Find thermal radiation intensity:
Relate W(f) to I(θ,φ,f) = “radiation intensity”
i.e.,
θ
∫ VOL ∫SURFACE
to
Unit area
Consider slab of
blackbody radiation:
vol. = δ,
contains Wδ[J Hz-1]
Let slab radiate without replacement: δ
[Jm-3 Hz-1]
[Wm-2 ster-1 Hz-1]
Wδ = ∫ V W(f) dV = ∫ I(f) dt dA dΩ [J Hz-1]
4πIoδ
8π 3 hf/kT
-1 = ∫ Io cos θ • δ/(c • cos θ) • 1 • dΩ = c
Wδ = δ 3 hf e
c
Io(f,θ,φ) = 2hf3 c2 ehf/kT -1
Wm-2 Hz-1 ster-1
Planck’s radiation law
20
2kT
Io(f,θ,φ) ≅ 2 in the limit hf <<kT
λ
Rayleigh-Jeans Law
log Io
Rayleigh-Jeans region
hf << kT
1000 To
100 To
10 To
To
log λ
21
Paradox:
blackbody
(radio limit)
KT
WHz-1
≠ f(λ)
single mode
2kT
Z0
λ2
antenna
2kT
how?
2
λ
Wm − 2Hz −1ster −1
= f (λ )
Answer: kT WHz-1 mode-1
Two polarizations
Number of modes/Hz•m2•ster
2kT
λ2
22
Note: In the radio limit: hfj...
hf j • n j = hf
1
e
hf kT
−1
[
]
≅ kT J mode ­ 1 [TEM line cavity]
Recall: kT/2 J/degree of freedom • m mechanical systems
sin ωt
orthogonal degrees of freedom, 2 per mode
cos ωt
Therefore we also obtain kT/2 J/degree of freedom
for thermal radiation
23
Thermal radiation from lossy lines:
dz
Zo
Zo
A
slightly lossy
→ kT e-αdz + emission = kT in equilibrium
P+ = kT →
(
)
∴ Emission = kT 1 − e − αdz ≅ kT (1 − [1 − αdz]) = kTα dz
24
What is thermal emission when not in equilibrium?
Answer: same, recall linearity of Maxwell’s equations
L
kTout = kTin
L
− ∫ α dz
− ∫ α dz
L
e o
+ k T( z)α( z)e z
dz
∫
o
T(z)
kTin
?
z
Tin
Tout = Tin
L
− ∫ αdz
e o
τMAX
+
τmax
∫
T( τ)e − τ dτ
0
Equation of radiative transfer
Tout
(
Tout = Tine − τ + Tline 1 − e − τ
)
Tline
25
Definition of a decibel:
dB gain ∆ 10 log10 P2 P1
e.g.:
0 dB if P2 = P1
10 dB if P2 = 10P1
20 dB if P2 = 100P1
26
Example of thermal noise from lossy line:
3°K
300°K
Tout(°K) = ?
2 dB loss
2
P
(e − τ = 2 = 10 10 = 0.63)
P1
−
Case 1:
τ = 0 ⇒ Tout = 3K
Case 2:
2 dB loss ⇒ Tout = 3 × 0.63 + 300(1 - 0.63) ≅ 113K
Case 3:
τ = ∞ ⇒ Tout = 300K
27
Thermal noise voltage (Johnson noise) in circuits
R
Zo = R
noise
R
R
Thermal noise
+
-
en
Zo = R
B
Hz
0°K
Bandpass filter
Watts dissipated in load R: kTB = (erms 2)2 R
erms (thermal noise) = 4kTBR volts (in B Hz)
Example: Amplifier, 50Ω input, B = 100 MHz, T = 300K
erms = 4 • 1.38 × 10 − 23 × 300 × 108 × 50 = 9.1µv
(9.1 mv if R = 50MΩ)
28
Shot Noise
For example, occurs in diodes:
i(t)
i(t)
i(t)
i(t)
0
t
single electron
S(f) WHz-1
t
0
S(f) = ei
B
f
If each charge moves independently,
arrivals are poisson distributed:
For example, let i = 1 ma, B = 108 Hz
We later show i 2AC = 2eiB = 2 × 1.6 • 10 −19 × 10 −3 × 108
Then i ACrms × 50Ω ≅ 9µV shot noise
29
Approximate derivation of shot noise
Sounds like falling shot
i(t) Poisson distributed
arrival times
anode
i(t)
cathode
B
0
f
i = ne
n ∆ avg. # electrons/sec
t
∆
Assume boxcar i(t)
i(t)
e/δ
e = electron charge
n >> B, by assumption
δ
t
30
Approximate derivation of shot noise
i(t)
i = ne
n avg. # electrons/sec
Assume boxcar i(t)
e/δ
e = electron charge
n >> B, by assumption
t
δ
i( t )
↓
φi ( τ) ↔ Φi ( f )
φi ( τ) ∆ E[i(t)i(t - τ)]
n(e δ )2 δ
2
2
2
⎛e δ ⎞
⎜ • ⎟ = i = (ne )
⎝δ ∆⎠
-δ
0
δ
τ
31
Approximate derivation of shot noise
φi ( τ) ∆ E[i(t)i(t - τ)]
i( t )
↓
n(e δ )2 δ
φi ( τ) ↔ Φi ( f )
2
2
2
⎛e δ ⎞
⎜ • ⎟ = i = (ne )
⎝δ ∆⎠
Φi ( f ) = Φi
DC
+ Φi
2
AC
( )
Φi(f)
i [watts, DC]
0
n e2 = ei Hz −1
f
B
[
∴ in B Hz : σi2 AC = 2Bei Amp 2
τ
δ
0
~1/δ Hz
]
32
Shot noise example
R = 5KΩ
B = 10 6 Hz
i = 1 ma
v = i R = 10
i(t)
+
­3
• 5K = 5 volts
-
R
- v(t) +
σi2 AC = 2Bei
vrms (shot ) = σi2 ACR = 2 • 106 • 1.6 × 10 −19 × 10 −3 5000 ≅ 0.1 mv
⎡vrms ( thermal ) = 4kTBR
⎤
⎢
⎥
−
23
6
⎢⎣≅ 4 × 1.38 × 10
× 300 × 10 × 5000 ≅ 0.01 mv ⎥⎦
33
Receiver Architecture
Professor David H. Staelin
Graphics: Scott Bressler
Receivers-A1
Uses of receivers
I. Power measurement
II. Finite set of transmitted signals is possible; which is it?
III. Infinite set possible; estimate one or more parameters,
e.g. arrival time, amplitude, Doppler, etc., e.g.
v(t)
τ
Design of waveform sets is part of our problem
A2
Measurement of noise power in B Hz
Noise voltage
Noisy
receiver
Estimate
average power
for τ sec
2
Simply compute average output power over τ sec: ∝ v ( t )
A3
Total power radiometer
B
PA = kTA
B
( )2
+
h(t)
display
f
0
PR = kTR
vi(t)
Voltage for TA+TR
vd(t)=vi2(t)
∞
v o ( t ) = ∫ v d ( t − τ)h( τ)dτ
0
vi(t)
t
vd(t)
t
vo(t)
t
t
A4
Total power radiometer
B
B
PA = kTA
( )2
+
display
h(t)
0
PR = kTR
vi(t)
vd(t)=vi2(t)
Voltage for TA+TR
∞
v o ( t ) = ∫ v d ( t − τ)h( τ)dτ
0
vo(t)
t
t
v o ∝ (TA + TR )
vi(t)
vo(t) compressed time
vrms
vo
t
0
t
A5