Norris Lecture 1 Flashcards
What are complex numbers?
The square root of -1 is denoted by i (sometimes by j=-i), and by definition i^2 = −1.
Complex numbers have a real and an imaginary part: z=a+ib. Where a is real and ib
imaginary. Imaginary numbers are a useful tool and very practical in the real world.
They can be used to provide a compact description of position coordinates in 2D as a
single number.
Consider two complex numbers: z1 = a1 + ib1;
z2 = a2 + ib2;
How does addition work in this context?
ADDITION i.e. z1 + z2
Addition: z1 + z2 = a1 + a2 + i(b1 + b2)
Consider two complex numbers: z1 = a1 + ib1;
z2 = a2 + ib2;
How does multiplication work in this context?
MULTIPLICATION z1z2
Multiplication: z1z2 = a1a2 − b1b2 + i (a2b1 + a1b2)
Consider two complex numbers: z1 = a1 + ib1;
z2 = a2 + ib2;
How does conjugation work in this context?
Conjugation:
z1∗ = a1 − ib1
hence
z1z1∗ = a1^2 + b1^2
When are two complex numbers equal?
if two complex numbers are equal then both their real and imaginary parts
must have the same value
Conjugation
Multiplying a number by its complex conjugate will always give a real value. This
is useful when confronted with a situation where the denominator is complex.
1/1 + i = (1/1 + i) × (1 − i/1 − i) = 1 − i/2
What is an Argand diagram?
Any complex number can be represented in a so-called Argand diagram as z=x+iy.
This representation is similar to standard Cartesian coordinates. We can also define
any point on the diagram in polar coordinates by r,θ instead of the standard coordinates
(x, y). Where r2 = x2 + y2 and tanθ = y/x. If we convert from polars to Cartesian then
we have: z=r(cos θ +isin θ ).
Why is it natural to describe circular motion using polar coordinates?
It is natural to describe circular motion using polar coordinates because whereas during
circular motion both x and y coordinates change constantly (normal coordinates), the
length of the vector does not change, only the angle (polar coordinates).
What is the angle change as function of time if the rotation is at a constant angular frequency?
φ (t) = ωt + θ
What is the locus of a point on the unit circle rotating at angular frequency ω,
and at coordinate (1,0) at t=0
this choice is equivalent to setting θ = 0 and the formula is z = cos (ωt) + i sin (ωt)
What is the background to understand z = cos (ωt) + i sin (ωt)?
The point moves with constant angular velocity, so that the angle φ (t) at a given
time will be the angular freuency ω (in units of radians per second) times the time t that
has passed plus the initial phase angle θ .
Look at the picture, at a time t, the real part of z (x) will equal the cosine of the
angle φ and the imaginary part of z (iy) will equal the sine of the angle. (Try drawing
a triangle if you do not see this immediately.)
What are different, yet equivalent ways to look at a phase shift?
- In an Argand diagram with two rotating points, then the instantaneous phase
difference is the difference in angular coordinate of the points at a given time.
You can think of this as a form of geometric phase difference. - The sine and cosine are essentially the same function, but they are shifted relative
to each other by a phase shift of 90 degrees (π/2). You can hence shift a sinusoidal function by adding a phase to it. This can be thought of as a trigonometric phase.
The second point can be understood by considering sin (φ + π/2) = sin (φ ) cos ( π/2) +cos (φ ) sin ( π/2) = cos φ
What is the general relationship between frequency and phase?
As shown in equation 1 the phase at a given time is given by
φ (t) = ωt
where we assume for simplicity that φ (t = 0) = 0 . If the frequency changes over time
then we have to integrate the frequency to obtain the total phase φ (t) = ∫( t - 0 ) ω (t′) dt′
What does Euler’s formula state?
Euler’s formula states that:
exp (ix) = cos x + i sin x, ∀x
and
exp (−ix) = cos x − i sin x
Non-examinable (deriving eulers formula)
Look up at page 4/5 (math for neuroimaging lecture notes) if interested
Use eulers formula to get cos(x)
cos x = (exp (ix) + exp (−ix))/2
Use eulers formula to get sin(x)
sin x = i (exp (−ix) − exp (ix))/2
what is de Moivre Theorem?
