expanded by Jozef Goetz, 2009 The McGraw-Hill Companies, Inc., 2006

Chapter 3 Signals PART III expanded by Jozef Goetz, 2009 The McGraw-Hill Companies, Inc., 2006 expanded by Jozef Goetz, 2009

3.4 Transmission Impairment Attenuation Distortion Noise

Figure 3.25Impairment types Attenuation = meansloss of energy Distortionmeansthesignalchanges its formorshape Noise is another cause of impairment

Data Rate of a Channel. • Theory: A perfect (noiseless) channel will still have a finite transmission capacity. • Introducing noise into a channel will further reduce the capacity of that channel. • Max Bit Ratefornoiseless channel(is rarely achieved): If bandwidth = Band the signal-power-to-noise-powerratio is SNR = S/N The ratio itself is not quoted, instead the quantity is given A ratio of 100 is 20 dB, ratio of 1000 is 30 dB SNRdB = 10 log SNR [decibels] = 10 log10 (S/N) [decibels]

Figure 3.21Attenuation Attenuation = meansloss of energy to overcome the resistance of the medium (converted to heat) We can use the decibel to measure the changes in the strength of a signal The decibel (dB) measures the relative strength of 2 signals where P1 and P2 are thepowers of a signal at points 1 and 2, respectively. If A < 0 a signal is attenuated If A > 0 a signal is amplified A = 10 log10 (P2/P1) [dB]

Example 3.26 • Imagine a signal travels through a transmission medium and its power is reduced to half. • This means that P2 = 1/2 P1. • In this case, the attenuation (loss of power) can be calculated as Solution A = 10 log10 (P2/P1)= 10 log10 (0.5P1/P1) = 10 log10 (0.5) = 10(–0.3) = –3 dB A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.

Example 3.27 • Imagine a signal travels through an amplifier and its power is increased ten times. • This means that P2 = 10 x P1. • In this case, the amplification (gain of power) can be calculated as A = 10 log10 (P2/P1) = 10 log10 (10P1/P1) = 10 log10 (10) = 10 (1) = 10 dB What is A when its power is increased 2 times?

Example 3.28 • One reason that engineers use the decibel to measure the changes inthestrengthof a signal is that decibel numberscan beadded (or subtracted) when we are talking about several points instead of just two (cascading). • In Figure 3.27 a signal travels a long distance from point 1 to point 4. • The signal is attenuated by the time it reaches point 2. • Between points 2 and 3, the signal is amplified. • Again, between points 3 and 4, the signal is attenuated. • We can find the resultantdecibel for the signal just by adding the decibel measurements between each set of points.

Figure 3.27Example 3.28 strength of signal = –3dB + 7dB – 3dB = +1 dB

Example 3.29 • Sometimes the decibel is used to measure signal power in milliwatts. • In this case, it is referred to as dBm and is calculated as dBm = 10 Pm, • where Pm is the power in milliwatts. • Find/calculate the power of a signal with dBm = −30. • Solution • We can calculate the power in the signal as

Example 3.30 The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km? Solution The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power as dB XXXXXX

Transmission of Light through Fiber • Attenuation of light through fiber in theinfrared region of the spectrum (visible light is 0.4 – 0.7 microns) • 3 wavelength bands(25,000 – 30,000 GHz wide) are used for optical communication. • They are centered at .85, 1.30, 1.55 microns • Last 2 haveattenuation property less than 5% loss per km (kilometer) • .85 microns band has higher attenuation but at that wavelength the laser and electronics can be made the samematerial (gallium arsenide) • Attenuation = 10 (R / T) [decibels] Where: T = transmitted power, R = received power e.g. a factor of 2 loss gives an attenuation of A = 10 log100.5 = -3 dB

Figure 3.28Distortion Distortionmeansthesignalchanges its formorshape. Each signal has its own propagation speed through a medium, so its own delay

Figure 3.29Noise is another cause of impairment • Noise types: • thermal • the random motion of electrons • induced noise • comes from e.g. motor and appliances • crosstalk • one wire on the other • impulse noise • comes from a signal with high energy in very short period of time: e.g. power lines, lightning • can corrupt the signal.

Example 3.31 The power of a signal is 10 mW (miliwatt) and the power of the noise is 1 μW; what are the values of SNR (signal-power-to-noise-powerratio) and SNRdB? Solution The values of SNR and SNRdB can be calculated as follows: μW

Example 3.32 The values of SNR and SNRdB for a noiseless channel are infinite We can never achieve this ratio in real life; it is an ideal.

