Summary of Signal Types
Pulses:
v(t)
↓
R(τ)
Typical Sets of Units
↔
V(f)
↔
↓
2
S(f) = ⏐V(f)⏐
↔
Vm[v]
[V]
↔
↓
2
Φm = ⏐Vm⏐
↔
↔
[V]
–1
↔ [VHz ]
↓
↓
2
–1 2
[ V s ] ↔ [VHz ]
–1
[ V ] ↔ [VHz ]
↓
↓
[J] ↔ [JHz–1]
1-ohm load
Periodic:
v(t)
↓
R(τ)
↔
[V]
↓
[ V2]
↔
↓
[ V2 ]
[?]
[V]
↔
[?]
↓
Φ(f)
↓
2
[V ]
↓
2
–1
↔ [ V Hz ]
[V] ↔
[V]
↓
↓
[W] ↔
[W]
1-ohm load
Random:
v(t)
↓
Φ(τ)
6.661 Fall 2001
Slide 1
[V] ↔
[?]
↓
↓
[ W ] ↔ [ W/Hz ]
G6
Probability Distributions
Gaussian Noise:
1
P{v } =
e
σ 2π
[ ]= ∫
Ev
2
p(v)
−( v / σ )2 / 2
∞
p( v ) v 2dv
−∞
σ
= σ2
v
0
Band-limited Gaussian white noise, e.g. N /2
o
Φ(f)
Single-sided
equivalent
Φ(f)
(
No WHz −1
f
[ ]
6.661 Fall 2001
Slide 2
)
No 2
E v 2 = σ2 = NoB
∴ σ = NoB
[WHz −1]
B
Power
spectral
f density
G7
Probability Distributions
Binomial Distribution:
Assume we have n bits, 0 or 1, where p{1} ≡ p, p{0} ≡ 1 − p
⎛n⎞ k
⎛ n⎞ ∆
n!
n −k
p{k 1' s} = ⎜ ⎟ p (1 − p )
where ⎜ ⎟
⎝k ⎠
⎝ k ⎠ (n − k ) ! k!
Note: There are n positions possible for the first “1,”
n – 1 for the second “1,” and a total of
n(n – 1)…(n – k + 1)/k! ways to arrange those
k “1’s” among the n available positions.
E[k ] = np =
n
∑ k p(k )
k =0
6.661 Fall 2001
Slide 3
G8
Probability Distributions
E[k ] = np =
Poisson Distribution:
n
∑ k p(k )
k =0
Assume we have n bits, 0 or 1, where p{1} ≡ p, p{0} ≡ 1 − p
∆
If n >> 1, p << 1: np = λ ≅ 1; variance = np (N
1-p) ≅ λ
λk − λ
then p{k} ≅ e
k!
Mean of k = λ = np
Variance of k = λ
Laplacian Distribution:
1 − 2 r / σo
p{r} =
e
2σo
6.661 Fall 2001
Slide 4
~0
1/ 2σo
0
σo / 2
⎛ Arises, for example, if r 2 = x 2 + y 2 , x 2 = y 2 ; xy = 0 ⎞
⎜
⎟
⎜ where x, y are Gaussian, variance σ2 , zero mean ⎟
⎝
⎠
o
G9
Receiver-Noise Processes
Receivers are limited by noise, many types
Thermal noise:
Cases:
6.661 Fall 2001
Slide 5
1-D (TEM transmission line)
3-D (Multimode waveguide)
Equation of radiative transfer (1-D)
RF and optical limits; IR case
2
Thermal noise, 1-D (TEM) case
Zo(Ω), c(ms-1)
Zo, T°K
lossless TEM line
Electromagnetic
energy
Heat
Heat
6.661 Fall 2001
Slide 6
3
Approach:
P­
power P+
closed container
very slightly lossy
D
z
1) Find average energy density W(f)[J/m Hz]
2) Find average power P+[W/Hz] power flow
6.661 Fall 2001
Slide 7
4
Find average energy density W(f)[Jm-1 Hz-1]
modes
W(f) =
Hz
photons
mode
energy
photon
1
D
hf [Joules] (h ≡ Planck’s constant)
f is frequency (Hz)
h = 6.6252 × 10-34 (J s)
6.661 Fall 2001
Slide 8
5
modes
W(f) =
Hz
Find modes/Hz:
photons
mode
energy
photon
1
D
m=1
V(z)
Resonator modes
z=0
z
m=3
D
m=2
2Dfm
2D
(vp = phase velocity)
= v
Therefore m =
λm
p
2D
dm
modes/Hz
=
vp
df
6.661 Fall 2001
Slide 9
6
Find photons/mode
∆
nj; (jth mode)
Photons obey Bose-Einstein statistics; therefore
any number can occupy each mode.
