Types of Digital Signals• Unit step signalu(n) ≡1, n ≥ 0,0, n < 0.• Unit impulse (unit sample)δ(n) ≡1, n = 0,0, n 6= 0.u(n) =nXm=−∞δ(m) summing,δ(n) = u(n) − u(n − 1) differencing.• Complex exponentials (cisoids)x(n) = A exp[j(ωn + θ)]obtained by sampling an analog cisoidxa(t) = A exp[j(Ω t + θ)],EE 524, Fall 2004, # 2 1i.e. x(n) = xa(nT ), where T is the sampling interval. Thus,ω = Ω T, or, equivalently, f =FFs(using Fs= 1/T , ω = 2πf, Ω = 2πF ),• Sinusoidsx(n) = A sin(ωn + θ)Useful properties:exp[j(ωn + θ)] = cos(ωn + θ) + j sin(ωn + θ),cos(ωn + θ) =exp[j(ωn + θ)] + exp[−j(ωn + θ)]2,sin(ωn + θ) =exp[j(ωn + θ)] − exp[−j(ωn + θ)]2j.EE 524, Fall 2004, # 2 2A sine wave as the projection of a complex phasor onto theimaginary axis:EE 524, Fall 2004, # 2 3Sampled vs. Analog Exponentials• Analog exponentials and (co)sinusoids are pe riodic with T =2π/Ω , discrete-time sinusoids are not necessarily periodic(although their values lie on a periodic envelope.)Periodicity condition: (also for sines and cosines)x(n) = x(n + N) =⇒ ejωn= ejω(n+N)=⇒ exp(jωN) = 1=⇒ ω =2πmNm integer, or f =mN(ω = 2πf).• For sampled exponentials, the frequency ω is expressed inradians, rather than radians/second.• Digital signals have ambiguity.EE 524, Fall 2004, # 2 4Ambiguity in Discrete-time SignalsAmbiguity Condition for Discrete-time Sinusoidssin(Ω1T ) = s in(Ω2T ), Ω16= Ω2⇒2πF1T = 2πF2T + 2πm, m = . . . , −2, −1, 1, 2, . . . ⇒|F1− F2| =mT= mFs, m = 1, 2, . . .EE 524, Fall 2004, # 2 5Example: lowpass signal (with spe ctrum |X(F )|2concentratedin the interval [−Fm, Fm]):Taking F1= Fmand F2= −Fm, it follows that there is noambiguity if the signal is sampled withFs=1T> 2Fm.where Fsis the sampling frequency. The above equation is aparticular form of the sampling theorem.• The frequency FN= 2Fmis referred to as the Nyquist rate.• Discrete-time signal ambiguity is often termed as the aliasingeffect.EE 524, Fall 2004, # 2 6Discrete-time Systemsy(n) = T [x(n)]where T [·] denotes the transformation (operator) that maps aninput sequence x(n) into an output sequence y(n).Linear system: a system is linear if it obeys the supe rpositionprinciple:The response of the system to the weighted sum of signals ≡corresponding weighted sum of the responses (outputs) of thesystem to each of the individual input signals. Mathematically:T [ax1(n) + bx2(n)] = aT [x1(n)] + bT [x2(n)]= ay1(n) + by2(n).-Linear system T [·]-ax1(n) + bx2(n) ay1(n) + by2(n)EE 524, Fall 2004, # 2 7Example: (Square-law device) Let y(n) = x2(n) (i.e. T [·] =(·)2). ThenT [x1(n) + x2(n)] = x21(n) + x22(n) + 2x1(n)x2(n)6= x21(n) + x22(n).Hence, the system is nonlinear!A time-invariant (or shift-invariant) system has input-outputproperties that do not change in time :if y(n) = T [x(n)] =⇒ y(n − k) = T [x(n − k)].Linear time-invariant (LTI) system is a syste m that is bothlinear and time-invariant [sometim es referred to as linear shift-invariant (LSI) system].EE 524, Fall 2004, # 2 8Discrete-time Signals via ShiftedImpulse Functionsx(n) =∞Xk=−∞x(k)δ(n − k).EE 524, Fall 