Figure 3.30Two cases of SNR: a high SNR and a low SNR

3.5 Data Rate Limits NoiselessChannel: Nyquist Bit Rate NoisyChannel: Shannon Capacity Using Both Limits

3.5 Data Rate Limits A very important consideration in data communications is how fast we can send data, inbits per second, over a channel. Data Rate depends on 3 factors: The bandwidth available The levels of signals we can use The quality (level of the noise) of the channel

Noiseless Channel: Nyquist Bit Rate Henry Nyquist proved that if an arbitrary signal has been run through a low-pass filterof bandwidth B, the filtered signal can be completely reconstructed by making only (at least) 2B [samples/sec] (data encoded rate) Nyquist theoretical maximum bit rate: NyquistBitRate [bps] = 2 B log2L = sampling rate [sample/sec]x log2L[#bits/sample] = 2B [1/sec] x log2L [bit] = 2Blog2L [bit/sec] • B – bandwidth • L - # of signal levels • Each signal may convey several bits – • for example if 8 voltages are possible per signal, then 3 bits are sent on every signal (per sample). • If the signal is BINARY (only two voltage levels), then the bit rate is equal to 2B (or the baud rate). • e.g. a noiselessB = 3-kHz channel cannot transmit binary (i.e. two-level) signals at a bit rateexceeding 2B = 6000 bps.

Note • Increasing the levels of a signal L may reduce the reliability of the system • b/c it is very unreliable to distinguish signals for high value of L e.g. 64 by the receiver

Example 3.33 • Does the Nyquist theorem bit rate agree with the intuitivebit rate described in baseband transmission? • Solution • They match when we have only two levels. We said, in baseband transmission, the • bit rate = 2 x B • if we use only the first harmonic in the worst case. • However, the Nyquist formula is more general than what we derived intuitively; • it can be applied • to baseband transmission and modulation • it can be applied when we have two or more levels of signals.

Example 3.34 Consider a noiseless channel with a bandwidthof3000 Hz transmitting a signal with two signal levels. The maximumbit rate can be calculated as BitRate [bps] = 2 B log2L BitRate = 2  3000[1/sec] log2 2 [bit] = 6000 [bps]

Example 3.35 Consider the same noiseless channel, transmitting a signal with 4 signal levels (for each level, we send 2 bits). The maximum bit rate can be calculated as: BitRate [bps] = 2 B log2L Bit Rate = 2 x 3000 [1/sec] x log2 4 [bit]= 12,000 [bps]

Example 3.36 We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need? Solution We can use the Nyquist formula as shown: • Since this result is nota power of 2, we need to • either increase the number of levels or • reduce the bit rate. • If we have 128 levels, the bit rate is 280 kbps. • If we have 64 levels, the bit rate is 240 kbps.

Noisy Channel: Shannon Capacity • Theory: A perfect (noiseless) channel will still have a finite transmissioncapacity. • Introducing noise (thermal or white noise) into a channel will further reduce the capacity of that channel. • Theoreticalmaximum data ratefor a noisy channel =Capacity C SNR - signal-to-noise ratio • the formula defines characteristic of the channel, not the method of transmission b/c there is no indication of the signal level • E.g. a channel of B =3000 Hz bandwidth with the a signal to thermal noise ratio of 3162 dB (typical for analog tel. system) can never transmit much more than 30,000 bps • No matter how many or how few signal levels are used and • No matter how often or how infrequently samples are taken C [bps]= B log2 (1 + SNR)

Example 3.37 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 [bps]= B log2 (1 + SNR) =B log2 (1 + 0)= B log2 (1) = B  0= 0

Example 3.38 We can calculate the theoretical highestbit 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 3162. For this channel the capacity is calculated as C = B log2 (1 + SNR) = 3000 [1/s] log2 (1 + 3162) [bit]= 3000 log2 (3163) [bit/s] C = 3000  11.62 = 34,860 bps This is the highest bit rate. If we want to send faster than this, we can either increase B or SNR

Example 3.39 The signal-to-noise ratio is often given in decibels. Assume that SNRdB = 36 and the channel bandwidth is 2 MHz. The theoretical channel capacity can be calculated as

Example 3.40 C = B log2 (1 + SNR) For practical purposes, when the SNR is very high, we can assume that SNR + 1 is almost the same as SNR. In these cases, the theoretical channel capacity can be simplified to For example, we can calculate the theoretical capacity of the previous example as

