Data and Signals

Data must be transformed to electronic signals (why?) Analog data: Continuous information, e.g., voice has a value at any time. Analog clock have values at any time! Digital data: Discrete state information. e.g., digital clock. Analog signal: They have infinitely many levels over a period of time. Digital signals: They have limited number of defined values. Periodic Signals: “Repeat a pattern every measurable time frame, “period” or “cycle”. (most used for analog signals) Non-Periodic (Aperiodic) Signals: No pattern or cycle. (most used for digital signals)

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Figure 3.1 Comparison of analog and digital signals

Periodic Analog Signals

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Figure 3.2 A sine wave

Frequency (f) = 1/T and T = 1/f

Phase: It is the position of waveform relative to time = 0.

Ex: The “sine” phase is 0, where as the “cos” phase = π/2

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Figure 3.3 Two signals with the same phase and frequency, but different amplitudes

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Figure 3.4 Two signals with the same amplitude and phase,but different frequencies

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Figure 3.5 Three sine waves with the same amplitude and frequency,but different phases

Wave length: It relates the frequency / period of a signal to its propagation

speed in the medium. In case of “light” over fiber: λ = c / f c = propagation speed of light signal. f = frequency of light signal. λ = Wave length of light signal.

Time vs. Frequency domain:The amplitude values are plotted versus time in case of time domain signals.

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Figure 3.7 The time-domain and frequency-domain plots of a sine wave

The amplitude is shown for each frequency component of a signal in the frequency domain. The above sin(6t) has one frequency of 6 Hz, with 5 volt peak amplitude.

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Figure 3.8 The time domain and frequency domain of three sine waves

Composite Signals: Every composite signal is made of many sine waves of different amp’s, freq’s, phases. (Fourier analysis)

A periodic composite signal can be decomposed into a number of

signals with discrete frequencies in the frequency domain

.

Whereas, a non-periodic composite signal is a group of sine waves with continuous frequencies in the frequency domain.

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Figure 3.9 A composite periodic signal

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Figure 3.10 Decomposition of a composite periodic signal in the time andfrequency domains

Whereas, a non-periodic composite signal is a group of sine waves with continuous frequencies in the frequency domain.

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Figure 3.11 The time and frequency domains of a nonperiodic signal

Band width: (non/composite signals). Refer fig 3.12 BWcs = fh – fl Hz

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Figure 3.12 The bandwidth of periodic and nonperiodic composite signals

Transmission Impairment

Causes:

1) Attenuation: Loss of energy as a function of the signal traveling distance and its power. Decibel: Unit measure of signal’s power loss or gain Power units dB = 10 log10 (Pdest/Psrc) Pdest: Signal Power at destination Psrc: Signal power at source 2) Distortion: Signal changes in form / shape. Composite signal components have different propagation speed through the medium; hence destination arrivals delays/phase shifts causing incorrect receiving of the original signal at the destination. 3) Noise:

i) Thermal (white) caused by electrons random movement in the medium (extra superimposed signals)

ii) Induced: Caused by appliances acting as a sending antenna with medium as receiving one

iii) Crosstalk: Mutual affect between wires acting as sending/ receiving antenna

iv) Impulse: caused by a power lines, lightning shaped as a “spike” (high frequency on very short time)

Signal to Noise Ratio (SNR):

SNR = avg. signal power / avg. noise power SNR plays a profound role in deciding the bit rate limit (max). It is always desirable to have it high (lower noise than signal power) for good reception of signals passing through noisy environment (low quality channel). It is measured (and always given) in decibel units (powers ratio) dB To convert to watt: (SNRdB/10) SNRwatt =10 A SNRdB = 40 is to be converted to watt --> 10(40/10) = 10000 watt

Digital Signals

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Figure 3.16 Two digital signals: one with two signal levels and the otherwith four signal levels

• Bit Rate: “non-periodic” digital signals are the most used in digital

data transfer, hence period or frequency are not used, instead we use “Bit rate” which is the number of bits Tx per sec.

• Bit length

•

= propagation speed * bit duration (distance one bit occupies on the Tx medium) Bit interval:

• Time to Tx a bit = 1/bit rate = s/b

Baud rate

: Number of distinct changes (in signal) over the medium per sec

Bit rate = Baud rate X number of bits per a signal change

In binary

signals encoding, bit rate = baud rate (why?)

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Figure 3.17 The time and frequency domains of periodic and nonperiodicdigital signals

Transmission of Digital Signals: 1) Baseband (digital/encoding): We use low-pass channel with its BW

starting from 0 (i.e., flow = 0), since we are Tx digital signal.

