Photon noise
black (no reflection)
σp (watts ) = nsσn2 hf τ
T°K
Area A
Ω Ster.
⎛⎜ number of states ⎞⎟
⎝ in cavity in Bτ ⎠
Calculation of the number of electromagnetic modes = m:
2kT
­1
­1
W Hz ­ 1m − 2ster −1
Recall, for hf << kT, kT W Hz • mode ⇒
λ2
2kT
2
2 ⎛ modes ⎞
= 2(f c ) ⎜
Therefore
kT =
⎟ ≠ f (T )
2
2
2
⎝ m • ster ⎠
λ
λ
Therefore
Lec 09.6- 1
1/11/01
m = 2(f c )2 AΩ propagation modes
L1
Photon noise
Therefore
m = 2(f c )2 AΩ modes
σp (watts ) = nsσn2 hf τ
Each mode has 2Bτ degrees of freedom
(2B samples sec-1 times τ sec) (Nyquist sampling)
Each energy state (ε = hf) has 2 degrees of freedom
(sin ωt, cos ωt)
Therefore number of states ns @ hf in Bτ:
⎛ degrees ⎞ ⎛ states ⎞
ns = (# modes in AΩ ) • ⎜
⎟
⎟•⎜
⎝ mode ⎠ ⎝ degree ⎠
= (m )(2Bτ )(1 2 ) = 2(f c )2 AΩ(2Bτ )(1 2 )
Lec 09.6- 2
1/11/01
L2
Photon noise
ns = 2(f c )2 AΩBτ
σp = nsσn2 hf τ
photons state = n =
σn2 = n + n
2
1
ehf kT − 1
2
2
⎛f⎞
Therefore σp = 2⎜ ⎟ AΩBτ n + n (hf )2 τ2 W " quantum limit"
⎝c⎠
If boxcar h(t) has τ = 0.5 sec, ≡ 1-Hz post-detection bandwidth,
yields units of W Hz -1 2 (NEPR )
2
2
⎛f⎞
NEPR (f ) = 4 AΩ⎜ ⎟ B n + n (hf )2 W Hz ­ 1 2
⎝c⎠
“Noise-equivalent power” due to radiation noise
Lec 09.6- 3
1/11/01
L3
Photon noise
2
2
⎛f⎞
(
)
NEPR f = 4 AΩ⎜ ⎟ B n + n (hf )2 W Hz ­ 1 2
⎝c⎠
“Noise-equivalent power” due to radiation noise
(B → ∞)
NEPR for a blackbody, all frequencies
∞
⎡
NEPR∞ = ⎢ 4AΩ ∫
⎢
0
⎣
(
hf 2 c
)
12
⎤
2⎞ ⎥
⎜ n + n ⎟ df ⎥
⎝
⎠
⎦
2⎛
12
σSB 4 ⎤
⎡
NEPR∞ = ⎢4 AΩ(4kT )
T ⎥
π
⎦
⎣
where n(f ) =
1
ehf kT − 1
[WHz −1 2 ]
Where σSB ≡ “Stefan-Boltzmann constant” = 5.67 × 10-8 Wm-2K-4
and Hz-1 refers to the detector output bandwidth
Lec 09.6- 4
1/11/01
L4
Photon noise
12
σSB 4 ⎤
⎡
NEPR∞ = ⎢4 AΩ(4kT )
T ⎥
π
⎣
⎦
WHz −1 2
Where σSB ≡ “Stefan-Boltzmann constant” = 5.67 × 10-8 Wm-2K-4
Recall: blackbodies radiate
σ
Pr = AΩ SB T 4 watts (small Ω ) , Pr = AσSB T 4 if Ω = 2π
π
Therefore:
NEPR∞ = [Pr 16kT ]1 2 WHz −1 2
for Ω = 2π
Therefore minimize T,A
NEPR∞ = 4 AσSBkT 5
A
Lec 09.6- 5
1/11/01
Ω = 4π
≅ 4 × 10 −16 WHz −1 2
for A = 10 ­ 6 , T = 4°K
L5
Bolometer noise analysis
Bolometers first
convert photons
to heat
Ω(ster)
Area A
hf
σp
R = f(T)
Rbias >> R
B (Hz) vo
Responsivity S
heat, Gt conductivity
Tb
cold bath
R, Rb produce Johnson noise
Radiated photons have shot noise, i.e. “radiation noise”
“Phonon noise” arises from shot noise in phonons
carrying heat to the cold bath
NEP = 4kTbR S2 + 16AΩkT5σSB π + 4kGt Tb2
Johnson noise
Lec 09.6- 6
1/11/01
(
W Hz-12
)
Photon shot noise Phonon shot noise
L6
Optical Superheterodynes
s(t) = signal
Radiometer
output
photodetector “mixer”
(photons s-1)
ωi.f . f
P photons/sec
(laser L.O.)
