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- Homework Statement
- (a) Determine the voltage required on the input of the amplifier to give

the maximum output of 85 dBμV.

(b) If the signal level from the aerial is 5 dBmV and the input noise level

is 20 dBμV, calculate the signal-to-noise ratio on the output of the

amplifier.

Values:

Bandwidth 40–862 MHz

Gain 20 dB

Noise Figure 6 dB

Max. Output 85 dBV

Input Impedance 75Ω

Output Impedance 75Ω

- Relevant Equations
- SNR=20Log(Vs/Vn)

SNR=10Log(Ps/Pn)

So pretty confident I understand part (a) however for part (b) I'm not sure if I have carried it out correctly if someone could give me a pointer?

(b)

5dBmV Input as a voltage:

5=20Log(V/1mV)

V=10

V=1.77827941mV

Then the noise level is 20dbμV so changing to a voltage:

20=20*Log(V/1μV)

V=10

V=10μV

Then I can find the SNR of the input using:

SNR

SNR

SNR

Then as I have the Noise figure from the data as F=6dB I can do the following:

F=SNR

SNR

SNR

SNR

(b)

5dBmV Input as a voltage:

5=20Log(V/1mV)

V=10

^{5/10}V=1.77827941mV

Then the noise level is 20dbμV so changing to a voltage:

20=20*Log(V/1μV)

V=10

^{20/20}V=10μV

Then I can find the SNR of the input using:

SNR

_{In}=20*Log(V_{p}/V_{n})SNR

_{In}=20*Log((1.77827941*10^{-3})/(10*10^{-6}))SNR

_{In}=45dBThen as I have the Noise figure from the data as F=6dB I can do the following:

F=SNR

_{In}/SNR_{Out}SNR

_{Out}=SNR_{In}/FSNR

_{Out}=45*6SNR

_{Out}=270dB**This seems far too high? I have found websites that say the formula is actually F = SNR**_{In}-SNR_{Out}however the learning materials say: