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Vadim26 [7]
4 years ago
9

What is advantage and disadvantage of flat top sampling in digital signal trasmission?​

Engineering
1 answer:
Gelneren [198K]4 years ago
5 0
The primary advantage of digital transmission over analog transmission is noise immunity. Digital signals are inherently less susceptible than analog signals to interference caused by noise because it is not necessary to evaluate the precise amplitude, frequency, or phase to ascertain its logic condition. A disadvantage is sampling error. Digital communications require greater bandwidth. The detection of digital signals requires the communications system to be synchronized.
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Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t
kodGreya [7K]

Answer:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

h = 20 \frac{W}{m^2 K} represent the heat transfer coefficient.

p represent the perimeter for this case and would be given by:

p = 2*0.05m +2*0.001m= 0.102m

k = 200 \frac{W}{m C} represent the thermal conductivity

w = 5cm =0.05 m represent the width

h = 1mm =0.001m represent the thickness

A= wh= 0.05m *0.001m = 0.00005 m^2

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}

Where m is a coefficient given by:

m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}

The value of x for this case represent the distance x =5 cm =0.05m

T_b =130 C represent the base temperature

T_{\infty}= 20 represent the temperature of the sorroundings or the ambient.

If we replace we have this:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

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Answer:

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Explanation:

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