In relation to a/c mains electricity, or to audio frequencies, 400 kHz is very high. In relation to radio waves it is fairly low.
The maxium frequency swing in FM is ± 75 kHz so 75 kHz x 60% = ± 45 kHz
The standard US AM band is 520 kHz-1,610 kHz. The mean would be at 1,065 kHz.
Solution Let fh is the highest frequency and fl is the lowest frequency. Bandwidth = fh - fl = 4000 - 40 KHz = 3960 KHz = 3.96 MHz
According to Niquest Theorem, it has to be more than twice the input frequency.
I came up with 14.8 khz, is this right?
BW = (1 MHz - 10 KHz) = (1,000 KHz - 10 KHz) = 990 KHz
Low Frequency band(30-300 kHz). It is located between VLF (Very Low Frequency) 3-30 kHz and MF (Medium Frequency) . In this band you will find Ham Radio, Time Beacons, and possibly submarines among other services.
freq range of rf is 3hz (extremely low frequency) to 300Ghz (extremely high frequency)
Assuming that the receiver uses a high-side local oscillator and an IF of 455 KHz, the image frequency is 910 KHz above. When tuned to 1600 KHz, the image frequency would be 2,510 KHz.
If 10 V input causes a frequency shift of 4 kHZ then 2,5v causes a freuency shift of 1 kHz. The input signal frequency of 1 kHz is irelevant.
Absorption of low frequency sound is weak. The main cause of sound attenuation in fresh water, and at high frequency in sea water (above 100 kHz) is viscosity. Important additional contributions at lower frequency in seawater are associated with the ionic relaxation of boric acid (up to c. 10 kHz)[6] and magnesium sulfate (c. 10 kHz-500 kHz).
Trovan 128 Khz and ISO 134 Khz
The maxium frequency swing in FM is ± 75 kHz so 75 kHz x 60% = ± 45 kHz
The standard US AM band is 520 kHz-1,610 kHz. The mean would be at 1,065 kHz.
kHz (kilohertz) is a unit of frequency, not a unit of speed.
20 khz.
38 kHz