APPENDIX A: NOISE MODELING WITH APPLICATIONS TO COMMUNICATIONS 1
Problem A.1
All parts of the problem are solved using the relation
p
Vrm s = 4kT RB
where
k = 1:38 10 23 J/K
B = 30 MHz = 3 107 Hz
a. For R = 10; 000 ohms and T = T0 = 290 K
p
Vrm s = 4 (1:38 10 23 ) (290) (104 ) (3 107 )
= 6:93 10 5 V rms
= 69:3 V rms
p
b. Vrm s is smaller than the result in part (a) by a factor of 10 = 3:16: Thus
Vrm s = 21:9 V rms
p
c. Vrm s is smaller than the result in part (a) by a factor of 100 = 10: Thus
Vrm s = 6:93 V rms
p
d. Each answer becomes smaller by factors of 2; 10 = 3:16; and 10, respectively.
Problem A.2
Use
eV
I = Is exp 1
kT
eV
We want I > 20Is or exp kT 1 > 20.
e 1:6 10 19
a. At T = 290 K, kT = 1:38 1023 290 = 40, so we have exp (40V ) > 21 giving
ln (21)
V > = 0:0761 volts
40
eV
i2rm s = 2eIB ' 2eBIs exp
kT
i2rm s eV
or = 2eIs exp
B kT
19 5
= 2 1:6 10 1:5 10 exp (40 0:0761)
22 2
= 1:0075 10 A /Hz
e
b. If T = 90 K, then kT = 129, and for I > 20Is , we need exp(129V ) > 21 or
ln (21) 2
V > = 2:36 10 volts
129
APPENDIX A: NOISE MODELING WITH APPLICATIONS TO COMMUNICATIONS 2
Thus
i2rm s 19 5
= 2 1:6 10 1:5 10 exp (129 0:0236)
B
= 1:0079 10 22
A2 /Hz
approximately as before.
Problem A.3
a. Use Nyquist’s formula to get the equivalent circuit of Req in parallel with RL , where Req is
given by
R3 (R1 + R2 )
Req =
R 1 + R 2 + R3
The noise equivalent circuit consists of a rms noise voltage, Veq , in series with Req and a rms noise
voltage, VL , in series with with RL with these series combinations being in parallel. The equivalent
noise voltages are
p
Veq = 4kT Req B
p
VL = 4kT RL B
The rms noise voltages across the parallel combination of Req and RL , by using superposition and
voltage division, are
Ve q RL VL Re q
V01 = and V02 =
Req + RL Req + RL
Adding noise powers to get V02 we obtain
2 2
Veq RL 2
VL2 Req
V02 = 2 + 2
(Req + RL ) (Req + RL )
(4kT B) RL Req
=
Req + RL
RL R3 (R1 + R2 )
= 4kT B
R 1 R 3 + R2 R 3 + R1 RL + R 2 R L + R 3 R L
Note that we could have considered the parallel combination of R3 and RL as an equivalent load
resistor and found the Thevenin equivalent. Let
R 3 RL
Rjj =
R3 + R L
The Thevenin equivalent resistance of the whole circuit is then
R3 RL
Rjj (R1 + R2 ) R3 +RL (R1 + R2 )
Req2 = = R3 RL
Rjj + R1 + R2 R3 +RL + R1 + R 2
RL R3 (R1 + R2 )
=
R 1 R 3 + R2 R3 + R1 RL + R2 RL + R 3 R L
APPENDIX A: NOISE MODELING WITH APPLICATIONS TO COMMUNICATIONS 3
and the mean-square output noise voltage is now
V02 = 4kT BReq2
which is the same result as obtained before.
b. With R1 = 2000 ; R2 = RL = 300 ; and R3 = 500 , we have
V02 RL R3 (R1 + R2 )
= 4kT B
B R1 R3 + R 2 R 3 + R 1 R L + R 2 R L + R3 RL
4 1:38 10 23 (290) (300) (500) (2000 + 300)
=
2000 (500) + 300 (500) + 2000 (300) + 300 (300) + 500 (300)
= 2:775 10 18
V2 /Hz
Problem A.4
Find the equivalent resistance for the R1 ; R2 ; R3 combination and set RL equal to this to get
R3 (R1 + R2 )
RL =
R1 + R2 + R 3
Problem A.5
Using Nyquist’s formula, we …nd the equivalent resistance looking back into the terminals with Vrm s
across them. It is
Req = 50 k k 20 k k (5 k + 10 k + 5 k)
= 50 k k 20 k k 20 k
= 50 k k 10 k
(50 k) (10 k)
=
50 k + 10 k
= 8; 333
Thus
2
Vrm s = 4kT Req B
23
= 4 1:38 10 (400) (8333) 2 106
= 3:68 10 10
V2
which gives
Vrm s = 19:18 V rms
Problem A.6
APPENDIX A: NOISE MODELING WITH APPLICATIONS TO COMMUNICATIONS 4
To …nd the noise …gure, we …rst determine the noise power due to a source at the output, then
due to the source and the network, and take the ratio of the latter to the former. Initally assume
unmatched conditions. The results are
2
R2 k RL
V02 due to RS , only
= (4kT RS B)
R S + R1 + R 2 k RL
2
R2 k RL
V02 due to R1 and R2
= (4kT R1 B)
R S + R 1 + R2 k RL
2
RL k (R1 + RS )
+ (4kT R2 B)
R2 + (R1 + RS ) k RL
2
R2 k RL
V02 due to RS ; R1 and R2
= [4kT (RS + R1 ) B]
RS + R1 + R 2 k R L
2
RL k (R1 + RS )
+ (4kT R2 B)
R2 + (R1 + RS ) k RL
The noise …gure is (after some simpli…cation)
2 2
R1 RL k (R1 + RS ) R S + R1 + R 2 k R L R2
F =1+ +
RS R2 + (R1 + RS ) k RL R2 k RL RS
In the above,
Ra Rb
Ra k Rb =
Ra + Rb
Note that the noise due to RL has been excluded because it belongs to the next stage. Since this is
a matching circuit, we want the input matched to the source and the output matched to the load.
