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UNO [17]
4 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]4 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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Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.
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Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30  + ( bicarbonate ion ) * 50/61

                                   = 15 * (50/30) + 120*( 50/61 )  = 123.361 mg/l  as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57  = 2.692*10^-5  moles/l

[ OH^- ]   = 2.692*10^-5  * 179/mole * 10^3 mg/g  = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3  = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

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