1 / 48

# cs412 introduction to computer networking telecommunication - PowerPoint PPT Presentation

CS412 Introduction to Computer Networking & Telecommunication. Theoretical Basis of Data Communication. Topics. Analog/Digital Signals Time and Frequency Domains Bandwidth and Channel Capacity Data Communication Measurements. Signals.

Related searches for cs412 introduction to computer networking telecommunication

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'cs412 introduction to computer networking telecommunication' - Patman

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### CS412 Introduction to Computer Networking & Telecommunication

Theoretical Basis of

Data Communication

Chi-Cheng Lin, Winona State University

• Analog/Digital Signals

• Time and Frequency Domains

• Bandwidth and Channel Capacity

• Data Communication Measurements

• Information must be transformed into electromagnetic signals to be transmitted

• Signal forms

• Analog or digital

• Periodic or aperiodic

• Analog signal

• Continuous waveform

• Can have a infinite number of values in a range

• Digital signal

• Discrete

• Can have only a limited number of values

• E.g., 0 or 1

Figure 3.1Comparison of analog and digital signals

• Periodical signal

• Contains continuously repeated pattern

• Period (T): amount of time needed for the pattern to complete

• Aperiodical signal

• Contains no repetitive signals

• Simple analog signal

• Sine wave

• 3 characteristics

1. Peak amplitude (A)

2. Frequency (f)

3. Phase ()

• Composite analog signal

• Composed of multiple sine waves

Figure 3.2A sine wave

Figure 3.3Amplitude

s(t): instantaneous amplitude

t

• Peak amplitude: highest intensity

• Frequency (f)

• Number of cycles/rate of change per second

• Measured in Hertz (Hz), KHz, MHz, GHz, …

• Period (T): amount of time it takes to complete one cycle

• f = 1/T

• Phase: position of the waveform relative to time 0

Figure 3.4Period and frequency

Figure 3.5Relationships between different phases

Figure 3.6Sine wave examples

Figure 3.6Sine wave examples (continued)

Figure 3.6Sine wave examples (continued)

• Changes in the three characteristics provides the basis for telecommunication

• Used by modems (later …)

• The sine waves shown previously are plotted in its time domain.

• An analog signal is best represented in the frequency domain.

Figure 3.7Time and frequency domains

• A composite signal can be decomposed into component sine waves - harmonics

• The decomposition is performed by FourierAnalysis

Signal with DC Component

The McGraw-Hill Companies, Inc., 1998

WCB/McGraw-Hill

Figure 3.8-3.10Square wave and the first three harmonics

Figure 3.11Frequency spectrum comparison

• Frequency spectrum

• Collection of all component frequencies it contains

• Bandwidth

• Width of frequency spectrum

Figure 3.13Bandwidth

If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is the bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.

Solution

B = fh-fl = 900 - 100 = 800 Hz

The spectrum has only five spikes, at 100, 300, 500, 700, and 900 (see Figure 13.4 )

Figure 3.14Example 3

A signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all integral frequencies of the same amplitude.

Solution

B = fh- fl

20 = 60 - fl

fl = 60 - 20 = 40 Hz

Figure 3.15Example 4

A signal has a spectrum with frequencies between 1000 and 2000 Hz (bandwidth of 1000 Hz). A medium can pass frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz). Can this signal faithfully pass through this medium?

Solution

The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies between 3000 and 4000 Hz; the signal is totally lost.

• 0s and 1s

• Bit interval and bit rate

• Bit interval: time required to send 1 bit

• Bit rate: #bit intervals in one second

A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval)

Solution

The bit interval is the inverse of the bit rate.

Bit interval = 1/ 2000 s = 0.000500 s = 0.000500 x 106ms = 500 ms

• A digital signal can be decomposed into an infinite number of simple sine waves (harmonics), each with a different amplitude, frequency, and phase

A digital signal is a composite signal with an infinite bandwidth.

• Significant spectrum

• Components required to reconstruct the digital signal

Harmonics of a Digital Signal

The McGraw-Hill Companies, Inc., 1998

WCB/McGraw-Hill

• (a) A binary signal and its root-mean-square Fourier amplitudes.

• (b) – (e) Successive approximations to the original signal.

Exact and Significant Spectrums

The McGraw-Hill Companies, Inc., 1998

WCB/McGraw-Hill

• Channel capacity

• Max. bit rate a transmission medium can transfer

• Nyquist theorem

• C = 2H log2V

where C: channel capacity (bit per second)

H: bandwidth (Hz)

V: signal levels (2 for binary)

• C is proportional to H

 Significant bandwidth puts a limit on channel capacity

Figure 3.18Digital versus analog

To transmit 6bps, we need a bandwidth = 3 - 0 = 3Hz

• Nyquist theorem is for noiseless (error-free) channels.

• Shannon Capacity

• C = H log2(1 + S/N)

where C: (noisy) channel capacity (bps)

H: bandwidth (Hz)

S/N: signal-to-noise ratio

dB = 10 log10S/N

• In practice, we have to apply both for determining the channel capacity.

Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as

BitRate = 2  3000  log2 2 = 6000 bps

Example 8

Consider the same noiseless channel, transmitting a signal with four signal levels (for each level, we send two bits). The maximum bit rate can be calculated as:

Bit Rate = 2 x 3000 x log2 4 = 12,000 bps

Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity is calculated as

C = B log2 (1 + S/N) = B log2 (1 + 0)= B log2 (1) = B  0 = 0

We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually 35dB, i.e., 3162. For this channel the capacity is calculated as

C = B log2 (1 + S/N) = 3000 log2 (1 + 3162) = 3000 log2 (3163)

C = 3000  11.62 = 34,860 bps

We have a channel with a 1 MHz bandwidth. The S/N for this channel is 63; what is the appropriate bit rate and signal level?

Solution

First, we use the Shannon formula to find our upper limit.

C = B log2 (1 + S/N) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps

Then we use the Nyquist formula to find the

number of signal levels.

4 Mbps = 2  1 MHz  log2L L = 4

• Throughput

• How fast data can pass through an entity

• Propagation speed

• Depends on medium and signal frequency

• Propagation time (propagation delay)

• Time required for one bit to travel from one point to another

• Wavelength

• Propagation speed = wavelength X frequency

Figure 3.25Throughput

Figure 3.26Propagation time

Figure 3.27Wavelength