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 ).
1 answer:
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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Answer:
Explanation:
The turbine at steady-state is modelled after the First Law of Thermodynamics:
The specific enthalpies at inlet and outlet are, respectively:
Inlet (Superheated Steam)
Outlet (Liquid-Vapor Mixture)
The power produced by the turbine is:
Answer:
The value of R is 10101
Explanation:
As per the given data
D = 1000100100
G = 100101
Redundant bit = 6-bits - 1-bit = 5-bits
No add fice zero to D
D = 100010010000000
Now calculate R as follow
R = D / G
R = 100010010000000 / 100101
R = 10101
Workings are attached with this question
Question
A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m.K), and the wire/sheath interface is characterized by a thermal contact resistance of Rtc = 3E-4m².K/W. The convection heat transfer coefficient at the outer surface of the sheath is 10 W/m²K, and the temperature of the ambient air is 20°C.
If the temperature of the insulation may not exceed 50°C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?
Answer:
a. 4.52W/m
b. 13mm
Explanation:
Given
Diameter of electrical wire = 2mm
Wire Thickness = 2-mm
Thermal Conductivity of Rubberized sheath (k = 0.13 W/m.K)
Thermal contact resistance = 3E-4m².K/W
Convection heat transfer coefficient at the outer surface of the sheath = 10 W/m²K,
Temperature of the ambient air = 20°C.
Maximum Allowable Sheet Temperature = 50°C.
From the thermal circuit (See attachment), we my write
E'q = q' = (Tin,i - T∞)/(R'cond + R'conv)
= (Tin,i - T∞)/(Ln (r in,o / r in,i)/2πk + (1/(2πr in,o h)))
Where r in,i = D/2
= 2mm/2
= 1 mm
= 0.001m
r in,o = r in,i + t = 0.003m
T in, i = Tmax = 50°C
Hence
q' = (50 - 20)/[(Ln (0.003/0.001)/(2π * 0.13) + 1/(2π * 0.003 * 10)]
= 30/[(Ln3/0.26π) + 1/0.06π)]
= 30/[(1.34) + 5.30)]
= 30/6.64
= 4.52W/m
The critical radius is unaffected by the constant resistance.
Hence
Critical Radius = k/h
= 0.13/10
= 0.013m
= 13mm
