Declaration of Conflict of Interest or
Relationship
Speaker Name: Greig Scott
Consultant for Boston Scientific. Our lab also receives funding from
General Electric Healthcare.
I have no conflicts of interest to disclose with regard to the subject matter
of this presentation
Theory of RF
Reciprocity
Greig Scott
Department of Electrical Engineering, Stanford,
Topics
• Intro to Reciprocity
• Faraday Law & Reciprocity
• Circular Polarization Mathematics
• Equivalent Sources
• Lorentz Reciprocity
• Applications
PMRIL
Stanford Electrical Engineering
What is Reciprocity?
Left hand
polarized
Response to a source is unchanged when source and
measurer swapped.
What is Reciprocity?
Left hand
polarized
Response to a source is unchanged when source and
measurer swapped.
Series Resistor Capacitor
J =I/A
A
I (ω )
I (ω )
d1
σ E1=J/σ
d2
Id 2
=
V
ε E2=J/jωε
jωεA
Id 2 d1
V=
jωεA
d2
σ E1=J1=0
ε E2=i/jωεA
+ V= I
j ωC
V=
I (ω )
I
+
j ωC -
I (ω )
Response is unchanged when source & observer are swapped
Assume ejωt so fields match electric circuit convention
Time Domain Convolution
I (ω )
+
V (ω ) =
-
I (t ) = δ (t )
R
I (ω )
1 + jωRC
H (ω )
h(t )
Fourier
transform
pair
I (t )
v(t ) = ∫ i (τ )h(t − τ )dτ
Frequency Domain Output: Product of input & filter response.
Time Domain Output: convolution of input and impulse response.
Spin Precession
dM
= γM × B
dt

ω = −γBo z
1H, 31P, 19F:
17O,
γ>0, negative (left hand) angular velocity
electrons : γ< 0, positive (right hand) angular velocity
Faraday Law & Reciprocity
M
B
I=1
dΦ
emf = ∫ E ⋅ dl = −
dt
d
emf = − ∫ B ⋅ Mdv
dt V
Quasistatic approximation of B. EMF imply loop contour integral.
Hoult & Richards, J Magn. Reson, 24:71, 1976
for vector potential proof: Haacke et al, Magnetic Resonance Imaging: Physical Principles and
Sequence Design, pg 97-99, 1999
Real Coils aren’t Simple Loops
How do I include matching capacitors, inductors etc?
Which loop do we use for contour integration?
Circular Polarization
a + = a x − ja y
a − = a x + ja y
ℜe[a + e jφ e jωt ] =
cos(ωt + φ )a x + sin( ωt + φ )a y
ℜe[a − e jφ e jωt ] =
cos(ωt + φ )a x − sin( ωt + φ )a y
y
y
φ x
φ
Right hand
x
Left hand
KEY PROPERTIES
a+ ⋅ a+ = 0
a− ⋅ a − = 0
∗
a+ ⋅ a − = 0
a+ ⋅ a − = 2
Quadrature Field
+
-90o
B
Transmit
¼ cycle
time delay
M
Receive
-1
-90o
¼ cycle
time delay
To receive, a quadrature coil must create a field rotating
opposite the direction of precession.
Chen, Hoult & Sank: Quadrature Detection Coils…J. Magn. Reson., 54:324,1983
Constructing Polarized Fields
Time Domain:
Η xy (t ) = hx cos(ωt + θ x )a x + hy cos(ωt + θ y )a y
Frequency Domain Phasor:
H xy (t ) = ℜe[H xy e
j ωt
]
where
Circularly Polarized Phasor:
(
jθ x
H xy = hx e a x + hy e a y
a − = a x + ja y
)
jθ y
a + = a x − ja y
(
)
1
1
jθ y
jθ y
jθ x
jθ x
Η xy = hx e + jhy e a + + hx e − jhy e a −
2
2
+ve/right hand
-ve/ left hand
In MRI (Hoult):
ℜe[( H x − jH y ) ∗ / 2] ⇒ x
ℑm[( H x − jH y ) ∗ / 2] ⇒ y
Imaginary part gives y component
In Electromagnetics :
ℜe[a − ( H x − jH y ) / 2] ⇒
1
(hx cos θ x + hy sin θ y )a x +
2
1
(hy cosθ y − hx sin θ x )a y
2
Extract x, y components from real
part of complex vector
The Principle of Reciprocity in Signal
Strength Calculations, D.I. Hoult,
Concepts Magn. Reson. 12:173,2000.
