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T wo diffrent manufacturers present their receiving systems as shown in

F ig. 16. 6. A potential customer wants to buy one of these receivers for his

particular application, where it is known that the input S/ N ratio is 12. 6 dB .

W hich receiver will provide the best output S/ N ratio?

Soluti on

F or each receiver, the overall noise factor must be calculated with ( 4. 78) and

then, applying F riiss definition, the output S/ N ratio can be evaluated.

16.2 Multistage Ampliers 547

Fig. 16.6. Block diagrams of the proposed receivers

However, in a preliminary step, all individual noise factors must be de-

duced from the noise gure values or from the noise temperatures.

Receiver 1. The waveguide has a length of 80 cm, hence its loss is

0 . 8(0 . 08) = 0 . 064 dB. Expressed as a ratio, this means L = 1 . 015. Accord-

ing to the theorem of Case Study 16.1, the noise factor of the waveguide is

F 1 =L=1 . 015.

The noise temperatures can be converted into noise factors with (4.76);

in this way, the block diagram can be redrawn as in Fig. 16.7, where the

notation is consistent.

Fig. 16.7. Equivalent block diagram of Receiver 1

Therefore

F REC1 =F 1 + F 2 −

+ F 3 −

G 1 G 2

G 1 G 2 G 3

F 4 −

G 1

=1 . 015 + 1 . 026

1 / 1 . 015

−

(1 / 1 . 015)10 +

−

(1 / 1 . 015)(10)(1000)

−

=1 . 447 or 1 . 6db .

12 . 9

548 16 Noise in Communication Systems

By means of Friis’s denition of the noise factor (expressed in decibels),

we deduce the output S/N ratio:

S o

N o

S i

N i −

F REC1 =12 . 6

−

1 . 6=11dB

Finally, Receiver 1 has an output S/N ratio

S o

N o

=11dB

REC1

Receiver 2. The block diagram, with all parameters expressed in consis-

tent units, is shown in Fig. 16.8. In this case, we can write

F REC2 =F 1 + F 2 −

+ F 3 −

G 1 G 2

G 1 G 2 G 3

G 1

=1 . 31 + 1 . 2

−

(0 . 76)(100) +

−

(0 . 76)(100)(100)

−

0 . 76

This yields:

F REC2 = 1 . 615 or 2 . 08 dB

and

S o

N o

S i

N i −

F REC2 =12 . 6

−

2 . 08 = 10 . 52 dB

The output S/N ratio of the second receiver is

S o

N o

REC2 = 10 . 52 dB

Fig. 16.8. Equivalent block diagram of the second receiver

Concluding Remark. The customer will very likely buy the rst receiver,

since its output S/N ratio is about 0.5 dB higher than that which would

be obtained under equivalent conditions with receiver 2. The result is not

surprising, as the loss of the waveguide situated at the front-end of the rst

receiver is smaller than that of the second receiver waveguide (recall that the

noise performance of the front-end stage is of paramount importance).

F 4 −

4 . 2

16.4 Receivers 557

at the antenna is dominant, it is not reasonable to buy a high-quality (less

noisy) receiver, because the improvement in overall performance is rather

insignicant (this statement is consistent with Rule 4 in Sect. 13.2.1).

For communication systems operating up to 30 MHz (LW, MW, and SW

bands), a receiver with a noise gure in the range 8–20 dB is quite ap-

propriate (given the usual values of the noise at the antenna). However,

above 30 MHz (where antenna noise drops o), a less noisy receiver con-

siderably improves system global performance.

CASE STUDY 16.14 [244]

A receiver has a preamplier connected to the amplier by a cable whose

length is 24 m. The performance of these components is given in Table 16.4.

Determine the minimum gain required in the preamplier if the global noise

gure of the system is not to exceed 6 dB.

Table 16.4. Performance of various components

Noise gure of the

preamplier

Cable loss

Noise gure of the

amplier

2.5 dB

0.2 dB/m

10 dB

Fig. 16.14. Block diagram of the receiver

Solution

The noise gure (NF) of all components must rst be expressed as a noise

factor (ratio):

NF 1 = 2.5 dB so that F 1 = 1.778

L 2 = 4.8 dB or 3.02 (as ratio); thus, G 2 = 1/L 2 = 1/3.02

NF 3 =10dBsothatF 3 = 100

NF = 6 dB so that F = 4.

A block diagram of this receiving system is shown in Fig. 16.14.

Making use of the theorem in Case Study 16.1, the noise factor of the

cable is equal to its attenuation (expressed as a ratio). Therefore, F 2 = 3.02.

558 16 Noise in Communication Systems

Expression (4.78) then yields the overall noise factor of several cascaded

two-ports, which in this case can be written

F=F 1 + F 2 −

+ F 3 −

G 1 G 2

G 1

From this, the preamplier gain G 1 (required to achieve an overall noise

factor of 7.94) becomes

G 1 =

(F 2 −

1) + F 3 −

−

F 1

G 2

(3 . 02 − 1) + 100

1 / 3 . 02

−

= 135 . 46

−

1 . 1778

Finally, the noise gure of the system does not exceed 6 dB, provided that

G 1 ≥

10 log(135 . 46) = 21 . 3dB

CASE STUDY 16.15 [244]

The input matching stage of a receiver has a noise temperature of 100 K and

a loss of 4 dB. It is followed by three identical IF stages, each with 10 dB gain,

a 6-MHz bandwidth, and a noise gure of 3 dB. With the system matched

throughout and operating at T = 290 K, determine

a) the overall noise gure;

b) the equivalent noise temperature of the receiver when connected to an

antenna with a noise temperature T a = 150 K;

c) the receiver sensitivity, if for high-quality communication an output

signal-to-noise ratio of at least 10 dB is required.

Solution

a) Applying (4.76), the noise factor of the input stage is

F 1 =1+ T e

T o

=1+ 100

290 =1 . 345

A block diagram is shown in Fig. 16.15, where the parameters are ex-

pressed as ratios.

The overall noise factor of the cascade is calculated by means of expression

(4.78):

F=F 1 + F 2 −

G 1 G 2

G 1 G 2 G 3

F 4 −

G 1

=1 . 345 +

1 / 2 . 51 +

−

10 / 2 . 51 +

−

100 / 2 . 51

−

+ F 3 −

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