Noise in Cascaded Amplifiers - MIT OpenCourseWare
Transcript of Noise in Cascaded Amplifiers - MIT OpenCourseWare
Noise in Cascaded Amplifiers
1 2 313 F2,G2 F1+2,GF1,G1 ≡
1213 33 11 Swhere
NS NS
F =∆+
⎟⎟ ⎠
⎞ ⎜⎜ ⎝
⎛ ==
22 11
12 NS NS
N
( ) o2223 FGN −+=
( )( )( )o221133 FGNS −+=∴
( ) ⎟⎟ ⎠
⎞ ⎜⎜ ⎝
⎛ − +=
1 2
1o1 G 1F
FkTS
1
2 1
33
11 G
1FF
NS NS
F −
+== +∴
L1
1+2
2 1 S G G
1 1 o F Recall G F kT
1 1 o kT 1 G F kT G
1 2 1 2 kT 1 F G G S G G
amplified from “2” excess from “2”
2 1
Noise in multiple cascade amplifiers
In general, 1F
G 1F
FF 21
3
1
2 1 … … +
− +
− +=
A B B Avs.
Note:
The better choice is not obvious. F1,2
and gain of the first amplifier, not just on F1
21
3 1FFF
− += +By extension:
1
2 1 G
1FFF
− += +
L2
formula noise cascade G G ,2,1
also depends on interstage impedance mismatches
2 1 3,2,1 G G 2 1
Noise in superheterodyne receivers
×
0 f f0
S1,N1 S2,N2
L.O. Gc,Fc Gi.f.,Fi.f.
“i.f.” “intermediate frequency”∆
( ) ( )S
iS G 1L
1
2
c c =∆∆
( )r 2 ot N i∆ =
3) ( ) orc1
o1
22
11 mixer LS
kTS NS NS
F ==∆
L3
fo fi.f.
0 S2 S1
“upper sideband”
“lower (rf) sideband”
i.f. passband fL.O.
f
1) “Conversion loss of mixer” SSB ., f.r
. f.
2) “Noise temperature ratio of mixer” .f. kT
r c t L kT t
Noise in superheterodyne receivers
3) ( ) orc1
o1
22
11 mixer LS
kTS NS NS
F ==∆
mixer
i mixerimixer G
1FFF
− += +
( ) ( )1tFL1FL r.f.ic.f.icrc −+=−+=
4)
( )2imixer −−≅+
L4
r c t L kT t
. f.amp. . f.
t L
dB 9 3 ~ 8 F e.g. amp. . f.
Basic receiver types-Amplification
vo(t)( )2 f
B
OR
OR
×
N1
h(t)1) Simple detector
2) RF Amplifier
3) Superheterodyne
Basic receiver types-Combinors
1) vo(t) Total power radiometer
2) vo(t)
3) × vo(t)
Multiplication (correlation) receiver
4) N M N-way combiner, N M<>
L I N E A R
N2
Dicke radiometer
Passive Multiport Networks
beamsplitter (4-port)
1)
RCVR+
adding interferometer
2)
3)
white light
(N-port)
diffraction grating
N detectors
N3
4) × 3-dB hybrid
L.O. Zo
λ/4
cancel add in phase
input
output
output
Passive Multiport Networks
5) “Magic tee”
2
1
3 4
Port 1 orthogonal to Port 2 Port 3 orthogonal to Port 4 All 4 ports can be matched
N4
LSB, (lower sideband)
USB, (upper sideband)
I.F.
noise
6) Frequency converter
LSB
f
I.F.
Nonlinear
USB
Linear Passive N-port Networks
Define exchangeable power = ( )⎪⎭⎪⎬⎫
⎪⎩
⎪⎨⎧
2 i
2 i
ib
ia
iia
ib j
Net power entering port “i” = 2 i
2 i ba −
Scattering matrix equation: b =
is defined only when the n-port is imbedded in a network
i and bi can be defined as some linear combinations of those for voltage and current
S Note:
( ) ( ) ooioooi VV2YV −=+== + P1
1 WHz or W port from
port toward
aS
1) The scattering matrix
2) The phases of a
Z 8 I Z b , Z 8 I Z a .g.e
Gain definition for N-port networks “Transducer gain”
AjDk 2
kj 2
j 2
kTkj PPSabG ==∆
Therefore available power from port K =
But FPA = fractional power emitted, if reciprocity applies.
i.e. fractional power absorbed (FPA) = 2 kk2
k
2 k
2 k S1
a
ba −=
−
( )2 kk
2 k S1b −
Therefore [ ] ( ) =−= 2
j 2
kk 2
kE aS1bG kj kjE2
kk
2 kj G S1
S =
−
“Exchangeable gain” (~”available gain”)
Note: available power out due to possible port-k mismatch
2 jE
E E
a
? P
P G
j
k kj
=∆ 2 kb≥
P2
Constraints on N-port networks
Lossless passive networks ∑∑ ==
=⇒ N
1i
2 i
N
1i
2 i ba
Reciprocity tSS =⇒
P3
Example of constrained N-port networks
bbaa tt ∗∗ =•⇒Losslessness
( ) ( ) SaaSbb tttt ∗∗∗∗ =••=•
Therefore ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡ ==
∗
100 010 001
ISSt ; if the combiner is lossless
Does an ideal power combiner exist?
