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Komunikasi Data
Erfan Rusdi S.Kom
A. Komunikasi Data
Istilah telekomunikasi berarti komunikasi dengan jarak. Kata data mengacu pada informasi yang ditampilkan apapun bentuk yang disetujui oleh pihak yg membuat dan menggunakan. Komunikasi data adalah pertukaran data diantara dua perangkat melalui berbagi bentuk media transmisi seperti kabel.
Topik yg didiskusikan pada bab ini:
- Komponen
- Representasi Data
- Aliran Data
Figure 1.1 Lima komponen dari komunikasi data
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Figure 1.2 Aliran Data flow (simplex, half-duplex, and full-duplex)
B. Jaringan
Sebuah jaringan adalah kumpulan perangkat (sering disebut sebagai nodes) yg dihubungkan oleh jalur komunikasi. Sebuah node dapat berupa komputer, printer, atau perangkat apapun yg mampu mengirim dan atau menerima data yang dibuat oleh node lainnya pada jaringan.
Topik yg didiskusikan pada bab ini:
- Proses terdistribusi
- Kriteria Jaringan
- Struktur Fisik
- Model Jaringan
- Jenis – jenis Jaringan
- Interkoneksi pada Jaringan : Internetwork
Figure 1.3 Tipe dari koneksi: point-to-point and multipoint
Figure 1.4 Jenis – jenis topologi
Figure 1.5 Topologi mesh (5 perangkat)
Figure 1.6 Topologi star
Figure 1.7 Topologi bus
Figure 1.8 Topologi ring
Figure 1.9 Topologi hybrid
Figure 1.10 Contoh LAN
Figure 1.11 WAN
Figure 1.12 Jaringan WAN dan LAN
C. Internet
Internet telah merevolusi berbagai aspek dlm hidup kita. Internet adalah sebuah sistem komunikasi yang membawa informasi ke ujung jari kita dan diorganisasikan untuk kita gunakan.
Topik yg didiskusikan pada bab ini:
- Sejarah singkat
- Internet Hari ini
Figure 1.13 Struktur Internet
D. Protocol dan Standard
Pada bagian ini, kita akan menjelaskan dua hal yang banyak digunakan : protocol dan standard. Pertama, kita jelaskan protocol, dimana sinonim dengan aturan. Kemudian kita diskusikan standard, dimana sinonim dengan aturan yang disepakati.
Topik yg didiskusikan pada bab ini:
- Protocol
- Standard
- Orgranisasi Standards
- Standar Internet
A. Tugas terlapis
Kita menggunakan konsep lapisan pada kehidupan kita sehari2. Sebagai contoh, mari kita lihat dua orang teman yang berkomunikasi melalui surat pos. Proses untuk mengirimkan surat ke teman akan menjadi komplek jika tidak ada layanan yg tersedia dari kantor pos.
Topik yg didiskusikan di bagian ini:
1.Pengirim, penerima dan pembawa Hirarki
Figure 2.1 Pekerjaan yg terlibat dlm mengirimkan surat
B. Osi Model
Berdiri pada thn 1947, the International Standards Organization (ISO) adalah badan multinational yg mengkhususkan pada persetujuan seluruh dunia pada standar internasional. Sebuah standar ISO yg membahas semua aspek pada jaringan data adalah Open Systems Interconnection (OSI) model. Model ini pertama kali dikenalkan pada akhir 1970.
Topik yg didiskusikan di bagian ini:
- ArsitekturBerlapis
- Proses Peer-to-Peer
- Encapsulation
Figure 2.2 7 lapisan OSI Model
Figure 2.3 Interaksi diantara lapisan pada OSI model
Figure 2.4 An exchange using the OSI model
C. Lapisan pada OSI Model
Pada bagian ini kita akan menjelaskan secara singkat fungsi setiap layer pada OSI model.
Topik yg didiskusikan di bagian ini:
- Physical Layer
- Data Link Layer
- Network Layer
- Transport Layer
- Session Layer
- Presentation Layer
- Application Layer
Figure 2.5 Physical layer
Physical layer bertanggung jawab untuk memindahkan bit – bit dari satu hop (node) ke hop berikutnya.
Figure 2.6 Data link layer
Data link layer bertanggung jawab untuk memindahkan frame dari satu hop (node) ke hop berikutnya.
Figure 2.7 Pengantaran dari hop-ke-hop
Figure 2.8 Network layer
Network layer bertanggung jawab untuk mengantarkan paket dari host sumber ke host tujuan.
Figure 2.9 Pengantaran dari sumber-ke-tujuan
Figure 2.10 Transport layer
Transport layer bertanggung jawa untuk mengantarkan sebuah pesan dari satu proses ke proses lainnya.
Figure 2.11 Pengantaran sebuah pesan dari process-ke-process yg tangguh
Figure 2.12 Session layer
Session layer bertanggung jawab untuk kontrol dialog dan sinkronisasi.
Figure 2.13 Presentation layer
Presentation layer bertanggung jawab untuk pengkodean, kompresi dan penyandian.
Figure 2.14 Application layer
Application layer bertanggung jawab untuk menyediakan layanan ke pengguna.
Figure 2.15 Kesimpulan dari masing2 layer
D. TCP/IP Protocol Suite
Layer2 pada TCP/IP protocol suite tidak sama persis dengan OSI model. TCP/IP protocol suite yg asli telah digambarkan memiliki 4 layer : host-to-network (network access), internet, transport, and application.
Topik yg didiskusikan di bagian ini:
- Network Access Layer
- Internet Layer
- Transport Layer
- Application Layer
Figure 2.16 TCP/IP and OSI model
E. Pengalamatan
4 tingkatan pada pengalamatn digunakan pada internet menggunakan protokol TCP/IP : physical, logical, port, and specific.
