Noise in Cascaded Amplifiers
F1,G1
≡
F2,G2
1
2
F1+ 2 ∆
3
F1+2,G1+2
1
3
S1 N1
where S3 = G1G2S1
S3 N3
⎛
S N ⎞
N2 = kToF1G1 ⎜⎜ Recall F1 = 1 1 ⎟⎟
S2 N2 ⎠
⎝
N3 = G2kToF1G1 + G2 (F2 − 1)kTo
amplified from “2”
excess from “2”
∴ S3 N3 = G1G2S1 (G1G2F1 + G2 (F2 − 1))(kTo )
= (S1 kTo )
∴
⎛
F − 1⎞
⎜ F1 + 2 ⎟
⎜
G1 ⎟⎠
⎝
S1 N1
F2 − 1
= F1 +
F1+ 2 =
S3 N3
G1
L1
Noise in multiple cascade amplifiers
F3 − 1
By extension: F1,2,3 = F1+ 2 +
G1G2
F2 − 1
F1+ 2 = F1 +
G1
In general,
F2 − 1 F3 − 1
+
+ … cascade noise formula
F1,2,… = F1 +
G1G2
G1
Note:
A
B
vs.
B
A
The better choice is not obvious.
F1,2 also depends on interstage impedance mismatches
and gain of the first amplifier, not just on F1
L2
Noise in superheterodyne receivers
S1,N1 S2,N2
×
0
fo
f
Gc,Fc
0 fi.f.
L.O.
“lower (rf) sideband”
f
Gi.f.,Fi.f.
“i.f.” ∆ “intermediate frequency”
“upper sideband”
S2
0
S1
i.f. passband
f
fL.O.
S2 (i.f .)
1
∆
∆
1) “Conversion loss of mixer” Lc
=
Gc S1(r.f ., SSB )
2) “Noise temperature ratio of mixer” ∆ t = N ( i.f.) kT
r
S1 kTo
S1 N1
3) Fmixer ∆
=
= Lc tr
S2 N2 (S1 Lc ) tr kTo
2
o
L3
Noise in superheterodyne receivers
S1 kTo
S1 N1
3) Fmixer ∆
=
= Lc tr
S2 N2 (S1 Lc ) tr kTo
Fi.f . − 1
4) Fmixer +i.f . amp. = Fmixer +
Gmixer
= L c tr + L c (Fi.f . − 1) = L c (Fi.f . + t r − 1)
e.g. Fmixer +i.f . amp. ≅ 2 − 8(~ 3 − 9 dB )
L4
Basic receiver types-Amplification
1) Simple detector
B
()
2
h(t)
vo(t)
f
OR
2) RF Amplifier
OR
3) Superheterodyne
×
N1
Basic receiver types-Combinors
1)
2)
vo(t)
Total power radiometer
vo(t)
Dicke radiometer
vo(t)
Multiplication (correlation)
receiver
3)
×
4)
N
L
I
N
E
A
R
M
N-way combiner, N >< M
N2
Passive Multiport Networks
1)
2)
beamsplitter
(4-port)
+
RCVR
adding interferometer
3)
N detectors
3-dB hybrid
4)
×
Zo
white light
diffraction grating
L.O.
cancel
add in phase
output
(N-port)
input
λ/4
output
N3
Passive Multiport Networks
5) “Magic tee”
6) Frequency converter
USB, (upper sideband)
2
I.F.
3
4
Nonlinear
noise
1
LSB, (lower sideband)
LSB
Port 1 orthogonal to Port 2
Port 3 orthogonal to Port 4
All 4 ports can be matched
USB
f
I.F.
N4
Linear Passive N-port Networks
⎧⎪ a 2 toward port i
⎫⎪
Define exchangeable power = ⎨ i 2
⎬
­1
⎪⎩ bi from port i W or WHz ⎪⎭
(
ai
)
i
bi
j
2
Net power entering port “i” = ai − bi
2
Scattering matrix equation: b = Sa
Note:
1) The scattering matrix S is defined only when the n-port
is imbedded in a network
2) The phases of ai and bi can be defined as some linear
combinations of those for voltage and current
e.g. ai = V + Yo 2 = (V + Zo I)
8Zo , bi = (V − Zo I)
8Zo
P1
Gain definition for N-port networks
“Transducer gain”
GTkj ∆ bk
2
aj
2
= Skj
2
= PDk PAj
“Exchangeable gain” (~”available gain”)
PE
?
2
GE ∆ k =
Note: available power out ≥ bk
2
kj
PE
aj
due to possible port-k mismatch
j
i.e. fractional power absorbed (FPA) =
ak
2
− bk
ak
2
= 1 − Skk
2
2
But FPA = fractional power emitted, if reciprocity applies.
Therefore available power from port K = bk
[ ] (1− S ) a
Therefore GE = bk
kj
2
kk
2
2
j
=
2
(1− S )
kk
Skj
2
2
1 − Skk
2
= GE
kj
P2
Constraints on N-port networks
N
Lossless passive networks ⇒ ∑ ai
i=1
Reciprocity ⇒ S = S
2
N
= ∑ bi
2
i =1
t
P3
Example of constrained N-port networks
Does an ideal power combiner exist?
kT1
⎡0 0 α⎤
We want S = ⎢ 0 0 α ⎥
⎥
⎢
k(T1 +T2)
⎣⎢α α β ⎥⎦
1
?
