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.MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.161 Signal Processing - Continuous and Discrete Parallel Derivations of the Z and Laplace Transforms 1 The following is a summary of the derivation of the Laplace and Z transforms from the continuous-and discrete-time Fourier transforms: 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 ex-ample 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 + jΩ so that we can write F (s) = F˜(jω|σ) = � 0 ∞ f(t)e−stdt F (s) is the one-sided Laplace Transform. Note that the Laplace variable s = σ + jΩ is expressed 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 ∞ fne−jnΩΔTF ∗(jΩ) = � n=0 (2) We note that for some sequences fn (for exam-ple the unit step sequence), the summation does not converge. (3) We introduce a weighted sequence {wn} = �fnr−n� and note lim 1 {wn} = {fn}r→ 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 ��fnr−n� e−jnΩΔTW ∗(jΩ) = F˜∗(jΩ r) = ∞ | n=0 ∞ jΩΔT �−n = � fn �re 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. D. Rowell October 22, 2008 1 1j W  { z } � Á { z } The Laplace Transform (contd.) (5) For a causal function f(t), the region of con-vergence (ROC) includes the s-plane to the right of all poles of F (jΩ). s: : s - p l a n e R O C The Z transform (contd.) (5) For a right-sided (causal) sequence {fn} the region of convergence (ROC) includes the z-plane at a radius greater than all of the poles of F (z). : z - p l a n e R O C : 1 . 0- 1 . 0 u n i t c i r c l e We note that the mapping between the s plane and the z plane is given by sΔT z = e and that the imaginary axis (s = jΩ) in the s-plane maps to the unit circle (z = ejΩΔT )inthe z-plane. Furthermore we note that the mapping of the unit circle in the z-plane to the imaginary axis in the s-plane is periodic with period 2π, and that the mapping of the jΩ axis to the unit circle produces aliasing for |Ω| >π/ΔT . If we define a normalized discrete-time frequency that is independent of ΔT , that is ω =ΩΔT ω ≤ π we can make the following comparisons: The Laplace Transform (contd.) (6) If the ROC includes the imaginary axis, the FT of f(t)is F (jΩ): F (jΩ) = F (s) |s=jΩ (7) The convolution theorem states � ∞ Lf(t) ⊗ g(t)= f(τ)g(t − τ )dτ ⇐⇒ F (s)G(s) −∞ (8) For an LTI system with transfer function H(s), the frequency response is H(s) |s=jΩ = H(jΩ) if the ROC includes the imaginary axis. The Z transform (contd.) (6) If the ROC includes the unit circle, the DFT of {fn}, n =0, 1,...,N − 1. is {Fm} where Fm = F (z) |z=ejωm = F (ejωm), where ωm =2πm/N for m =0, 1,...,N − 1. (7) The convolution theorem states ∞ Z{fn}⊗{gn} = fmgn−m ⇐⇒ F (z)G(z) m=−∞ (8) For a discrete LSI system with transfer func-tion H(z), the frequency response is H(z) | jω = H(ejω) |ω|≤πz=e if the ROC includes the unit circle. 2� � �  { z } � � � � The Laplace Transform (contd.) (9) The transfer function bmsm + bm−1sm−1 + ...+ b1s + b0H(s)= ansn + an−1sn−1 + ...+ a1s + a0 is derived from the ordinary differential equation dny dn−1y dy ann + an−1 n−1 ...+ a1 + a0ydt dt dt dmf df = bm + ...+ b1 + b0f dtm dt (10) Poles of H(s) in the rh-plane indicate insta-bility in the continuous-time system. (11) The frequency response H(jω) may be in-terpreted geometrically from the poles and zeros of H(s) according to the following diagram: xx s - p l a n e s j W w o r r qq 1 12 2 q q jj 1 1 2 2 then H(jΩo)= H(s)| and s=jΩo m | H(jΩ)| = K� i=1 qi n i=1 ri n n = H(jΩ) φi − θi i=1 i=1 F(12) If f(t) ⇐⇒ F (s) then F−sτ F (s),f(t − τ ) ⇐⇒ e which is the delay property of the Laplace trans-form. The Z transform (contd.) (9) The transfer function bmz−m + bm−1z−(m−1) + ...+ b1z−1 + b0H(z)= anz−n + an−1z−(n−1) + ...+ a1z−1 + a0 is derived from the difference equation a0yk + a1yk−1 + ...+ an−1yk−(n−1) + a0yk−n = b0fk + b1fk−1 + ...+ bmfk−m (10) Poles of H(z) outside the unit circle indicate instability in the discrete-time system. (11) The frequency response H(jω), (ω =Ω/ΔT ) may be interpreted geometrically from the poles and zeros of H(z) according to the following dia-gram: x x z - p l a n e W or r qq 1 12 2 q q jj 1 1 2 2 0p u n i t c i r c l e Á { z } then H(ejω)= H(z)| jω and z=e � � � m � H(ejω)� = K� in =1 qi i=1 ri n n H(ejω)= φi − θi i=1 i=1 Z(12) If { fn} ⇐⇒ F (z) then Z{ fn−m} ⇐⇒ z −mF (z), which is the delay (shift) property of the Z trans-form.