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(a)
To Find: The wavelength of the AM signals.
(a)
Answer to Problem 38P
The wavelength of the AM signals of 550 kHz is 545 m and wavelength of the AM signals of 1600 kHz is 187.5 m .
Explanation of Solution
Given:
Lower frequency of AM signals, f1 = 550 kHz
Higher value of frequency of AM signals, f2 = 1600 kHz
Speed of wave, v = 3×108 m/s
Formula used:
The expression for wave speed.
v = f×λ
Here, λ is the wavelength of the AM Signal.
v is speed.
f is frequency of signal.
Thus, wavelength can be written as:
λ = vf
Calculation:
Substitute the given values in the above equation,
For 550 kHz , Wavelength
λ1 = 3×108550×103 = 545 m
For 1600 kHz , Wavelength
λ2=3×1081600×103=187.5 m
Conclusion:
The wave length of the AM signals of 550 kHz is 545 m and wave length of the AM signals of 1600 kHz is 187.5 m .
(b)
To Find: The wavelength of the FM signals.
(b)
Answer to Problem 38P
The wave length of the FM signals of 88 MHz is 3.41 m and wave length of 108 MHz FM signals is 2.78 m .
Explanation of Solution
Given:
Lower frequency of AM signals, f1 = 550 kHz
Higher value of frequency of AM signals, f2 = 1600 kHz
Speed of wave, v = 3×108 m/s
Formula used:
The expression for wave speed.
v = f×λ
Here, λ is the wavelength of the AM signals. Thus, wavelength can be written as:
λ = vf
Calculation:
Substitute the given values in the above equation,
For 88 MHz , Wave length λ1 = 3×10888×106 = 3.41 m
For 108 MHz , Wave length λ2 = 3×108108×106 = 2.78 m
Conclusion:
Thus, the wave length of the FM signals of 88 MHz is 3.41 m and wave length of 108 MHz FM signals is 2.78 m .
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