Ask question

Ask question

All categories

3 years ago

6

The output S/N at thereceiver must be greater than 40 dB. The audio signal has zero mean, maximum amplitude of 1, power of ½ Wan

d bandwidth of 15 kHz. The power spectral density of white noise N0/2 = 10-10W/Hz and the power loss in the channel is 50 dB. Determine the transmit power required and the bandwidth needed.

1 answer:

abruzzese [7]3 years ago

3 0

Given that,

The output signal at the receiver must be greater than 40 dB.

Maximum amplitude = 1

Bandwidth = 15 kHz

The power spectral density of white noise is

$\dfrac{N}{2}=10^{-10}\ W/Hz$

Power loss in channel= 50 dB

Suppose, Using DSB modulation

We need to calculate the power required

Using formula of power

$P_{L}_{dB}=10\log(P_{L})$

Put the value into the formula

$50=10\log(P_{L})$

$P_{L}=10^{5}\ W$

For DSB modulation,

Figure of merit = 1

We need to calculate the input signal

Using formula of FOM

$FOM=\dfrac{\dfrac{S_{o}}{N_{o}}}{\dfrac{S_{i}}{N_{i}}}$

$1=\dfrac{\dfrac{S_{o}}{N_{o}}}{\dfrac{S_{i}}{N_{i}}}$

$\dfrac{S_{i}}{N_{i}W}=\dfrac{S_{o}}{N_{o}}$

Put the value into the formula

$\dfrac{S_{i}}{2\times10^{-10}\times15\times10^{3}}$

$\dfrac{S_{i}}{30\times10^{-7}}$

$S_{i}$

$S_{i}=30\times10^{-3}$

We need to calculate the transmit power

Using formula of power transmit

$S_{i}=\dfrac{P_{t}}{P_{L}}$

$P_{t}=S_{i}\times P_{L}$

Put the value into the formula

$P_{t}=30\times10^{-3}\times10^{5}$

$P_{t}=3\ kW$

We need to calculate the needed bandwidth

Using formula of bandwidth for DSB modulation

$bandwidth=2W$

Put the value into the formula

$bandwidth =2\times15$

$bandwidth = 30\ kHz$

Hence, The transmit power is 3 kW.

The needed bandwidth is 30 kHz.

You might be interested in

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75

Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process $T_{3}$ = 2042.56 K

(b). The value of pressure at the end of heat addition process $P_{3}$ = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle $E_{otto}$ = 0.4478

(d). The value of mean effective pressure of the cycle $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

Explanation:

Compression ratio $r_{p}$ = 8

Initial pressure $P_{1}$ = 95 k pa

Initial temperature $T_{1}$ = 278 °c = 551 K

Final pressure $P_{2}$ = 8 × $P_{1}$ = 8 × 95 = 760 k pa

Final temperature $T_{2}$ = $T_{1}$ × $r_{p} ^{\frac{\gamma - 1}{\gamma} }$

Final temperature $T_{2}$ = 551 × $8 ^{\frac{1.4 - 1}{1.4} }$

Final temperature $T_{2}$ = 998 K

Heat transferred at constant volume Q = 750 $\frac{KJ}{kg}$

(a). We know that Heat transferred at constant volume $Q_{S}$ = $m C_{v} ( T_{3} - T_{2} )$

⇒ 1 × 0.718 × ( $T_{3}$ - 998 ) = 750

⇒ $T_{3}$ = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ $\frac{P_{3} }{P_{2} }$ = $\frac{T_{3} }{T_{2} }$

⇒ $P_{3}$ = $\frac{2042.56}{998}$ × 760

⇒ $P_{3}$ = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle $Q_{R} = m C_{v} ( T_{4} - T_{1} )$

For the compression and expansion process,

⇒ $\frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }$

⇒ $\frac{2042.56}{998} = \frac{T_{4} }{551}$

⇒ $T_{4}$ = 1127.7 K

Heat rejected $Q_{R}$ = 1 × 0.718 × ( 1127.7 - 551)

⇒ $Q_{R}$ = 414.07 $\frac{KJ}{kg}$

Net heat interaction from the cycle $Q_{net}$ = $Q_{S}$ - $Q_{R}$

Put the values of $Q_{S}$ & $Q_{R}$ we get,

⇒ $Q_{net}$ = 750 - 414.07

⇒ $Q_{net}$ = 335.93 $\frac{KJ}{kg}$

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ $Q_{net}$ = $W_{net}$

⇒ $W_{net}$ = 335.93 $\frac{KJ}{kg}$

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

$E_{otto} = 1- \frac{T_{1} }{T_{2} }$

Put the values of $T_{1}$ & $T_{2}$ in the above formula we get,

$E_{otto} = 1- \frac{551 }{998 }$

⇒ $E_{otto}$ = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure $P_{m}$ :-

We know that mean effective pressure of the Otto cycle is given by

$P_{m}$ = $\frac{W_{net} }{V_{s} }$ ---------- (1)

where $V_{s}$ is the swept volume.

