When Hailey participated in a tree plantation drive, she wore her most comfortable loungewear. However, the same attire was not
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Answer:
a.
y[n] = x[n] x[n-1] x[n+1]
(i) Memory-less - It is not memory-less because the given system is depend on past or future values.
(ii) Causal - It is non-casual because the present value of output depend on the future value of input.
(iii) Invertible - It is invertible and the inverse of the given system is
(iv) Stable - It is stable because for all the bounded input, output is bounded.
(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.
b.
y[n] = cos(x[n])
(i) Memory-less - It is memory-less because the given system is not depend on past or future values.
(ii) Causal - It is casual because the present value of output does not depend on the future value of input.
(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .
For example - for x[n] = 0 , y[n] = cos(0) = 1
for x[n] = 2 , y[n] = cos(2
) = 1
(iv) Stable - It is stable because for all the bounded input, output is bounded.
(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.
Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