Reciprocity Tool Kit
• Impressed Electric Current Source
• Impressed Magnetic Current Source
• Impressed Magnetic Dipole
• Lorentz Reciprocity Theorem
• Rumsey Reaction Integral
Electric Current Source Ji
Ji is an impressed current element independent of field
J i = Iδ (r − ro )a z
δz
Induces E, H
Ji
+
V
-
J=σE
D=εE
∇ × H = (σ + jωε ) E + J i
− ∇ × E = jωµH
E
E
H
B=µH
∫ E ⋅ J dv = ∫ E ⋅ Idzδ (r)dv = I ∫ E ⋅ dz = − IV
i
V
V
Can compute a voltage across Ji from a field E!
N Port Impedance Matrix
Ib
Ia
-
+
J
c
i
V = ZI
-
J
d
i
+
+
J
a
i
-
J
b
i
+
-
Ic
Id
Wherever we impress a current, we create current source port
b
a
a b
E
⋅
J
dv
=
−
I
V
i
∫
V
Electric fields integrated over this source reduce to a port
voltage.
Magnetic Current Source Ki
Ki = impressed magnetic current independent of field
δz
K i = jωµ o Mδ (r − ro )a z
E
Induces E, H
∇ × H = (σ + jωε ) E
− ∇ × E = jωµH + K i
Ki
J=σE
D=εE
H
B=µH
A magnetic current element can be physically created by a
time varying magnetic dipole.
K i ~ jωµ o M i
Ki
Voltage Source Concept
Va
c
i
Vd
+
-
Vc
-
K
I = YV
-
+
K
+
d
i
+
K
a
i
-
K
b
i
Vb
Magnetic current looping a wire creates a voltage source port
Magnetic
dipole toroid
K
∫H
b
⋅ K dv = − I V
a
i
b
V
wire
A loop of K induces an emf in the wire like a transformer.
a
Lorentz Reciprocity Theorem
Exp. A: electric current source Jia,
magnet current source Kia
Exp. B: electric current source Jib,
magnet current source Kib
∇ × H a = (σ + jωε ) E a + J ia
a
a
a
− ∇ × E = jωµH + K i
∇ × H b = (σ + jωε ) E b + J ib
b
b
b
− ∇ × E = jωµH + K i
J ib
J ia
K ia
K ib
×
(
E
H
−
E
×
H
)
⋅
ndS
=
0
∫
a
S
b
b
a
Reaction in Reciprocity
Exp. A: electric current source Jia,
magnet current source Kia
Exp. B: electric current source Jib,
magnet current source Kib
∇ × H a = (σ + jωε ) E a + J ia
a
a
a
− ∇ × E = jωµH + K i
∇ × H b = (σ + jωε ) E b + J ib
b
b
b
− ∇ × E = jωµH + K i
J ib
J ia
K ia
∫ (E
V
K ib
a
⋅ J − H ⋅ K )dv = ∫ ( E ⋅ J − H ⋅ K )dv
b
i
Reaction: <a,b>
a
b
i
b
V
a
i
b
<b,a>
Beware! σ, ε, µ are actually tensors and must be symmetric.
a
i
NMR Reciprocity Case
Exp. A: electric current filament Jia
Exp. B: magnetic current Kib=jωµoM
jωµ o M ib → E b , H b
J ia → H a , E a
Unit current
I(ω)
J
V
a+
i
b -
dM ib
K = µo
dt
b
i
jω ↔
d
dt
If symmetric
σ, ε, µ
− V b I a = ∫ E b ⋅ J ia dv = − ∫ H a ⋅ jωµ o M ib dv
V coil voltage
V
Vesselle et al, IEEE Trans. Biomed. Eng. 42, 497,1995
Ibrahim, T., Magn. Reson. Med. 54, 677, 2005
0
+ stuff
NMR Reciprocity Case
Exp. A: electric current filament Jia
Exp. B: magnetic current Kib=jωµoM
jωµ o M ib → E b , H b
J ia → H a , E a
− V b I a = − ∫ H a ⋅ jωµ o M ib dv
V
Unit current
I(ω)
J
V
a+
i
b -
dM ib
K = µo
dt
b
i
(
)
H a ⋅ M ib = H +a a + + H −a a − ⋅ m+ a +
(H +a + ⋅ m+a + ) = 0
j 2ωµ o
V =
H − m+ dv
∫
Ia V
b
Field rotating opposite precession determines sensitivity
Time Domain Reciprocity
Let a unit current impulse δ(t)
generate the time varying
field Hxy(t).