Let signal s( t ) =
vm(t)
( )2
h(t)τ
vo
communication signal
2 ∆
2S cos ωs t ∝ E(t ) , s (t) = S
local oscillator p( t ) =
2P cos ωo t
Both are coherent lasers
D = dark photons/sec
Mixer output (L.O. ≡ 0) = ηS + D counts/sec, where η ≡ quantum efficiency
( 2So cos ωs t ) (volts)
= constant [η(S + P + SP cos ωi.f .t ) + D] where ωi.f. = ωs − ωo
v m ( t ) = constant • η
Lec 09.6- 7
1/11/01
2
We want P >> S, η SP >> D
N1
Optical Superheterodyne CNR
(
= constant [η(S + P +
vm ( t ) = constant • η 2So cos ωst
)2
(volts)
) ]
SP cos ωi.f .t + D where ωi.f. = ωs − ωo
We want P >> S, η SP >> D
∆
Let " constant" = 1 so v m ( t ) units are counts sec
v mix ( t ) ≅ v m [signal] + v m [noise]
v m [signal] ≅ η SP cos ωi.f .t
Conveys information in S(t), ωi.f .
⎛ counts sec ⎞ ⎡
v m [rms noise] ≅ 2ηP ⎜
⎟ ⎢=
Hz
⎝
⎠⎣
(≅
2D if D >> ηP )
v o = η2SP cos 2 ωi.f .t
Lec 09.6- 8
1/11/01
⎤
ηP
, τ = 0.5 sec for Hz ⎥
τ
⎦
<
Conveys information in s(t) for ω ~
2π
<< ωi.f .
τ
N2
Optical Superheterodyne CNR
∆
Let " constant" = 1 so v m ( t ) units are counts sec
v mix ( t ) ≅ v m [signal] + v m [noise]
v m [signal] ≅ η SP cos ωi.f .t
Conveys information in S(t), ωi.f .
⎤
ηP
⎛ counts sec ⎞ ⎡
τ
=
0.5
sec
for
v m [rms noise] ≅ 2ηP ⎜
=
,
Hz
⎟⎢
⎥
τ
Hz
⎝
⎠⎣
⎦
(≅
2D if D >> ηP )
v o = η2SP cos 2 ωi.f .t
vo
rms noise
=
(
2ηPB
<
Conveys information in s(t) for ω ~
)
2
2π
<< ωi.f .
