Discrete-Time David Johns and Ken Martin University of Toronto ([email protected]) ([email protected]) University of Toronto 1 of 40 © D. Johns, K. Martin, 1997 Overview of Some Signal Spectra st() xn()= x ()nT c yn() ys()t ysh()t xc()t xs()t convert to convert to analog yc()t discrete-time DSP impulse hold low-pass sequence train filter Conceptual x ()t y ()t sh xn()= xc()nT yn() sh xc()t sample D/A analog A/D y ()t and DSP converter low-pass c converter hold with hold filter Typical Implementation University of Toronto 2 of 40 © D. Johns, K. Martin, 1997 Example Signal Spectra τxs()t A Xc()f xc()t f t fo f τ → 0 s A X ()f --- s T xn() f f o fs 2fs 4567 n A 0 12 3 --- X()ω T ω ()2πfo ⁄ fs 2π 4π xsh()t A X ()f t sh f f o fs 2fs time frequency University of Toronto 3 of 40 © D. Johns, K. Martin, 1997 Example Signal Spectra • Xs()f has same spectra as Xc()f but repeats every fs (assuming no aliasing occurs). • X()ω has same spectra as Xs()f freq axis normalized. sinx • Spectra for X ()f equals X ()f multiplied by ---------- sh s x response — in effect, filtering out high frequency images. University of Toronto 4 of 40 © D. Johns, K. Martin, 1997 Laplace Transform & Discrete-Time xc()t (all pulses) τxs()t nT t 2T 3T T τ (single pulse at nT) τxsn()t • xs()t scaled by τ such that the area under the pulse at nT equals the value of xc()nT . • In other words, at tn= T, we have x ()nT x ()nT = ---------------c (1) s τ University of Toronto 5 of 40 © D. Johns, K. Martin, 1997 Laplace Transform & Discrete-Time • Thus as τ → 0, height of xs()t at time nT goes to ∞ and so we plot τxs()t instead. • Define ϑ()t to be the step function, 1 ()t ≥ 0 ϑ()t ≡ (2) 0 ()t < 0 • then single-pulse signal, xsn()t , can be written as x ()nT x ()t = ---------------c []ϑ()tnT– – ϑ()tnT– – τ (3) sn τ and the entire signal xs()t as ∞ xs()t = ∑ xsn()t (4) n = –∞ University of Toronto 6 of 40 © D. Johns, K. Martin, 1997 Laplace Transform & Discrete-Time • Above signals are defined for all time — we can find Laplace transforms of these signals. • The Laplace transform Xsn()s for xsn()t is –sτ 1 1 – e –snT X ()s = --------------------- x ()nT e (5) sn τs c and X()s is simply a linear combination of xsn()t , which results in ∞ –sτ 1 1 – e –snT X ()s = --------------------- x ()nT e (6) s τs ∑ c n = –∞ University of Toronto 7 of 40 © D. Johns, K. Martin, 1997 Laplace Transform & Discrete-Time 2 x x • Using the expansion e = 1 ++x ----- +…, when τ → 0, 2! the term before the summation in (6) goes to unity. • Therefore, as τ → 0, ∞ –snT Xs()s = ∑ xc()nT e (7) n = –∞ • This Laplace transform only depends on sample points, xc()nT which in turn depends on the relative sampling-rate, T. University of Toronto 8 of 40 © D. Johns, K. Martin, 1997 Spectra of Discrete-Time Signals • xs()t spectra can be found by replacing sj= ω in (7) • However, a more intuitive approach is ... • Define a periodic pulse train, st() as ∞ st() = ∑ δ()tnT– (8) n = –∞ where δ()t is the unit impulse function. • Then xs()t can be written as xs()t = xc()t st() (9) 1 X ()jω = ------ X ()jω ⊗ Sj()ω (10) s 2π c where ⊗ denotes convolution. University of Toronto 9 of 40 © D. Johns, K. Martin, 1997 Spectra of Discrete-Time Signals • Since the Fourier transform of a periodic impulse train is another periodic impulse train we have ∞ 2π 2π Sj()ω = ------ δω()– k------ (11) T ∑ T k = –∞ • Thus, the spectra Xs()jω is found to be ∞ 1 jk2π X ()jω = --- X ()jω – ----------- (12) s T ∑ c T k = –∞ • or equivalently, ∞ 1 X ()f = --- X ()j2πfjk– 2πf (13) s T ∑ c s k = –∞ University of Toronto 10 of 40 © D. Johns, K. Martin, 1997 Spectra for Discrete-Time Signals • The spectra for the sampled signal xs()t equals a sum of shifted spectra of xc()t . • No aliasing will occur if Xc()jω is bandlimited to fs ⁄ 2 . • Note that xs()t can not exist is practice as it would require an infinite amount of power (seen by integrating Xs()f over all frequencies). University of Toronto 11 of 40 © D. Johns, K. Martin, 1997 Spectra Example xn() X ()f , X()ω x ()t s c τxs()t t f 0.25 0.5 1Hz 4Hz 8Hz ()n ()ω ()1 ()2 ()0.5π ()2π ()4π xn() Xs2()f , X2()ω xc()t τxs()t f t 1Hz 12Hz 24Hz 0.25 0.5 ()n π ()3 ()6 --- ()2π ()4π ()ω 6 