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:
(a) Calculate the field, armature, and load currents versus load = 306A
(b) Determine the terminal voltage at no-load and at rated load conditions = 300V
(c) Calculate the voltage regulation of the generator. Use the no-load voltage as the base value = 6.67%
(d) Plot the terminal voltage as a function of the load. Determine the load that corresponds to a 5% voltage drop using the no-load voltage as the base = 294.74V
Explanation:
CHECK THE ATTACHED FILES FOR DETAILED EXPLANATION.
Answer:
Explanation:
Given
Initial Thickness=45 mm
Final thickness=20 mm
Roll diameter=600 mm
Radius(R)=300 mm
coefficient of friction between rolls and strip (
)=0.15
maximum draft![(d_{max})=\nu ^2R](https://tex.z-dn.net/?f=%28d_%7Bmax%7D%29%3D%5Cnu%20%5E2R)
![=0.15^2\times 300=6.75 mm](https://tex.z-dn.net/?f=%3D0.15%5E2%5Ctimes%20300%3D6.75%20mm)
Minimum no of passes![=\frac{45-20}{6.75}=3.70\approx 4](https://tex.z-dn.net/?f=%3D%5Cfrac%7B45-20%7D%7B6.75%7D%3D3.70%5Capprox%204)
(b)draft per each pass
![d=\frac{Initial\ Thickness-Final\ Thickness}{min.\ no.\ of\ passes}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7BInitial%5C%20Thickness-Final%5C%20Thickness%7D%7Bmin.%5C%20no.%5C%20of%5C%20passes%7D)
![d=\frac{45-20}{4}=6.25 mm](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B45-20%7D%7B4%7D%3D6.25%20mm)
Answer:
a) 89300
b) ![8.93*10^-^8](https://tex.z-dn.net/?f=%208.93%2A10%5E-%5E8%20)
Explanation:
To find the Reynolds number, we use the formula:
Re = VD / v
Therefore, For part (a)
we are given:
D= 0.1
V= 10 * 0.1(m/s) = 1m/s
![v= 1.12*10^-^6 m^2/s](https://tex.z-dn.net/?f=%20v%3D%201.12%2A10%5E-%5E6%20m%5E2%2Fs%20)
Substituting the figures in the formula, we have:
![R_e = [(1m/s) (0.1m)] / 1.12*10^-^6 m^2/s](https://tex.z-dn.net/?f=%20R_e%20%3D%20%5B%281m%2Fs%29%20%280.1m%29%5D%20%2F%201.12%2A10%5E-%5E6%20m%5E2%2Fs%20)
Re = 89,300
For part (b)
We are given:
;
;
![v = 1.12*10^-^6 m^2/s](https://tex.z-dn.net/?f=%20v%20%3D%201.12%2A10%5E-%5E6%20m%5E2%2Fs%20)
Therefore,
;
![R_e = 8.93*10^-^8](https://tex.z-dn.net/?f=%20R_e%20%3D%208.93%2A10%5E-%5E8%20)