2) 3.58

Figure 3.18 Baseband transmission

Since a digital signal is a composite of (infinite) analog signal with infinite BW (frequency components), the idle case is to have low-pass channel with infinite bandwidth (not in real life!) that must start from frequency component zero and have a very wide BW channel, in the frequency spectrum.(why?

)

Remember: Due to the filtering effect of the channel, The loss of low and high frequency components of the transmitted digital signal will result in the loss of the flat part and the sharp edges (respectively) of the received digital signal.

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Figure 3.19 Bandwidths of two low-pass channels

Yet, we can still send only the signal’s frequency components with significant amplitudes within some acceptable threshold

for minimum distortion and receive a reasonably acceptable digital signal at the receiver. As the bit rate increases, the signal significant BW (BWss) increases, hence we need a wider medium low-pass wider BW (BWm)

For safe digital signal reception: BWm >= BWss

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Figure 3.20 Baseband transmission using a dedicated medium

In fig 3.20, the input signal is Non-Periodic digital (user transmitted burst data hence it has continuous BW (0 to ∞). The channel BW filters out the low and high frequency component of the input signal, hence the receives gets a distorted output signal. In case of the channel passing enough signal BWss then we are OK, otherwise bad signal reception!

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Figure 3.21 Rough approximation of a digital signal using the first harmonicfor worst case

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Figure 3.22 Simulating a digital signal with first three harmonics

Fig. 3.22: As you capture more (low and high frequencies)

harmonics you get better digital signal representation.

2) Broadband (analog/modulation): Digital signal analog signal If digital signals are so so so desirable for many reasons (easier manipulation of refreshing for long channels, compressing for BW

If the available channel BW is scarce, hence we are saving, securing via encryption, ...) WHY do we ever need analog?

forced to to use a bandpass channel (Fig. 3.23) with its flow = f1 is non-zero value. Consequently, We do not have enough BW for digital transmission! Then we need to convert our digital signal to analog

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Figure 3.23 Bandwidth of a bandpass channel

signal (modulation process), as shown in fig. 3.24

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Figure 3.24 Modulation of a digital signal for transmission on a bandpasschannel

Since we are modulating our digital data into analog signals (sin wave carriers with different amplitude/frequencies/phases) we need not to start from the zero in the spectrum, but anywhere we select our carrier frequency (no need to have infinite number of frequency components), hence band-pass channel instead of low-pass as in the case of baseband digital Tx.

SNRdB = 10log10 SNR where Signal to Noise Ratio (given in dB always):

to convert to watts: SNRwatt = Signal power watt / Noise power watt

= 10(SNRdB

/10)

Noise Free channel: Max Bit Rate ––– CNyq = 2 BWch * log2 L -----------(1)

Nyquist Capacity (max channel bit rate)

Where L is the number of signal levels Since there is no upper limit on the value of L, we assume that the channel is noise free for our analysis, otherwise large L will make no sense over noisy channel!

Yet, if we choose a value for L, regardless of the channel noise, then in practice, we might be right or wrong in such choice of the value of L. Hence, we need another measure to judge our choice as depicted in the next theorem for Shannon:

Noisy Channel: Max Bit Rate ––– CShan = BWch log2 (1 + SNRwatt) ---(2)

Shannon Capacity

Both limits are important CNyq to decide the encoding bit/level and CShan for the actual noisy environment limitation. In case of noisy

channel, obtaining CShan from (2) and plugging it in lhs of (1) will guide us in the practical and correct choice of encoding (bit/level). Otherwise, if we made a wrong choice of encoding (not following the above) , then we might end up with an upper bound CNyq that is much lower than a valid CShan.

Example: What is the max bit-rate capacity of a 4000 Hz binary

CNyq = 2*4000 * log2 2 = 8000 b/s

channel (2 levels encoding) has a SNR of 30 dB?

CShan = 4000 * log2 (1+ 10(30/10)) = 4000 * 9.9672 ≈ 39869 b/s Even though, the CShan is much higher than the CNyq , yet we are forced to the lowest value of the two obtained capacities (i.e., 8000 b/s), as a result of bad encoding decision of only two levels!

Performance of Networks

1) Throughput: The actual speed of data traveling a link (notice, the BW of a link is its potential speed) Throughput (T) <= BW Example: bad encoding might waste BW in case of Ethernet ME. Therefore TME < BW

2) Delay (Latency): L = propagation + Tx + queuing + processing

Propagation delay = Distance (m) / propagation speed (m/s) Tx delay = message size (bit) / data-rate (bit/sec)

3) Jitter: Packets of data will encounter different delays. (not very good for real time multimedia applications!)