y1 = 0.6 811) 0.3 y2 =0.5 (1)=0.2 h1=0.4 812) = 0 h2=0.3 12) = 0...

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y1 = 0.6 811) 0.3 y2 =0.5 (1)=0.2 h1=0.4 812) = 0 h2=0.3 12) = 0 o Let the learning rate be 0.1, the old connection weight fr

Ο Ο Let the learning rate be 0.1, the old connection weight from y1 to h1 is 1, what is the new weight? Ο Ο Ο 1.24 Ο 1.024 Ο

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Input Learning Old connetion ate 0.1 Aficlden :Chi, h) ut : (५), ५) h2M 0.3 enisting auigut+leroningrae adea New weigut 1l ttO.1 o.6x0.4 1t0.024 - 1.0a4 (Option A

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y1 = 0.6 811) 0.3 y2 =0.5 (1)=0.2 h1=0.4 812) = 0 h2=0.3 12) = 0...

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y1 = 0.6 811) 0.3 y2 =0.5 (1)=0.2 h1=0.4 812) = 0 h2=0.3 12) = 0...

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By application of the impulse-invariant transformation of H (s), the following transfer functions have been obtained. Find H (s), assuming T = 0.1 s: H1(z) = 2z (z − e−0.2)(z − e−0.1) H2(z) = z − e−0.6 (z − e−0.5)(z − e−0.4) H3(z) = z (z − 0.9)(z − 0.3)

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y1 = 0.6 811) 0.3 y2 =0.5 (1)=0.2 h1=0.4 812) = 0 h2=0.3 12) = 0...

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Add Answer to: y1 = 0.6 811) 0.3 y2 =0.5 (1)=0.2 h1=0.4 812) = 0 h2=0.3 12) = 0...

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By application of the impulse-invariant transformation of H (s), the following transfer functions have been obtained. Find H (s), assuming T = 0.1 s: H1(z) = 2z (z − e−0.2)(z − e−0.1) H2(z) = z − e−0.6 (z − e−0.5)(z − e−0.4) H3(z) = z (z − 0.9)(z − 0.3)

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Add Answer to:
y1 = 0.6 811) 0.3 y2 =0.5 (1)=0.2 h1=0.4 812) = 0 h2=0.3 12) = 0...