answersLogoWhite

0


Best Answer

df

User Avatar

Wiki User

11y ago
This answer is:
User Avatar
More answers
User Avatar

Wiki User

11y ago

250hz

This answer is:
User Avatar

User Avatar

JOYCE JON BETRICE FE...

Lvl 2
3y ago

2khz?

This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: What is the minimum sample rate for a signal frequency of 4kHz?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

What is the lowpass filter cutoff frequency that must be used to decimate to reduce the sampling rate from 8kHz to 4kHz?

2kHz - That's the nyquist frequency at a sample frequency of 4kHz.


A telephone line has 4 KHz?

Most phone lines sample at 8kHz, which means any signal beyond the 4kHz mark will be down sampled to the point of significant data loss and returned as noise, so yes, most phone lines only operate with 4kHz of bandwidth.


What is the highest frequency that can normally be transmitted by a conventional telephone system?

4000 Hz (4kHz)


On cubase 4 can you remove vocals from cds?

No you cannot. But may be u can cut the vocal frequency range which is between 1-4khz but ul also be deleting other instruments that plays in that wavelength.


What is the bandwidth of a analog voice channel?

Bandwidth is defined as a frequency span - the difference between a high frequency and a lower frequency. For the low end voice it depends if its male bass bariton tenor or female alto and soprano. A bass voice goes down to 100 Hz. The harmonics go up to 15 kHz. So the bandwidth for voices is arround 15 kHz.


Difference between channel and bandwidth?

a channel is actually a path through which a signal of a particular frequency travels and bandwidth is the capacity of that path it tells about the number or range of frequencies which a path can carry


How many cycles of a 4KHz sound wave will occur in 0.4seconds?

1,600 (4,000 x 0.4 = 1,600)


A communication channel with additive white gaussian noise has a bandwidth of 4kHz and SNR of 31dB. Its channel capacity is?

20kbps


How you can Find the maximum data rate of a noiseless 4-kHz channel-using Analog encoding eg QPSK with 2 bits per sample?

for 4KHz then for noisy channel using Shannon theorem, sampling rate will be 8K samples/sec. So with 2 bit encoding, 2 bits are sent per sample. So the data rates is 8000 samples / sec * 2 bits = 16000bits / sec = 16kbps.


What is mean by full carrier?

At this point we're talking about amplitude modulation. If you're transmitting with "29.000" MHz displaying on your radio's frequency display and your bandwidth is set to 4 kHz, you are transmitting useful intelligence from 28.996MHz to 29.004MHz--4KHz on each side of the carrier--in "full carrier" or "double sideband" mode. You wouldn't actually transmit this for a couple of reasons. The most important reason is that using filters in your radio to remove either the signal below 29MHz (called "upper sideband" or USB operation) or the signal above it (called "lower sideband" or LSB) allows another operator to use the piece of spectrum you relinquished. The other advantage is SSB makes more efficient use of your amplifier, so you can talk to people farther away from you and they can hear you better. So..."full carrier" means double sideband, and then there's single sideband.


What is the frequency range of most people with good hearing?

The maximum human hearing range is 20Hz to 20,000Hz. 20Hz is actually more felt than heard, though a trained musician can identify the pitch at 20Hz. Only very young children and dogs can clearly hear a pitch at 20kHz, generally, and every human's ability to hear very high frequencies gradually shifts downward with age - that is normal. Even the very elderly, barring other hearing problems, can still generally hear from 40Hz to 4kHz as a minimum - about six and one-half octaves. Generally, a healthy adult can hear clearly in a range of 30-15,000Hz into their sixth decade of life.


A noiseless 4 kHz channel is sampled every 1 msec What is the maximum data rate?

by Nyquist theorem max data rate = 2H log2 V bits /sec so H= 4KHz.. and v=2 ( say binary is transmitting ) therefore max data rate = 2 * 4 *103 *log22 bps max data rate = 8000 bps Bcoz it is sampling per 1ms so