2004, # 2 9Response of LTI SystemLet h(n) be the response of the system to δ(n). Due to thetime-invariance property, the response to δ(n − k) is simplyh(n − k). Thusy(n) = T [x(n)] = T∞Xk=−∞x(k)δ(n − k)=∞Xk=−∞x(k)T [δ(n − k)]=∞Xk=−∞x(k)h(n − k)= {x(n)} ? {h(n)} convolution sum.The sequence {h(n)} ≡ impulse response of LTI system.EE 524, Fall 2004, # 2 10Convolution: PropertiesAn important property of convolution:{x(n)} ? {h(n)} =∞Xk=−∞x(k)h(n − k)=∞Xk=−∞h(k)x(n − k)= {h(n)} ? {x(n)},i.e. the order in which two sequences are convolved isunimportant!Other properties:{x(n)} ? [{h1(n)} ? {h2(n)}] associativity= [{x(n)} ? {h1(n)}] ? {h2(n)}.{x(n)} ? [{h1(n)} + {h2(n)}] distributivity= {x(n)} ? {h1(n)} + {x(n)} ? {h2(n)}.EE 524, Fall 2004, # 2 11Stability of LTI SystemsAn LTI system is stable if and only if∞Xk=−∞|h(k)| < ∞,Proof: (absolute summability ⇒ stability) Let the inputx(n) be bounded so that |x(n)| ≤ Mx< ∞, ∀n ∈ [−∞, ∞].Then|y(n)| =∞Xk=−∞h(k)x(n − k)≤∞Xk=−∞|h(k)||x(n − k)|≤ Mx∞Xk=−∞|h(k)| ⇒ |y(n)| < ∞ if∞Xk=−∞|h(k)| < ∞.=⇒ Now, it remains to prove that, ifP∞k=−∞|h(k)| = ∞,then a bounded input can be found for which the output is notbounded. Considerx(n) =(h∗(−n)|h∗(−n)|, h(n) 6= 0,0, h(n) = 0.y(0) =∞Xk=−∞h(k)x(−k) =∞Xk=−∞|h(k)| =⇒ifP∞k=−∞|h(k)| = ∞, the output sequence is unbounded.EE 524, Fall 2004, # 2 12Causality of LTI SystemsDefinition. A system is causal if the output does notanticipate future values of the input, i.e. if the output atany time depends only on values of the input up to that time.Thus, a causal system is a system whose output y(n) dependsonly on {. . . , x(n − 2), x(n − 1), x(n)}.Consequence: A system y(n) = T [x(n)] is causal if wheneverx1(n) = x2(n) for all n ≤ n0then y1(n) = y2(n) for alln ≤ n0, where y1(n) = T [x1(n)], y2(n) = T [x2(n)].Comments:• All real-time physical systems are causal, because time onlymoves forward. (Imagine that you own a noncausal systemwhose output depends on tomorrow’s stock price.)• Causality does not apply to spatially-varying signals. (Wecan move both left and right, up and down.)• Causality does not apply to systems processing recordedsignals (e.g. taped sports games vs. live broadcasts).Proposition. An LTI s yste m is causal if and only if its impulseresponse h(n) = 0 for n < 0.Proof. From the definition of a causal system:y(n) =∞Xk=−∞h(k)x(n − k)EE 524, Fall 2004, # 2 13=∞Xk=0h(k)x(n − k).Obviously, this equation is valid ifP−1k=−∞h(k)x(n − k) = 0for all x(n − k) =⇒ h(n) = 0 for n < 0. The other directionis obvious. 2If h(n) 6= 0 for n < 0, system is noncausal.h(n) = 0 , n < −1,h(−1) 6= 0=⇒y(n) =∞Xk=0h(k)x(n − k) + h(−1)x(n + 1) =⇒y(n) depends on x(n + 1) =⇒ noncausal system!EE 524, Fall 2004, # 2 14Example:An LTI system withh(n) = anu(n) =an, n ≥ 0,0, n < 0.• Since h(n) = 0 for n < 0, the system is causal.• To decide on stability, we must compute the sumS =∞Xk=−∞|h(k)| =∞Xk=0|a|k=11−|a|, |a| < 1,∞, |a| ≥ 1.Thus, the system is stable only for |a| < 1.EE 524, Fall 2004, # 2 15Linear Constant-Coefficient Difference(LCCD)