Topic: Communication Systems
The minimum input signal that gives output signal to noise ratio of 10 dB in a system that has an input impedance equal to $50~\Omega$, a noise figure of 4 dB and bandwidth of 100 kHz is
- A. $0.32~\mu V$
- B. $1.41~\mu V$
- C. $0.245~\mu V$
- D. None of above
Answer: C. $0.245~\mu V$
Explanation: $R_{in}=50~\Omega$, N.F. = 4 dB, $(\frac{S}{N})_{o}=10dB$, B.W. = 100 kHz. $(\frac{S}{N})_{i}=10~log_{10}N.F + 10~log_{10}(\frac{S}{N})_{o}$. $10~log_{10}S-10~log_{10}N=4+10$, where $N=KTB$. $10~log_{10}N=10~log_{10}KT+10~log_{10}B = -174~dBm+10log_{10}10^{5} = -174+50=-124$. $10~log S=-124+14 = -110 \Rightarrow S=10^{-11}$. $\frac{V^{2}}{R_{in}}=10^{-11} \Rightarrow V=0.245~\mu V$.