I = δ (t )
H xy (t )
The receive signal is convolution
of the field impulse response with
the time varying magnetization
+
V (t )
µo
dM
dt
-
dM i (τ )
V (t ) = ∫ H (t − τ ) ⋅ µo
dτ
dτ
0
t
Reciprocal Media
B = µH
scalar
 Bx   µ x
B  =  0
 y 
 Bz   0
µy
0
0 H x 
0   H y 
µ z   H z 
Principal axes align x,y,z
z
z
y
x
0
 Bx   µ xx
 B  = µ
 y   xy
 Bz   µ xz
µ xy
µ yy
µ yz
Principal axes arbitrary
z
y
x
µ xz   H x 

µ yz   H y 
µ zz   H z 
y
x
T
a
b
b
a
[
µ
]
=
[
µ
]
H
µ
H
H
µ
H
⋅
[
]
−
⋅
[
]
=0
If
then
Reciprocal Media have symmetric material property tensors.
Reciprocity is satisfied if material is reciprocal.
Circuit Reciprocity
v1
i1 (ω )
+
-
v2 (ω )
i1
 z11
z
 21
R + 1
v
 1 
j ωC
=
v   1
 2
j ωC

z12 

z 22 
v2

jωC   i1 
1
 i2 
j ωC 
1
z12 = z 21
In reciprocal circuits, Zmn = Znm for n!=m
i2
When Can Reciprocity Fail?
Left hand
polarized
Left hand
polarized
Cyclotron motion qEarth
magnetic field
ionosphere
Hmm? What if I’ve got charged ions moving in a magnetic
field? Or electron spin acting like a gyroscope, or NMR?
Gyrotropic Media
 Dx   ε t
 D  =  − jη
t
 y 
 Dz   0
jη t
εt
0
0   Ex 
0   E y 
ε z   E z 
Magnetized plasma (electrons
undergo cyclotron motion
 Bx   µ
 B  =  − jκ
 y 
 Bz   0
jκ
µ
0
0 H x 
0   H y 
µ o   H z 
Magnetized ferrite
T
a
b
b
a
[
µ
]
≠
[
µ
]
If
then H ⋅ [ µ ]H ≠ H ⋅ [ µ ]H
Material tensors not symmetric so reciprocity is not satisfied.
Foundations for Microwave Engineering, Robert E Collin
Bloch equation describes electron motion in
ferrites!
Phenomena & Applications
• Surface coil sensitivity asymmetry
• Guide wire artifact patterns
• Reversed Polarization
• RF Current Density Imaging
• Electric Properties Tomography
Transmit & Receive Asymmetry
7T
Different Excitation & Reception
Distributions with a Single Loop TransmitReceive Surface Coil near a head-sized
spherical phantom at 300 MHz, C.M. Collins
et al, MRM, 47, 1026, fig 2, 2002
3T
Reciprocity & Gyrotropism in magnetic
resonance transduction, James Tropp,
Phys. Rev. A, 74, 062103, fig 3, 2006
Guidewire Artifacts
15° flip
experiment
Ross Venook
simulation
Transmit and Receive coupling of a guidewire to a body coil
creates two distinct null locations
Reversed RF Polarization
Forward Polarization
Reversed Polarization
• MR signal is created only by one circular polarization
• This has been exploited for wireless catheter tracking
Haydar Celik et al., MRM 58:1224, 2007.
Pacemaker Lead
Forward Polarization
Reversed Polarization
Linear polarized fields generated by conducting structures
Electric Properties Tomography
∇ 2 H q = jωµ o (σ + jωε ) H q
∇ 2 H + = jωµ o (σ + jωε ) H +
q=x,y,z
∇ 2 H − = jωµ o (σ + jωε ) H −
CHALLENGE: Independently measure H+ and HU Katscher et al, Proc
ISMRM 14, 3035, 2006
σ
ε ε
U Katscher et al, Proc
ISMRM 15, 1774, 2007
raw
Summary
• Lorentz Reciprocity Theorem central
to deriving NMR signal detection.