Topik yg didiskusikan di bagian ini:
- Physical Addresses
- Logical Addresses
- Port Addresses
- Specific Addresses
Figure 2.17 Pengalamatan pada TCP/IP
Figure 2.18 Relationship of layers and addresses in TCP/IP
Figure 2.19 Physical addresses
Figure 2.20 IP addresses
Physical addresses akan berubah dari satu hop ke hop lainnya, Tetapi logical address biasanya tetap sama
Figure 2.21 Port addresses
A. Analog dan Digital
Data dapat berupa analog atau digital. Istilah analog data menunjuk pada informasi yg terus menerus; digital data menunjuk pada informasi yg memiliki keadaan diskrit. Analog data memiliki nilai kontinu. Digital data memiliki nilai diskrit.
Topics discussed in this section:
- Analog and Digital Data
- Analog and Digital Signals
- Periodic and Nonperiodic Signals
Signals dapat berupa analog atau digital. Sinyal analog dapat memiliki range jumlah nilai yg tak terhingga; sinyal digital signal hanya dapat memiliki jumlah nilai terbatas.
Figure 3.1 Comparison of analog and digital signals
In data communications, we commonly use periodic analog signals and nonperiodic digital signals.
B. Periodic Analog Signals
Periodic analog signals can be classified as simple or composite. A simple periodic analog signal, a sine wave, cannot be decomposed into simpler signals. A composite periodic analog signal is composed of multiple sine waves.
Topics discussed in this section
- Sine Wave
- Wavelength
- Time and Frequency Domain
- Composite Signals
- Bandwidth
Figure 3.2 A sine wave
We discuss a mathematical approach to sine waves in Appendix C.
Example 3.1
The power in your house can be represented by a sine wave with a peak amplitude of 155 to 170 V. However, it is common knowledge that the voltage of the power in U.S. homes is 110 to 120 V. This discrepancy is due to the fact that these are root mean square (rms) values. The signal is squared and then the average amplitude is calculated. The peak value is equal to 2½ × rms value.
Figure 3.3 Two signals with the same phase and frequency, but different amplitudes
Example 3.2
The voltage of a battery is a constant; this constant value can be considered a sine wave, as we will see later. For example, the peak value of an AA battery is normally1.5 V.
Frequency and period are the inverse of each other.
Figure 3.4 Two signals with the same amplitude and phase, but different frequencies
Table 3.1 Units of period and frequency
Example 3.3
The power we use at home has a frequency of 60 Hz. The period of this sine wave can be determined as follows:
Example 3.4
From Table 3.1 we find the equivalents of 1 ms (1 ms is 10−3 s) and 1 s (1 s is 106 μs). We make the following substitutions:.
Example 3.5
The period of a signal is 100 ms. What is its frequency in kilohertz?
First we change 100 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10−3 kHz).
- Frequency is the rate of change with respect to time.
- Change in a short span of time means high frequency.
- Change over a long span of time means low frequency.
- If a signal does not change at all, its frequency is zero.
- If a signal changes instantaneously, its frequency is infinite.
- Phase describes the position of the waveform relative to time 0.
Figure 3.5 Three sine waves with the same amplitude and frequency, but different phases
Example 3.6
A sine wave is offset 1/6 cycle with respect to time 0. What is its phase in degrees and radians?
We know that 1 complete cycle is 360°. Therefore, 1/6 cycle is
Figure 3.6 Wavelength and period
Figure 3.7 The time-domain and frequency-domain plots of a sine wave
A complete sine wave in the time domain can be represented by one single spike in the frequency domain.
Example 3.7
The frequency domain is more compact and useful when we are dealing with more than one sine wave. For example, Figure 3.8 shows three sine waves, each with different amplitude and frequency. All can be represented by three spikes in the frequency domain.
Figure 3.8 The time domain and frequency domain of three sine waves
- A single-frequency sine wave is not useful in data communications; we need to send a composite signal, a signal made of many simple sine waves.
- According to Fourier analysis, any composite signal is a combination of simple sine waves with different frequencies, amplitudes, and phases.
- Fourier analysis is discussed in Appendix C.
- If the composite signal is periodic, the decomposition gives a series of signals with discrete frequencies;
- if the composite signal is nonperiodic, the decomposition gives a combination of sine waves with continuous frequencies.
Example 3.8
Figure 3.9 shows a periodic composite signal with frequency f. This type of signal is not typical of those found in data communications. We can consider it to be three alarm systems, each with a different frequency. The analysis of this signal can give us a good understanding of how to decompose signals.
Figure 3.9 A composite periodic signal
Figure 3.10 Decomposition of a composite periodic signal in the time and frequency domains
Example 3.9
Figure 3.11 shows a nonperiodic composite signal. It can be the signal created by a microphone or a telephone set when a word or two is pronounced. In this case, the composite signal cannot be periodic, because that implies that we are repeating the same word or words with exactly the same tone.
Figure 3.11 The time and frequency domains of a nonperiodic signal
The bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal.
Figure 3.12 The bandwidth of periodic and nonperiodic composite signals
Example 3.10
If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then
The spectrum has only five spikes, at 100, 300, 500, 700, and 900 Hz (see Figure 3.13).
Figure 3.13 The bandwidth for Example 3.10
Example 3.11
A periodic 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 frequencies of the same amplitude.
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then
The spectrum contains all integer frequencies. We show this by a series of spikes (see Figure 3.14).
Figure 3.14 The bandwidth for Example 3.11
Example 3.12
A nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 V. The two extreme frequencies have an amplitude of 0. Draw the frequency domain of the signal.
The lowest frequency must be at 40 kHz and the highest at 240 kHz. Figure 3.15 shows the frequency domain and the bandwidth.
Figure 3.15 The bandwidth for Example 3.12
Example 3.13
An example of a nonperiodic composite signal is the signal propagated by an AM radio station. In the United States, each AM radio station is assigned a 10-kHz bandwidth. The total bandwidth dedicated to AM radio ranges from 530 to 1700 kHz. We will show the rationale behind this 10-kHz bandwidth in Chapter 5.
