Answer:
the state of the circuit is a function of the voltage level. The interpretation is up to the user.
Explanation:
A binary digital circuit adopts one of two states, depending on whether the voltage level is above or below some threshold that depends on the design of the circuit. Within each state, the voltage may have some typical range. When the voltage is near the threshold, the state of the circuit may actually be "indeterminate".
The internal/output voltage is a function of the state of the circuit. The interpretation of that voltage as a true/false or 1/0 or other meaning is up to the user of the circuit.
The circuit interprets a given input voltage as intending to convey a particular input signal state according to the circuit specifications. Input voltages near the threshold between states may cause unexpected or even destructive results.
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In order to conserve space, some digital circuits use more than 2 different voltage levels to signify more than 2 different states.
The question is not complete. We are supposed to find the average value of v_o.
Answer:
v_o,avg = 0.441V
Explanation:
Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;
3/(T/4) = 0.7/t1
So, 12/T = 0.7/t1
So, t1 = 0.7T/12
t1 = 0.0583 T
Also, from symmetry of triangles,
t2 = T/2 - t1
So, t2 = T/2 - 0.0583 T
t2 = 0.4417T
Average of voltage output is;
v_o,avg = (1/T) x Area under small triangle
v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)
v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)
v_o,avg = (1/T) x 2.3 x 0.1917T
T will cancel out to give;
v_o,avg = 0.441V
Answer:
(a) The mean time to fail is 9491.22 hours
The standard deviation time to fail is 9491.22 hours
(b) 0.5905
(c) 3.915 × 10⁻¹²
(d) 2.63 × 10⁻⁵
Explanation:
(a) We put time to fail = t
∴ For an exponential distribution, we have f(t) = 
Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

e^(1000·λ) - 0.1·e^(1000·λ) = 1
0.9·e^(1000·λ) = 1
1000·λ = ㏑(1/0.9)
λ = 1.054 × 10⁻⁴
Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours
The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours
b) Here we have to integrate from 5000 to ∞ as follows;
![p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [ -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905](https://tex.z-dn.net/?f=p%28t%3E5000%29%20%3D%20%5Cint%5Climits%5E%7B%5Cinfty%7D_%7B5000%7D%20%7B%5Clambda%20e%5E%7B-%5Clambda%20t%7D%7D%20%5C%2C%20dt%20%3D%5Cleft%20%5B%20%20-e%5E%7B%5Clambda%20t%7D%5Cright%20%5D_%7B5000%7D%5E%7B%5Cinfty%7D%20%3D%20e%5E%7B5000%20%5Clambda%7D%20%3D%200.5905)
(c) The Poisson distribution is presented as follows;

p(x = 3) = 3.915 × 10⁻¹²
d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours
The Cumulative Distribution Function is given as follows;
p( t ≤ 1/4)
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AnsweCalculate a heat pump's coefficient of performance. ... Heat pumps, air conditioners, and refrigerators utilize heat transfer from cold to hot.Explanation: