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UNO [17]
3 years ago
15

Design an Armstrong indirect FM modulator to generate an FM signal with a carrier frequency 98.1 MHz and a frequency deviation △

f = 75 kHz. A narrow-band FM generator is available at a carrier frequency 100 kHz and a frequency deviation △f = 10 Hz. The stock room also has an oscillator with an adjustable frequency in the range of 10 MHz to 11 MHz. There are also plenty of frequency doublers, triplers, and quintuplers (which means the frequency multiplication ratio must be integers like 2n × 3 m × 5 k ).

Engineering
1 answer:
Bess [88]3 years ago
3 0

Answer:

See explaination

Explanation:

In the Armstrong method of FM generation, the phase of the carrier is directly modulated in the combing network through summation, generating indirect frequency modulation.

Very high frequency stability is achieved through Armstrong method since the crystal oscillator is used as carrier frequency generator.

Please kindly check attachment for the step by step solution of the given problem.

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3 years ago
Tech a says the higher the numarical gear ratio (4:1), the more torque that will be applied to the wheels. Tech b says that the
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Answer:

Tech A

Explanation:

The amount of energy required to apply the same force with a 1:1 ratio is divided into 4, so you can apply 4 times as much force than a 1:1 ratio. efficiency and speed come into play here, but assuming the machine powering the gear can run at a unlimited RPM, 4:1 will have more force and a slower output speed than a 2:1 ratio.

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John has just graduated from State University. He owes $35,000 in college loans, but he does not have a job yet. The college loa
IRISSAK [1]

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= 35000\times (\frac{F}{P}, 4 \ percent,5 )

= 35000\times 1.216 7

= 42584.50

Now,

time (t) will be:

= 5\times 12

= 60 \ monthly \ payments

So, monthly payment will be:

= 42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )

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6 0
3 years ago
Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat
Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 =  0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = \frac{m}{\rho}

V = Av = \frac{\pi}{4} * d^2 * v1

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V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = \frac{V}{v} = \frac{0.30787}{0.11418}

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = \frac{v2}{A2}

v2 = \frac{0.3705}{\frac{\pi}{4} * 0.28^2}

v2 = 6.017 m/s

7 0
3 years ago
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