Answer:
Output voltage equation is ![V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)](https://tex.z-dn.net/?f=V_%7Bout%7D%20%28t%29%20%3D%202.72%20%5Ccos%20%282%5Cpi%20%2857000%29t%20%2B18.3%29)
Explanation:
Given:
dc gain
dB
Input signal ![V_{in} (t) = 0.18 \cos (2\pi (57000)t +18.3)](https://tex.z-dn.net/?f=V_%7Bin%7D%20%28t%29%20%3D%200.18%20%5Ccos%20%282%5Cpi%20%2857000%29t%20%2B18.3%29)
Now convert gain,
![A = 10^{\frac{23.6}{20} } = 15.13](https://tex.z-dn.net/?f=A%20%3D%2010%5E%7B%5Cfrac%7B23.6%7D%7B20%7D%20%7D%20%3D%2015.13)
DC gain at frequency
is given by,
![A = \frac{V_{out} }{V_{in} }](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7BV_%7Bout%7D%20%7D%7BV_%7Bin%7D%20%7D)
![V_{out} =AV_{in}](https://tex.z-dn.net/?f=V_%7Bout%7D%20%3DAV_%7Bin%7D)
![V_{out} = 15.13 \times 0.18 \cos (2\pi (57000)t +18.3)](https://tex.z-dn.net/?f=V_%7Bout%7D%20%3D%2015.13%20%5Ctimes%20%20%200.18%20%5Ccos%20%282%5Cpi%20%2857000%29t%20%2B18.3%29)
At zero frequency above equation is written as,
![V_{out} = 2.72 \times \cos 18.3](https://tex.z-dn.net/?f=V_%7Bout%7D%20%3D%202.72%20%5Ctimes%20%5Ccos%2018.3)
![V_{out} = 2.72](https://tex.z-dn.net/?f=V_%7Bout%7D%20%3D%202.72)
Now we write output voltage as input voltage,
![V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)](https://tex.z-dn.net/?f=V_%7Bout%7D%20%28t%29%20%3D%202.72%20%5Ccos%20%282%5Cpi%20%2857000%29t%20%2B18.3%29)
Therefore, output voltage equation is ![V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)](https://tex.z-dn.net/?f=V_%7Bout%7D%20%28t%29%20%3D%202.72%20%5Ccos%20%282%5Cpi%20%2857000%29t%20%2B18.3%29)
Answer:
b. 10A
Explanation:
Using the formula, E= k × r×I
200= 0.5 ×2000×0.02×I
200=20×I
Dividing with 20
I = 200/20= 10A
Answer:
a) ![\frac{Ws}{Es} = \frac{200}{1+1.2s}](https://tex.z-dn.net/?f=%5Cfrac%7BWs%7D%7BEs%7D%20%20%3D%20%5Cfrac%7B200%7D%7B1%2B1.2s%7D)
b) attached below
c) type zero system
d) k > ![\frac{g}{200}](https://tex.z-dn.net/?f=%5Cfrac%7Bg%7D%7B200%7D)
e) The gain K increases above % error as the steady state speed increases
Explanation:
Given data:
Motor voltage = 12 v
steady state speed = 200 rad/s
time taken to reach 63.2% = 1.2 seconds
<u>a) The transfer function of the motor from voltage to speed</u>
let ;
be the transfer function of a motor
when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec
hence the transfer function of the motor from voltage to speed
= ![\frac{Ws}{Es} = \frac{200}{1+1.2s}](https://tex.z-dn.net/?f=%5Cfrac%7BWs%7D%7BEs%7D%20%20%3D%20%5Cfrac%7B200%7D%7B1%2B1.2s%7D)
<u>b) draw the block diagram of the system with plant controller and the feedback path </u>
attached below is the remaining part of the detailed solution
c) The system is a type-zero system because the pole at the origin is zero
d) ) k > ![\frac{g}{200}](https://tex.z-dn.net/?f=%5Cfrac%7Bg%7D%7B200%7D)
Answer:
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