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eduard
3 years ago
9

A signal containing both a 5k Hz and a 10k Hz component is passed through a low-pass filter with a cutoff frequency of 4k Hz. Wh

ich component(s) will be attenuated? g
Engineering
1 answer:
abruzzese [7]3 years ago
3 0

Answer:

The 5 kHz signal will be less attenuated and the 10 kHz will be completely attenuated.

Low Pass Filter:

The purpose of a filter is to remove the undesired frequency components in a signal. A low-pass filter is one in which we allow low frequencies components to pass through it and block the higher frequency components. The frequency  where it starts to attenuate the signal is called cut-off frequency.

Explanation:

We have a signal that contains 5 kHz and 10 kHz frequency component.

The cut-off frequency is 4 kHz which means the filter will attenuate the frequency components of 4 kHz or greater than that.

So, in theory both 5 kHz and 10 kHz frequency components will be attenuated.

But filters are not ideal and they usually have a little margin also called roll off which means it will allow some of the frequency components greater than cut-off frequency of 4 kHz.

Conclusion:

So to conclude, the 5 kHz signal will be less attenuated and the 10 kHz will be completely attenuated.

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What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?
Yakvenalex [24]

Answer:

Activation\ Energy=2.5\times 10^{-19}\ J

Explanation:

Using the expression shown below as:

N_v=N\times e^{-\frac {Q_v}{k\times T}

Where,

N_v is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

{Q_v} is the activation energy

T is the temperature

Given that:

N_v=2.3\times 10^{13}

N = 10 moles

1 mole = 6.023\times 10^{23}

So,

N = 10\times 6.023\times 10^{23}=6.023\times 10^{24}

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (425 + 273.15) K = 698.15 K  

T = 698.15 K

Applying the values as:

2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

Q_v=2.5\times 10^{-19}\ J

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A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate
olga2289 [7]

Answer:

1) The probability of at least 1 defective is approximately 45.621%

2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%

Explanation:

The given parameters are;

The defective rate of the device = 3%

Therefore, the probability that a selected device will be defective, p = 3/100

The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;

The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97

The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927

The probability of at least 1 = 1 - The probability of 0 defective in 20

∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621

The probability of at least 1 defective ≈ 0.45621 = 45.621%

2) The probability of at least 1 defective in a shipment, p ≈ 0.45621

Therefore, the probability of not exactly 1 defective = q = 1 - p

∴ q ≈ 1 - 0.45621 = 0.54379

The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;

P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212

Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%

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3 years ago
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