Example 3.14
Another example of a nonperiodic composite signal is the signal propagated by an FM radio station. In the United States, each FM radio station is assigned a 200-kHz bandwidth. The total bandwidth dedicated to FM radio ranges from 88 to 108 MHz. We will show the rationale behind this 200-kHz bandwidth in Chapter 5.
Example 3.15
Another example of a nonperiodic composite signal is the signal received by an old-fashioned analog black-and-white TV. A TV screen is made up of pixels. If we assume a resolution of 525 × 700, we have 367,500 pixels per screen. If we scan the screen 30 times per second, this is 367,500 × 30 = 11,025,000 pixels per second. The worst-case scenario is alternating black and white pixels. We can send 2 pixels per cycle. Therefore, we need 11,025,000 / 2 = 5,512,500 cycles per second, or Hz. The bandwidth needed is 5.5125 MHz.
C. Digital Signals
In addition to being represented by an analog signal, information can also be represented by a digital signal. For example, a 1 can be encoded as a positive voltage and a 0 as zero voltage. A digital signal can have more than two levels. In this case, we can send more than 1 bit for each level.
Topics discussed in this section:
- Bit Rate
- Bit Length
- Digital Signal as a Composite Analog Signal
- Application Layer
Figure 3.16 Two digital signals: one with two signal levels and the other with four signal levels
Appendix C reviews information about exponential and logarithmic functions.
Example 3.16
A digital signal has eight levels. How many bits are needed per level? We calculate the number of bits from the formula
Each signal level is represented by 3 bits.
Example 3.17
A digital signal has nine levels. How many bits are needed per level? We calculate the number of bits by using the formula. Each signal level is represented by 3.17 bits. However, this answer is not realistic. The number of bits sent per level needs to be an integer as well as a power of 2. For this example, 4 bits can represent one level.
Example 3.18
Assume we need to download text documents at the rate of 100 pages per minute. What is the required bit rate of the channel?
A page is an average of 24 lines with 80 characters in each line. If we assume that one character requires 8 bits, the bit rate is
Example 3.19
A digitized voice channel, as we will see in Chapter 4, is made by digitizing a 4-kHz bandwidth analog voice signal. We need to sample the signal at twice the highest frequency (two samples per hertz). We assume that each sample requires 8 bits. What is the required bit rate?
The bit rate can be calculated as
Example 3.20
What is the bit rate for high-definition TV (HDTV)?
HDTV uses digital signals to broadcast high quality video signals. The HDTV screen is normally a ratio of 16 : 9. There are 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second. Twenty-four bits represents one color pixel.
The TV stations reduce this rate to 20 to 40 Mbps through compression.
Figure 3.17 The time and frequency domains of periodic and nonperiodic digital signals
Figure 3.18 Baseband transmission
A digital signal is a composite analog signal with an infinite bandwidth.
Figure 3.19 Bandwidths of two low-pass channels
Figure 3.20 Baseband transmission using a dedicated medium
Baseband transmission of a digital signal that preserves the shape of the digital signal is possible only if we have a low-pass channel with an infinite or very wide bandwidth.
Example 3.21
An example of a dedicated channel where the entire bandwidth of the medium is used as one single channel is a LAN. Almost every wired LAN today uses a dedicated channel for two stations communicating with each other. In a bus topology LAN with multipoint connections, only two stations can communicate with each other at each moment in time (timesharing); the other stations need to refrain from sending data. In a star topology LAN, the entire channel between each station and the hub is used for communication between these two entities. We study LANs in Chapter 14.
Figure 3.21 Rough approximation of a digital signal using the first harmonic for worst case
Figure 3.22 Simulating a digital signal with first three harmonics
In baseband transmission, the required bandwidth is proportional to the bit rate;if we need to send bits faster, we need more bandwidth.
Table 3.2 Bandwidth requirements
Example 3.22
What is the required bandwidth of a low-pass channel if we need to send 1 Mbps by using baseband transmission?
The answer depends on the accuracy desired.
a. The minimum bandwidth, is B = bit rate /2, or 500 kHz.
b. A better solution is to use the first and the third harmonics with B = 3 × 500 kHz = 1.5 MHz.
c. Still a better solution is to use the first, third, and fifth harmonics with B = 5 × 500 kHz = 2.5 MHz.
Example 3.22
We have a low-pass channel with bandwidth 100 kHz. What is the maximum bit rate of this channel?
The maximum bit rate can be achieved if we use the first harmonic. The bit rate is 2 times the available bandwidth, or 200 kbps.
Figure 3.23 Bandwidth of a bandpass channel
If the available channel is a bandpass channel, we cannot send the digital signal directly to the channel; we need to convert the digital signal to an analog signal before transmission.
Figure 3.24 Modulation of a digital signal for transmission on a bandpass channel
Example 3.24
An example of broadband transmission using modulation is the sending of computer data through a telephone subscriber line, the line connecting a resident to the central telephone office. These lines are designed to carry voice with a limited bandwidth. The channel is considered a bandpass channel. We convert the digital signal from the computer to an analog signal, and send the analog signal. We can install two converters to change the digital signal to analog and vice versa at the receiving end. The converter, in this case, is called a modem which we discuss in detail in Chapter 5.
Example 3.25
A second example is the digital cellular telephone. For better reception, digital cellular phones convert the analog voice signal to a digital signal (see Chapter 16). Although the bandwidth allocated to a company providing digital cellular phone service is very wide, we still cannot send the digital signal without conversion. The reason is that we only have a bandpass channel available between caller and callee. We need to convert the digitized voice to a composite analog signal before sending.
D. Transmission Impairment
Signals travel through transmission media, which are not perfect. The imperfection causes signal impairment. This means that the signal at the beginning of the medium is not the same as the signal at the end of the medium. What is sent is not what is received. Three causes of impairment are attenuation, distortion, and noise.
Topics discussed in this section:
- Attenuation
- Distortion
- Noise
Figure 3.25 Causes of impairment
Figure 3.26 Attenuation
Example 3.26
Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as
A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.