Matching at the input requires that
R2 RL
RS = Rin = R1 + R2 k RL = R1 +
R2 + RL
and matching at the output requires that
R2 (R1 + RS )
RL = Rout = R2 k (R1 + RS ) =
R1 + R 2 + R S
Next, these expressions are substituted back into the expression for F . After some simpli…cation,
this gives
2 2
R1 2RL RS (R1 + R2 + RS ) = (RS R1 ) R2
F =1+ + 2 2
RS R2 (R1 + RS + RL ) + RL (R1 + R2 + RS ) RS
Note that if R1 >> R2 we then have matched conditions of RL = R2 and RS = R1 : Then, the
noise …gure simpli…es to
R1
F = 2 + 16
R2
APPENDIX A: NOISE MODELING WITH APPLICATIONS TO COMMUNICATIONS 5
Note that the simple resistive pad matching circuit is very poor from the standpoint of noise. The
equivalent noise temperature is found by using
Te = T0 (F 1)
R1
= T0 1 + 16
R2
Problem A.7
a. The important relationships are
T el
Fl = 1 +
T0
Tel = T0 (Fl 1)
T e2 T e3
T e0 = T e1 + +
Ga1 Ga1 Ga2
Completion of the table gives
Ampl. No. F T ei Gai
1 not needed 300 K 10 dB = 10
2 6 dB 864.5 K 30 dB = 1000
3 11 dB 3360.9 K 30 dB = 1000
Therefore,
864:5 3360:9
T e0 = 300 + +
10 (10) (1000)
= 386:8 K
Hence,
T e0
F0 = 1+
T0
= 2:33 = 3:68 dB
b. With ampli…ers 1 and 2 interchanged
300 3360:9
T e0 = 864:5 + +
10 (10) (1000)
= 865:14 K
This gives a noise …gure of
865:14
F0 = 1+
290
= 3:98 = 6 dB
APPENDIX A: NOISE MODELING WITH APPLICATIONS TO COMMUNICATIONS 6
c. See part (a) for the noise temperatures.
d. For B = 50 kHz, TS = 1000 K, and an overall gain of Ga = 107 , we have, for the con…guration
of part (a)
Pna; out = Ga k (T0 + Te0 ) B
= 107 1:38 10 23 (1000 + 386:8) 5 104
9
= 9:57 10 watts
We desire
Psa; out Psa; out
= 104 = 9
Pna; out 9:57 10
which gives
5
Psa; out = 9:57 10 watts
For part (b), we have
Pna; out = 107 1:38 10 23
(1000 + 865:14) 5 104
8
= 1:29 10 watts
Once again, we desire
Psa; out Psa; out
= 104 = 8
Pna; out 1:29 10
which gives
4
Psa; out = 1:29 10 watts
and
Psa; out 11
Psa; in = = 1:29 10 watts
Ga
Problem A.8
a. The noise …gure of the cascade is
F2 1 F 1
Foverall = F1 + =L+ = LF
Ga1 (1=L)
b. For two identical attenuator-ampli…er stages
F 1 L 1 F 1
Foverall = L + + + = 2LF 1 2LF; L >> 1
(1=L) (1=L) L (1=L) L (1=L)
c. Generalizing, for N stages we have
Foverall NFL
APPENDIX A: NOISE MODELING WITH APPLICATIONS TO COMMUNICATIONS 7
Problem A.9
a. The data for this problem is
Stage Fi Gi
1 (preamp) 2 dB = 3.01 G1
2 (mixer) 8 dB = 6.31 1.5 dB = 1.41
3 (ampli…er) 5 dB = 3.16 30 dB = 1000
The overall noise …gure is
F2 1 F3 1
F = F1 + +
G1 G1 G2
which gives
6:31 1 3:16 1
5 dB = 3:16 1:58 + +
G1 1:41G1
or
5:31 (2:16=1:41)
G1 = 2:39 = 2:79 dB
1:58
b. First, assume that G1 = 15 dB = 31:62: Then
6:31 1 3:16 1
F = 1:58 + +
31:62 (1:41) (31:62)
= 1:8 = 2:54 dB
Then
Tes = T0 (F 1)
= 290 (1:8 1)
= 230:95 K
and
Toverall = Tes + Ta
= 230:95 + 300
= 530:95 K
Now use G1 as found in part (a):
F = 3:16
Tes = 290 (3:16 1) = 626:4 K
Toverall = 300 + 626:4 = 926:4 K
c. For G1 = 15 dB = 31:62; Ga = 31:62 (1:41) (1000) = 4:46 104 : Thus
Pna; out = Ga kToverall B
= 4:46 104 1:38 10 23
(530:9) 109
9
= 3:27 10 watts
APPENDIX A: NOISE MODELING WITH APPLICATIONS TO COMMUNICATIONS 8
For G1 = 3:79 dB = 2:39; Ga = 2:39 (1:41) (1000) = 3:37 103 : Thus
Pna; out = 3:37 103 1:38 10 23
(926:4) 107
10
= 4:31 10 watts
Note that for the second case, we get less noise power out even wth a larger Toverall . This is due
to the lower gain of stage 1, which more than compensates for the larger input noise power.
