Dr. Clincy Professor of CS Chapter 4 Handout #4 Dr. Clincy Professor of CS Dr. Clincy Lecture 2

Four Data-to-Signal Cases DATA SIGNAL D D A A D Already covered Will cover today Dr. Clincy Lecture 2

ANALOG-TO-DIGITAL CONVERSION We have seen in Chapter 3 that a digital signal is superior to an analog signal. The tendency today is to change an analog signal to digital data. In this section we describe two techniques, pulse code modulation and delta modulation.

Components of PCM encoder PCM – Pulse Code Modulation 1st: analog signal is sampled 2nd: sampled signal is quantized 3rd: quantized values are encoded as bit streams (or codes) Analog signal is sampled every Ts seconds Sample rate is fs = 1/Ts

Three different sampling methods for PCM High-speed switch used – able to retain the shape of the signal Ideal but complex Sample-and-hold method that creates flat-top samples by using a circuit Sampling process also called pulse amplitude modulation (PAM)

Recovery of a sampled sine wave for different sampling rates Nyquist theorem states that the sampling rate must be at least 2 times the highest frequency of the signal Catches the essence of the signal Doesn’t improve the case Doesn’t capture the essence of the signal

Quantization and encoding of a sampled signal actual-amplitude/D actual amplitude Quantization steps: Determine Vmin and Vmax Divide range into L zones, each of height D D = [Vmin - Vmax]/L 3. Assign quantized values of 0 to L-1 to midpoint of each zone 4. Map the sample value to a quantized value Norm. Actual Error between actual and nornalized Quant. value for code Code that represents the voltage level Assume sample amplitudes between -20V and +20V Let L = 8 (levels) – therefore, D = [20 - -20]/8 = 5 Quantization error can contribute to Shannon’s SNR: SNRdB = 6.02nb + 1.76 where nb is bits per sample Bit rate = sampling rate x # of bits per sample = fs x nb

Components of a PCM decoder PCM decoder recovers the original signal Smooths out the staircase signal What is the minimum bandwidth of the filter the digitized signal will need ? Bmin = c x nb x 2 x Banalog x 1/r (nb = # bits per sample) If 1/r=1 and c=1/2, Bmin=nb x Banalog If the data rate and number of signal levels are fixed, minimum bandwidth is Bmin = N / [2 x log2 L]

The process of delta modulation PCM is more complex than Delta modulation PCM finds the amplitude of the signal; delta modulation simply finds the change in the signal from the previous sample Delta modulation doesn’t use codes – bits are sent one after another Positive changes are encoded as 1; negative changes are encoded as 0

TRANSMISSION MODES The transmission of binary data across a link can be accomplished in either parallel or serial mode. In parallel mode, multiple bits are sent with each clock tick. In serial mode, 1 bit is sent with each clock tick. While there is only one way to send parallel data, there are three subclasses of serial transmission: asynch, syn and iso approaches (asynchronous, synchronous, and isochronous.)

Parallel transmission

Serial transmission

Asynchronous transmission In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1s) at the end of each byte. There may be a gap between each byte. Asynchronous here means “asynchronous at the byte level,” but the bits are still synchronized; their durations are the same.

Synchronous transmission In synchronous transmission, we send bits one after another without start or stop bits or gaps. It is the responsibility of the receiver to group the bits.

Isochronous Transmission For realtime audio and video, uneven delays between frames is not acceptable – so synchronous transmission doesn’t work well The entire stream of bits must be synchronized – this is isochronous transmission Isochronous transmission guarantees data at a fixed rate

Dr. Clincy Professor of CS Chapter 5 Handout #4 Dr. Clincy Professor of CS Dr. Clincy Lecture 2

Digital-to-analog conversion Based on the digital data, the Modulator changes characteristics of the “controllable” analog signal (bandpass analog signal) on the transmitter side to represent the digital data Demodulator interprets the analog signal in re-creating the digital data on the receiver side Terminology: “modulating digital data into an analog signal” The analog signal we can control ? Sine Wave, Carrier Signal, Periodic Signal Dr. Clincy Lecture

Types of digital-to-analog conversion Change amplitude to represent a bit Change frequency to represent a bit Change phase to represent a bit Combination of changing both amplitude and phase to represent a set of bits Dr. Clincy Lecture

Recall For digital transmission, bit rate (data rate) and signal rate (baud rate) relationship was S = N x 1/r where r = # of data elements per signal element and N is the data rate in bps (and S is the signaling or baud rate) For analog, r = log2L where L is the type of signal (versus level) Dr. Clincy Lecture

Example An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate. Solution In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from Dr. Clincy Lecture

Example 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? Solution 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. Dr. Clincy Lecture

Binary amplitude shift keying changing the original amplitude Explain this not changing the original amplitude Bandwidth (B) is proportional to the signal rate (S) and depending on the modulation and filtering process, the required bandwidth can range between S to 2S (where middle bandwidth is fc). The value of d relates to the modulation and filtering process B = (1 + d) x S Dr. Clincy Lecture

Example 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? Solution 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). Dr. Clincy Lecture S = N * 1/r

Binary frequency shift keying changing the original frequency Use two different carrier frequencies, f1 and f2, for 0 and 1 Explain this not changing the original frequency Dr. Clincy Lecture

Example 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? Solution The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means The difference (delta) between the two frequencies Dr. Clincy Lecture

Binary phase shift keying changing the original phase Explain this not changing the original phase Dr. Clincy Lecture

QPSK and its implementation QPSK – Quadrature Phase Shift Keying Use 2 bits in each signal element – decreases baud rate and bandwidth Uses 4 possible phases (versus 2) 2 composite signals are created Because the 2 signals are using the same bandwidth – each signal has ½ bandwidth Dr. Clincy Lecture

Example Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0. Solution 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. B = (1 + d) x S Dr. Clincy Lecture

Concept of a constellation diagram Helps define the amplitude and phase of a signal element the amplitude of the 2nd carrier Peak Amplitude Phase This is the amplitude given one carrier Only use 1 carrier and phase is static and 2 amplitude levels Only use 1 carrier and 1 amplitude and 2 phases (0o and 180o) Uses 2 carriers and 1 amplitude and 4 phases (45o, 135o, -45o, -135o) Dr. Clincy Lecture Binary Phase Shift Keying – uses 2 different phases and the same amplitude

Constellation diagrams for some QAMs QAM – Quadrature Amplitude Modulation For QPSK, we only changed the phase For QAM, we change both the phase and amplitude Has a 0 amplitude and a positive amplitude (with 2 carriers) Has a negative amplitude and a positive amplitude (with 2 carriers) Has 2 positive amplitudes (with 2 carriers) Has 4 negative levels and 4 positive levels (with 2 carriers) Dr. Clincy Lecture