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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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6 0
3 years ago
Read 2 more answers
Gtjffs
grandymaker [24]

the required documents is 3000

4 0
2 years ago
The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be
lakkis [162]

Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = \frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}       .....................1

Q = \frac{T \infty - Tso}{\frac{1}{hA}}                         .....................2

now we compare both equation 1 and 2 and put here value

\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}            

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

3 0
2 years ago
The output voltage of a power supply is normally distributed with mean 12 V and standard deviation 0.11 V. If the upper and lowe
podryga [215]

Answer:

82.62%

Explanation:

The z score is a score used in statistics to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean\ and\ \sigma=standard\ deviation.\\\\Given \ that\ \mu=12V, \sigma=0.11V.\\\\For\ x11.85V:\\\\z=\frac{11.85-12}{0.11} =-1.36\\\\

From the normal distribution table, P(11.85 < x < 12.15) = P(-1.36 < z < 1.36) = P(z < 1.36) - P(z < -1.36) = 0.9131-0.0869 = 0.8262 = 82.62%

4 0
2 years ago
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