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## Digital Communications Tutorial

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**Digital Communications Tutorial**Cognitive Radio Communications @ Virginia Tech NSF Research Experiences for Undergraduates (REU) Site Ratchaneekorn (Kay) Thamvichai tkay@vt.edu**Outline**• Introduction • Fourier Transform • Sampling • Pulse Amplitude Modulation (PAM) • InterSymbol Interference (ISI) • Digital Bandpass Modulation**Analog vs. Digital**• Transmitted bits can be detected and regenerated, so noise does not propagate additively. • More signal processing techniques are available to improve system performance: source coding, channel (error-correction) coding, equalization, encryption, filtering,… • Digital ICs are inexpensive to manufacture • Digital communications permits integration of voice, video, and data on a single system (ISDN) • Implementation of various algorithms can be done by software instead of hardware • Security is easier to implement.**Fourier Transform**• F() is the continuous-time Fourier transform of f(t). • The Fourier transformation F(ω) is the frequency domain representation of the original function f(t). It describes which frequencies are present in the original function.**Example 1:**x(t) t=0 Ex A: Find the Fourier Transform of x(t) = (t) Ex B: Find the Fourier Transform of x(t) = 0.5cos(500pt)**sinc(x)**1 x -3p -2p -p 0 p 2p 3p sinc function sinc(x) = sin(x) x • even function • zero crossings at • Amplitude decreases proportionally to 1/x**Ex D: Pulsed Cosine:**cos(w0t)rec(t/T) <=> (T/2) sinc(w-w0)T + sinc(w+w0)T 2 2**Linear Time-Invariant (LTI) system**h(t) Convolution: y(t) = x(t)*h(t) Its Fourier Transform: Y(ω) = X(ω)H(ω) where H(ω) is a frequency response or a transfer function of a system h(t).**Ideal filters**• A filter is used to eliminate unwanted parts of the frequency spectrum of a signal. • A filter is LTI system with an impulse response h(t). • The output y(t) of a filter can be founded in time domain using a convolution. • However, it is easier to do it in a frequency domain: Y(ω) = X(ω)H(ω)**Low Pass Filterwith a cutoff frequency wc**High Pass Filter**Example 2: Given x(t) = cos(500pt)cos(1000pt), find an**impulse response h(t) of a low-pass filter that passes the low frequency component of the signal. x(t) y(t) = low freq. component of x(t) Low-pass filter h(t)**H(w)**Y(w) = H(w)X3(w) = p/2[d(w-500p) + d(w+500p)] => y(t) = 0.5cos(500pt) H(w ) = rect(w/2000p) => h(t) = 1000sinc(1000t)**Outline**• Introduction • Fourier Transform • Sampling • Pulse Amplitude Modulation (PAM) • InterSymbol Interference (ISI) • Digital Bandpass Modulation**Sampling Continuous-Time signals**Sampling – generating of an ordered number of sequence by taking values of f(t) as specified instants of time i.e. f(t1), f(t2), f(t3), … where tm are instants at which sampling occurs. Sampling operation is implemented in hardware by an analog-to-digital converter (ADC) – electronic device used to sample physical voltage signals. In most cases, continuous-time signals are sampled at equal increments of time. The sample increment, called sample period, is usually denoted as Ts.**Impulse sampling**Define the continuous time impulse train as: p(t) is an infinite train of continuous time impulse functions, spaced Ts seconds apart.**Let x(t) be a continuous time signal we wish to sample. We**will model sampling as multiplying a signal x(t) by p(t).**Sampling Theorem**let P(ω) be a Fourier Transform of p(t), X(ω) be a Fourier Transform of x(t), Xs(ω) be a Fourier Transform of xs (t), Since xs(t) = x(t)p(t) by a multiplication property (Fourier Transform),**where Ck are the Fourier Series coefficients of the**periodic signal.**We see that an impulse train in time, p(t), has a Fourier**Transform that is an impulse train in frequency, P(w). • The spacing between impulses in time is Ts, and the spacing between impulses in frequency is ω0 = 2p/Ts. Note: If we increase the spacing in time between impulses, this will decrease the spacing between impulses in frequency, and vice versa.**Spectrum of a sampled signal**replicated scaled versions of X(w), spaced every w0 apart in frequency**Time-domain**Frequency-domain ω0 = 2p/Ts**If w0-wc < wc , ALIASING (overlap area) occurs**If w0-wc ≥ wc , Note: if ω0 - ωc ≥ ωc orω0 ≥ 2ωc, there is no aliasing**Sampling Theorem**Let x(t) be a band-limited signal with X(ω) = 0 for |ω| > ωc. Then x(t) is uniquely determined by its samples x(nTs), n = 0, ±1, ± 2, … if ω0 ≥ 2ωc where ω0 = 2p/Ts. • This is how to choose a sampling frequency (fs = 1/Ts) or period (Ts) such that an original continuous-time signal x(t) can be recovered from a sampled version xs(t). => a sampling rate (ω0) MUST be at least twice the highest frequency (ωc) of a signal to avoid aliasing problem.