d. A transmission line with loss L = 2 dB connects the antenna to the preamp. We …rst …nd
TS for the transmission line/preamp/mixer/amp chain:
F1 1 F2 1 F3 1
FS = FTL + + + ;
GTL GTL G1 GTL G1 G2
where
2=10
GTL = 1=L = 10 = 0:631 and FTL = L = 102=10 = 1:58
Assume two cases for G1 : 15 dB and 6.37 dB. First, for G1 = 15 dB = 31:6, we have
1:58 1 6:31 1 3:16 1
FS = 1:58 + + +
0:631 (0:631) (31:6) (0:631) (31:6) (1:41)
= 2:84
This gives
TS = 290 (2:84 1) = 534 K
and
Toverall = 534 + 300 = 834 K
Now, for G1 = 3:79 dB = 2:39, we have
1:58 1 6:31 1 3:16 1
FS = 1:58 + + +
0:631 (0:631) (2:39) (0:631) (2:39) (1:41)
= 7:04
This gives
TS = 290 (7:04 1) = 1752 K
and
Toverall = 1752 + 300 = 2052 K
Problem A.10
a. (a) Using
Pna; out = Ga kTS B
with the given values yields
6
Pna; out = 7:45 10 watts
APPENDIX A: NOISE MODELING WITH APPLICATIONS TO COMMUNICATIONS 9
b. We want
Psa; out
= 105
Pna; out
or
Psa; out = 105 7:45 10 6
= 0:745 watts
This gives
Psa; out 8
Psa; in = = 0:745 10 watts
Ga
= 51:28 dBm
APPENDIX A: NOISE MODELING WITH APPLICATIONS TO COMMUNICATIONS 10
Problem A.11
a. For A = 1 dB, Y = 1:259 and the e¤ective noise temperature is
600 (1:259) (300)
Te = = 858:3 K
1:259 1
For A = 1:5 dB, Y = 1:413 and the e¤ective noise temperature is
600 (1:413) (300)
Te = = 426:4 K
1:413 1
For A = 2 dB, Y = 1:585 and the e¤ective noise temperature is
600 (1:585) (300)
Te = = 212:8 K
1:585 1
b. These values can be converted to noise …gure using
Te
F =1+
T0
With T0 = 290 K, we get the following values: (1) For A = 1 dB; F = 5:98 dB; (2) For
A = 1:5 dB; F = 3:938 dB; (3) For A = 2 dB; F = 2:39 dB.
Problem A.12
a. Using the data given, we can determine the following:
= 0:039 m
2
= 202:4 dB
4 d
GT = 39:2 dB
PT GT = 74:2 dBW
This gives
PS = 202:4 + 74:2 + 6 5= 127:2 dBW
b. Using Pn = kTe B for the noise power, we get
Te
Pn = 10 log10 kT0 B
T0
Te
= 10 log10 [kT0 ] + 10 log10 + 10 log10 (B)
T0
1000
= 174 + 10 log10 + 10 log10 106
290
= 108:6 dBm
= 138:6 dBW
APPENDIX A: NOISE MODELING WITH APPLICATIONS TO COMMUNICATIONS 11
c.
PS
= 127:2 ( 138:6)
Pn dB
= 11:4 dB
= 101:14 = 13:8 ratio
d. Assuming the SNR = z = Eb =N0 = 13:8; we get the results for various digital signaling
techniques given in the table below:
Modulation type Error probability
p
BPSK Q 2z = 7:4 10 8
1 z
DPSK 2e = 5:06 10 7
1 z=2
Noncoh. FSK 2e = 5:03 10 4
QPSK Same as BPSK
