MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.V + C 1 Massachusetts Institute of Technology Department of Mechanical Engineering 2.161 Signal Processing - Continuous and Discrete Fall Term 2008 Lecture 131 Reading: • Proakis & Manolakis, Chapter 3 (The z-transform) • Oppenheim, Schafer & Buck, Chapter 3 (The z-transform) Intro duction to Time-Domain Digital Signal Processing Consider a continuous-time filter C o n t i n u o u s f ( t ) s y s t e m y ( t ) ( h ( t ) , H ( s )) such as simple first-order RC high-pass filter: - R v o described by a transfer function H(s)= RCs RCs +1 . The ODE describing the system is τ dy dt + y = τ df dt where τ = RC is the time constant. Our task is to derive a simple discrete-time equivalent of this prototype filter based on samples of the input f(t) taken at intervals ΔT . fn a l g DoSr i tPh m yn 1copyright cD.Rowell 2008 13–1� � � If we use a backwards-difference numerical approximation to the derivatives, that is dx (x(nΔT ) − x((n − 1)ΔT )≈ dt ΔT and adopt the notation y = y(nΔT ), and let a = τ/ΔT ,n a(yn − yn−1)+ yn = a(fn − fn−1) and solving for yn a a a yn = yn−1 + fn − fn−11+ a 1+ a 1+ a which is a first-order difference equation, and is the computational formula for a sample-by-sample implementation of digital high-pass filter derived from the continuous prototype above. Note that • The “fidelity” of the approximation depends on ΔT , and becomes more accurate when ΔT  τ . • At each step the output is a linear combination of the present and/or past samples of the output and input. This is a recursive system because the computation of the current output depends on prior values of the output. In general, regardless of the design method used, a LTI digital filter implementation will be of a similar form, that is N M yn = aiyn−i + bifn−i i=1 i=0 where the ai and bi are constant coefficients. Then as in the simple example above, the current output is a weighted combination of past values of the output, and current and past values of the input. • If ai ≡ 0 for i =1 ...N, so that M � yn = bifn−i i=0 The output is simply a weighted sum of the current and prior inputs. Such a filter is a non-recursive filter with a finite-impulse-response (FIR), and is known as a moving average (MA) filter, or an all-zero filter. • If bi ≡ 0 for i =1 ...M, so that N yn = aiyn−i + b0fn i=0 only the current input value is used. This filter is a recursive filter with an infinite-impulse-response (IIR), and is known as an auto-regressive (AR) filter, or an all-pole filter. 13–2� � � � � 2 • With the full difference equation N M yn = aiyn−i + bifn−i i=1 i=0 the filter is a recursive filter with an infinite-impulse response (IIR), and is known as an auto-regressive moving-average (ARMA) filter. The Discrete-time Convolution Sum For a continuous system C o n t i n u o u sf ( t ) s y s t e m y ( t )( h ( t ) , H ( s )) the output y(t), in response to an input f(t), is given by the convolution integral: � ∞ y(t)= f(τ )h(t − τ)dτ 0 where h(t) is the system impulse response. For a LTI discrete-time system, such as defined by a difference equation, we define the pulse response sequence {h(n)} as the response to a unit-pulse input sequence {δn}, where 1 n =0 δn = 0 otherwise. { d n } { hn } fn a l g Do Sr i tPh m yn. . . . - 2 - 1 0 1 2. . . . n. . . .- 2 - 1 0 1 2. . . . n If the input sequence {fn} is written as a sum of weighted and shifted pulses, that is ∞ fn = fkδn−k k=−∞ then by superposition the output will be a sequence of similarly weighted and shifted pulse responses ∞ yn = fkhn−k k=−∞ which defines the convolution sum, which is analogous to the convolution integral of the continuous system. The z-Transform The z-transform in discrete-time system analysis and design serves the same role as the Laplace transform in continuous systems. We begin here with a parallel development of both the z and Laplace transforms from the Fourier transforms. 13–3 3� � � � � The Laplace Transform (1) We begin with causal f(t) and find its Fourier transform (Note that because f(t)is causal, the integral has limits of 0 and ∞): � ∞ F (jΩ) = f(t)e −jΩtdt 0 (2) We note that for some functions f(t) (for example the unit step function), the Fourier integral does not converge. (3) We introduce a weighted function w(t)= f(t)e −σt and note lim w(t)= f(t)σ→0 The effect of the exponential weighting by e−σt is to allow convergence of the integral for a much broader range of functions f(t). (4) We take the Fourier transform of w(t) � ∞ W (jΩ) = F˜(jΩ|σ)= � f(t)e −σt � e −jΩtdt 0 � ∞ = f(t)e −(σ+jΩ)dt 0 and define the complex variable s = σ + jΩso that we can write � ∞ F (s)= F˜(jω|σ)= f(t)e −stdt 0 F (s) is the one-sided Laplace Transform. Note that the Laplace variable s = σ + jΩ is ex-pressed in Cartesian form. The Z transform (1) We sample f(t) at intervals ΔT to produce f∗(t). We take its Fourier transform (and use the sifting property of δ(t)) to produce ∞ F ∗ (jΩ) = fne −jnΩΔT n=0 (2) We note that for some sequences fn (for example the unit step sequence), the summa-tion does not converge. (3) We introduce a weighted sequence {wn} = fnr −n and note lim {wn} = {fn}r→1 The effect of the exponential weighting by r−n is to allow convergence of the summation for a much broader range of sequences fn. (4) We take the Fourier transform of wn ∞ W ∗ (jΩ) = F˜∗ (jΩ|r)= � fnr −n � e −jnΩΔT n=0 ∞ = � fn � rejΩΔT �−n n=0 and define the complex variable z = rejΩΔT so that we can write ∞ F (z)= F˜∗ (jΩ|r)= fnz −n n=0 F (z) is the one-sided Z-transform. Note that z = rejΩΔT is expressed in polar form. 13–4s j W Â { z } Á { z } � � The Laplace Transform (contd.) (5) For a causal function f(t), the region of convergence (ROC) includes the s-plane to the right of all poles of F (jΩ). : : s - p l a n e R O C (6) If the ROC includes the imaginary