- Given, input has a signal power of S(f)=2μW and noise power N(f)=323nW . Hence the required signal to noise ratio SNR=N(f)S(f) , Now substituting the values in this equation, SNR=323×10−91×10−6=0.3231
- Noise ratio =So/NoSi/Ni=53.1=0.62 and NF=10log10(F)=10log10(0.62)
- We know that Te=To(F−1) but To=295K Hence, Te=295(0.62−1)
- Average noise power NP=KBTB
⇒NB=1.3×10−23j/k×(75+460)95×33×1000W
=12750.8×10−20W
=1.75×10−16W
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