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sergejj [24]
1 year ago
8

Which of these analysis methods describes neural computing? a. historical if-then-else cases are used to recognize patterns O b.

a mathematical procedure predicts the value of a dependent variable based on a single independent variable ca specialized set of algorithms sorts through data and forms statistical rules about relationships among the items d. historical data is examined for patterns that are then used to make predictions
Engineering
1 answer:
Lera25 [3.4K]1 year ago
3 0

Historical data is examined for patterns that are then used to make predictions is one of the analysis methods that describe neural computing

What is neural computing?

A neural network is an artificial intelligence technique that instructs computers to analyze data in a manner modeled after the human brain. It is a kind of artificial intelligence technique known as deep learning that makes use of interconnected neurons or nodes in a layered structure to mimic the human brain.

Historical data is nothing but the existing network data which is stored for the predicting in future

In the context of neural networks, the word "pattern" refers to a collection of activations over a group of units (neurons).

Hence to conclude neural netwoks almost describes the patterns

To know more on neural networks follow this link

brainly.com/question/27371893

#SPJ1

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The boost converter of Fig. 6-8 has parameter Vs 20 V, D 0.6, R 12.5 , L 10 H, C 40 F, and the switching frequency is 200 kHz. (
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Answer:

a) the output voltage is 50 V

b)

- the average inductor current is 10 A

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Explanation:

Given the data in the question;

a) the output voltage

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given that; V_s = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

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V₀ = 50 V

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b)

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I_L = V_s / ( 1 - D )²R

given that R = 12.5 Ω, V_s = 20 V, D = 0.6

we substitute

I_L = 20 / (( 1 - 0.6 )² × 12.5)

I_L = 20 / (( 0.4)² × 12.5)

I_L = 20 / ( 0.16 × 12.5 )

I_L = 20 / 2

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- the maximum inductor current

I_{Lmax = [V_s / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V_s = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

I_{Lmax = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

I_{Lmax = [20 / 2 ] + [ 60 / 20 ]    

I_{Lmax = 10 + 3

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Therefore, the maximum inductor current is 13 A

- The minimum inductor current

I_{Lmax = [V_s / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V_s = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

I_{Lmin = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

I_{Lmin = [20 / 2 ] -[ 60 / 20 ]    

I_{Lmin = 10 - 3

I_{Lmin  = 7 A

Therefore, the maximum inductor current is 7 A

 

c)  the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

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under ideal components; diode current = output current

hence the diode current will be;

I_D = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

I_D = 50 / 12.5

I_D = 4 A

Therefore, the average current in the diode under ideal components is 4 A

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