As we know that (exp x)^2 = exp (2x)
We can also derive de Moivre’s theorem that
(cos θ + i sin θ )^n = exp (inθ ) = cos nθ + i sin nθ
Which is a generally useful way of working out formulae for cosines and sines of multiple angles cos 2θ etc
What is the Extension to de Moivre Theorem
By extension consider
exp (i (A + B)) = exp (iA) exp (iB)
Hence
cos (A + B) + i sin (A + B) = (cos (A) + i sin (A)) (cos (B) + i sin (B))
Equating real and imaginary parts
cos (A + B) = cos (A) cos (B) − sin (A) sin (B)
and
sin (A + B) = cos (A) sin (B) + sin (A) cos (B)
We shall use these formulae later in the course.
What is a concise notation for circular motion?
At the start of the lecture we worked out that circular motion could be described
by:
z = cos (ωt) + i sin (ωt)
Now we see that this can be written as
z = exp (iωt)
This is a very important function as we shall see when we look at Fourier transforms.
If we have a non-zero initial phase then:
φ (t) = ωt + θ
Hence
z = exp (iφ (t)) = exp (i (ωt + θ )) = exp (iωt) exp (iθ )
How does rotation in MRI work?
The fundamental equation of nuclear magnetic resonance is ω = γB
This gives us the precession frequency as a function of field strength (γ = 2.67 × 108
rad T−1s−1 for protons).
If we only consider the static field then we term the frequency the Larmor frequency
ω0 = γB0
Strictly, as B is a vector then ω should also be a vector, but we normally dispense with
this.
What is the rotating frame of reference?
- We often need to consider the magnetisation orientation while discounting the Larmor frequency. For example when we want to know the phase of the mag-
netisation. - In the laboratory frame of reference the magnetisation vector will rotate at the
Larmor frequency and it is difficult to define a meaningful stationary axis to use as a point of reference. - If you consider a frame of reference rotating at the Larmor frequency then the magnetisation will be stationary and the phase will be easy to define. The x-axis
is then determined by the orientation of B1-vector during the excitation pulse. - If we apply a 90° along the x-axis then it will generate magnetisation along the y-axis. If we apply the B1-field along the y-axis then the magnetisation will be along the negative x-axis.
Magnetic field gradients
If we apply a magnetic field gradient then the total field at a coordinate x will be given by
ω (t) = γB0 + γxGx (t)
We often are just interested in differences in frequency to the Larmor frequency and
can write
ω (t) = γxGx (t)
If we consider gradients in all three directions then we can write
ω (x, y, z) = γ (xGx + yGy + zGz)
if we define a position vector r, and a gradient vector G as
r = ( x )
( y )
( z )
G = ( Gx )
( Gy )
( Gz )
Then we can write
ω (r) = γr.G
This allows us to determine the frequency at any position (r) for any combination of
applied gradients (G)
Effect of gradients: What are the three main ways in which we can use magnetic field gradients?
There are three main ways in which we can use magnetic field gradients:
- The instantaneous frequency will be proportional to the position coordinate
(ω (r) = γr.G). - After a gradient pulse has been applied the signal phase will be proportional to
the position coordinate
(φ (t) = ∫ (t-0) ω (t′) dt′ = ∫ (t-0) γr.G (t′) dt′) - The difference in phase as a function of position will reduce the signal intensity,
and if the gradient is strong enough it will even completely dephase the signal
How can the signal intensity be calculated?
The total signal will be given by the integral over the spatial domain we are inter-
ested in.
In general S = ∫ρ (r) exp (−iφ ) dr
Where S is the signal intensity, and ρ (r) describes the density distribution of the spins.
If you assume that the spin density is uniform (for example imaging a water phantom)
then you can set ρ (r) = 1, so in a specific instance you can calculate the total signal
by inserting the correct gradient value and limits of integration
