DC -16-17,18-19.pptx
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(2) Recap of Lecture 15 Types of Digital-To-Digital Encoding Unipolar Encoding Polar Encoding Bipolar Encoding .
(3) Overview of Lecture 16 . Analog-to-Digital Conversion Pulse. Amplitude Modulation (PAM). Pulse. Code Modulation (PCM) Quantization Binary Encoding Digital-To-Digital Conversion.
(4) Conversion Methods.
(5) Types of Digital-to-Digital Encoding Digital/Digital Encoding. Unipolar. Polar. Bipolar.
(6) Analog-to-Digital Conversion.
(7) Analog-to-Digital Conversion . The most common technique to change an analog signal to digital data (digitization) is called pulse code modulation (PCM). A PCM encoder has three processes, as shown. 1. The analog signal is sampled. 2. The sampled signal is quantized. 3. The quantized values are encoded as streams of bits..
(8) Pulse Code Modulation (PCM) . Modifies pulses created by PAM. . Four Separate Processes: PAM Quantization Binary Encoding Digital/Digital Encoding.
(9) Pulse Code Modulation (PCM).
(10) Sampling Pulse Amplitude Modulation (PAM) . First step in Analog-to-Digital Conversion. . This technique takes an Analog signal, Samples it, and Generates a series of Pulses based on the results of Sampling.
(11) Sampling Pulse Amplitude Modulation (PAM) Sampling The first step in PCM is sampling. The analog signal is sampled every Ts , where Ts is the sample interval or period. The inverse of the sampling interval is called the sampling rate or sampling frequency .
(12) Sampling Pulse Amplitude Modulation (PAM) . There are three sampling methodsideal, natural, and flat-top-as.
(13) Sampling Pulse Amplitude Modulation (PAM) In ideal sampling, pulses from the analog signal are sampled. This is an ideal sampling method and cannot be easily implemented. In natural sampling, a high-speed switch is turned on for only the small period of time when the sampling occurs. The result is a sequence of samples that retains the shape of the analog signal. The most common sampling method, called sample and hold, however, creates flat-top samples by using a circuit..
(14) Pulse Amplitude Modulation (PAM) The sampling process is sometimes referred to as pulse amplitude modulation (PAM).
(15) Sampling Rate . How many samples are sufficient?. . Nyquist theorem: The sampling rate must be at least twice the highest frequency.
(16) Sampling Rate.
(17) Sampling Rate Example 4.6 For an intuitive example of the Nyquist theorem, let us sample a simple sine wave at three sampling rates: fs = 4f (2 times the Nyquist rate )'/s = 2f (Nyquist rate), and f s =f (one-half the Nyquist rate). Figure 4.24 shows the sampling and the subsequent recovery of the signal..
(18) Sampling Rate.
(19) Quantization Quantization is a method of assigning integral values in a specific range to the sampled instances.
(20) Quantization The following are the steps in quantization: 1. We assume that the original analog signal has instantaneous amplitudes between Vmin and Vmax' 2. We divide the range into L zones, each of height ~ (delta) 3. We assign quantized values of 0 to L - I to the midpoint of each zone. 4. We approximate the value of the sample amplitude to the quantized values..
(21) The following are the steps in quantization: 1. We assume that the original analog signal has instantaneous amplitudes between Vmin and Vmax' 2. We divide the range into L zones, each of height ~ (delta) 3. We assign quantized values of 0 to L - I to the midpoint of each zone. 4. We approximate the value of the sample amplitude to the quantized values..
(22) Quantization Quantization Levels The choice of L, the number of levels, depends on the range of the amplitudes of the analog signal and how accurately we need to recover the signal. If the amplitude of a signal fluctuates between two values only, we need only two levels; if the signal, like voice, has many amplitude values, we need more quantization levels. In audio digitizing, L is normally chosen to be 256; in video it is normally thousands. Choosing.
(23) Quantization Quantization Error One important issue is the error created in the quantization process. Quantization is an approximation process. The input values to the quantizer are the real values; the output values are the approximated values. The output values are chosen to be the middle value in the zone. If the input value is also at the middle of the zone, there is no quantization error; otherwise, there is an error. In the previous example, the normalized amplitude of the third sample is 3.24, but the normalized quantized value is 3.50. This means that there is an error of +0.26. The value of the error for any sample is.
(24) Quantization.
(25) Binary Encoding.
(26) Result of PCM.
(27) Full PCM Process.
(28) Bit Rate BitRate SamplingRate No.ofbits / sample.