Total energy fixed; combinations favor more likely distributions
nj =
∞
Σ n pj(n)
n=0
pj(n) ∆ p{n photons in state j}
pj(n) = Q e-nWj/kT , “Boltzmann distribution”
∞
where
Σ
n=0
6.661 Fall 2001
Slide 10
pj(n) ≡ 1, Wj
∆
hfj, Q = constant
7
pj(n) = Q e-nWj/kT , “Boltzmann distribution”
∞
where
Σ
n=0
∞
pj(n) ≡ 1, Wj
∞
Σ pj(n) = Q • Σ
n=0
-Wj/kT
e
∆
hfj, Q = constant
n
Q
=
n=0
-Wj/kT
1-e
∞
Recall
Σ xn = 1/(1 - x)
if x < 1
n=0
Therefore Q = 1 - e-Wj/kT
6.661 Fall 2001
Slide 11
8
pj(n) = Q e-nWj/kT , “Boltzmann distribution”
Where Q = 1 - e-Wj/kT
Therefore
-Wj/kT
pj(n) = 1 - e
∞
nj =
6.661 Fall 2001
Slide 12
Σ
n=0
-nWj/kT
e
-Wj/kT
n pj(n) = 1 - e
∞
Σ
-Wj/kT
n
ne
n=0
9
nj =
∞
∑ n p j (n) = ⎛⎜⎝1 − e
− W j kT ⎞ ∞
⎛
⎟ ∑ n⎜ e
⎠n=0 ⎝
n= 0
∞
Recall
d
n
Σ n x = x dx
n=0
∞
Σ
n=0
-Wj/kT
So nj = 1 - e
xn
e
− W j kT ⎞n
⎟
⎠
-1
d
x
(1 - x) =
= x
2
dx
1-x
-Wj/kT
-Wj/kT
2
1-e
[
W j kT
⎛
nj = 1 ⎜e
− 1⎟⎞ photons/mode W j = hf j
⎝
⎠
6.661 Fall 2001
Slide 13
]
10
Solution - Average Energy Density Jm-1Hz-1
W(f) =
modes
Hz
photons
mode
1
2D
(hf)
W(f) =
W
/kT
vp e j
-1
W(f) = W+ + W­ = 2W+
1
=
D
energy
photon
1
D
2hf
vp e
hf/kT
Jm-1Hz-1
-1
(powers and energies superimpose
if waves are “orthogonal”)
W+ = forward-moving energy density
6.661 Fall 2001
Slide 14
11
Solution - Thermal power in TEM line:
1
2D
(hf)
W(f) =
W
/kT
vp e j
-1
1
=
D
2hf
vp e
hf/kT
Jm-1Hz-1
-1
W(f) = W+ + W­ = 2W+
P+ WHz-1 = vgW+ = vgW/2
If the TEM line is non-dispersive, then vp = vg and
P+(f) WHz
6.661 Fall 2001
Slide 15
-1
=
hf
e hf/kT - 1
12
Recall ex = 1 + x + x2/2! + … ≈ 1 + x for x << 1
[
P+ ( f ) WH z−1
] = ehf kT − 1 ≅ kT for hf << kT
hf
“Rayleigh-Jeans limit”
P ≅ kTB watts in uniform bandwidth B(Hz)
P+(f)
Rayleigh-Jeans region (radio vs. optical)
kT
0
B
fo
f(Hz)
hfo ≅ kT, so fo = kT/h ≅ 20 • T(°K) GHz
Planck’s constant:
Boltzmann’s constant:
6.661 Fall 2001
Slide 16
h ≅ 6.6 × 10-34 [J sec]
k = 1.38 × 10-23 [J/°K]
13
Problem: Find thermal radiation intensity I(watts/Hz • m2 • ster)
Approach:
Assume closed container, very
slightly lossy, filled with photons
1) Find energy density spectrum W(f)[J/m3 Hz]
2) Relate W(f) to I(f)
I(f)
e.g. antenna
Intercepting
surface
x
b
W(f)
z
a
6.661 Fall 2001
Slide 17
d
y
14
First:
Find energy density spectrum W(f)[Jm-3Hz-1]
modes
W(f) =
Hz
waveguide
modes
Ez≡ 0
2
6.661 Fall 2001
Slide 18
= a,
2
1
vol.
hf
Hz≡ 0
m=4
b
nλx
energy
photon
TEm,n, TMm,n 1 ehf/kT -1
x
mλy
photons
mode
z
=b
n=3
λx
d
a
λy
a y
b
15
Begin finding modes/Hz
2
Claim:
2
⎛ cm ⎞
⎛ cn ⎞
⎛ cp ⎞
fm,n,p = ⎜
⎟ +⎜ ⎟ +⎜ ⎟
⎝ 2a ⎠
⎝ 2b ⎠
⎝ 2d ⎠
2
Recall wave eqn: ∇2 + ω2µε E = 0
y
∂2
∂2
∂2
+ 2 + 2
2
∂x
∂y
∂z
λo
λy
k
wave
direction
- jkxx - jkyy - jkzz
E = Eoe
[kx = 2π/λx]
2π
⇒ k 2x + k 2y + k 2z = k 2o = ω2µoεo =
λo
6.661 Fall 2001
Slide 19
λx
x
2
16
Uniform plane wave
E = Eoe- jkxx - jkyy - jkzz
(∇ 2 + ω2µε)E = 0
[kx = 2π/λx]
2π
k 2x + k 2y + k 2z = k 2o = ω2µoεo =
λo
y
2
λo
λy
k
wave
direction
λx
2
1/λ X
x
+ 1/λ 2y + 1/λ 2z = 1/λ 2o = (f/c)2
Therefore
λy
Note: m = a ⇒ (m/2a)2 + (n/2b)2 + (p/2d)2 = (f/c)2 = QED
2
Next, use this relation to find modes/Hz
6.661 Fall 2001
Slide 20
17
Find modes/Hz:
fm,n,p =
cm
2a
2
+
cn
2b
2
cp
+
2d
2
p(c/2d)
# modes in df shell =
= vol. of shell × 2/vol. cell
2
df
1
0
1
m(c/2a)
TE + TM
n(c/2b)
0
1
2
3
4πf2df • 2
=
8
c
c
c
•
•
2a 2b 2c
8πf2 V df ⇒ modes df
= 3 ol
Hz
c
abd
6.661 Fall 2001
Slide 21
18
Find energy density spectrum W(f):
modes
W(f) =
Hz
photons
mode
energy
photon
1
vol.
1
8π hf3
8πf2
W(f) = 3 V hf/kT
• hf •1/V = 3 hf/kT [Jm-3Hz-1]
-1
c e
c
e
-1
6.661 Fall 2001
Slide 22
19