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:
Answer is attached.
Explanation:
A NMOS is a n-channel MOSFET or Metal Oxide
Semiconductor Field Effect Transistor. This type
of transistor might be an enhancement or
depletion type nMOS transistor designed using
layers of Metal-oxide, Silicon-oxide and Silicon
fabricated on a substrate.
Answer:
(N-1) × (L/2R) = (N-1)/2
Explanation:
let L is length of packet
R is rate
N is number of packets
then
first packet arrived with 0 delay
Second packet arrived at = L/R
Third packet arrived at = 2L/R
Nth packet arrived at = (n-1)L/R
Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R
Now
L / R = (1000) / (10^6 ) s = 1 ms
L/2R = 0.5 ms
average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2
the average queuing delay of a packet = 0 ( put N=1)
Answer: 383.22K
Explanation:
L = 3m, w = 1.5m
Area A = 3 x 1.5 = 4.5m2
Q' = 750W/m2 (heat from sun) ,
& = 0.87
Q = &Q' = 0. 87x750 = 652.5W/m2
E = QA = 652.5 x 4.5 = 2936.25W
T(sur) = 300K, T(panel) = ?
Using E = §€A(T^4(panel) - T^4(sur))
§ = Stefan constant = 5.7x10^-8
€ = emmisivity = 0.85
2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)
T(panel) = 383.22K
See image for further details.