Books
in black and white
Main menu
Share a book About us Home
Books
Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics
Ads

Theory of Linear and nonlinear circuits - Engberg J.

Engberg J. Theory of Linear and nonlinear circuits - Wiley & sons , 1995. - 154 p.
ISBN 0-47-94825
Download (direct link): noisetheoryoflinearandnonlinear1995.pdf
Previous << 1 .. 19 20 21 22 23 24 < 25 > 26 27 28 29 30 31 .. 85 >> Next

r? , — 1 , ^e,3 — 1 , Fc,m - 1 гЛ
= Fe, 1 H--------p----- + y;—у !-•■•+ J, j, P- fo.5)
^e,l beiiUe,2
m F • - 1
= ^i + EfprTT- (5.6)
i=2 11; = 1е,;
In order to see how the signs behave the combination of two stages is examined. From Table 4.2 it is seen that Tee has the same sign as the source resistance Rs-The formula for Tce of two stages in cascade is (Equation 5.3):
T — т 4. ^-e-2
1 ее — ee,l "r
^«,1
This gives the signs as shown in Table 5.1.
Tee, 1 Ge,l Fe,2 Tee
> 0 > 0 > 0 > 0
> 0 < 0 < 0 > 0
< 0 > 0 < 0 < 0
< 0 < 0 > 0 < 0
- — !o-*- *•“ г г...- two two ports 1
It is seen from Table 5.1 that the sign for Tee for a two-stage amplifier is the same as the sign of the source immittance. Taking these two two-ports as a single two-port and then adding another two-port; the combination of three two ports also
5.1. Two-ports in cascade
73
follows the sign of the source immittance. This procedure can be repeated as often as necessary, so
for Rs > 0 then T,. > 0 and Fs - I > 0
and for Rs < 0 then Tee < 0 and Fe - I < 0
Example 5.1 An antenna amplifier consists of two identical stages each with the following data:
Yn = 10 + j 2.1 mS YV2 = 0.50 -j 0.86 mS
5-21 = 19-j30mS Y-22 = 1.0 + j3.0mS
Rn = 25 P. Gn = 4.8 mS
Yy = 2.0 + j 7.5 mS
The source admittance is the antenna admittance which is I5 = 20 mS.
The noise factor of the antenna amplifier is computed in the following way.
Fe, 1 = 1 + ^ (G„ + Rn\Ys + K7|2)
= 1 + ^ (4.8 + 0.025 |20 + 2 + j7.5|2) = 1.92
\Y2i\2 Gs
Ge,i =
Re[{(rn + Ys)Y22 - Yl2Y21}(Yn + Ysyj (192 + 302)20
Re [{(10 + j 2.1 + 20)(1.0 + j 3.0) - (0.50 - j 0.86)(19 - j 30)}( 10 + j 2.1 + 20)-= 17.3
Yl2 y21
Yont,\ — Yn. —
Yn + YS
f0.50 - 7 0-cSfiV 19 — ? 30 >
i,U + J 'j'° 10 + j 2.1 + 20 = 1.61 + j 4.00 mS
Fe,2 = 1 + (G„ + Rn \Y0ut,l + У/Г
1 + (4.8 + 0.02511.61 + j 4.00 4- 2.0 + j 7.512) = 6-23
1.01 '
... г -|2''
F, . - I 6.23 - !
i m
G> Д i < .0
It is seen that the second stage has a rather high noise factor, but because of the also rather high gain of the first stage it does not spoil the combined noise factor very much. If a lossless matching network was added between the two amplifier stages in order to
74
5. Noise measure and graphic representations
transform the output admittance of the first stage to the optimum source admittance for minimum noise factor for the second stage the combined noise factor can be determined. Instead of using Fei2 above the F.i2in,n is used-.
Fe,2m!n = 1 + 2 + yjRnGn + (RnG^j
= 1 + 2 ^0.025 x 2.0 + ^0.025 x 4.S + (0.025 x 2.0)2)
The second stage source admittance is computed by Equation (4.17) to 14 As the exchangeable power gain is unchanged
Fe = 1.92 + -■— - = 1.96
1 I .6
Example 5.2 An antenna with impedance 50 Л and noise temperature T„ = 2900 К is connected to an amplifier via a 50 Q. cable. This cable has a loss of 1.76 dB and its temperature can be taken to be the standard temperature, To (17 °C). The noise factor of the amplifier is 1,8.
The 50 fi cable has a loss of 1.76 dB and as its temperature is To [K] It is seen from Example 3.2 that its noise factor is 1.50 ~ 1.76 dB and the gain is 1/1.50. For the combination of cable and amplifier, the result is
Fe = F'tcMe + ~ 1 = 1.50 + = 2.7
^e,cable T~5Q
T« = (Fc - 1)T0 = (2.7 - 1)290 = 493 К
From this the operating noise temperature for the antenna - cable - amplifier combination referred to the antenna terminal is
Teop = Ta + T« = 2900 + 493 = 3393 ~ -3400 К
It should be noted that a rise in noise factor from 1.5 to 2.7 - which often should be avoided - is of no importance, when another noise source - here the antenna noise - is dominating.
5.2 The noiso measure
Take two two-ports with extended noise factors FeA and 2 and exchangeable power gains GcA and Ge,2 and fmd out which cascade combination is best - the one with two-port one first or the one with two-port two first T,pt the result be that
= 1.8 - j 7.5 mS.
5.2. The noise measure
Fe 12 < Ff2\. From Equation (5.5) it follows, for a passive source and С-д < 0 or Ge,i > i and Gs 2 < 0 or Ge2 > 1, that
p I ~ 1 r1 , ■f’e.l - 1
e, 1 i ------p;------ ^ I z,! T- ---^----------
v-'£.l ^«,2
- 1 - ctrj < lftj “ ,) (‘ ~ TT,_
Fe,l - 1 Fe.2 - 1
1 - r^- 1 - r1-
^«,1 ^е,2
It is seen tliat for two two-ports with a positive source immittance and an amplification which is either negative or greater than one, the choice should be the two-port with the smaUer extended noise measure, Me, first, where the extended noise measure is defined by
F„ - 1
Me = f—j- (5.7)
О?
If the source is chosen to be negative the same procedure shows that the negative noise measure closest to zero is the best. Haus and Adler [1] have extended the definition to the general case where Fe and Ge may differ from the conventional F and Ga and denoted this extended definition of noise measure by Mc.
In the discussion of the noise performance of amplifiers, it turns out to be important to bear in mind the algebraic sign that Me assumes under various physical conditions. These are summarized in Table 5.2.
Previous << 1 .. 19 20 21 22 23 24 < 25 > 26 27 28 29 30 31 .. 85 >> Next