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 ![=\frac{Am^2}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7BAm%5E2%7D%7B2%7D)
From the question we are given that Amplitude Am = 10mV
substituting this value into the power formula
Power of the signal
μW
From the question we where given that the signal to noise ratio is
![(\frac{S}{N})_dB = 5dB](https://tex.z-dn.net/?f=%28%5Cfrac%7BS%7D%7BN%7D%29_dB%20%3D%205dB)
Note: The dB of a value means the same thing as 10 log of the value
![10log(\frac{S}{N} ) = 5](https://tex.z-dn.net/?f=10log%28%5Cfrac%7BS%7D%7BN%7D%20%29%20%3D%205)
![\frac{S}{N} = 10^{0.5} =3.16](https://tex.z-dn.net/?f=%5Cfrac%7BS%7D%7BN%7D%20%20%3D%2010%5E%7B0.5%7D%20%3D3.16)
Now to obtain noise power we make it the subject in the above equation
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 = ![G_1G_2G_3 = 10^{-6}*10^{-4}*1 = 10^2](https://tex.z-dn.net/?f=G_1G_2G_3%20%3D%2010%5E%7B-6%7D%2A10%5E%7B-4%7D%2A1%20%3D%2010%5E2)
Now that we have gotten this we can now compute the output signal power gain denoted by ![S_o](https://tex.z-dn.net/?f=S_o)
![S_o = P_m *(G_1G_2G_3)](https://tex.z-dn.net/?f=S_o%20%20%3D%20P_m%20%2A%28G_1G_2G_3%29)
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
![N_o = N *(G_4G_5G_6) = 15.3*10^{-6} * 10 * 10^{0.01} * 10 = 16.12*10^{-4}W](https://tex.z-dn.net/?f=N_o%20%3D%20N%20%2A%28G_4G_5G_6%29%20%3D%2015.3%2A10%5E%7B-6%7D%20%2A%2010%20%2A%2010%5E%7B0.01%7D%20%2A%2010%20%3D%2016.12%2A10%5E%7B-4%7DW)
Output signal to noise ratio (S/N) =![\frac{50*10^{-4}}{16.16*10^{-4}} =\frac{50}{16.16}](https://tex.z-dn.net/?f=%5Cfrac%7B50%2A10%5E%7B-4%7D%7D%7B16.16%2A10%5E%7B-4%7D%7D%20%3D%5Cfrac%7B50%7D%7B16.16%7D)
![(\frac{S}{N}) _dB = 10log(\frac{50}{16.6} ) = 4.905dB](https://tex.z-dn.net/?f=%28%5Cfrac%7BS%7D%7BN%7D%29%20_dB%20%3D%2010log%28%5Cfrac%7B50%7D%7B16.6%7D%20%29%20%3D%204.905dB)
Answer:
Explanation:
A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a compressive strength of more than
1300 KN/m2. The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. Suppose compressive strength for specimens of this mixture is normally distributed with ? = 68. Let ? denote the true average compressive strength.
(b) Let
X
denote the sample average compressive strength for n = 11 randomly selected specimens. Consider the test procedure with test statistic X and rejection region x ? 1331.26. What is the probability distribution of the test statistic when H0 is true? (Round your standard deviation to three decimal places.)
(c) What is the probability distribution of the test statistic when
? = 1350? (Round your standard deviation to three decimal places.)
Using the test procedure of part (b), what is the probability that the mixture will be judged unsatisfactory when in fact
? = 1350 (a type II error)? (Round your answer to four decimal places.)
(d) How would you change the test procedure of part (b) to obtain a test with significance level 0.05? (Round your answer to two decimal places.) Replace 1331.26 KN/m2 wit
What impact would this change have on the error probability of part (c)? (Round your answer to four decimal places.)
The probability that the mixture will be judged unsatisfactory when in fact ? = 1350 will change to .
(e) Consider the standardized test statistic
Z = (X ? 1300)/(?/n).
What are the values of Z corresponding to the rejection region of part (b)? (Round your answer to two decimal places.)
Answer:
a. 60V
b. 77.46V
c. 600 W
d. there would be no effect when this is done
Explanation:
a.
to get value of average output
Vo = VsD
Vs = 100V
D = 0.6
Vo = 100 V * 0.6
Vo = 60V
Therefore 60V is the value of average voltage
b.
The rms voltage:
= Vs√D
= 100 x √0.6
= 100 x 0.7746
= 77.46V
the rms voltage is Therefore 77.46V
c.
value of average power:
= (RMS)²/R
R = 10
when we put into formula above,
= (77.46)²/10
= 6000.05/10
= 600.005 W
= 600W
d.
If switching frequency is changes to be 2khz it would actually have no effect on the average voltage, average power or rms voltage.