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Sigmoid units:
Always remember that the function inside units take as input the weighted sum, S and of the values coming from the units connected to it. However the function inside sigmoid units calculates the following value that given a real-valued input S: is as
Whereas e is the base of natural logarithms, e = 2.718...
Where we plot the output from sigmoid units given various weighted sums as input that looks remarkably like a step function as:
Obviously by getting a differentiable function that looks like the step function was the whole point of the exercise. Moreover, not only is this function differentiable but there the derivative is fairly simply expressed in terms of the function itself as:
So note that the output values for the σ function range between but never make it to 0 and 1. It means that e-S is never negative so the denominator of the fraction tends to 0 as S gets very big in the negative direction and then tends to 1 as it gets very big in the positive direction. In fact this tendency happens fairly quickly as: in the middle ground between 0 and 1 is rarely seen means the sharp or near step in the function. Just because of it looking like a step function that we can think of it firing and not-firing as in a perceptron as: if a positive real is input and the output will generally be close to +1 then a negative real is input the output will generally be close to -1.
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how to breadboARD THE 4 BIT COMPARATOR
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Simplify the given expressions using Boolean postulates Y = (A + B)(A‾ + C)(B + C) Ans. Y = (A + B)(A‾ + C)(B + C) = (A A‾ + AC + B A‾ + BC) (B + C) = (AC + B A‾ + BC) (B + C)
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