# Nyquist Bandwidth

Please reference this figure for the question:

11. Considering only the Nyquist bandwidth limitation. Determine the bandwidth required of the media (at Point D) given the following:

a. 20-analog voice signals digitized using standard G.711 PCM; a TDM multiplexer with an added output overhead of 15%; and a modem that employs QPSK modulation.

b. Repeat part "a" for 16-ary QAM modulation.

c. Repeat part "b" using 16-ary QAM modulation. Each standard G.711 PCM digitized voice signal is compressed using a 4:1 ratio before going into the multiplexer. In addition the multiplexer is inserting a 1-kbps control signal for each input in addition to the general overhead.

Â© BrainMass Inc. brainmass.com December 24, 2021, 8:00 pm ad1c9bdddfhttps://brainmass.com/computer-science/algorithms/nyquist-bandwidth-237000

## SOLUTION This solution is **FREE** courtesy of BrainMass!

Nyquist Bandwidth simplified for second time. Please check attachment.

Please reference this figure for the question:

11. Considering only the Nyquist bandwidth limitation. Determine the bandwidth required of the media (at Point D) given the following:

a. 20-analog voice signals digitized using standard G.711 PCM; a TDM multiplexer with an added output overhead of 15%; and a modem that employs QPSK modulation.

b. Repeat part "a" for 16-ary QAM modulation.

c. Repeat part "b" using 16-ary QAM modulation. Each standard G.711 PCM digitized voice signal is compressed using a 4:1 ratio before going into the multiplexer. In addition the multiplexer is inserting a 1-kbps control signal for each input in addition to the general overhead.

Answer

G.711 (PCM: Pulse code modulation) is an international standard and widely used in the conversion of analog voice signals for use in digital transmission networks. The G.711 quality and characteristics are widely used as a reference point when new or improved algorithms are used in testing. Two sub-methods exist, mu-law (US) and A-law (non-US). G.711 has a 64 kbits/second data rate.

Here 8 kHz sampling rate and an 8 bit encoding is taken into the consideration.

So the digitized analog signal using G.711 PCM We get

220 samples. . ( here n=20;)

here data rate=64 kbits/second;

So the resultant bits are 220*10%;

here fc =8kHz;

n=220*10%;

S is the signal power and N is the noise power.

S=resultant bits;

N=1;

Just put the value in the above formula and Bandwidth will be available.

Here is the description is given for the better understanding. All the topics related to the question is discussed here and the mathematical calculation is done on the basis of discussion.

A time division multiplexing (TDM) facility is presented for time division multiplexing at least two inter-system channel (ISC) data streams to create a TDM multiplexed data stream. The TDM multiplexed data stream can be forwarded across the network over a single wavelength of a wavelength division multiplexing (WDM) network. The TDM multiplex data stream may be wavelength division multiplexed with one or more other TDM multiplexed data stream. Different protocols are presented for maintaining disparity balance within the ISC data streams depending upon whether a given sequence in one of the data streams is an idle sequence, frame sequence or continuous sequence.

Here QPSK modulation is used. In Quadrature Phase Shift Keying (QPSK) modulation, a cosine carrier is varied in phase while keeping a constant amplitude and frequency. In QPSK, information in conveyed through phase variations. In each time period, the phase can change once. Since there are four possible phases, there are 2 bits of information conveyed within each time slot. The rate of change (baud) in this signal determines the signal bandwidth, but the throughput or bit rate for QPSK is twice the baud rate.

. .

b)

The error-rate performance is close to that of 16-QAM â€” it is only about 0.5 dB better â€” but its data rate is only three-quarters that of 16-QAM. Thus 8-PSK is often omitted from standards and, as seen above, schemes tend to 'jump' from QPSK to 16-QAM (8-QAM is possible but difficult to implement).

In case of 16-QAM the signal to Noise ratio or SNR is

SNR=

Es =Signal Power,

N0 =Noise Power,

So the digitized analog signal using G.711 PCM We get

220 samples. . ( here n=20;)

here data rate=64 kbits/second;

So the resultant bits are 220*10%;

here fc =8kHz;

n=220*10%;

Please just put the value in the above formula and output bandwidth(BW) will be available.

c) Initially the data rate is divided into 4:1 ratio. Then the individual data that is transmitted over the media is multiplied with the overhead.

So the digitized analog signal using G.711 PCM using 16-ary QAM modulation ,

We get 220 samples. . ( here n=20;)

here data rate=64 kbits/second;

So the resultant bits are 220*10%;

So the data rate for each multiplexed media

= 220*10%*4/5;

Now when the data is passing from the multiplexer the amount of data =(220*10%*4/5+1024)byte;

Here 8 kHz sampling rate and an 8 bit encoding is taken into the consideration.

here fc =8kHz;

n=220*10%;

SNR=

S is the signal power and N is the noise power.

S=resultant bits;

N=1;

Just put the value in the above formula and Bandwidth(BW) will be available.

https://brainmass.com/computer-science/algorithms/nyquist-bandwidth-237000