Answer:
The answer is 4.905 dB
Explanation:
Let say that that signal is sinusoidal i.e Am sin(wt)
Hence the power of the signal [tex]=\frac{Am^2}{2}[/tex]
From the question we are given that Amplitude Am = 10mV
substituting this value into the power formula
Power of the signal [tex]=\frac{(10*10^{-3})^{2}}{2}= 50[/tex]μW
From the question we where given that the signal to noise ratio is
[tex](\frac{S}{N})_dB = 5dB[/tex]
Note: The dB of a value means the same thing as 10 log of the value
[tex]10log(\frac{S}{N} ) = 5[/tex]
[tex]\frac{S}{N} = 10^{0.5} =3.16[/tex]
Now to obtain noise power we make it the subject in the above equation
[tex]Noise \ Power \ N = \frac{Signal \ Power (P_m)}{3.16} = \frac{50*10^{-6}}{3.16} =15.3[/tex]μW
Now to obtain the overall signal gain we multiply the individual gain for the frequency that we are considering i.e 1KHz as our signal
Overall signal gain = [tex]G_1G_2G_3 = 10^{-6}*10^{-4}*1 = 10^2[/tex]
Now that we have gotten this we can now compute the output signal power gain denoted by [tex]S_o[/tex]
[tex]S_o = P_m *(G_1G_2G_3)[/tex]
[tex]= 50*10^{-6} *10^2 = 50*10^{-4}[/tex]W
Now to obtain the overall signal gain we multiply the individual gain for the frequency that we are considering i.e 10KHz as our noise
[tex]N_o = N *(G_4G_5G_6) = 15.3*10^{-6} * 10 * 10^{0.01} * 10 = 16.12*10^{-4}W[/tex]
Output signal to noise ratio (S/N) =[tex]\frac{50*10^{-4}}{16.16*10^{-4}} =\frac{50}{16.16}[/tex]
[tex](\frac{S}{N}) _dB = 10log(\frac{50}{16.6} ) = 4.905dB[/tex]