τ
Bτ for P >> S, P >> D
[
Define CNR “Carrier-to-Noise Ratio” for v m ( t ) = η2SP 4ηPB
]
Bτ = ηS τ 16B
We assume τ > 1/B so τ provides additional noise smoothing
Lec 09.6- 9
1/11/01
CNR ~
< η Sτ 4
(best we could do is CNR ≤ 1
for 4 photons/bit)
N3
Optical Superheterodynes, Comparisons
1) Radio total-power radiometer:
CNR = TA ∆TRMS = TA
Bτ TR for TR >> TA
radio expression
Optical superheterodyne CNR = η
S τ 16B = η(kTAB hf ) τ 16B
N
photons sec
­ 1
Therefore “TR” = 4hf/kη if CNR (radio) = CNR (optical)
This is 4 times radio quantum limit if i.f. noise etc. is negligible
(Note: PA ≠ kTAB in optical)
Thus optical superheterodynes can approach quantum limit
Lec 09.6- 10
1/11/01
N4
Optical Superheterodynes, Comparisons
2) Optical non-superheterodyne if D >> S; then
v o sig = ηS (gain normalized )
v onoise = 2DB
CNR = ηS τ 2D
Bτ
versus CNRS.H. = ηS τ 16B
Therefore a superheterodyne is better if Bi.f . < D 8
i.e. the worse D is, the higher B can be before L.O. shot
noise dominates (assuming no mixer or i.f. excess noise)
Lec 09.6- 11
1/11/01
N5
Antennas
Basic Characterization
Professor David H. Staelin
Lec 09.6- 12
1/11/01
Uses of Antennas
In
Antennas
Out
Signal
Processor
Transducer
Electromagnetic
Environment
Transducer
Antennas couple electromagnetic radiation and transmission lines
for transmission and reception
We have studied: hf << kT
hf >> kT
hf ≈ kT
All bands use antennas
Lec 09.6- 13
1/11/01
Radio
Optical
IR
A1
Antennas – Characterization
y
Side Lobes
G(f,θ,φ)
PT
Transmitter
Power
θ,φ
[
]
PTR [W ] =
∫4π PdΩ
z
0
P(f , θ, φ ) W ster -1 radiated power
Main Lobe
⇓
" total radiated power"
Radiation efficiency:
ηR ∆ PTR PT
Gain (over isotropic):
P(f, Θ, φ )
∆
= ηRD(f , θ, φ )
G(f , θ, φ )
PT 4π
Directivity (over isotropic):
P(f, θ, φ )
∆
D(f , θ, φ )
PTR 4π
Antenna pattern:
Lec 09.6- 14
1/11/01
t (f , θ, φ ) ∆
G(f, θ, φ ) ∆ D(f, θ, φ )
≤1
Go
Do
A2
Antenna Example
(
)
P Wm − 2 = G(θ, φ ) •
PT
4πR
2
P(Wm-2)
Target
R(meters)
Moon
3 × 108
10-5
Jupiter
1012
10-12
Antares
3 × 1016
10-21
MIT “Haystack” antenna @ λ =1 CM; Go ≅ 73 dB
Assume it radiates 1–MWatt radar pulses
Assume kTB ≅ 1.4 × 10-23 × 10°K × 1Hz ≅ 10-22
Watts(say TR ≅ 10° and we use 1–Hz CW radar)
Then P(Wm-2) received on Antares is comparable
to receiver noise power kTB
Lec 09.6- 15
1/11/01
A3
Receiving Properties of Antennas
Characterized by Effective Area A(f, θ, φ):
I(f,θ,φ)
Pr(f)
Power spectral density received:
Pr (f ) = A (f , θ, φ ) • [I(f , θ, φ ) • ∆Ω]∆f (W )
[(
[m2 ]
“flux density” S Wm −2Hz −1
)]
∆f, ∆Ω are source bandwidth, solid angle
Lec 09.6- 16
1/11/01
Recall: Radiation intensity I(f,θ,φ) received
from blackbody at temperature T is:
2kT
I(f , θ, φ ) =
Wm − 2Hz −1ster −1
λ2
[
]
A4
Receiving System Example
Recall – 1-MW radar on Antares ⇒ 10-21 W/m2 on earth (GT = 73 dB)
Radio
Antares
Earth
R = 3 × 1016 m
Received power = A (f, θ, φ ) • [I(f, θ, φ )dΩ] • B = 10 −17 W from Antares
say 10 4 m2
S
10 −21Wm −2Hz −1
Suppose kTB = 10-22W (recall above) then SNR = 105
Audio at 104Hz ⇒ SNR = 10 (commercial opportunity?)
Lec 09.6- 17
1/11/01
A5
Relation Between A(f, θ,φ) and G(f, θ,φ)
Go
Ao
G(θ,φ)
0
A(θ,φ)
θ
0
θ
G(θ, φ ) A (θ, φ )
=
We later prove (using reciprocity) that
Go
Ao
Lec 09.6- 18
1/11/01
C1
Receiving Properties Deduced from
Reciprocity and Thermodynamics
dΩ
“Principle of detailed balance”
TB(f, θ, φ)
P(f, θ, φ)
T°K
Zo
Zo
Reciprocity + thermal equilibrium says that within dΩ,
power out = power in
[
]
Antenna radiates P(f, θ, φ ) W Hz -1ster −1 into dΩ, so
power out = P(f, θ, φ )dΩdf = (kTdf 4π )GdΩ(watts )
Antenna receives from dΩ:
power in =
Lec 09.6- 19
1/11/01
1 2kTB (θ, φ )
df dΩ A (f, θ, φ )[watts ]
2
2
λ
One
polarization
Wm-2Hz-1ster-1
C2
Receiving Properties Deduced from
Reciprocity and Thermodynamics
Antenna radiates P(f, θ,φ) [W Hz-1 ster-1] into dΩ, so
power out = P(f, θ,φ) dΩ df = (kTdf/4π) GdΩ (watts)
Antenna receives from dΩ:
1 2kTB (θ, φ )
power in =
df dΩ A (f, θ, φ )[watts ]
2
One
λ2
polarization
Wm-2Hz-1ster-1
In thermal equilibrium T = TB(f, θ, φ); then equating
radiation and reception (detailed balance) yields
G(f , θ, φ ) =
4π
2
λ
A (f , θ, φ )
This assumes hf << kT and that powers superimpose, i.e., that the
TB (θ1, φ1 ) signal E(t ) is uncorrelated with that for TB (θ2 , φ2 )
Lec 09.6- 20
1/11/01
C3
Antennas Used to Provide a Radio Link
Pt
Wm −2 at receiver
Pr
Pr =
r
Gain = GT
Gr =
4π
Ar
Pt
4πr
2
• Gt • A r Watts
isotropic
m2 effective area
λ2
Note: Pr → ∞ as r →0!, so this relation requires r > r minimum
Let Pr = Pt at rmin and At = Ar = D2 (m2) [D ≅ aperture diameter in practice]
Then
Gt A r
4 πr 2min
= 1=
A t Ar
λ2r 2min
(
=
D4
λ2r 2min
Therefore rmin = D2 λ in practice we want r > 2D2 λ
)
This zone where r ~
> 2D2 λ is called the " far field" of the aperture
Lec 09.6- 21
1/11/01
C4
Definition of Antenna Temperature TA(°K)
(
kTA W Hz
(m2 ) (Wm −2Hz−1ster −1)
­1
) = ∫4π A(θ, φ) I(θ, φ)dΩ
Received power
spectral density
for a specific polarization
2kTB 1
Since I =
• for thermal radiation, single polarization
2
2
λ
Therefore
Lec 09.6- 22
1/11/01
TA =
1
2
∫ A (θ, φ)TB (θ, φ ) dΩ
λ
1
=
G(θ, φ )TB (θ, φ ) dΩ
∫
4π
For TB
uncorrelated
in angle
C5
Observing Small Thermal Sources TB(θ,φ,f)
Limiting case:
TA S
Go TBΩS A o TBΩS
1
G(θ, φ ) TB (θ, φ ) dΩ =
=
=
∫
Ω
4π
4π S
λ2
Go/2
Go
θ, φ
small source, << θB
TB, Ωs
average source TB↑ ↑ source solid angle
TAS is due to source (assume zero background)
Physical interpretation (for θS << θB):
Define Ω A , " beam solid angle" so that GoΩ A ∆ ∫
4π
Go
ΩA
GdΩ = 4πηr
Go = 4πηr Ω A
⎛Ω
⎞
Then TA S ∆ ⎜⎜ S ηr ⎟⎟TB
⎝ ΩA ⎠
Coupling coefficient ≤ ηr
Ωs, TBS
TAS
Lec 09.6- 23
1/11/01
ΩA
Geometric coupling ratio =
ΩS
ΩA
C6
Ways to Characterize Small Thermal Sources
Limiting case:
TA S
Go TBΩS A oTBΩS
1
G(θ, φ ) TB (θ, φ ) dΩ =
=
=
∫
Ω
4π S
4π
λ2
1. TB (θ, φ, f ) (for each of 2 polarizations)
2. TBS average brightness temperatur e
[
]
3. S(f ) Wm − 2Hz −1 = ∫
ΩS
I(f , θ, φ ) dΩ ≠ f(antennas )
if source small, ΩS << ΩA
Units of S : 1 " jansky" = 10 ­ 26 Wm ­ 2Hz −1
Lec 09.6- 24
1/11/01
C7