$V_{s}$ = $V_{1} - V_{2}$ ---------- ( 2 )

From ideal gas equation $P_{1}$ $V_{1}$ = m × R × $T_{1}$

Put all the values in above formula we get,

⇒ 95 × $V_{1}$ = 1 × 0.287 × 551

⇒ $V_{1}$ = 0.6 $m^{3}$

From the same ideal gas equation

$P_{2}$ $V_{2}$ = m × R × $T_{2}$

⇒ 760 × $V_{2}$ = 1 × 0.287 × 998

⇒ $V_{2}$ = 0.377 $m^{3}$

Thus swept volume $V_{s}$ = 0.6 - 0.377

⇒ $V_{s}$ = 0.223 $m^{3}$

Thus from equation 1 the mean effective pressure

⇒ $P_{m}$ = $\frac{335.93}{0.223}$

⇒ $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

This is the value of mean effective pressure of the cycle.

4 0

3 years ago

An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a

DedPeter [7]

Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

b) the annual cost of heat loss from line is $12904.25

Explanation:

a)

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

so

A = π × 0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

b)

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶ × 18413.588/0.9 × ( 3600 × 24 × 365 )

A = $12904.25

Therefore the annual cost of heat loss from line is $12904.25

4 0

3 years ago

A compressor receives air at 290 K, 95 kPa and shaft work of 5.5 kW from a gasoline engine. It should deliver a mass flow rate o

aev [14]

Answer:

P2 = 3.9 MPa

Explanation:

Given that

T₁ = 290 K

P₁ = 95 KPa

Power P = 5.5 KW

mass flow rate = 0.01 kg/s

solution

with the help of table A5

here air specific heat and adiabatic exponent is

Cp = 1.004 kJ/kg K

and k = 1.4

so

work rate will be

W = m × Cp × (T2 - T1) ..........................1

here T2 = W ÷ ( m × Cp) + T1

so T2 = 5.5 ÷ ( 0.001 × 1.004 ) + 290

T2 = 838 k

so final pressure will be here

P2 = P1 × $(\frac{T2}{T1})^\frac{k}{k-1}$ ..............2

P2 = 95 × $(\frac{838}{290})^\frac{1.4}{1.4-1}$

P2 = 3.9 MPa

3 0

3 years ago

For a metal that has an electrical conductivity of 6.1 × 107 (Ω∙m)–1, what is the resistance of a wire that is 4.3 mm in diamete

kotegsom [21]

Answer: (C) 9.14 . 10⁻³ Ω

Explanation:

The resistance of a resistor, is proportional to his length and inversely proportional to his area, being the proportionality constant a property of the material, called resistivity.

The resistivity is defined as the inverse of the electrical conductivity, which depends on the number of charge carriers and the mobility of these carriers, which is different for each material.

So, we can calculate the resistance as follows:

R = 1/σ . L / A, where:

σ = electrical conductivity, l= length of the wire , A = wire cross-section (assumed circular).

Replacing by the values, we can calculate R as follows:

R = 1/6.1. 10⁷ (Ω.m) . 8.1 m. / π (0.0043)² m / 4 = 9.14 . 10⁻³ Ω

7 0

3 years ago

According to the basic speed law, if conditions make it unsafe to follow posted speed limits, you should:____.

lapo4ka [179]

According to the basic speed law, if conditions make it unsafe to follow posted speed limits, you should: Reduce your speed to less than the maximum posted speed limit.

<h3>What is the most speed limit?</h3>

The highest published speed limit in the country is 85 mph (137 km/h) and can be located only on Texas State Highway 130, a toll road that bypasses the Austin metropolitan area for long-distance traffic.

<h3>What is the 'rule'? </h3>

The 'rule' itself is quite straightforward: if the speed limit is (for example) 30mph, the rule notes that you won't get a speeding ticket unless you are going 10% plus 2 mph faster than the limit.

To learn more about speed limit, refer

brainly.com/question/19567226

#SPJ4

7 0

2 years ago