Example 3.27
A signal travels through an amplifier, and its power is increased 10 times. This means that P2 = 10P1 . In this case, the amplification (gain of power) can be calculated as
Example 3.28
One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are measuring several points (cascading) instead of just two. In Figure 3.27 a signal travels from point 1 to point 4. In this case, the decibel value can be calculated as
Figure 3.27 Decibels for Example 3.28
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 log10 Pm , where Pm is the power in milliwatts. Calculate the power of a signal with dBm = −30.
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?
The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power as
Figure 3.28 Distortion
Figure 3.29 Noise
Example 3.31
The power of a signal is 10 mW and the power of the noise is 1 μW; what are the values of SNR and SNRdB ?
The values of SNR and SNRdB can be calculated as follows:
Example 3.32
The values of SNR and SNRdB for a noiseless channel are
We can never achieve this ratio in real life; it is an ideal.
Figure 3.30 Two cases of SNR: a high SNR and a low SNR
E. Data Rate Limits
A very important consideration in data communications is how fast we can send data, in bits per second, over a channel. Data rate depends on three factors:
1. The bandwidth available
2. The level of the signals we use
3. The quality of the channel (the level of noise)
Topics discussed in this section:
- Noiseless Channel: Nyquist Bit Rate
- Noisy Channel: Shannon Capacity
- Using Both Limits
Increasing the levels of a signal may reduce the reliability of the system.
Example 3.33
Does the Nyquist theorem bit rate agree with the intuitive bit rate described in baseband transmission?
They match when we have only two levels. We said, in baseband transmission, the bit rate is 2 times the bandwidth 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. Also, it can be applied when we have two or more levels of signals.
Example 3.34
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
Example 3.35
Consider the same noiseless channel transmitting a signal with four signal levels (for each level, we send 2 bits). The maximum bit rate can be calculated as
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?
We can use the Nyquist formula as shown:
Since this result is not a 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.
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 C is calculated as
This means that the capacity of this channel is zero regardless of the bandwidth. In other words, we cannot receive any data through this channel.
Example 3.38
We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000. The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as
This means that the highest bit rate for a telephone line is 34.860 kbps. If we want to send data faster than this, we can either increase the bandwidth of the line or improve the signal-to-noise ratio.
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
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
We have a channel with a 1-MHz bandwidth. The SNR for this channel is 63. What are the appropriate bit rate and signal level?
First, we use the Shannon formula to find the upper limit.
Example 3.41 (continued)
The Shannon formula gives us 6 Mbps, the upper limit. For better performance we choose something lower, 4 Mbps, for example. Then we use the Nyquist formula to find the number of signal levels.
The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal levels we need.
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:
- Bandwidth
- Throughput
- Latency (Delay)
- Bandwidth-Delay Product
In networking, we use the term bandwidth in two contexts.
Ø The first, bandwidth in hertz, refers to the range of frequencies in a composite signal or the range of frequencies that a channel can pass.
Ø The second, bandwidth in bits persecond, refers to the speed of bit transmission in a channel or link.
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 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.
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?
We can calculate the throughput as
The throughput is almost one-fifth of the bandwidth in this case.
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 × 108 m/s in cable.
We can calculate the propagation time as
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.
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 × 108 m/s.
We can calculate the propagation and transmission time as shown on the next slide:
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.
We can calculate the propagation and transmission times as shown on the next slide.
Example 3.47 (continued)
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.
Figure 3.31 Filling the link with bits for case 1
Example 3.48
We can think about the link between two points as a pipe. The cross section of the pipe represents 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.
Figure 3.32 Filling the link with bits in case 2
The bandwidth-delay product defines the number of bits that can fill the link.
Figure 3.33 Concept of bandwidth-delay product
A. Digital-To-Analog Conversion
Digital-to-analog conversion is the process of changing one of the characteristics of an analog signal based on the information in digital data.
Topics discussed in this section:
- Aspects of Digital-to-Analog Conversion
- Amplitude Shift Keying
- Frequency Shift Keying
- Phase Shift Keying
- Quadrature Amplitude Modulation
Figure 5.1 Digital-to-analog conversion
Figure 5.2 Types of digital-to-analog conversion
Bit rate is the number of bits per second. Baud rate is the number of signalelements per second.
In the analog transmission of digital data, the baud rate is less than or equal to the bit rate.
Example 5.1
An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate.
In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from
Example 5.2
An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do we need?
In this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L.
Figure 5.3 Binary amplitude shift keying
Figure 5.4 Implementation of binary ASK
Example 5.3
We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 1?
The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at fc = 250 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1).
Example 5.4
In data communications, we normally use full-duplex links with communication in both directions. We need to divide the bandwidth into two with two carrier frequencies, as shown in Figure 5.5. The figure shows the positions of two carrier frequencies and the bandwidths. The available bandwidth for each direction is now 50 kHz, which leaves us with a data rate of 25 kbps in each direction.
Figure 5.5 Bandwidth of full-duplex ASK used in Example 5.4
Figure 5.6 Binary frequency shift keying
Example 5.5
We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What should be the carrier frequency and the bit rate if we modulated our data by using FSK with d = 1?
This problem is similar to Example 5.3, but we are modulating by using FSK. The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means
Figure 5.7 Bandwidth of MFSK used in Example 5.6
Example 5.6
We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwidth.
We can have L = 23 = 8. The baud rate is S = 3 MHz/3 = 1000 Mbaud. This means that the carrier frequencies must be 1 MHz apart (2Δf = 1 MHz). The bandwidth is B = 8 × 1000 = 8000. Figure 5.8 shows the allocation of frequencies and bandwidth.
Figure 5.8 Bandwidth of MFSK used in Example 5.6
Figure 5.9 Binary phase shift keying
Figure 5.10 Implementation of BASK
Figure 5.11 QPSK and its implementation
Example 5.7
Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0.
For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz.
Figure 5.12 Concept of a constellation diagram
Example 5.8
Show the constellation diagrams for an ASK (OOK), BPSK, and QPSK signals.
