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Marizza181 [45]
2 years ago
4

6.58 In World War II, voice radio scramblers used modulation schemes to distort a signal so that enemy forces could not understa

nd it unless it was demodulated properly. In one scheme, a signal x (t), bandlimited to 4 kHz, is modulated to generate an output signal (a) Plot the range of the spectrum of y(t). (b) Describe why the scrambled signal's spectrum is ''distorted.'' (c) Show that x(t) can be recovered from y(t) using

Engineering
1 answer:
jonny [76]2 years ago
3 0

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

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A system consisting of 3 lb of water vapor in a piston–cylinder assembly, initially at 350°F and a volume of 71.7 ft3, is expand
Alla [95]

Answer:

isobaric expansion = 281.09 Btu

isothermal compression= 72 Btu

Explanation:

The first law of thermodynamics is:

Q_{AB}=W_{AB}+deltaU_{AB}

where:

Q=heat transferred

W= work

U=internal energy  

W_{AB}=P*(V_{B}-V_{A})

U_{AB}=n*C_{v}(T_{2}-T_{1})

P=pressure, V= volume, T= temperature, n =  moles, Cv= specific heat at constant volume.

In a isobaric process heat transferred is:

Q=P*(V_{B}-V_{A})+n*C_{v}(T_{2}-T_{1})

For an isothermal process (T2-T1 = 0) so

Q=P*(V_{B}-V_{A})= W_{AB}

From the data we know that the energy transferred to the system in the isothermal compression by work was 72 Btu that is the heat transferred to the system.

For the first process

Q=P*(V_{B}-V_{A})+n*C_{v}(T_{2}-T_{1})

we have to properties at the beginning of the process : temperature (350°F) and specific volume (V/mass)

specific-volume=\frac{71.7 ft^{3}}{3Lb}=23.9\frac{ft^{3}}{Lb}

we use this information in the appropriate unit to find the pressure in thermodynamic tables.

T1= 176°C

v1= 1.49 m^3/kg

P=1.37 bar

in the second state we have

P=1.37 bar =137000Pa

v_{2}=\frac{85.38ft^{3}}{3Lb}= 28.46\frac{ft^{3}}{Lb}

with thee properties we check in the thermodynamic tables

T2= 255°C

n=mass/Mw = 3Lb*\frac {1kg}{2.2Lb}*\frac{1000gr}{1kg}*\frac{1mol}{18gr}=75.75 mol

we usually find Cp on tables for water but from the Mayer relation we have:

C_{v}=C_{p}+R

Cp for water vapor is: 33.12 J/mol*K

R=8.314 J/mol*K

Cv= 41.434 J/mol*K

replacing in the equation for Q

Q=137000 Pa*(2.41m^{3}-2.030m^{3})+75.75mol*41.434\frac{ J}{mol*K}*(528.15-449.81 K)=296569J

296569J =281.09 Btu

5 0
2 years ago
An annealed copper strip of 228 mm wide and 25 mm thick being rolled to a thickness of 20 mm, in one pass. The roll radius is 30
sdas [7]

Answer:

The roll force is 1.59 MN

The power required in this operation is 644.96 kW

Explanation:

Given;

width of the annealed copper, w = 228 m

thickness of the copper, h₀ = 25 mm

final thickness, hf = 20 mm

roll radius, R = 300 mm

The roll force is given by;

F = LwY_{avg}

where;

w is the width of the annealed copper

Y_{avg} is average true stress of the strip in the roll gap

L is length of arc in contact, and for frictionless situation it is given as;

L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm

Now, determine the average true stress, Y_{avg}, for the annealed copper;

The absolute value of the true strain, ε = ln(25/20)

ε = 0.223

from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.

Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa

Finally determine the roll force;

F = LwY_{avg}

F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F =   1.59 \ MN

The power required in this operation is given by;

P = \frac{2\pi FLN}{60}\\\\P =  \frac{2\pi (1.59*10^6)(0.03873)(100)}{60}\\\\P = 644955.2 \ W\\\\P = 644.96 \ kW

5 0
2 years ago
The legend that Benjamin Franklin flew a kite as a storm approached is only a legend—he was neither stupid nor suicidal. Suppose
Delicious77 [7]

Answer: 0.93 mA

Explanation:

In order to calculate the current passing through the water layer, as we have the potential difference between the ends of the string as a given, assuming that we can apply Ohm’s law, we need to calculate the resistance of the water layer.

We can express the resistance as follows:

R = ρ.L/A

In order to calculate the area A, we can assume that the string is a cylinder with a circular cross-section, so the Area of the water layer can be written as follows:

A= π(r22 – r12) = π( (0.0025)2-(0.002)2 ) m2 = 7.07 . 10-6 m2

Replacing by the values, we get R as follows:

R = 1.4 1010 Ω

Applying Ohm’s Law, and solving for the current I:

I = V/R = 130 106 V / 1.4 1010 Ω = 0.93 mA

7 0
2 years ago
What properties should the head of a carpenter’s hammer possess? How would you manufacture a hammer head?
BabaBlast [244]

Properties of Carpenter's hammer possess

Explanation:

1.The head of a carpenter's hammer should possess the impact resistance, so that the chips do not peel off the striking face while working.

2.The hammer head should also be very hard, so that it does not deform while driving or eradicate any nails in wood.

3.Carpenter's hammer is used to impact smaller areas of an object.It can drive nails in the wood,can crush  the rock and shape the metal.It is not suitable for heavy work.

How hammer head is manufactured :

1.Hammer head is produced by metal forging process.

2.In this process metal is heated and this molten metal is placed in the cavities said to be dies.

3.One die is fixed and another die is movable.Ram forces the two dies under the forces which gives the metal desired shape.

4.The third process is repeated for several times.

5 0
2 years ago
An electric field is expressed in rectangular coordinates by E = 6x2ax + 6y ay +4az V/m.Find:a) VMN if point M and N are specifi
Fittoniya [83]

Answer:

a.) -147V

b.) -120V

c.) 51V

Explanation:

a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).

b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.

c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.

Honestly, these things take practice to get used to. It's really hard to explain this.

3 0
2 years ago
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