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laila [671]
3 years ago
6

You want to capture and digitize an analog signal that has a frequency bandwidth of 5500 Hz. You select a sampling rate of 5500

samples per second. What happens to your digitized signal? a. You should be fine since your sampling rate is the same as the analog signal’s bandwidth b. You cannot sample an analog signal since it is infinite and continuous c. Your digital signal will experience aliasing because you did not follow Nyquist sampling rate requirements d. None of the above
Physics
1 answer:
VladimirAG [237]3 years ago
6 0

Answer:

c.

Explanation:

When you take samples of an analog signal, you must take samples fast enough so then you can recover the original signal, just passing the digitized signal through a low-pass filter.

The Nyquist criteria states that in order to be able to recover the original signal completely, you must take samples at a rate greater than 2 times the highest frequency component of the signal.

In our case, we should sample the analog signal at a rate > 2*5500 Hz = 11000 Hz.

If we sampled the 5500 Hz signal (assuming be a sinusoid just for simplicity) two times each cycle, we could be so unfortunate that the samples fall exactly when the signal crosses by zero, so the digitized signal would be only a train of zeros.

So, if we sample the signal only one time in each cycle, clearly we will not  be able  to  recover the signal, and the digital signal will experience aliasing, due we are not following Nyquist sample rate  requirements, as stated  in the  option c.

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Explanation:

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A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th
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Answer:

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Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

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v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}

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Hence,

T = \mu\cdot v^{2}

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\rho = \dfrac{m}{V}

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m=\rho\cdot V

If the length is denoted by l, then

\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}

T = \dfrac{\rho\cdot V}{l} v^{2}

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The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

V = \pi \dfrac{d^{2}}{4} l

T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2

T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2

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