Figure 5.13 shows the three constellation diagrams.
Figure 5.13 Three constellation diagrams
Quadrature amplitude modulation is a combination of ASK and PSK.
Figure 5.14 Constellation diagrams for some QAMs
B. Analog and Digital
Analog-to-analog conversion is the representation of analog information by an analog signal. One may ask why we need to modulate an analog signal; it is already analog. Modulation is needed if the medium is bandpass in nature or if only a bandpass channel is available to us.
Topics discussed in this section:
- Amplitude Modulation
- Frequency Modulation
- Phase Modulation
Figure 5.15 Types of analog-to-analog modulation
Figure 5.16 Amplitude modulation
The total bandwidth required for AMcan be determinedfrom the bandwidth of the audio
signal: BAM = 2B.
Figure 5.17 AM band allocation
The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BFM = 2(1 + β)B.
Figure 5.18 Frequency modulation
Figure 5.19 FM band allocation
Figure 5.20 Phase modulation
The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal:BPM = 2(1 + β)B.
Figure 7.1 Transmission medium and physical layer
Figure 7.2 Classes of transmission media
A. Guided Media
Guided media, which are those that provide a conduit from one device to another, include twisted-pair cable, coaxial cable, and fiber-optic cable.
Topics discussed in this section:
- Twisted-Pair Cable
- Coaxial Cable
- Fiber-Optic Cable
Figure 7.3 Twisted-pair cable
Figure 7.4 UTP and STP cables
Table 7.1 Categories of unshielded twisted-pair cables
Figure 7.5 UTP connector
Figure 7.6 UTP performance
Figure 7.7 Coaxial cable
Table 7.2 Categories of coaxial cables
Figure 7.8 BNC connectors
Figure 7.9 Coaxial cable performance
Figure 7.10 Bending of light ray
Figure 7.11 Optical fiber
Figure 7.12 Propagation modes
Figure 7.13 Modes
Table 7.3 Fiber types
Figure 7.14 Fiber construction
Figure 7.15 Fiber-optic cable connectors
Figure 7.16 Optical fiber performance
B. Unguided Media: Wireless
Unguided media transport electromagnetic waves without using a physical conductor. This type of communication is often referred to as wireless communication.
Topics discussed in this section:
1. Radio Waves
2. Microwaves
3. Infrared
Figure 7.17 Electromagnetic spectrum for wireless communication
Figure 7.18 Propagation methods
Table 7.4 Bands
Figure 7.19 Wireless transmission waves
Figure 7.20 Omnidirectional antenna
Radio waves are used for multicast communications, such as radio and television, and paging systems.
Figure 7.21 Unidirectional antennas
Microwaves are used for unicast communication such as cellular telephones, satellite networks,
and wireless LANs.
Infrared signals can be used for short-range communication in a closed area using line-of-sight propagation.
Figure 8.1 Switched network
Figure 8.2 Taxonomy of switched networks
A. Circuit-Switched Network
A circuit-switched network consists of a set of switches connected by physical links. A connection between two stations is a dedicated path made of one or more links. However, each connection uses only one dedicated channel on each link. Each link is normally divided into n channels by using FDM or TDM.
Topics discussed in this section:
- Three Phases
- Efficiency
- Delay
- Circuit-Switched Technology in Telephone Networks
A circuit-switched network is made of a set of switches connected by physical links, in which each link is divided into n channels.
Figure 8.3 A trivial circuit-switched network
In circuit switching, the resources need to be reserved during the setup phase;
the resources remain dedicated for the entire duration of data transfer until the teardown phase.
Example 8.1
As a trivial example, let us use a circuit-switched network to connect eight telephones in a small area. Communication is through 4-kHz voice channels. We assume that each link uses FDM to connect a maximum of two voice channels. The bandwidth of each link is then 8 kHz. Figure 8.4 shows the situation. Telephone 1 is connected to telephone 7; 2 to 5; 3 to 8; and 4 to 6. Of course the situation may change when new connections are made. The switch controls the connections.
Figure 8.4 Circuit-switched network used in Example 8.1
Example 8.2
As another example, consider a circuit-switched network that connects computers in two remote offices of a private company. The offices are connected using a T-1 line leased from a communication service provider. There are two 4 × 8 (4 inputs and 8 outputs) switches in this network. For each switch, four output ports are folded into the input ports to allow communication between computers in the same office. Four other output ports allow communication between the two offices. Figure 8.5 shows the situation.
Figure 8.5 Circuit-switched network used in Example 8.2
Figure 8.6 Delay in a circuit-switched network
Switching at the physical layer in the traditional telephone network uses the circuit-switching approach.
B. Datagram Networks
In data communications, we need to send messages from one end system to another. If the message is going to pass through a packet-switched network, it needs to be divided into packets of fixed or variable size. The size of the packet is determined by the network and the governing protocol.
Topics discussed in this section:
- Routing Table
- Efficiency
- Delay
- Datagram Networks in the Internet
In a packet-switched network, there is no resource reservation;resources are allocated on demand.
Figure 8.7 A datagram network with four switches (routers)
Figure 8.8 Routing table in a datagram network
A switch in a datagram network uses a routing table that is based on the destination address.
The destination address in the header of a packet in a datagram network remains the same during the entire journey of the packet.
Figure 8.9 Delay in a datagram network
Switching in the Internet is done by using the datagram approach to packet switching at
the network layer.
C. Virtual-Circuit Networks
A virtual-circuit network is a cross between a circuit-switched network and a datagram network. It has some characteristics of both.
Topics discussed in this section:
- Addressing
- Three Phases
- Efficiency
- Delay
- Circuit-Switched Technology in WANs
Figure 8.10 Virtual-circuit network
Figure 8.11 Virtual-circuit identifier
Figure 8.12 Switch and tables in a virtual-circuit network
Figure 8.13 Source-to-destination data transfer in a virtual-circuit network
Figure 8.14 Setup request in a virtual-circuit network
Figure 8.15 Setup acknowledgment in a virtual-circuit network
In virtual-circuit switching, all packets belonging to the same source and
destination travel the same path; but the packets may arrive at the destination with different delays
if resource allocation is on demand.
