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borishaifa [10]
1 year ago
6

The output signal from an analogue Internet of Things (IoT) sensor is sampled every 235 µs to convert it into a digital represen

tation. What is the corresponding sampling rate expressed in kHz?
According to the Sampling Theorem (Section 3.3.1), for this sampling rate value, what approximately could be the highest frequency present in the sensor signal, in kHz, assuming the sensor’s lowest frequency is very close to zero?

If each sample is now quantised into 1024 levels, what will be the resulting device output bitrate in kbps?

Give your answer in scientific notation to one decimal place.

Hint: you need to determine the number of bits per sample which allow for 1024 quantisation levels (see Section 2.4, Block 1 and Section 3.3.2, Block 3).

ii.The above IoT sensor is now replaced by a new sensor which has a lowest frequency very close to zero, and a maximum frequency of 3 kHz. Briefly explain how the sampling period of 235 µs used in Part (i) will need to be modified to avoid aliasing in the new sensor’s sampled signal
Physics
1 answer:
ElenaW [278]1 year ago
4 0

In order to find the sampling rate per second for 1024 levels and the resulting device output bitrate in kbps, you would get 10kbps

<h3>Calculations and Parameters</h3>

First, convert from base 10 to base 2

log₂(1024)=log₁₀(1024)/log₁₀(2)

= 3.01029995664/0.301029995664

= 10kbps

Hence, we can note that your question is incomplete so I gave you a general overview to get a better understanding of the concept.

Read more about bit rates here:

brainly.com/question/12977416

#SPJ1

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Which of the following statements is true? (A) The measured value for the pressure of a gas in a container is almost independent
Lana71 [14]

Answer:

The correct answer to the question is

Both A and B are true

Explanation:

The particles of a gas are free to move to occupy the entire volume in which they are placed due to the smallerinter molecular forces holding them together hence due to the face that pressure is a measure of the Force per unit area that is Pressure P = ( Force F)/ (Area A) then the force per unit area, exerted on the all of the container by the gaseous particles which are colliding with each other and with the walss of the container is fairly constant through out the surface oof the container

In the case of the liquid which are held on together by more stronger forces, the force per nit area exerted by the liquid particle is transmitted from one particle to the next until it reaches the container's surface. Then remembering that the force of gravity on the liquid is acting in one direction (that is downwards) the sum of the fprce due to the weight incrreases as we progress deaper into the liquid hence the pressure increases per unit depth

5 0
2 years ago
Compare these two collisions of a PE student with a wall.
Stolb23 [73]

1) The variable that is different in the two cases is \Delta t, the duration of the collision

2) The change in momentum is the same in the two cases

3) The impulse is the same in the two cases

4) Case B will experience a greater force

Explanation:

1)

The variable that is different in the two cases is \Delta t, the duration of the collision.

In fact, in the first case the wall is padded: this means that the collision will be "softer" and therefore will last longer, so the duration of the collision, \Delta t, will be larger.

In the second case instead, the wall is unpadded: this means that the collision is "harder" and so it will last less time, therefore the duration of the collision \Delta t will be smaller.

2)

The change in momentum in the two cases is the same.

In fact, the change in momentum is given by:

\Delta p = m(v-u)

where:

m is the mass of the student

u is the initial velocity

v is the final velocity

In both cases, we have:

m = 75 kg

u = 8 m/s

v = 0 (they both comes to rest)

Therefore, the change in momentum is

\Delta p = (75)(0-8)=-600 kg m/s

3)

The impulse in the two cases is the same.

In fact, impulse is defined as the product of force applied, F, and duration of the collision, \Delta t:

J=F \Delta t

However, the force can be rewritten as product of mass (m) and acceleration (a), according to Newton's second law:

F=ma

So the impulse is

J=ma\Delta t

The acceleration can be rewritten as rate of change of velocity:

a=\frac{\Delta v}{\Delta t}

So the impulse becomes

J=m\frac{\Delta v}{\Delta t}\Delta t = m\Delta v

So, the impulse is equal to the change in momentum: and since in the two cases the change in momentum is the same, the impulse is the same as well.

4)

The force in the collision is related to the impulse by

J=F\Delta t

where

J is the impulse

F is the force

\Delta t is the duration of the collision

The equation can be rewritten as

F=\frac{J}{\Delta t}

In the two situations described in the problem (A and B), we already said that the impulse is the same (because the change in momentum is the same). However, in case A (padded wall) the time \Delta t is longer, while in case B (unpadded wall) the time \Delta t is shorter: since the force F is inversely proportional to the duration of the collision, this means that in case B the student will experience a greatest force compared to case A.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

3 0
3 years ago
A man pushes a box along a flat, frictionless surface using a force of 500 N. The box was moved a distance of 2.5 m. The actual
andrey2020 [161]

Answer:

Workdone = 1250Nm

Explanation:

<u>Given the following data;</u>

Force, F = 500N

Distance, d = 2.5m

Workdone is given by the formula;

Workdone = force * distance

Substituting into the equation, we have

Workdone = 500 * 2.5  

Workdone = 1250Nm

Therefore, the actual work done by the worker is 1250 Newton-meter.

4 0
3 years ago
The average surface temperature of Mars is estimated to be -55°C. What is this temperature on the Fahrenheit scale? -99°F -63°F
Mamont248 [21]
-55 C to F is -67.

I hope this helps you!

Brainliest answer is always appreciated!
4 0
3 years ago
A student climbs to the top of the gym and drops a baseball to the ground below. A physics student standing nearby has a motion
sladkih [1.3K]

Answer:

The baseball was falling during 1.733 seconds.

Explanation:

The baseball experimented a free fall, which is a particular case of uniformly accelerated motion, in which object is accelerated by gravity. In this case, we need to determine the time spent by the ball before hitting the ground (t), measured in seconds, which is determined by the following kinematic formula:

t = \frac{v-v_{o}}{g} (1)

Where:

v_{o}, v - Initial and final speeds of the baseball, measured in meters per second.

g - Gravitational acceleration, measured in meters per square second.

If we know that v_{o} = 0\,\frac{m}{s}, v = 17\,\frac{m}{s} and g = 9.807\,\frac{m}{s^{2}}, then the time spent by the baseball is:

t = \frac{17\,\frac{m}{s}-0\,\frac{m}{s}  }{9.807\,\frac{m}{s^{2}} }

t = 1.733\,s

The baseball was falling during 1.733 seconds.

6 0
2 years ago
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