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natita [175]
3 years ago
14

An analog baseband audio signal with a bandwidth of 4kHz is transmitted through a transmission channel with additive white noise

. The channel is assumed to be distortionless, and the power spectral density of white noise, No/2 is 10 WHz. An RC low-pass filter with a 3-dB bandwidth of 8 kHz is used at the receiver to limit the output noise power. Calculate the output noise power.
Engineering
1 answer:
siniylev [52]3 years ago
8 0

Answer:

2k20

Explanation:

4k ✈

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Carbon dioxide flows at a rate of 1.5 ft3 /s from a 3-in. pipe in which the pressure and temperature are 20 psi (gage) and 120 °
Monica [59]

Answer:

the absolute pressure in the smaller pipe = 19.63 psi

Explanation:

Let A be the diameter of the first pipe = 3 inches

Let B be the diameter of the second pipe.  = 1.5 inches

To feet (ft) ; we have

Diameter of the pipe A D_1 = (\dfrac{3}{12})ft = 0.25  \ ft

Diameter of pipe B  D_1 = (\dfrac{1.5}{12})ft = 0.125  \ ft

Temperature T = 120° F = (120+ 460)°R

= 580 ° R

The pressure gage to atmospheric pressure ; we have:

P_{Absolute }=P _{Atm} + P_{guage}

where;

atmospheric pressure = 1.47 psi

pressure gage = 20 psi

P_{Absolute }=(1.47+20)psi

P_{Absolute }=34.7 \ psi

To lb/ft²; we have:

P_{Absolute }=(34.7 *144 ) lb/ft^2

P_{Absolute }= 4.998.6 fb/ft²

The density of carbon dioxide can be calculated by using the relation

\rho = \dfrac{P}{RT}

\rho = \dfrac{4996.8}{(1130 \ lb /slug ^0 R)*(580{^0} R)}

\rho = 7.64*10^{-3}\ slug /ft^3

Formula for calculating cross sectional area is

A = \dfrac{\pi}{4}D

For diameter of pipe D_1 = 0.025

A₁ = \dfrac{\pi}{4}*0.25^2

A₁ = 0.04909 ft²

For diameter of pipe D_2 - 0.0125

A₂ =\dfrac{\pi}{4}*0.125^2

A₂ = 0.012227 ft²

Using the continuity equation to determine the velocities V₁ and V₂ respectively.

For V₁

Q = A₁V₁

V₁ = Q₁/ A₁

V₁ = 1.5/0.04909

V₁ = 30.557 ft/s

For V₂

Q = A₂V₂

V₂= Q₂/ A₂

V₂ = 1.5/0.04909

V₂ = 30.557 ft/s

Finally; using Bernoulli's Equation to the flow of the carbon dioxide from the larger pipe to the smaller pipe ; we have:

p_1 + \dfrac{\rho V_1^2}{2}+\gamma Z_1= p_2 + \dfrac{\rho V_2^2}{2}+\gamma Z_2

Since the pipe is horizontal then;

\gamma Z_1= \gamma Z_2

So;

p_1 + \dfrac{\rho V_1^2}{2}= p_2 + \dfrac{\rho V_2^2}{2}

p_2 =p_1 +\dfrac{1}{2}  \rho(V_1^2-V_2^2)

p_2 =4996.8+\dfrac{1}{2}  *7.624*10^{-3}(30.557^2-122.23^2)

p_2 =4943.41 \ lb/ft^2

To psi;

p_2 =\dfrac{4943.41 }{144}psi

p_2 =34.33 \ psi gage

The absolute pressure in the smaller pipe can be calculated as:

p_2 _{absolute} = 34.33 - 14.7

p_2 _{absolute} = 19.63 \ \  absolute

Hence, the absolute pressure in the smaller pipe = 19.63 psi

7 0
4 years ago
What is the IMA of a fixed pulley ​
Amanda [17]

Answer:

Not sure

Explanation:

Looking now

4 0
3 years ago
Read 2 more answers
100000000000x1000000000000=?
zhannawk [14.2K]

Answer:

Thy answer to your very sophisticated question is 1E23

Explanation:

IT JUST IS! Dont ask any questions

mAsquErade, mAsquerade tHat iS mY naME

7 0
2 years ago
Read 2 more answers
A steam power plant with a power output of 230 MW consumes coal at a rate of 60 tons/h. If the heating value of the coal is 30,0
NARA [144]

Answer:

\eta =46\%

Explanation:

Hello!

In this case, we compute the heat output from coal, given its heating value and the mass flow:

Q_H=60\frac{tons}{h}*\frac{1000kg}{1ton}*\frac{1h}{3600s}*\frac{30,000kJ}{kg}\\\\Q_H=500,000\frac{kJ}{s}*\frac{1MJ}{1000J} =500MW

Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:

\eta =\frac{230MW}{500MW}*100\% \\\\\eta =46\%

Best regards!

7 0
3 years ago
Im passed due someone help meeeeeee
vovangra [49]

Answer:

how are supposed to help when you can't do anything?

8 0
3 years ago
Read 2 more answers
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