2
symmetry
kT2
3
t∗
t∗
Losslessness ⇒ a • a = b b
t∗
( ) • (S • a) = a
b • b = Sa
Therefore
t∗
t∗
t∗
S Sa
⎡ 1 0 0⎤
S S = I = ⎢0 1 0⎥ ; if the combiner is lossless
⎥
⎢
⎢⎣0 0 1⎥⎦
t∗
P4
Example of constrained N-port networks
⎡ 1 0 0⎤
Therefore S S = I = ⎢0 1 0⎥ ; if the container is lossless
⎥
⎢
⎢⎣0 0 1⎥⎦
⎡0 0 α⎤
⎢0 0 α⎥
S
=
⎡ α2 α2
⎥
⎢
α∗ β ⎤
⎢ 2
⎥
⎣⎢α α β ⎥⎦
t∗
2
∗
Test: S S = ⎢ α
α β ⎥
α
⎢ ∗
2⎥
2
∗
⎢⎣αβ αβ 2 α + β ⎥⎦
t∗
2
2
2
If α = 1, then 2 α + β ≠ 1
Therefore constraints (8-14) and (8-15) cannot be
satisfied simultaneously for this system
Ideal matched 2-input-port combiners are impossible
P5
Another N-port network example
Can we match all 3 ports simultaneously?
1
3
2
(Note: S is defined only when imbedded in network)
For 3 matched ports:
t∗
Does S S = I ?
⎡ 0 α β⎤
S = ⎢α 0 γ ⎥
⎢
⎥
⎢⎣ β γ 0⎥⎦
(lossless passive constraint)
P6
Matched 3-port example
⎡ 0 α β⎤
S = ⎢α 0 γ ⎥
⎢
⎥
⎣⎢ β γ 0⎥⎦
For 3 matched ports:
t∗
Does S S = I ?
(lossless passive constraint)
⎡α 2 + β2
⎢
t∗
S S = ⎢ βγ∗
⎢
∗
α
γ
⎣⎢
β∗ γ
2
α +γ
αβ∗
α∗ γ
2
⎤
⎥ ?
∗
α β ⎥ = I
2
2⎥
β +γ ⎥
⎦
∗
∗
∗
If β γ = α γ = α β = 0 , then 2 of (α,β,γ) and at least
t∗
one diagonal element of S S = 0
Therefore not possible to match all 3 ports simultaneously
P7
Lossless passive reciprocal symmetric 4-port network
Ports 1, 2 are isolated; also 3, 4
1
3
2
4
We can show ∆φ for
1
3
versus
symmetry axis, all ports matched
1
4
paths is unique
using losslessness, reciprocity, and symmetry
P8
Mixer
+ vThsignal
+
S(t ) + (t) = VTh
vo(t)
Rs
i(t)
RL
-
local
oscillator
diode:
i
io = VTh
i ≅ v 2 for diode
d
v d ( t ) ≅ v sin ωo t + v s sin ωs t
+ small high - order terms
(
2
v o (t ) = iRL ∝ v Th
RL
)
2
= k o + k1v Th + k 2v Th
+ k 3 v 3Th + ...
( RL + RS ) >> i(t)
load line i =
RL + RS
vd
0
diode i,vTh(t)
VTh − v d
VTh
dominant
R1
Mixer
2
(∝ v R )
3
2
d L
v o (t) = iRL = k o + k1v Th + k 2 v Th + k 3 v Th + ...
dominant
Spectral content of detector output:
Vo ( f )
0
fif = fs − fo
image
fo fs
2fo 2fs
fo + fs
3fo
3fs
f
Pure product (frequency component for pure s(t) • (t))
(Note: v 4 ⇒ 2fi.f . , can be in i.f. passband)
d
Normally:
a 4-port model suffices
to find F,TSSB , " TR " :
signal
1
4
image
2
3
output
internal
noise
R2
Mixer Noise Figure Using 4-port Model
TR = TSSB = (F − 1)To = To (Lc tr − 1)
signal
1
4
for single sideband (SSB) operation
image
2
3
FSBB =
S1 N1
=
S 4 N4 ⎡ S S 2 ⎤
⎢ 1 41 2 ⎥
⎢⎣1 − S 44 ⎥⎦
output
internal
noise
S1 kTo
⎡kT S 2 + kT S 2 + kT S 2 ⎤
2 42
3 43
⎥
⎢ o 41
2
⎥⎦
1 − S 44
⎣⎢
[Note: “S”, signifies signal in port 1, Skj is a scattering matrix element]
2
2
TSSB
T3 S 43
T2 S 42
Simplifying, FSSB = 1 +
+
= 1+
2
2
To
To S
To S
41
41
Therefore
TSSB = T2
S 42
2
S 41
2
+ T3
S 43
2
S 41
2
R3
Double-sideband Receiver
Therefore TSSB = T2
S 42
2
S 41
2
+ T3
S 43
2 signal
S 41
2
image
(
2
It follows that
TDSB = T3 S 43
Often suggested:
2
[S
2
2
4
2
3
output
internal
noise
Both ports 1 and 2 are signal, so
S 4 = kTo S 41 + S 42
1
) (1− S )
41 + S 42
44
2
2
]
FSSB ≅ 2FDSB
R4
Double-sideband Receiver
Therefore TSSB = T2
S 42
2
S 41
2
+ T3
Often suggested:
2
Let S 41 = S 42
2
S 43
2 signal
S 41
2
image
1
4
2
3
FSSB ≅ 2FDSB
output
internal
noise
and To = T1 = T2 ; then
2
TDSB
1 T3 S 43
FDSB = 1 +
= 1+
2 To S 2
To
41
2
T3 S 43
TSSB
FSSB = 1 +
= 2+
= 2FDSB
2
To S
To
41
R5