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
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
Note: The dB of a value means the same thing as 10 log of the value
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 =
Now that we have gotten this we can now compute the output signal power gain denoted by
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
Output signal to noise ratio (S/N) =
Answer:
a) it is periodic
N = (20/3)k = 20 { for K =3}
b) it is Non-Periodic.
N = ∞
c) x(n) is periodic
N = LCM ( 5, 20 )
Explanation:
We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.
then the period of the signal is given as
N = ( 2π/w₀)K
k is least integer for which N is also integer
Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2
now
a) cos(2π(0.15)n)
w₀ = 2π(0.15)
Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3
so, it is periodic
N = (20/3)k = 20 { for K =3}
b) cos(2n);
w₀ = 2
Now, 2π/w₀ = 2π/2) = π
so, it is Non-Periodic.
N = ∞
c) cos(π0.3n) + cos(π0.4n)
x(n) = x1(n) + x2(n)
x1(n) = cos(π0.3n)
x2(n) = cos(π0.4n)
so
w₀ = π0.3
2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3
∴ N1 = 20
AND
w₀ = π0.4
2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5
∴ N² = 5
so, x(n) is periodic
N = LCM ( 5, 20 )
Answer:
(a) T₂ =747.5 and K= 474.5 °C (b) 330.178 kJ/kg
Explanation:
Solution
T₁ = 35°C = 308
the first step to take is to Use the Table A-17: Ideal gas properties for air:
Now,
At T₁ = 308 K
V₁ = 217.67 + [308-305/310-305] (221.25 -217.67)
So,
V₁ =219.818 kJ/kg
Thus,
Vr₁ = 596 + [308-305/310-305] (572.3 - 596)
= 581.78
so,
Vr₂/Vr₁ = 1/10
Vr₂ =58.178
Applying Table A-17, at Vr₂ = 58.178
Then,
(a) T₂ = 740 + [58.178 - 59.82/57.63 -59.82] (750 -740)
T₂ = 747.5 and K = 474.5 °C
V₂ =544.02 + [58.178 - 59.82/57.63 -59.82] (551.99 - 544.02)
so,
V₂ =549.996 kJ/kg
Hence,
(b) q - w = v₂ -v₁
= 0 -w = 549.996- 219.818
w = 330.178 kJ/kg
Why not simply include each step's setup instructions within it? is it due to our desire to maintain DRY (Don't Repeat Yourself) code?
<h3>What is the purpose of unit testing?</h3>
Program testing is known as "unit testing" involves testing individual software components. When developing an application, unit testing is done on the software product. An individual component could be a technique or a specific function.
Unit testing's primary goal is to separate written code for testing to see if it functions as intended. Unit testing is a crucial stage in the development process because, when done properly, it can aid in finding early code issues that could be more challenging to identify in subsequent testing phases.
The core of the testing process consists of unit testing and functional testing. The primary distinction between the two is that during the development cycle, the developer conducts unit testing. The tester does functional testing at the system testing level.
To learn more about unit testing refers to:
brainly.com/question/22900395
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