Question 10.2: NOISE ANALYSIS OF A WIRELESS RECEIVER The block diagram of a...
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 N_{i} = kT_{A}B, where T_{A} = 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 T_{0}, with a characteristic impedance of 50 Ω, and an IF bandwidth of 10 MHz.
We first perform the required conversions from dB to numerical values:
G_{a}= 10 dB = 10 G_{ f} = −1.0 dB = 0.79 G_{m} = −3.0 dB = 0.5
F_{a} = 2 dB = 1.58 F_{ f} = 1 dB = 1.26 F_{m} = 4 dB = 2.51
Next, use (10.23) to find the overall noise figure of the system:
F_{cas}=F_{1}+\frac{F_{2}-1}{G_{1}}+\frac{F_{3}-1}{G_{1}G_{2}}+….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.
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 :
T_{e}=(F-1)T_{0}=(1.80-1)(290)=232 K.The overall gain of the system is G = (10)(0.79)(0.5) = 3.95. Then we can find the output noise power as
N_{o}=k(T_{A}+T{e})BG=(1.38\times 10^{-23})(150+232)(10 \times 10^{6})(3.95)=2.08 \times 10^{-13} W=-96.8 dBm.For an output SNR of 20 dB = 100, the input signal power must be
S_{i}=\frac{S_{o}}{G}=\frac{S_{o}}{N_{o}}\frac{N_{o}}{G}=100\frac{2.08\times 10^{-13}}{3.95}=5.27 \times 10^{-12} W=-82.8 dBm,For a 50 Ω system impedance, this corresponds to an input signal voltage of
V_{i}=\sqrt{Z_{o}S_{i}} = \sqrt{(50)(5.27 \times 10^{-12})}=1.62\times 10^{ -5} V=16.2 \mu V (rms)Note: It may be tempting to compute the output noise power from the definition of the noise figure, as
N_{o}=N_{i}F(\frac{S_{o}}{S_{i}} )=N_{i}FG=kT_{A}BFG=(13.38\times 10^{-23})(150)(10\times 10^{6})(1.8)(3.95)=1.47\times 10^{-13} W
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 kT_{0}B, while this problem involves an input noise of kT_{A}B, with T_{A} = 150 K ≠ T_{0}. 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.