**To recover x(t) from its sampled version xs(t), we use a low**pass filter (reconstruction filter) to recover the center island of Xs(w):**Ex: Given a signal x(t) with Fourier Transform with cutoff**frequency ωc as shown: Given three different pulse trains with periods Draw the sampled spectrum in each case. Which case(s) experiences aliasing?**Aliasing Phenomenon**• Sampling theorem: the signal is strictly band-limited (wc). However, in practice, no information-bearing signal is strictly band-limited. • Aliasing is the phenomenon of a high-frequency component in the spectrum of the signal seemingly taking on the identify of a lower frequency in the spectrum of its sampled version. • To prevent the effects of aliasing in practice • Prior to sampling : a low-pass anti-alias filter is used to attenuate those high-frequency components of a message signal that are not essential to the information being conveyed by the signal. • The filtered signal is sampled at a rate slightly higher than the Nyquist rate.**Example: Why 44.1 kHz for Audio CDs?**• Sound is audible in 20 Hz to 20 kHz range: fmax = 20 kHz and the Nyquist rate 2fmax = 40 kHz • What is the extra 10% of the bandwidth used? Rolloff from passband to stopband in the magnitude response of the anti-aliasing filter. • Okay, 44 kHz makes sense. Why 44.1 kHz? At the time the choice was made, only recorders capable of storing such high rates were VCRs. • NTSC: 60-Hz video (30 frames/s) - 490 lines per frame or 245 lines per field, 3 audio samples per line the sampling rate is 60 X 245 X 3 = 44.1 KHz**Outline**• Introduction • Fourier Transform • Sampling • Pulse Amplitude Modulation (PAM) • InterSymbol Interference (ISI) • Digital Bandpass Modulation**Pulse-Amplitude Modulation (PAM)**• The amplitude of regularly spaced pulses are varied in proportion to the corresponding sample values of a continuous message signal. • Two operations involved in the generation of the PAM signal • Instantaneous sampling of the message signal m(t) every Ts seconds, • Lengthening the duration of each sample, so that it occupies some finite value T.**Sample-and-Hold Filter : Analysis**• The PAM signal is • The h(t) is a standard rectangular pulse of unit amplitude and duration • The instantaneously sampled version of m(t) is**To modify mδ(t) so as to assume the same form as the PAM**signal: • The PAM signal s(t) is mathematically equivalent to the convolution of mδ(t) , the instantaneously sampled version of m(t), and the pulse h(t). • Its Fourier Transform:**One benefit of PAM**It enables the simultaneous transmission of multiple signals using time-division multiplexing (TDM). User 1 User 2**Quantization Process**• Amplitude quantization: The process of transforming the sample amplitude m(nTs) of a baseband signal m(t) at time t=nTs into a discrete amplitude v(nTs) taken from a finite set of possible levels. It will be represented by binary number(s)**Outline**• Introduction • Fourier Transform • Sampling • Pulse Amplitude Modulation (PAM) • MATLAB! • InterSymbol Interference (ISI) • Digital Bandpass Modulation**Baseband Transmission of Digital Data**• The transmission of digital data over a physical communication channel is limited by two unavoidable factors • Intersymbol interference • Channel noise**The level-encoded signal and the discrete PAM signal are**• The transmitted signal is • The channel output is • The output from the receive-filter is**The InterSymbol Interference (ISI) Problem**We may express the receive-filter output as the modified PAM signal where After sampling:**ISI (cont.)**Define where E is the transmitted signal energy / bit (symbol). What we desire is However, from Residual phenomenon, intersymbol interference (ISI)**Pulse-shaping**• Given the channel transfer function, determine the transmit-pulse spectrum and receive-filter transfer function so as to satisfy two basic requirements: • Intersymbol interference (ISI) is reduced to zero. • Transmission bandwidth is conserved.**The Nyquist Channel**• The optimum solution for zero ISI at the minimum transmission bandwidth possible in a noise-free environment • For zero ISI, it is necessary for the overall pulse shape p(t) and the inverse Fourier transform of the pulse spectrum P(f) to satisfy the condition**The overall pulse spectrum is defined by the optimum**brick-wall function: • The brick-wall spectrum defines B0 as the minimum transmission bandwidth for zero intersymbol interference. • The optimum pulse shape is the impulse response of an ideal low-pass channel with an amplitude response Popt(f) in the passband and a bandwidth B0**Symbol 2**Symbol 1 Symbol 3