Figure 8.16 Delay in a virtual-circuit network
Switching at the data link layer in a switched WAN is normally implemented by using
virtual-circuit techniques.
D. Structure Of A Switch
We use switches in circuit-switched and packet-switched networks. In this section, we discuss the structures of the switches used in each type of network.
Topics discussed in this section:
- Structure of Circuit Switches
- Structure of Packet Switches
Figure 8.17 Crossbar switch with three inputs and four outputs
Figure 8.18 Multistage switch
In a three-stage switch, the total number of crosspoints is 2kN + k(N/n)2 which is much smaller than the number of crosspoints in a single-stage switch (N2).
Example 8.3
Design a three-stage, 200 × 200 switch (N = 200) with k = 4 and n = 20.
In the first stage we have N/n or 10 crossbars, each of size 20 × 4. In the second stage, we have 4 crossbars, each of size 10 × 10. In the third stage, we have 10 crossbars, each of size 4 × 20. The total number of crosspoints is 2kN + k(N/n)2, or 2000 crosspoints. This is 5 percent of the number of crosspoints in a single-stage switch (200 × 200 = 40,000).
According to the Clos criterion:
n = (N/2)1/2
k > 2n – 1
Crosspoints ≥ 4N [(2N)1/2 – 1]
Example 8.4
Redesign the previous three-stage, 200 × 200 switch, using the Clos criteria with a minimum number of crosspoints.
We let n = (200/2)1/2, or n = 10. We calculate k = 2n − 1 = 19. In the first stage, we have 200/10, or 20, crossbars, each with 10 × 19 crosspoints. In the second stage, we have 19 crossbars, each with 10 × 10 crosspoints. In the third stage, we have 20 crossbars each with 19 × 10 crosspoints. The total number of crosspoints is 20(10 × 19) + 19(10 × 10) + 20(19 ×10) = 9500.
Figure 8.19 Time-slot interchange
Figure 8.20 Time-space-time switch
Figure 8.21 Packet switch components
Figure 8.22 Input port
Figure 8.23 Output port
Figure 8.24 A banyan switch
Figure 8.25 Examples of routing in a banyan switch
Figure 8.26 Batcher-banyan switch
Bandwidth utilization is the wise use of available bandwidth to achieve specific goals.
Efficiency can be achieved by multiplexing; privacy and anti-jamming can be achieved by spreading.
A. Multiplexing
Whenever the bandwidth of a medium linking two devices is greater than the bandwidth needs of the devices, the link can be shared. Multiplexing is the set of techniques that allows the simultaneous transmission of multiple signals across a single data link. As data and telecommunications use increases, so does traffic.
Topics discussed in this section:
- Frequency-Division Multiplexing
- Wavelength-Division Multiplexing
- Synchronous Time-Division Multiplexing
- Statistical Time-Division Multiplexing
Figure 6.1 Dividing a link into channels
Figure 6.2 Categories of multiplexing
Figure 6.3 Frequency-division multiplexing
FDM is an analog multiplexing technique that combines analog signals.
Figure 6.4 FDM process
Figure 6.5 FDM demultiplexing example
Example 6.1
Assume that a voice channel occupies a bandwidth of 4 kHz. We need to combine three voice channels into a link with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the configuration, using the frequency domain. Assume there are no guard bands.
We shift (modulate) each of the three voice channels to a different bandwidth, as shown in Figure 6.6. We use the 20- to 24-kHz bandwidth for the first channel, the 24- to 28-kHz bandwidth for the second channel, and the 28- to 32-kHz bandwidth for the third one. Then we combine them as shown in Figure 6.6.
Figure 6.6 Example 6.1
Example 6.2
Five channels, each with a 100-kHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 kHz between the channels to prevent interference?
For five channels, we need at least four guard bands. This means that the required bandwidth is at least 5 × 100 + 4 × 10 = 540 kHz, as shown in Figure 6.7.
Figure 6.7 Example 6.2
Example 6.3
Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration, using FDM.
The satellite channel is analog. We divide it into four channels, each channel having a 250-kHz bandwidth. Each digital channel of 1 Mbps is modulated such that each 4 bits is modulated to 1 Hz. One solution is 16-QAM modulation. Figure 6.8 shows one possible configuration.
Figure 6.8 Example 6.3
Figure 6.9 Analog hierarchy
Example 6.4
The Advanced Mobile Phone System (AMPS) uses two bands. The first band of 824 to 849 MHz is used for sending, and 869 to 894 MHz is used for receiving. Each user has a bandwidth of 30 kHz in each direction. How many people can use their cellular phones simultaneously?
Each band is 25 MHz. If we divide 25 MHz by 30 kHz, we get 833.33. In reality, the band is divided into 832 channels. Of these, 42 channels are used for control, which means only 790 channels are available for cellular phone users.
Figure 6.10 Wavelength-division multiplexing
WDM is an analog multiplexing technique to combine optical signals.
Figure 6.11 Prisms in wavelength-division multiplexing and demultiplexing
Figure 6.12 TDM
TDM is a digital multiplexing technique for combining several low-rate
channels into one high-rate one.
Figure 6.13 Synchronous time–division multiplexing
In synchronous TDM, the data rate of the link is n times faster, and the unit duration is n times shorter.
Example 6.5
In Figure 6.13, the data rate for each input connection is 3 kbps. If 1 bit at a time is multiplexed (a unit is 1 bit), what is the duration of (a) each input slot, (b) each output slot, and (c) each frame?
We can answer the questions as follows:
1. The data rate of each input connection is 1 kbps. This means that the bit duration is 1/1000 s or 1 ms. The duration of the input time slot is 1 ms (same as bit duration).