University of Toronto 12 of 40 © D. Johns, K. Martin, 1997 Z-Transform • The z-transform is merely a shorthand notation for (7). • Specifically, defining sT ze≡ (14) • we can write ∞ –n Xz()≡ ∑ xc()nT z (15) n = –∞ • where X()z is called the z-transform of the samples xc()nT . University of Toronto 13 of 40 © D. Johns, K. Martin, 1997 Z-Transform • 2 properties of the z-transform are: –k — If xn()↔ Xz(), then xn()– k ↔ z Xz() — Convolution in the time domain is equivalent to multiplication in the frequency domain. X()z is not a function of the sampling-rate! • A 1Hz signal sampled at 10Hz has the same transform as a similar 1kHz signal sampled at 10kHz • X()z is only related to the numbers, xc()nT while Xs()s is the Laplace transform of the signal xs()t as τ → 0. • Think of the series of numbers as having a sample- rate normalized to T = 1 (i.e. f's = 1Hz). University of Toronto 14 of 40 © D. Johns, K. Martin, 1997 Z-Transform • Such a normalization results in 2πf Xs()f = X()-------- (16) fs or equivalently, a frequency scaling of 2πf ω = -------- (17) fs • Thus, discrete-time signals have ω in units of radians/sample. • Continuous-time signals have frequency units of cycles/ second (hertz) or radians/second. University of Toronto 15 of 40 © D. Johns, K. Martin, 1997 Example Sinusoidal Signals xn() xn() 10 15 1 5 n 1 5 10 15 n 0 rad/sample= 0 cycles/sample π ⁄ 8 rad/sample = 116⁄ cycles/sample xn() xn() 5 15 15 1 10 n 1 5 10 n π ⁄ 4 rad/sample = 18⁄ cycles/sample π ⁄ 2 rad/sample = 14⁄ cycles/sample University of Toronto 16 of 40 © D. Johns, K. Martin, 1997 Example Sinusoidal Signals • A continuous-time sinusoidal signal of 1kHz when sampled at 4kHz will change by π ⁄ 2 radians between each sample. • Such a discrete-time signal is defined to have a frequency of π ⁄ 2 rad/sample. • Note that discrete-time signals are not unique since the addition of 2π will result in the same signal. • For example, a discrete-time signal having a frequency of π ⁄ 4 rad/sample is identical to that of 9π ⁄ 4 rad/sample. • Normally discrete-time signals are defined to have frequency components only between –π and π rad/sample. University of Toronto 17 of 40 © D. Johns, K. Martin, 1997 Downsampling n n L 0 4 8 0 1 2 L = 4 ω ω π 4π --- 2π 0 ------ 2π 6 6 • Keep every L‘th sample and throw away L – 1 samples. • It expands the original spectra by L. • For aliasing not to occur, original signal must be bandlimited to π ⁄ L. University of Toronto 18 of 40 © D. Johns, K. Martin, 1997 Upsampling n n L 0 1 2 0 4 8 L = 4 ω 4π π 0 ------ 2π --- π 2π 6 6 • Insert L – 1 zero values between samples • The frequency axis is scaled by L such that 2π now occurs where L2π occurred in the original signal. • No worry about aliasing here. University of Toronto 19 of 40 © D. Johns, K. Martin, 1997 Discrete-Time Filters un() Hz() yn() (yn() equals hn()if un()is an impulse) (discrete-time filter) • An input series of numbers is applied to a discrete- time filter to create an output series of numbers. • This filtering of discrete-time signals is most easily visualized with the shorthand notation of z-transforms. Transfer-Functions • Similar to those for continuous-time filters except instead of polynomials in “s ”, polynomials in “z ” are obtained. University of Toronto 20 of 40 © D. Johns, K. Martin, 1997 Cont-Time Transfer-Function • Low-pass continuous-time filter, Hc()s , 4 H ()s = -------------------------- (18) c 2 s ++2s 4 • The poles are the roots of the denominator polynomial • Poles: – 1.0 ± 1.7321j for this example. • Zeros: Defined to have two zeros at ∞ since the den poly is two orders higher than the numerator poly. University of Toronto 21 of 40 © D. Johns, K. Martin, 1997 Cont-Time Frequency Response high-frequency s-plane jω∞= jω dc (poles) jω = 0 • Poles and zeros plotted in the s-plane. • Substitution sj= ω is equivalent to finding the magnitude and phase of vectors from a point along the jω axis to all the poles and zeros. University of Toronto 22 of 40 © D. Johns, K. Martin, 1997 Discrete-Time Transfer-Function 0.05 Hz()= -------------------------------------- (19) 2 z – 1.6z + 0.65 • Poles: 0.8± 0.1j in the z-plane and two zeros are again at ∞.