Voice comes in 8000 Hz frequency, so 16000 samples required each seconds.
1 sample if of 8 bits.
16000 sample if of 128000bits.
these bits is send per second.
so bandwidth required is 128kbs for single voice.
for 24 it is 24 x 128 kbps
= 3072 kbps
= 3.1 mbps
4,000 Hz is the standard transmission frequency for voice.
4,000 Hz * 2 = 8,000 samples per second (sps)
8,000 sps * 8 bits per second (bps) = 64,000 bps
24 * 64,000 bps = 1,536,000 bps
A very usefull advantage is the exchange of SNR(signal to noise ratio) with Bandwidth... as on increasing the bandwidth the power required for transmission get reduced to a great extent.. is given by the formula SNR2 ~ (SNR1) B1/B2 AS we can see on increasing the bandwidth the SNR is reduced greatly
to shift the frequency of information signal ,at the frequency domain to a higher frequency ...so the information can be transmitted to the receiver .
Disadvantage of Analog Communication: 1)Analog Communication systems are costlier than digital communication system. 2)Less Security in Analog Communication 3)More prone to noise interference. Advantage of Analog Communication: 1)Transmission bandwidth required is less. 2)No need of synchronization. Advantage of Digital Communication: 1)Digital Communication systems are simpler and cheaper. 2)Using data encryption only permitted user are allowed to detect the transmitted data. 3)Since in Digital Communication, channel coding is used, therefore a large amount of errors may be detected 4)Since transmitted signal is digital in nature,large amount of noise tolerance may be tolerated Disadvantage of Digital Communication: 1)More transmission bandwidth required. 2)Digital Communication needs synchronization in case of synchronous modulation.
1.Find the Fourier Transform of the pulse used to transmit data over the channel. 2.Determine the bitrate of the signal by the modulation format (QPSK for example has 2bits/symbol so 1 symbol per second would equate to 2 bits/s) 3.The first null in the Fourier transform is the required bandwidth (~0.75 x bitrate in optical communications, depends on channel) 4. Divide bits/s by the required bandwidth to find the spectral efficiency.
It's 1920Bytes. As the bandwidth (f2-f1) is 20-16=4Hz. The sampling frequency should be twice the bandwidth (2B). the sampling frequency is 8Hz= 8 samples= 8 cycles/second. 4 minutes=(4*60) =240 Seconds 1sample->1 Byte 1Second->8 Samples 240 Seconds->(240*8) = 1920 Samples
How much bandwidth required for e-gov application?
required larger bandwidth
In Amplitude Modulation (AM), specifically in the case of Double Sideband Suppressed Carrier (DSB-SC) or Full Carrier (DSB-FC) AM, the required bandwidth is twice the bandwidth of the modulating signal. If the modulating signal has a bandwidth of B Hz, the bandwidth required for AM would be 2B Hz. This is because both the upper and lower sidebands of the carrier wave are utilized in the modulation process, each consuming bandwidth equivalent to the original signal.
QoS
transmitted light is required for the purpose of examining internal features of gemstone inclusion.
There is no information available as to how much bandwidth Essex FM online radio takes. It is not as much as required, but how much it takes up is what the worry is for. Most internet providers have a limit as to how much bandwidth you are allowed to use in a month.
600 Hz
A very usefull advantage is the exchange of SNR(signal to noise ratio) with Bandwidth... as on increasing the bandwidth the power required for transmission get reduced to a great extent.. is given by the formula SNR2 ~ (SNR1) B1/B2 AS we can see on increasing the bandwidth the SNR is reduced greatly
to shift the frequency of information signal ,at the frequency domain to a higher frequency ...so the information can be transmitted to the receiver .
no of sources: 5 bandwidth required for each source= 400 Hz no of guard times= 5 bandwidth of each guard time = 200 Hz minimum bandwidth = 5 *400 + 5*200 Hz
Use Nyquist and Shannon Heartly theorem to solve this Nyquist theorem says that Channel Capacity C = 2 * Bandwidth * log2 (Number of Signal levels) Shannon Heartly theorem says that Channel Capacity C = Bandwidth * log2( 1 + SNR) Important points to consider while solving Bandwidth is expressed in Hz SNR is expressed in dB it must be converted using dB value = 10 log10(SNR) (10 dB = 10, 20 dB = 100, 30 dB = 1000 etc..)
With full double sideband AM the bandwidth of the modulated signal is twice that of the baseband information signal. With suppressed carrier single sideband AM the bandwidth of the modulated signal is identical to that of the baseband information signal. With vestigial sideband AM the bandwidth of the modulated signal is somewhere between the above two cases, depending on how much of the vestigial sideband is included.