(29) Summary . Analog-to-Digital Conversion. . Pulse Code Modulation (PCM) Pulse. (PAM). Amplitude Modulation. Quantization Binary. Encoding Digital-To-Digital Conversion.
(30) Suggested Reading . Section 5.2, “Data Communications and Networking” 2nd Edition by Behrouz A. Forouzan.
(31) DATA COMMUNICATION. Lecture-17.
(32) Recap of Lecture 16 Analog-To-Digital Conversion Pulse Code Modulation (PCM) Pulse Amplitude Modulation (PAM) Quantization Binary Encoding Digital-to-Digital Conversion .
(33) Overview of Lecture 17 Digital-to Analog Conversion Bit Rate and Baud Rate Carrier Signals Amplitude Shift Keying (ASK) Frequency Shift Keying (FSK) Phase Shift Keying (PSK) Quadrature Amplitude Modulation (QAM) .
(34) Digital To Analog Conversion . Process of changing one of the characteristics of an analog signal based on the info in a digital signal. . Digital data must be modulated on an analog signal that has been manipulated to look like two distinct values corresponding to binary 1 to.
(35) Digital To Analog Conversion.
(36) Variation in Characteristics of Sine Wave . A sine wave is defined by 3 characteristics: Amplitude Frequency Phase. . By changing one aspect of a simple electrical signal back & forth,we can use it to represent digital data.
(37) Mechanisms for Modulating Digital Data to Analog Signals.
(38) Bit Rate & Baud Rate . Bit rate:no. of bits transmitted during one second. . Baud rate:no. of signal units per second that are required to represent that bit.
(39) Analogy for Bit rate &Baud Rate . In transportation a Baud is analogous to a Car a Bit is analogous to a Passenger. . If1000 cars can go from one point to another carrying only one passenger(only driver),than 1000 passengers are transported.
(40) Analogy for Bit rate &Baud Rate . However, if each car carries four passengers, then 4000 passengers are transported. . Note that the Number of Cars (Bauds), not the Numbers of Passengers (Bits) determines the traffic and therefore the need for wider highways.
(41) Example 5.6 . An analog signal carries 4 bits in each signal element.If 1000 signal elements are sent per second, find the Baud Rate and Bit Rate?. . Solution: Baud Rate= Number of Signal Elements Baud Rate =1000 bauds/second Bit Rate=Baud Rate * Number of bits per signal element.
(42) Carrier Signals The sending device produces a high frequency signal, that acts as a basis for the information signal. This base signal is called the Carrier Signal or Carrier Frequency.
(43) Amplitude Shift Keying (ASK) . The amplitude of the Carrier signal is varied to represent binary 1 or 0. . Both frequency and phase remain constant, while the amplitude changes.
(44) Amplitude Shift Keying (ASK).
(45) Effect Of Noise on ASK . Highly susceptible to noise interference. . ASK relies solely on Amplitude for recognition. . Noise usually affects the amplitude.
(46) On-Off- Keying (OOK) . A popular ASK Technique. . In OOK, one of the bit values is represented by no voltage. . The advantage is the reduction in the amount of energy required to transmit Information.
(47) Bandwidth for ASK (Figure).
(48) Bandwidth for ASK (Figure) . Bandwidth requirements for ASK are calculated using the formula BW = (1+d)*Nbaud.
(49) Example 5.8 . Find minimum bandwidth required for an ASK signal TX at 2000 bps. TX. Mode is half duplex. . Solution: In ASK, Baud Rate= Bit Rate Therefore, Baud Rate = 2000 Also ASK requires a minimum bandwidth equal to its Baud Rate Therefore Minimum BW = 2000 Hz.
(50) Summary Digital-to Analog Conversion Bit Rate and Baud Rate Carrier Signals Amplitude Shift Keying (ASK) .
(51) Suggested Reading . Section 5.3, “Data Communications and Networking” 2nd Edition by Behrouz A. Forouzan.
(52) DATA COMMUNICATION. Lecture-18.
(53) Recap of Lecture 17 Digital-to Analog Conversion Bit Rate and Baud Rate Carrier Signals Amplitude Shift Keying (ASK) .
(54) Overview of Lecture 18 Digital-to Analog Conversion Frequency Shift Keying (FSK) Phase Shift Keying (PSK) Quadrature Amplitude Modulation (QAM) .
(55) Frequency Shift Keying (FSK) . Frequency of signal is varied to represent binary 1 or 0. . Both peak amplitude and phase remains constant.
(56) Frequency Shift Keying (FSK).
(57) Effect of Noise on FSK . Avoids most of the Noise problems of ASK. . Receiving device is looking for specific frequency changes over a given number of periods, it can ignore voltage spikes.