2. The duration of each output time slot is one-third of the input time slot. This means that the duration of the output time slot is 1/3 ms.
3. Each frame carries three output time slots. So the duration of a frame is 3 × 1/3 ms, or 1 ms. The duration of a frame is the same as the duration of an input unit.
Example 6.6
Figure 6.14 shows synchronous TDM with a data stream for each input and one data stream for the output. The unit of data is 1 bit. Find (a) the input bit duration, (b) the output bit duration, (c) the output bit rate, and (d) the output frame rate.
We can answer the questions as follows:
1. The input bit duration is the inverse of the bit rate 1/1 Mbps = 1 μs.
2. The output bit duration is one-fourth of the input bit duration, or ¼ μs.
3. The output bit rate is the inverse of the output bit duration or 1/(4μs) or 4 Mbps. This can also be deduced from the fact that the output rate is 4 times as fast as any input rate; so the output rate = 4 × 1 Mbps = 4 Mbps.
4. The frame rate is always the same as any input rate. So the frame rate is 1,000,000 frames per second. Because we are sending 4 bits in each frame, we can verify the result of the previous question by multiplying the frame rate by the number of bits per frame.
Figure 6.14 Example 6.6
Example 6.7
Four 1-kbps connections are multiplexed together. A unit is 1 bit. Find (a) the duration of 1 bit before multiplexing, (b) the transmission rate of the link, (c) the duration of a time slot, and (d) the duration of a frame.
We can answer the questions as follows:
1. The duration of 1 bit before multiplexing is 1 / 1 kbps, or 0.001 s (1 ms).
2. The rate of the link is 4 times the rate of a connection, or 4 kbps.
3. The duration of each time slot is one-fourth of the duration of each bit before multiplexing, or 1/4 ms or 250 μs. Note that we can also calculate this from the data rate of the link, 4 kbps. The bit duration is the inverse of the data rate, or 1/4 kbps or 250 μs.
4. The duration of a frame is always the same as the duration of a unit before multiplexing, or 1 ms. We can also calculate this in another way. Each frame in this case has four time slots. So the duration of a frame is 4 times 250 μs, or 1 ms.
Figure 6.15 Interleaving
Example 6.8
Four channels are multiplexed using TDM. If each channel sends 100 bytes /s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bit rate for the link.
The multiplexer is shown in Figure 6.16. Each frame carries 1 byte from each channel; the size of each frame, therefore, is 4 bytes, or 32 bits. Because each channel is sending 100 bytes/s and a frame carries 1 byte from each channel, the frame rate must be 100 frames per second. The bit rate is 100 × 32, or 3200 bps.
Figure 6.16 Example 6.8
Example 6.9
A multiplexer combines four 100-kbps channels using a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate? What is the frame duration? What is the bit rate? What is the bit duration?
Figure 6.17 shows the output for four arbitrary inputs. The link carries 50,000 frames per second. The frame duration is therefore 1/50,000 s or 20 μs. The frame rate is 50,000 frames per second, and each frame carries 8 bits; the bit rate is 50,000 × 8 = 400,000 bits or 400 kbps. The bit duration is 1/400,000 s, or 2.5 μs.
Figure 6.17 Example 6.9
Figure 6.18 Empty slots
Figure 6.19 Multilevel multiplexing
Figure 6.20 Multiple-slot multiplexing
Figure 6.21 Pulse stuffing
Figure 6.22 Framing bits
Example 6.10
We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (a) the data rate of each source, (b) the duration of each character in each source, (c) the frame rate, (d) the duration of each frame, (e) the number of bits in each frame, and (f) the data rate of the link.
We can answer the questions as follows:
1. The data rate of each source is 250 × 8 = 2000 bps = 2 kbps.
2. Each source sends 250 characters per second; therefore, the duration of a character is 1/250 s, or
4 ms.
3. Each frame has one character from each source, which means the link needs to send 250 frames per second to keep the transmission rate of each source.
4. The duration of each frame is 1/250 s, or 4 ms. Note that the duration of each frame is the same as the duration of each character coming from each source.
5. Each frame carries 4 characters and 1 extra synchronizing bit. This means that each frame is
4 × 8 + 1 = 33 bits.
Example 6.11
Two channels, one with a bit rate of 100 kbps and another with a bit rate of 200 kbps, are to be multiplexed. How this can be achieved? What is the frame rate? What is the frame duration? What is the bit rate of the link?
We can allocate one slot to the first channel and two slots to the second channel. Each frame carries 3 bits. The frame rate is 100,000 frames per second because it carries 1 bit from the first channel. The bit rate is 100,000 frames/s × 3 bits per frame, or 300 kbps.
Figure 6.23 Digital hierarchy
Table 6.1 DS and T line rates
Figure 6.24 T-1 line for multiplexing telephone lines
Figure 6.25 T-1 frame structure
Table 6.2 E line rates
Figure 6.26 TDM slot comparison
B. Spread Spectrum
In spread spectrum (SS), we combine signals from different sources to fit into a larger bandwidth, but our goals are to prevent eavesdropping and jamming. To achieve these goals, spread spectrum techniques add redundancy.
Topics discussed in this section:
1. Frequency Hopping Spread Spectrum (FHSS)
2. Direct Sequence Spread Spectrum Synchronous (DSSS)
Figure 6.27 Spread spectrum
Figure 6.28 Frequency hopping spread spectrum (FHSS)
Figure 6.29 Frequency selection in FHSS
Figure 6.30 FHSS cycles
Figure 6.31 Bandwidth sharing
Figure 6.32 DSSS
Figure 6.33 DSSS example


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Diana Butler Bass: Christianity After Religion
Posted by Carol on November 15, 2012 · 4 Comments
As we read about the rise of the spiritual but not religious, how do we respond? Do we think of it as a threat? A challenge? Or do we resonate with the category?
What are the SBNR's telling us about the church? How should the church be responding?
Join us for an insightful conversation with Diana Butler Bass, as she reflects on these questions with us.