? 2 3
1kT1
kT2 symmetry
k(T1 +T2) ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
βαα
αα
= 00 00
SWe want
P4
aS aS
Example of constrained N-port networks
⎥ ⎥ ⎥
⎦
⎤
⎢ ⎢ ⎢
⎣
⎡
βαα
α
α
= 00 00
S
Therefore ⎥ ⎥ ⎥
⎦
⎤
⎢ ⎢ ⎢
⎣
⎡ ==
∗
100 010 001
ISSt ; if the container is lossless
Test:
⎥ ⎥ ⎥ ⎥
⎦
⎤
⎢ ⎢ ⎢ ⎢
⎣
⎡
α
αα
αα
= ∗∗
∗
∗ ∗
22
22
22
t
2 SS
1If 222 ≠α=α
Therefore constraints (8-14) and (8-15) cannot be satisfied simultaneously for this system
P5
β + β α β α
β α
β α
2 then ,1 β +
Ideal matched 2-input-port combiners are impossible
Another N-port network example Can we match all 3 ports simultaneously?
2
3
1
(Note: is defined only when imbedded in network)S
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
γβ
γα
βα
=
0 0
0 SFor 3 matched ports:
SSt =
∗ Does (lossless passive constraint)
P6
? I
⎥ ⎥ ⎥
⎦
⎤
⎢ ⎢ ⎢
⎣
⎡
γβ
γα
βα
=
0 0
0 SFor 3 matched ports:
SSt =
∗ Does (lossless passive constraint)
Matched 3-port example
ISS ?
22
22
22
t =
⎥ ⎥ ⎥ ⎥
⎦
⎤
⎢ ⎢ ⎢ ⎢
⎣
⎡
β
α
α
= ∗∗
∗∗
∗∗ ∗
Therefore not possible to match all 3 ports simultaneously P7
0∗∗∗ If , then 2 of (α ,β ,γ ) and at least one diagonal element of 0SSt
= ∗
? I
γ + β α γ α
β α γ + γ β
γ α γ β β +
= β α = γ α = γ β
Lossless passive reciprocal symmetric 4-port network
31
42
symmetry axis, all ports matched
Ports 1, 2 are isolated; also 3, 4
We can show ∆φ for 31 41versus paths is unique
using losslessness, reciprocity, and symmetry
P8
R1
vTh+ -
+ = Th) (t) V
signal
local oscillator
i(t) Rs
+
-RL
vo(t)
0 VTh
diode:
diode i,vTh(t)
( )= +o L Si V R R i(t)
2 dvi ≅
load line i = −
+ Th d
L S
V v R R
i
vd
tsinvtsinv)t(v ssod ω+ω≅
termsorder-+
( )= ∝ 2 o L L) iR v R
= + + + +2 3 o 2 3 Thk
dominant
Mixer
S(t
>> Th
diode for
high small
Th v (t
1 Th Th k v k v k v ...
R2
( )= = + + + + ∝ 2 3 2
o L o 1 Th 2 3 Th d Lv (t) iR k k v k v
dominant Spectral content of detector output:
ofsfiff −=
)f(oV image
of sf o ssfof +
o sf0
Pure product (frequency component for pure s(t) • (t))
i4 dv ⇒(Note: , can be in i.f. passband)
1
2 3
4 output
internal noise
signal
image
Normally: a 4-port model suffices to find F,T RSSB
Mixer Th k v ... v R
f 2 f 2 f 3 f 3
. f.f 2
: " T " ,
1
2 3
4 output
internal noise
signal
image
R3
( )1TT)1F(TT rcooSSBR −=−==
for single sideband (SSB) operation
⎥ ⎥ ⎦
⎤
⎢ ⎢ ⎣
⎡
−
++
⎥ ⎥ ⎦
⎤
⎢ ⎢ ⎣
⎡
−
==
2 44
2 433
2 422
2 41o
2 44
2 411
o1
44 11
SBB
S1S1
kTS NS NSF
[Note: “S”, signifies signal in port 1, Skj is a scattering matrix element]
o
SSB 2
41
2 43
o
3 2
41
2 42
o 2
SSB T T1
S
S T T
S
S T T1F +=++=Simplifying,
2 41
2 43
32 41
2 42
2SSB S
ST
S
STT +=Therefore
Mixer Noise Figure Using 4-port Model
t L
S kT S kT S kT S S
2 41
2 43
32 41
2 42
2SSB S
ST
S
STT +=Therefore
Both ports 1 and 2 are signal, so
( ) ( )2 44
2 42
2 41o4 S1SSkTS −+=
1
2 3
4 output
internal noise
signal
image
Often suggested: DSBSSBF ≅
R4
It follows that [ ]SST 2
42 2
41 2
433DSB +=
Double-sideband Receiver
F2
S T
DSB2 41
2 43
o
3
o
SSBSSB
S
S T T2
T T1F =+=+=
then;TTS 21o 2
42 2
41 ===
2 41
2 43
o
3
o
DSBDSB
S
S T T
2 11
T T1F +=+=
Often suggested: DSBSSBF ≅
2 41
2 43
32 41
2 42
2SSB S
ST
S
STT +=
1
2 3
4 output
internal noise
signal
image Therefore
R5
Double-sideband Receiver
F2
T and S Let
F2