(58) Bandwidth of FSK.
(59) Example 5.11 . Find the minimum BW for an FSK signal transmitted at 2000 bps. TX is in half duplex mode and carrier must be separated by 3000 Hz. . Solution:. . For FSK,. . BW= 2000 + (fc1 – fco) = 2000 + 3000 = 5000 Hz. BW=Baud Rate + (fc1 – fco).
(60) Phase Shift Keying (PSK) . In PSK, phase of carrier is varied to represent binary 1 or 0. . Both peak amplitude and frequency remains constant as the phase changes.
(61) Phase Shift Keying (PSK).
(62) 2 PSK.
(63) Effect of Noise on PSK . PSK is not susceptible to the noise degradation that affects ASK, nor to the bandwidth limitations of FSK. . Smaller variations in signal can be detected reliably by the receiver.
(64) 4 PSK.
(65) 4 PSK.
(66) BW for PSK . Minimum bandwidth required for PSK transmission is the same as ASK. . PSK bit rate using the same BW can be two or more times greater.
(67) 8 PSK.
(68) Limitations of PSK . PSK is limited by the ability of the equipment to distinguish small differences in phase. . This factor limits its potential bit rate.
(69) Quadrature Amplitude Modulation (QAM) . Combination of ASK and PSK. . ‘x’ variation in phase and ‘y variations in amplitude result into a total of x * y variations.
(70) Quadrature Amplitude Modulation (QAM).
(71) 8 QAM.
(72) 16 QAM.
(73) Digital 16-QAM with example constellation points..
(74) Example 5.15 . A constellation diagram consists of eight equally spaced points on a a circle. If bit rate is 4800 bps, what is the Baud Rate?. . Solution: Constellation indicates 8 PSK with the points 45 degree apart Baud Rate= 4800 / 3 = 1600 baud.
(75) Summary Digital-to Analog Conversion Frequency Shift Keying (FSK) Phase Shift Keying (PSK) Quadrature Amplitude Modulation (QAM) .
(76) Suggested Reading . Section 5.3, “Data Communications and Networking” 2nd Edition by Behrouz A. Forouzan.
(77) DATA COMMUNICATION. Lecture-19.
(78) Recap of Lecture 18 Digital-to Analog Conversion Bit Rate and Baud Rate Carrier Signals Ask FSK PSK QAM Bit and Baud Rate Comparison .
(79) Overview of Lecture 19 Analog-to Analog Conversion Amplitude Modulation Frequency Modulation Phase Modulation .
(80) Analog To Analog Conversion . Representation of Analog information by an Analog signal. . For Example: Radio.
(81) Analog To Analog Conversion.
(82) Analog To Analog Conversion Methods.
(83) Amplitude Modulation (AM) . Amplitude of carrier signal is changed according to the amplitude of modulating signal. . Frequency and phase of the carrier remain the same.
(84) Amplitude Modulation (AM).
(85) AM Bandwidth . Bandwidth of AM signal (modulated signal) = 2 * bandwidth of modulating signal. . Significant spectrum of AM audio = 5 KHz 10 KHz bandwidth for an AM station.
(86) AM Bandwidth.
(87) AM Band Allocation.
(88) Example 5.18 . We have an audio signal with a BW of 4 KHz. What is the BW needed, if we modulate the signal using AM?. . Solution: AM signal requires twice the BW of original signal BW = 2 * 4 KHz = 8 KHz.
(89) Frequency Modulation (FM) . Frequency of carrier signal is changed according to the amplitude of modulating signal. . Amplitude and Phase of the carrier signal remain constant.
(90) Frequency Modulation (FM).
(91) FM Bandwidth . Bandwidth of FM signal (modulated signal) = 10 * bandwidth of modulating signal. . Significant spectrum of FM audio = 15 KHz Minimum 150 KHz bandwidth.
(92) FM Bandwidth.
(93) FM Band Allocation.
(94) Example 5.19 . We have an Audio signal with a BW of 4 MHz. What is the BW needed if we modulate the signal using FM?. . Solution: BW = 10 * 4 MHz = 40 MHz.
(95) Phase modulation (PM) . Simpler hardware requirements. . Phase is modulated with the amplitude. . Amplitude & Frequency of the carrier signal remain constant.
(96) Conversion Methods.
(97) Summary Analog-to Analog Conversion Amplitude Modulation Frequency Modulation Phase Modulation .
(98) Suggested Reading . Section 5.4, “Data Communications and Networking” 2nd Edition by Behrouz A. Forouzan.
(99)
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