subset: commoncrawl


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By Wayne Veysey | Chief Correspondent
Everton have made Roberto Martinez their No.1 target to replace Manchester United-bound David Moyes, Goal.com can reveal.
The Wigan manager is the Merseysiders' standout choice to take over from the Scot, who is poised to bring an end to his 11-year reign at Goodison Park.
Everton have been preparing for the possibility of Moyes's departure at the end of the season for months given his failure to commit to negotiations to extend a contract that expires this summer.
Chairman Bill Kenwright is also an admirer of Porto manager Vitor Pereira, who was a guest of the club at a Premier League match over the Christmas period, but the Toffees will move first for Martinez if, as expected, Moyes takes over from Sir Alex Ferguson at Old Trafford.
|Roberto Martinez is 13/8 with Bet365 to be the next Everton manager
Although Kenwright claimed in an interview in April that no plans are in place for what would be the club's first change of manager in 11 years, it is understood that a shortlist of possible successors has been compiled, with Martinez's name at the top.
The 39-year-old Spaniard is much admired by Everton chiefs for maintaining Wigan's Premier League status since joining the north-west club from Swansea City in the summer of 2009.
He will lead the Latics out in Saturday's FA Cup final against Manchester City on Saturday but the club are on the brink of losing the top-flight spot that they have held for nine seasons following Tuesday's 3-2 home defeat to the Swans.
It was a defeat that leaves Wigan three points adrift of safety with a trip to Arsenal and a home fixture against Aston Villa to come.
Martinez held talks with Liverpool last summer before rejecting the opportunity to succeed Kenny Dalglish after refusing to work under a planned tier of executives with shared control of recruiting players.
He is believed to regard Everton as an attractive post, given the foundations put in place by Moyes in recent years.


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Nobody who lives in the District of Columbia — or follows housing issues — needs to be told that rents are too damn high here. Nor that they consume an inordinate portion of low-income residents' budgets.
A just-released study by the Urban Institute is nonetheless newsworthy because it provides many and diverse figures on our affordable housing situation, along with details on our homeless population and its needs — met and unmet.
The full study covers not only the District, but other jurisdictions in the Washington metro area. So we get comprehensive figures and interesting opportunities for comparisons.
As is always the case, however, the figures for the District understate affordability problems because they're based on the median income for the entire area.
For the 2009-11 period covered by the housing portion of the study, that was $106,100 for a family of four. By way of rough comparison, the median income for four-person D.C. families was $84,400 last year.
But we've got to go with what we've got. So here are a few of the many things one can extract about what the study labels housing security in the District. As you'll see, it might more appropriately be labeled housing insecurity for the lowest-income residents.
Housing Burdens
The Urban Institute, like most analysts, uses the U.S. Department of Housing and Urban Development's affordability measures.
HUD sets 30% of household income as the affordability cut-off. A household that pays more is said to have a housing-cost burden. A household that pays more than half its income has a severe housing-cost burden.
Slightly more than half of all District households were, to some degree, cost-burdened — and 28% severely so. But housing-cost burdens were vastly more common for the District's 63,700 or so extremely low-income households, i.e., those with incomes at or below 30% of the area median.
All but 16% of them paid more than 30% of their income for housing — generally rent, plus basic utilities, though 18% were classified as homeowners.
And nearly two-thirds (66%) had a severe housing-cost burden. This is nearly three times greater than the percent for very low-income households, i.e., those in the next income tier.
Rental Housing Availability
The rental housing market was — and still is — extremely tight. Of the total rent units the Urban Institute identified, only 8% were vacant during the 2009-11 period.
So the old law of supply and demand helps explain the housing-cost burdens for lower-income residents, as well as the cost burdens for some much better-off households.
Only 26% of the units rented for less than $800 a month — roughly what an extremely low-income family of four could afford.
But the story is more complicated. About a third of these units were occupied by higher-income households. And only 0.9% of them were vacant.
So the rental housing market was shy 22,100 units that extremely low-income families could have lived in without a cost burden.
More units affordable for very low-income households were occupied by those with higher incomes. But because the District has more such units — and because more were vacant — the Urban Institute finds no shortage.
Subsidized Housing
In 2012, HUD subsidized roughly 33,900 housing units in the District. Housing Choice (formerly Section 8) vouchers accounted for 41% — some of them vouchers awarded to developers so they could charge affordable rents and some given directly to eligible households, which could then rent on the open market.
Public housing accounted for an additional 25% of the affordable units. Subsidies for the remaining 11,600 units came from a mix of programs. It's not clear that all these units were affordable for the District's lowest-income households.
What is clear is that there were far more extremely-low income households than HUD-subsidized units — and that the District's own voucher program fell far short of closing the gap.
Looking only at renter households, the Urban Institute reports 43 subsidized units for every 100 extremely low-income households during the 2009-11 period. This, recall, is before HUD's budget got hit by sequestration.
What's Missing
As informative — and depressing — as all these numbers are, they tell only part of the story. We need also to consider where the affordable units were.
As the Urban Institute says, "they may not be in neighborhoods of opportunity that were transit accessible, close to jobs, or had amenities like grocery stores." For the District, this is probably more apt now as gentrification has spread.
We need also to consider whether the affordable units were livable. The recent Washington Post exposé of conditions at Park Southern tells us that some surely weren't. Leaks, mold, rotting dead birds on the stairwell, etc.
Not a unique case, by any means, as a recent NPR story indicates.
What Now?
It would be nice to end this long post with a policy solution. The best I can do isn't good enough.
Clearly, as the DC Fiscal Policy Institute says, we need to invest more in affordable housing. Like the Urban Institute, it also says we should increase the total number of housing units, since this could relieve the demand pressures that are driving up costs.
The"we" here ought to be the federal government, as well as our local government and private sources. But it almost surely won't be any time soon — even if the House doesn't altogether get its way on what the HUD budget should be.
We need also to help extremely and very low-income households join the higher income tiers. An obviously large and varied agenda here.