Example 3.41 Using both formulas • We have a channel with a 1 MHz bandwidth. The SNRfor this channel is 63; • what is the appropriate maxbit rate and signal level? Solution First, we use the Shannon formula to find our upper limit. C = B log2 (1 + SNR) = 106[1/s] log2 (1 + 63) [bit] = 106 log2 (64) [bit/s] = 6 Mbps Then we use the Nyquist formula to find the number of signal levels. BitRate [bps] = 2 B log2L For better performance we choose 4 Mbps, the closest power of 2 but less than 6 Mbps (see C) 4 [Mbps]= 2  1 [MHz] log2L[bit] L = 4

Note • The Shannon capacity gives us the upper limit of data rate; • the Nyquist formula tells us how many signal levels we need. C [bps]= B log2 (1 + SNR) BitRate [bps] = 2 B log2L

3-6 PERFORMANCE • One important issue in networking is the performance of the network—how good is it? • We discuss quality of service, an overall measurement of network performance, in greater detail in Chapter 24. • In this section, we introduce terms that we need for future chapters. Topics discussed in this section: BandwidthThroughputLatency (Delay) Bandwidth-Delay Product

Note • In networking, we use the term bandwidth in two contexts. • ❏The first, bandwidth in hertz, refers to • the range of frequencies in a composite signal or • the range of frequenciesthat a channel can pass. • ❏ The second, bandwidth in bitsper second, refers to the speed of bit transmission in a channel or link.

Note An increase inbandwidth in hertz means an increase in bandwidth in bits/second

Example 3.42 The bandwidth of a subscriber line is 4 kHz for voice or data. The bandwidth of this line for data transmission can be up to 56,000 bps(b/c of 2 B log2L) using a sophisticated modem to change the digital signal to analog. Example 3.43 If the telephone company improves the quality of the line and increases the bandwidth to 8 kHz, we can send 112,000 bps by using the same technology as mentioned in Example 3.42.

Throughput 1. Throughput the measurement of how fast data can pass through an entity The bandwidth is a potential measurement of a link, but throughput is an actual measurement.

Example 3.44 A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network? Solution We can calculate the throughput as The throughput is almost one-fifth of the bandwidth in this case.

Propagation 2. Propagationtime measures the distance a signal can travel through a mediumin 1 sec In a vacuum, light is propagated with a speed 3 x 10 ^ 8 m/s = 300,000 km/s Distance = Propagation speed x Propagation time D = S x T

Example 3.45 What is the propagation time if the distance between the two points is 12,000 km? Assume the propagation speed to be 2.4 × 10^8 m/s in cable. Solution We can calculate the propagation time as Distance = Propagation speed x Propagation time The example shows that a bit can go over the Atlantic Ocean in only 50 ms if there is a direct cable between the source and the destination.

Transmissiontime • 3. Transmissiontime measures the time required • for transmission of a message • A time between the first bit leaving the sender and • the last bit arriving at the receiver. Transmission time = Message size / Bandwidth or Message size = Bandwidth x Transmission time In a vacuum, light is propagated with a speed 3 x 10 ^ 8 m/s = 300,000 km/s

Example 3.46 • What are the propagation time and the transmission time for a 2.5-kbyte message (an e-mail) if the bandwidth of the network is 1 Gbps? • Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 10^8 m/s. • Solution • We can calculate the propagation and transmission time

Example 3.46 (continued) • Note that in this case, because the message is short and the bandwidth is high, the dominant factor is the propagation time, not the transmission time. • The transmission time can be ignored.

Example 3.47 • What are the propagation time and the transmission time for a 5-Mbyte message (an image) if the bandwidth of the network is 1 Mbps? • Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s. • Solution

Example 3.47 (continued) Note that in this case, because the message is very long and the bandwidth is not very high, the dominant factor is the transmission time, not the propagation time. The propagation time can be ignored.

4. Delay (Latency) Delaydefines how long it takes for an entire msg to completely arrive at the destination from the time the 1st bit is sent out from the source. 4. Delay = Propagation time + Transmission time + Queuing time + Processing delay Queuing time – the time needed to hold the message before processing

Figure 3.31Filling the link with bits for case 1 Delaydefines how long it takes for an entire msg to completely arrive at the destination from the time the 1st bit is sent out from the source. Product 1 x 5 is the maximum bits can fill the link.

Figure 3.32Filling the link with bits in case 2 5 5 p.93 correct it

Note 5. The bandwidth-delay product defines the number of bits that can fill the link.

Concept of bandwidth-delay product We can think about the link between two points as a pipe. The cross section of the piperepresents the bandwidth, and the length of the pipe represents the delay. We can say the volume of the pipe defines the bandwidth-delay product, as shown in Figure 3.33.

expanded by Jozef Goetz, 2009 The McGraw-Hill Companies, Inc., 2006