Answer:
a) periodic (N = 1)
b) not periodic
c) not periodic
d) periodic (N = 8)
e) periodic (N = 16)
Explanation:
For function to be a periodic: f(n) = f(n+N)
![a) x[n]=sin(\frac{8\pi}{2}n+1)\\\\sin(\frac{8\pi}{2}n+1)=sin(4\pi n+1)](https://tex.z-dn.net/?f=a%29%20x%5Bn%5D%3Dsin%28%5Cfrac%7B8%5Cpi%7D%7B2%7Dn%2B1%29%5C%5C%5C%5Csin%28%5Cfrac%7B8%5Cpi%7D%7B2%7Dn%2B1%29%3Dsin%284%5Cpi%20n%2B1%29)
It is periodic with fundamental period N = 1
![b) x[n]=cos(\frac{n}{8} -\pi)\\\\\frac{1}{8} N=2\pi k](https://tex.z-dn.net/?f=b%29%20x%5Bn%5D%3Dcos%28%5Cfrac%7Bn%7D%7B8%7D%20-%5Cpi%29%5C%5C%5C%5C%5Cfrac%7B1%7D%7B8%7D%20N%3D2%5Cpi%20k)
N must be integer. So it is nor periodic
![c) x[n]=cos(\frac{\pi}{8} n^2)\\\\cos(\frac{\pi}{8} (n+N)^2)=cos(\frac{\pi}{8} (n^2+N^2+2nN)\\\\N^2 = 16 \:\:or\:\:2nN=16](https://tex.z-dn.net/?f=c%29%20x%5Bn%5D%3Dcos%28%5Cfrac%7B%5Cpi%7D%7B8%7D%20n%5E2%29%5C%5C%5C%5Ccos%28%5Cfrac%7B%5Cpi%7D%7B8%7D%20%28n%2BN%29%5E2%29%3Dcos%28%5Cfrac%7B%5Cpi%7D%7B8%7D%20%28n%5E2%2BN%5E2%2B2nN%29%5C%5C%5C%5CN%5E2%20%3D%2016%20%5C%3A%5C%3Aor%5C%3A%5C%3A2nN%3D16)
Since N is dependent to n. So it is not periodic.
![d) x[n]=cos(\frac{\pi }{2} n) cos(\frac{\pi }{4} n)\\\\x[n] = \frac{1}{2} cos(\frac{3\pi }{4} n) + \frac{1}{2} cos(\frac{\pi }{4} n)\\\\N_1=8\:\:and\:\:N_2=8\\](https://tex.z-dn.net/?f=d%29%20x%5Bn%5D%3Dcos%28%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20n%29%20cos%28%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%20n%29%5C%5C%5C%5Cx%5Bn%5D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20cos%28%5Cfrac%7B3%5Cpi%20%7D%7B4%7D%20n%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20cos%28%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20n%29%5C%5C%5C%5CN_1%3D8%5C%3A%5C%3Aand%5C%3A%5C%3AN_2%3D8%5C%5C)
So it is periodic with fundamental period N = 8.
![e) x[n]=2cos(\frac{\pi }{4} n)+sin(\frac{\pi }{8} n)-2cos(\frac{\pi }{2} n+\frac{\pi }{6} )\\\\N_1=8\:\:and\:\:N_2=16\:\:and\:\:N_3=4](https://tex.z-dn.net/?f=e%29%20x%5Bn%5D%3D2cos%28%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%20n%29%2Bsin%28%5Cfrac%7B%5Cpi%20%7D%7B8%7D%20n%29-2cos%28%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20n%2B%5Cfrac%7B%5Cpi%20%7D%7B6%7D%20%29%5C%5C%5C%5CN_1%3D8%5C%3A%5C%3Aand%5C%3A%5C%3AN_2%3D16%5C%3A%5C%3Aand%5C%3A%5C%3AN_3%3D4)
So it is periodic with N = 16.
Answer:
Class of fit:
Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).
Here minimum shaft diameter will be greater than the maximum hole diameter.
Medium Drive Force Fits are FN 2 Fits.
As per standard ANSI B4.1 :
Desired Tolerance: FN 2
Tolerance TZone: H7S6
Max Shaft Diameter: 3.0029
Min Shaft Diameter: 3.0022
Max Hole Diameter:3.0012
Min Hole Diameter: 3.0000
Max Interference: 0.0029
Min Interference: 0.0010
Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.
Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.
Explanation:
The statements regarding the refrigeration system as expressed by the technicians are correct. Hence, the correct statement is Option c.
<h3>What is a refrigeration system?</h3>
Refrigeration systems are a crucial process in the industry and home applications as they carry out cooling or keep the room temperature at a preferred value.
The missing information in the question is:
A.technician A only
B.technician B only
C.both A and B
D.neither A or B
Hence, The statements regarding the refrigeration system as expressed by the technicians are correct. the correct statement is Option c.
Learn more about Refrigeration systems:
brainly.com/question/12135611
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Answer:
a) 254.6 GPa
b) 140.86 GPa
Explanation:
a) Considering the expression of rule of mixtures for upper-bound and calculating the modulus of elasticity for upper bound;
Ec(u) = EmVm + EpVp
To calculate the volume fraction of matrix, 0.63 is substituted for Vp in the equation below,
Vm + Vp = 1
Vm = 1 - 0.63
Vm = 0.37
In the first equation,
Where
Em = 68 GPa, Ep = 380 GPa, Vm = 0.37 and Vp = 0.63,
The modulus of elasticity upper-bound is,
Ec(u) = EmVm + EpVp
Ec(u) = (68 x 0.37) + (380 x 0.63)
Ec(u) = 254.6 GPa.
b) Considering the express of rule of mixtures for lower bound;
Ec(l) = (EmEp)/(VmEp + VpEm)
Substituting values into the equation,
Ec(l) = (68 x 380)/(0.37 x 380) + (0.63 x 68)
Ec(l) = 25840/183.44
Ec(l) = 140.86 GPa
Answer:
Ceramic coating offers good protection to the car's surface. The nano-coating can protect the car from most scratches, dirt and chemical contaminants. Ceramic coating also doesn't have any side-effects to the original paint. Ceramic coating also lasts longer than regular paint.
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<em><u>Hope the helps.. </u></em></h2>
