When buttons or switches are pressed by humans for arbitrary periods of time, we need to convert a signal level to a pulse. In t
he following FSM, whenever the input L goes from low to high, the level-to-pulse converter produces a single pulse P, which is one clock period wide. Thus it is a synchronous rising-edge detector. Write a Verilog HDL for the following FSM. L=0 L- Level to pulse 00 Low input, Waiting for risg P0 01 Edge Detected! P=1 / 11 High input, Waiting for fall P=0 LEO CLK converter 1:0NoteDiagram cannot be explained
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Answer:
D. N= 11. 22 rad/s (CW)
Explanation:
Given that
Form factor R = 8
Speed of sun gear = 5 rad/s (CW)
Speed of ring gear = 12 rad/s (CW)
Lets take speed of carrier gear is N
From Algebraic method ,the relationship between speed and form factor given as follows
here negative sign means that ring and sun gear rotates in opposite direction
Lets take CW as positive and ACW as negative.
Now by putting the values
N= 11. 22 rad/s (CW)
So the speed of carrier gear is 11.22 rad/s clockwise.
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;
(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) .