Overview of Discrete-Time Fourier Transform Topics Handy Equations - PowerPoint PPT Presentation

Overview of Discrete-Time Fourier Transform Topics Handy Equations • Handy equations and limits ∞ 1 � a n = | a | < 1 1 − a, • Definition n =0 • Low- and high- discrete-time frequencies N − 1 1 − a N � a n • Convergence issues = 1 − a n =0 • DTFT of complex and real sinusoids N − 1 a M − a N • Relationship to LTI systems � a n = 1 − a • DTFT of pulse signals n = M ∞ a • DTFT of periodic signals � na n = 1 − a, | a | < 1 • Relationship to DT Fourier series n =0 N a ( N +0 . 5) − a − ( N +0 . 5) • Impulse trains in time and frequency � a n = a 0 . 5 − a − 0 . 5 n = − N You should be able to prove all of these. J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 1 J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 2 Handy Limits Orthogonality Defined Two non-periodic power signals x 1 [ n ] and x 2 [ n ] are orthogonal if and + ∞ Ω( N ± 1 � � sin 2 ) only if � lim = +2 π δ (Ω − 2 πℓ ) sin(Ω 1 N 2 ) N →∞ 1 � ℓ = −∞ x 1 [ n ] x ∗ lim 2 [ n ] = 0 2 N + 1 N →∞ + ∞ � Ω( N ± 1 � cos 2 ) n = − N � lim = ± 2 π δ (Ω − π − 2 πℓ ) cos(Ω 1 2 ) N →∞ ℓ = −∞ � Ω( N ± 1 � cos 2 ) lim = 0 sin(Ω 1 2 ) N →∞ Ω( N ± 1 � � sin 2 ) lim = 0 cos(Ω 1 2 ) N →∞ • First is roughly analogous to a sinc function • All are periodic functions of frequency Ω with fundamental period of 2 π J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 3 J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 4

Orthogonality of Complex Sinusoids Importance of Orthogonality Suppose that we know a signal is composed of a linear combination of Consider two (possibly non-harmonic) complex sinusoids non-harmonic complex sinusoids x 1 [ n ] = e j Ω 1 n x 2 [ n ] = e j Ω 2 n � π x [ n ] = 1 X (e j Ω ) e j Ω n dΩ Are they orthogonal? 2 π − π How do we solve for the coefficients X (e j Ω ) ? N N 1 1 � x 1 [ n ] x ∗ � e j Ω 1 n e − j Ω 2 n lim 2 [ n ] = lim 2 N + 1 2 N + 1 N N →∞ N →∞ n = − N n = − N � x [ n ]e − j Ω o n lim N N →∞ 1 n = − N � e j (Ω 1 − Ω 2 ) n = lim � 1 � π 2 N + 1 N N →∞ � X (e j Ω ) e j Ω n dΩ n = − N � e − j Ω o n = lim 2 π � N →∞ − π 1 Ω 1 − Ω 2 = 2 πℓ n = − N = � π � N � 0 Otherwise = 1 1 � X (e j Ω ) e j Ω n e − j Ω o n lim dΩ 2 π 2 N + 1 N →∞ − π n = − N J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 5 J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 6 Workspace Definition + ∞ � π � N � = 1 X (e j Ω ) = � x [ n ] e − j Ω n � F { x [ n ] } = X (e j Ω ) e j (Ω − Ω o ) n lim dΩ 2 π N →∞ − π n = −∞ n = − N � π x [ n ] = 1 � e j (Ω − Ω o )( N +0 . 5) − e − j (Ω − Ω o )( N +0 . 5) X (e j Ω ) e j Ω n dΩ = 1 � � F − 1 � X (e j Ω ) � = X (e j Ω ) lim dΩ 2 π e j (Ω − Ω o )0 . 5 − e − j (Ω − Ω o )0 . 5 2 π 2 π N →∞ − π � π � � = 1 sin[(Ω − Ω o )( N + 0 . 5)] FT X (e j Ω ) ⇒ X (e j Ω ) lim dΩ • Denote relationship as x [ n ] ⇐ 2 π sin[(Ω − Ω o )0 . 5] N →∞ − π • Why use this odd notation for the transform? � π ∞ = 1 X (e j Ω ) 2 π � δ (Ω − Ω o ± 2 πℓ ) dΩ • Wouldn’t X (Ω) be simpler than X (e j Ω ) ? 2 π − π ℓ = −∞ • Answer: this awkward notation is consistent with the z -transform � π X (e j Ω ) δ (Ω − Ω o ) dΩ = + ∞ � x [ n ] z − n X (e j Ω ) = X ( z ) | z =e j Ω X ( z ) = − π = X (e j Ω o ) n = −∞ • This also enables us to distinguish between the DT & CT Fourier transforms J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 7 J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 8

Mean Squared Error Observations + ∞ x [ n ] = 1 � + ∞ X (e j Ω ) e j Ω n dΩ X (e j Ω ) = � x [ n ]e − j Ω n � X (e j Ω ) = x [ n ] e − j Ω n F { x [ n ] } = 2 π 2 π n = −∞ n = −∞ 1 � X (e j Ω ) e j Ω n dΩ x [ n ] ˆ = • Called the analysis and synthesis equations, respectively 2 π 2 π • Recall that e j Ω n = e j (Ω+ ℓ 2 π ) n , for any pair of integers ℓ and n + ∞ � x [ n ] | 2 MSE = | x [ n ] − ˆ • Thus, X (e j Ω ) is a periodic function of Ω with a fundamental n = −∞ period of 2 π • Unlike the DT Fourier series, the frequency Ω is continuous • Like the Fourier series, it can be shown that X (e j Ω ) minimizes the • Thus the DT synthesis integral can be taken over any continuous MSE over all possible functions of Ω interval of length 2 π • Like the DTFS, the error converges to zero • Note: this isn’t in the text J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 9 J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 10 Comments Discrete-Time Harmonics + ∞ Equivalence of Discrete−Time Harmonics x [ n ] = 1 � X (e j Ω ) e j Ω n dΩ 1 X (e j Ω ) = � x [ n ]e − j Ω n 0.0 π 2 π 0 2 π n = −∞ −1 1 • X (e j Ω ) describes the frequency content of the signal x [ n ] 0.2 π 0 −1 • x [ n ] can be thought of as being composed of a continuum of 1 0.4 π frequencies 0 −1 • X (e j Ω ) represents the density of the component at frequency Ω 1 0.6 π 0 −1 1 0.8 π 0 −1 1 1.0 π 0 −1 −10 −8 −6 −4 −2 0 2 4 6 8 10 J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 11 J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 12

MATLAB Code Discrete-Time Frequency Concepts • Recall that e j (Ω+ ℓ 2 π ) n = e j Ω n function [] = Harmonics(); close all; n = -10:10; • If seemingly very high-frequency discrete-time signals, t = -10:0.01:10; cos ((Ω + ℓ 2 π ) n ) , are equal to low-frequency discrete-time signals, w = [0:0.2:1]*pi; nw = length(w); cos(Ω n ) , what does low- and high-frequency mean in FigureSet(1,’LTX’); discrete-time? for cnt = 1:length(w), subplot(nw,1,cnt); • Note that the units of Ω are radians per sample h = plot([min(t) max(t)],[0 0],’k:’,t,cos(t*w(cnt)),’b’,t,cos(t*(w(cnt)+2*pi)),’r’); hold on; h = stem(n,cos(n*w(cnt))); • A sinusoid with a frequency of 0.1 radians per sample is the same set(h(1),’Marker’,’.’); set(h(1),’MarkerSize’,5); as one with a frequency of ( 0 . 1 + 2 π ) radians per sample set(h,’Color’,’k’); hold off; • Recall that cos( πn ) = ( − 1) n ylabel(sprintf(’%3.1f \\pi’,w(cnt)/pi)); ylim([-1.05 1.05]); box off; • No DT signal can oscillate “faster” between two samples if cnt==1, title(’Equivalence of Discrete-Time Harmonics’); end; • No DT signal can oscillate “slower” than 0 radians per sample if cnt~=nw, set(gca,’XTickLabel’,’’); • Thus end; end; – Ω = π = ℓ ( π + 2 π ) is the highest perceivable DT frequency AxisSet(8); print -depsc Harmonics; – Ω = 0 = ℓ (2 π ) is the lowest perceivable frequency J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 13 J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 14 Discrete-Time Frequency Concepts Continued Example 4: Unit Impulse Find the Fourier transform of x [ n ] = δ [ n ] . X (e j Ω ) 1 Ω − 4 π − 3 π − 2 π − π π 2 π 3 π 4 π 0 X (e j Ω ) 1 Ω − π π − 4 π − 3 π − 2 π 0 2 π 3 π 4 π • Low frequencies are those that are near 0 • High frequencies are those near ± π • Intermediate frequencies are those in between • Note that the highest frequency, π radians per sample is equal to 0.5 cycles per sample • We will encounter this concept again when we discuss sampling J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 15 J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 16

Convergence Example 5: Inverse of Impulse Train Sketch the following impulse train and find the inverse Fourier + ∞ x [ n ] = 1 � X (e j Ω )e j Ω n dΩ transform. X (e j Ω ) = � x [ n ]e − j Ω n ∞ 2 π 2 π � X (e j Ω ) = 2 π n = −∞ δ (Ω − Ω 0 − 2 πℓ ) ℓ = −∞ • Sufficient conditions for the convergence of the discrete-time Fourier transform of a bounded discrete-time signal: (any one of the following are sufficient) – Finite duration: There exists an N such that x [ n ] = 0 for | n | > N ∞ � – Absolutely summable: | x [ n ] | < ∞ ∞ n = −∞ | x [ n ] | 2 < ∞ � – Finite energy: n = −∞ • The synthesis equation always converges • There is no Gibb’s phenomenon in the time domain J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 17 J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 18 Example 5: Workspace Example 6: Constant Find the Fourier transform of x [ n ] = 1 . J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 19 J. McNames Portland State University ECE 223 DT Fourier Transform Ver. 1.23 20