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Savatey [412]
3 years ago
7

Problem 2 (a) A sinusoid with a frequency of 2Hz is applied to a sampler/zero-order hold combination. The sampling rate is 10Hz.

List all the frequencies present in the output that are less than 50Hz. (b) Repeat part (a) if the input sinusoid has a frequency of 8 Hz.
Engineering
1 answer:
madreJ [45]3 years ago
3 0

Answer & Explanation:

(a) Frequency of 2Hz is applied to a sampler/zero-order hold combination

The sampling rate is 10Hz

List of all the frequencies present in the output that are less than 50Hz.

Adding:

fs + fm = 10 + 2

= 12 Hz

2fs + fm = 2 * 10 + 2

= 22 Hz

3fs + fm = 3 * 10 + 2

= 32 Hz

4fs + fm = 4 * 10 + 2

= 42 Hz

Subtracting:

fs - fm = 10 - 2

= 8 Hz

2fs - fm = 2 * 10 - 2

= 18 Hz

3fs - fm = 3 * 10 - 2

= 28 Hz

4fs - fm = 4 * 10 - 2

= 38 Hz

5fs - fm = 5 * 10 - 2

= 48 Hz

(b) Frequency of 8Hz is applied to a sampler/zero-order hold combination

The sampling rate is 10Hz

List of all the frequencies present in the output that are less than 50Hz.

Adding:

fs + fm = 10 + 8

= 18 Hz

2fs + fm = 2 * 10 + 8

= 28 Hz

3fs + fm = 3 * 10 + 8

= 38 Hz

Subtracting:

fs - fm = 10 - 8

= 2 Hz

2fs - fm = 2 * 10 - 8

= 12 Hz

3fs - fm = 3 * 10 - 8

= 22 Hz

4fs - fm = 4 * 10 - 8

= 32 Hz

5fs - fm = 5 * 10 - 8

= 42 Hz

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Answer:

Option D

Explanation:

Given information

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Considering that the volume of embankment is inversely proportional to the dry unit weight

\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}

Therefore, V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}

V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy

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(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since 1 yd^{3}= 27 ft^{3}

The embankment requires water of  6.78*27*440000= 80546400 lb

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Explanation:

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Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

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Answer:

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