NOISE ANALYSIS OF A WIRELESS RECEIVER The block diagram of a wireless receiver front-end is shown in Figure 10.10. Compute the overall noise figure of this subsystem. If the input noise power from a feeding antenna is Ni = kTAB, where TA = 150 K, find the output noise power in dBm. If we require

Question

NOISE ANALYSIS OF A WIRELESS RECEIVER

The block diagram of a wireless receiver front-end is shown in Figure 10.10. Compute the overall noise figure of this subsystem. If the input noise power from a feeding antenna is [latex]N_{i} = kT_{A}B[/latex], where [latex]T_{A}[/latex] = 150 K, find the output noise power in dBm. If we require a minimum signal-to-noise ratio (SNR) of 20 dB at the output of the receiver, what is the minimum signal voltage that should be applied at the receiver input? Assume the system is at temperature [latex]T_{0}[/latex], with a characteristic impedance of 50 Ω, and an IF bandwidth of 10 MHz.

Accepted Answer

We first perform the required conversions from dB to numerical values:

[latex]G_{a}[/latex]= 10 dB = 10 [latex]G_{ f}[/latex] = −1.0 dB = 0.79 [latex]G_{m} [/latex]= −3.0 dB = 0.5
[latex]F_{a}[/latex] = 2 dB = 1.58 [latex]F_{ f}[/latex] = 1 dB = 1.26 [latex]F_{m}[/latex] = 4 dB = 2.51

Next, use (10.23) to find the overall noise figure of the system:

[latex]F_{cas}=F_{1}+\frac{F_{2}-1}{G_{1}}+\frac{F_{3}-1}{G_{1}G_{2}}+.... [/latex]

[latex]F=F_{a}+\frac{F_{f}-1}{G_{a}}+\frac{F_{m}-1}{G_{a} G_{f}}=1.58+\frac{(1.26-1)}{10}+\frac{(2.51-1)}{(10)(0.79)}=1.80=2.55 dB. [/latex]

The best way to compute the output noise power is to use noise temperatures. From (10.12), the equivalent noise temperature of the overall system is :

[latex]T_{e}=(F-1)T_{0}=(1.80-1)(290)=232 K.[/latex]

The overall gain of the system is G = (10)(0.79)(0.5) = 3.95. Then we can find the output noise power as

[latex]N_{o}=k(T_{A}+T{e})BG=(1.38 imes 10^{-23})(150+232)(10 imes 10^{6})(3.95)=2.08 imes 10^{-13} W=-96.8 dBm.[/latex]

For an output SNR of 20 dB = 100, the input signal power must be

[latex]S_{i}=\frac{S_{o}}{G}=\frac{S_{o}}{N_{o}}\frac{N_{o}}{G}=100\frac{2.08 imes 10^{-13}}{3.95}=5.27 imes 10^{-12} W=-82.8 dBm, [/latex]

For a 50 Ω system impedance, this corresponds to an input signal voltage of

[latex]V_{i}=\sqrt{Z_{o}S_{i}} = \sqrt{(50)(5.27 imes 10^{-12})}=1.62 imes 10^{ -5} V=16.2 \mu V (rms)[/latex]

Note: It may be tempting to compute the output noise power from the definition of the noise figure, as

[latex]N_{o}=N_{i}F(\frac{S_{o}}{S_{i}} )=N_{i}FG=kT_{A}BFG[/latex]

[latex]=(13.38 imes 10^{-23})(150)(10 imes 10^{6})(1.8)(3.95)=1.47 imes 10^{-13} W[/latex]

This is an incorrect result! The reason for the disparity with the earlier result is that the definition of noise figure assumes an input noise level of [latex]kT_{0}B[/latex], while this problem involves an input noise of [latex]kT_{A}B[/latex], with [latex]T_{A}[/latex] = 150 K ≠ [latex]T_{0}[/latex]. This is a common error, and suggests that when computing absolute noise power it is often safer to use noise temperatures to avoid this confusion.

Question 10.2: NOISE ANALYSIS OF A WIRELESS RECEIVER The block diagram of a...

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