1Hz is unit of frequency,which is equals to one cycle per second........................
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The Nyquist frequency for a signal with a maximum bandwidth of 1 KHz is 500 Hz, however that will lead to aliasing unless perfect filters are available. The Nyquist rate for a signal with a maximum bandwidth of 1 KHz is 2 KHz, so the answer to the question is 2 KHz, or 500 microseconds.
According to the Nyquist theorem, the minimum sampling rate must be at least twice the maximum frequency of the input signal to avoid aliasing. Therefore, if the input frequency is 3 kHz, the minimum sampling rate should be 6 kHz.
Electricity, Heat, Natural obstacles during daylight hours
frequency means number of cycles per second. here 1 cycle takes .029 mili sec or .000029 sec so the number of cycles in 1 sec is 1/.000029 = 34482.75 Hz = 34.5 KHz
38 kHz
The period of a waveform is the reciprocal of its frequency. For a clock waveform with a frequency of 500 kHz, the period can be calculated as 1 / 500 kHz = 2 microseconds.
BW = (1 MHz - 10 KHz) = (1,000 KHz - 10 KHz) = 990 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.
Period = reciprocal of frequency = 1 / (500) = 0.002 second
The Nyquist frequency for a signal with a maximum bandwidth of 1 KHz is 500 Hz, however that will lead to aliasing unless perfect filters are available. The Nyquist rate for a signal with a maximum bandwidth of 1 KHz is 2 KHz, so the answer to the question is 2 KHz, or 500 microseconds.
To calculate the time of the cycle you just invert the Hz value. Hz = 1 / T, Where T is the time of the cycle in seconds so a 10,000Hz signal has a time of each cycle of: 0.0001 seconds.
Since period is the reciprocal of frequency, i.e Period = 1/frequency: Frequency = 4 kHz = 4000 cycles/sec Period = 1/(4000 cycles/sec) = 1 sec/4000 cycles = 0.00025 sec/cycle
Period = 1 / (frequency) = 1 / 500 = 0.002 second = 2 milliseconds
If the intelligence signal striking a microphone was doubled in frequency from 1 kHz to 2 kHz with constant amplitude, (fc) would change from 1 kHz to 2 kHz. Because the intelligence amplitude was not changed, however, the amount of frequency deviation above and below fc will remain the same. On the other hand, if the 1 kHz intelligence frequency were kept the same but its amplitude were doubled, the rate of deviation above and below fc would remain at 1 kHz, but the amount of frequency deviation would double.
The minimum sample rate required to record a frequency of 96 kHz is 192 kHz. This is because according to the Nyquist theorem, the minimum sampling rate must be at least twice the highest frequency in order to accurately reconstruct the original signal. So for a frequency of 96 kHz, the minimum required sampling rate is double, which equals 192 kHz.
The pulse repetition period is the inverse of the pulse repetition frequency. Therefore, if the pulse repetition frequency is 1 kHz, the pulse repetition period would be 1 millisecond (1/1000 seconds).
According to the Nyquist theorem, the minimum sampling rate must be at least twice the maximum frequency of the input signal to avoid aliasing. Therefore, if the input frequency is 3 kHz, the minimum sampling rate should be 6 kHz.