Question #167823

  1. Assume that an RF amplifier’s input has a signal power of 2μW and a noise power of 323nW, what is the signal-to-noise ratio? (5 points)
  2. Assume that the RF amplifier in problem number 1 has a signal-to-noise ratio of 5 at the output. Calculate the noise ratio and noise figure. (5 points)
  3. Given the noise ratio calculated in problem number 2, what is its equivalent noise temperature? (5 points)
  4. What is the average noise power of a device operating at temperature 75oF with a bandwidth of 33 kHz? (10 points)

Expert's answer

  1. Given, input has a signal power of S(f)=2μWS(f)=2μW and noise power N(f)=323nWN(f)=323nW . Hence the required signal to noise ratio SNR=S(f)N(f)SNR=\frac{S(f)}{N(f)} , Now substituting the values in this equation, SNR=1×10−6323×10−9=10.323SNR=\frac{1\times 10^{-6}}{323\times 10^{-9}}=\frac{1}{0.323}
  2. Noise ratio =Si/NiSo/No=3.15=0.62=\frac{S_i/N_i}{S_o/N_o}=\frac{3.1}{5}=0.62 and NF=10log⁡10(F)=10log⁡10(0.62)NF=10\log_{10}(F)= 10\log_{10}(0.62)
  3. We know that Te=To(F−1)T_e=T_o(F-1) but To=295KT_o=295K Hence, Te=295(0.62−1)T_e=295(0.62-1)
  4. Average noise power NP=KBTBN_P=K_BTB

⇒NB=1.3×10−23j/k×(75+460)59×33×1000W\Rightarrow N_B=1.3 × 10^{−23} j/ k\times (75+460)\frac{5}{9}\times 33\times 1000W

=12750.8×10−20W=12750.8\times 10^{-20} W

=1.75×10−16W=1.75\times 10^{-16}W


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