• Impressed current dipole creates H.
• Magnetic dipole induces V at current
dipole location.
• Rotating phasor a+ “reacts” with a• Time domain convolution requires
time reversal of one field.
Problems
• I need final freq domain formula showing H+
and m-, to go with time domain convolution,
and show H- dot m- gives 0.
• Want time domain to show as picture the
applied impulse and then the voltage response.
• See if can make picture of B- and B+ (just as in
my paper!)
• Photocopy the Lorentz proof from a textbook.
• Can I get a dyadic green’s function plot?
• Add impedance definitions, power definitions &
computation
NMR Reciprocity Case
Exp. A: electric current filament Jia
a
dD
∇× H a = J a +
+ J ia
dt
a
dB
a
−∇× E =
dt
Unit current
I(ω)
J
V
a+
i
b -
Exp. B: magnetic current Kib=jωµoM
b
dD
∇× H b = J b +
b dt
b
dM
dB
i
− ∇ × Eb =
+ µo
dt
dt
dM ib
K = µo
dt
b
i
b
dM
b
a
a
i
(
)
µ
⋅
=
−
⋅
dv
E
J
dv
H
i
o
∫V
∫V
dt
0
+ stuff
coil voltage
Warning: Not time domain YET! Assume
d
= jω
dt
Reciprocity Theorems
• Response to a source is unchanged when
source and measurer swapped
• Relate response at one source due to a
second source to the response at the
second source due to the first source.
PMRIL
Stanford Electrical Engineering
What is Reciprocity?
Left hand
polarized
Left hand
polarized
Cyclotron motion qEarth
magnetic field
ionosphere
Response to a source is unchanged when source and
measurer swapped.
Hmm? What if I’ve got charged ions moving in a magnetic
field? Or electron spin acting like a gyroscope, or NMR?
Time Domain Interpretation
H + (t ) = cos(ωt )a x + sin( ωt )a y
m− (t ) = cos(ωt )a x − sin( ωt )a y
H − (t ) = cos(ωt )a x − sin( ωt )a y
m−a − ⋅ H −a − = 0
t
y (t ) = ∫ m− (τ ) ⋅ H − (t − τ )dτ → 0
0
Time convolution is zero for same sense of rotation.
m−a − ⋅ H + a + ≠ 0
t
y (t ) = ∫ m− (τ ) ⋅ H + (t − τ )dτ → cos(ωt )
0
Time convolution yields cos(ωt) for opposing rotation.
Ignored “Stuff”
a
b
a
 a dD b



dB
dB
dD
b
a
b
b
a
a
b
⋅
−
⋅
+
⋅
−
⋅
+
⋅
−
⋅
∫  E dt E dt dv ∫ E J E J dv ∫  H dt H dt dv
[
D = [ε ]E
(
jω E a ⋅ [ε ]E b − E b ⋅ [ε ]E a
0
]
J = [σ ]E
) (E
a
⋅ [σ ]E b − E b ⋅ [σ ]E a
0
B = [ µ ]H
)
(
jω H a ⋅ [ µ ] H b − H b ⋅ [ µ ] H a
0
If [ε], [σ], [µ] are symmetric tensors, all terms are zero!
Reciprocal Media have symmetric material property tensors.
)
NMR Reciprocity Case
Exp. A: electric current filament Jia
Exp. B: magnetic current Kib=jωµoM
∇ × H a = (σ + jωε ) E a + J ia
∇ × H b = (σ + jωε ) E b
− ∇ × E b = jωµH b + jωµ o M ib
− ∇ × E a = jωµH a
Unit current
I(ω)
J
V
a+
i
b -
dM ib
K = µo
dt
b
i
d
jω ↔
dt
− V b I a = ∫ E b ⋅ J ia dv = − ∫ H a ⋅ jωµ o M ib dv
V coil voltage
V
Vesselle et al, IEEE Trans. Biomed. Eng. 42, 497,1995
Ibrahim, T., Magn. Reson. Med. 54, 677, 2005
If symmetric
σ, ε, µ
0
+ stuff
Rotating Frame Components
Scott et al, Trans. Med. Imag. 14, 515, 1995