! (Because contrast is determined by center of k-space.)

Note, can only use this trick once.

Now, fast imaging:

Rapid acquisiton w/ refocused echoes (RARE). Fast-spin echo (FSE). Multi-echo sequence: Put 180's very close together (5-10 msec) so T2 differences small.

Note, apply our dephasing lobe positively before 180, equivalent to negative gradient after pulse.

Apply phase encoding w/ +/-.

Can do this multiple times.

Advantage: timing.

Limits, T1 of tissue.

Problem, lose signal w/ each successive echo.

Say 16 lines of k-space. Use conjugate symmetry so only 8.

Say use double FSE. Get cyclic intensity variation. Result in intensity w/ bright point in image.

They do not try to correct, but do every 4 lines interlaced so reduction smooth.

What frequencies are we talking about?

Say grad 4000Hz/cm. See dephasing for big sphere. Don't see it for a point.

So, center of raw data space has low-spatial frequencies. Edges give details (high spatial frequencies). Things fall off rapidly as you go to the outer edges of k-space.

Back to FSE:

Possible to modulate contrast by determining where center of k-space lies.

Te(effective) = time from 90 to center of k-space

Handwave: Center of k-space determines contrast. Hence definition of effective te.

All T2s use FSE so faster. However, contrast behavior is not simple T2 weighted. Have spatially dependent contrast. Low-spatial frequency = >te contrast. High-frequency => te+/- something contrast. Minimize cycling of lines by interleaving.

Diffusion, will look at later.

Reality: Don't have perfect gradients.

How make gradient? Spool of wire, run e- current, produce b-field linear w/ position. Coils have inductance. Stores energy in B-field that counteracts tendency to change current. Resist changes in current measured by L, inductance. V = LdI/dt = EMF which counteracts. To get 0 ramp time, need infinite V.

Typically, we put constant voltage, so current changes linearly. Get it in .2 to 0.5 msec.

Regarding k-space, we have to make sure the gradient-time product results in even divisions so we get linear sampling of k-space.

For EPI this is very important, because we can sample unevenly in time.