Frequency-Domain Analysis Fourier Series Consider a continuous complex signal x(t) ∈ [−T/2,T/2]. Represent x(t) using an arbitrary orthonormal basis ϕn(t): ∞ X x(t) = αnϕn(t) n=−∞ Orthonormality condition: Z T/2 1 ∗ ϕn(t)ϕk(t)dt = δ(n − k). T −T/2 ∗ Multiplying the above expansion with ϕk(t) and integrating over the interval, we obtain ∞ 1 Z T/2 1 Z T/2 X x(t)ϕ∗(t)dt = α ϕ (t)ϕ∗(t)dt T k T n n k −T/2 −T/2 n=−∞ ∞ ! X 1 Z T/2 = α ϕ (t)ϕ∗(t)dt n T n k n=−∞ −T/2 ∞ X = αnδ(n − k) = αk. n=−∞ EE 524, Fall 2004, # 3 1 Thus, the coefficients of expansion are given by Z T/2 1 ∗ αk = x(t)ϕk(t)dt. T −T/2 Proposition. The functions ϕn(t) = exp(j2πnt/T ) are orthonormal at the interval [−T/2,T/2]. Proof. T/2 T/2 1 Z 1 Z 2π(n−k) ∗ j T t ϕn(t)ϕk(t)dt = e dt T −T/2 T −T/2 sin[π(n − k)] = = δ(n − k). π(n − k) 2 Thus, we can take exponential functions ϕn(t) = exp(j2πnt/T ) as orthonormal basis =⇒ we obtain Fourier series. Fourier series for a periodic signal x(t) = x(t + T ): ∞ X j 2πnt x(t) = Xne T n=−∞ Z T/2 1 −j 2πnt Xn = x(t)e T dt. T −T/2 EE 524, Fall 2004, # 3 2 Fourier coefficients can be viewed as a signal spectrum: 2πn X ∼ X(Ω ), where Ω = ⇒ n n n T Fourier series can be applied to analyze signal spectrum! Also, this interpretation implies that periodic signals have discrete spectrum. Example: Periodic sequence of rectangles: Z T/2 Z τ/2 1 −j 2πnt 1 −j 2πnt Xn = x(t)e T dt = Ae T dt T −T/2 T −τ/2 τ Aτ sin(πnT ) = · τ real coefficients. T πnT EE 524, Fall 2004, # 3 3 Remarks: • In general, Fourier coefficients are complex-valued, ∗ • For real signals, X−n = Xn. • Alternative expressions exist for trigonometric Fourier series, exploiting summation of sine and cosine functions. EE 524, Fall 2004, # 3 4 Convergence of Fourier Series Dirichlet conditions: Dirichlet conditions are met for most of the signals encountered in the real world. Still, convergence has some interesting characteristics: N X j 2πnt xN (t) = Xne T . n=−N As N → ∞, xN (t) exhibits Gibbs’ phenomenon at points of discontinuity. Under the Dirichlet conditions: EE 524, Fall 2004, # 3 5 • The Fourier series = x(t) at points where x(t) is continuous, • The Fourier series = “midpoint” at points of discontinuity. Demo: Fourier series for continuous-time square wave (Gibbs’ phenomenon). http://www.jhu.edu/~signals/fourier2/index.html. EE 524, Fall 2004, # 3 6 Review of Continuous-time Fourier Transform What about Fourier representations of nonperiodic continuous- time signals? Assuming a finite-energy signal and T → ∞ in the Fouries series, we get limT →∞ Xn = 0. Trick: To preserve the Fourier coefficients from disappearing as T → ∞, introduce Z T/2 −j 2πnt Xen = TXn = x(t)e T dt. −T/2 EE 524, Fall 2004, # 3 7 Transition to Fourier transform: X(Ω) = lim Xen T →∞ Z T/2 −j 2πnt = lim x(t)e T dt T →∞ −T/2 Z ∞ = x(t)e−jΩtdt, −∞ where the “discrete” frequency 2πn/T becomes the continuous frequency Ω. Transition to inverse Fourier transform: ∞ ∞ X j 2πnt X Xen j 2πnt x(t) = lim Xne T = lim e T T →∞ T →∞ T n=−∞ n=−∞ 1 Z ∞ 2π 2πn = X(Ω)ejΩtdΩ ⇐ dΩ = , Ω = . 2π −∞ T T EE 524, Fall 2004, # 3 8 Continuous-time Fourier transform (CTFT): Z ∞ X(Ω) = x(t)e−jΩtdt, −∞ 1 Z ∞ x(t) = X(Ω)ejΩtdΩ. 2π −∞ Example: Finite-energy rectangular signal: Z ∞ X(Ω) = x(t)e−jΩtdt −∞ Z τ/2 = Ae−jΩtdt −τ/2 sin(Ωτ/2) = Aτ real spectrum. Ωτ/2 EE 524, Fall 2004, # 3 9 Remarks: • In general, Fourier spectrum is complex-valued, • For real signals, X(−Ω) = X∗(Ω). EE 524, Fall 2004, # 3 10 Dirac Delta Function Definition: ∞, t = 0, Z ∞ δ(t) = , δ(t)dt = 1. 0, t 6= 0 −∞ Do not confuse continuous-time δ(t) with discrete-time δ(n)! Sifting property: Z ∞ f(t)δ(t − τ)dt = f(τ). −∞ The spectrum of δ(t − t0): Z ∞ X(Ω) = x(t)e−jΩtdt −∞ Z ∞ −jΩt = δ(t − t0)e dt −∞ = e−jΩt0. Delta function in time domain: 1 Z ∞ δ(t) = ejΩtdΩ. 2π −∞ EE 524, Fall 2004, # 3 11 Delta function in frequency domain: 1 Z ∞ ∞, Ω = 0, δ(Ω) = e−jΩtdt = . 2π −∞ 0, Ω 6= 0 For signal x(t) = AejΩ0t we have Z ∞ X(Ω) = x(t)e−jΩtdt −∞ Z ∞ = A e−j(Ω−Ω0)tdt −∞ = A 2π δ(Ω − Ω0). EE 524, Fall 2004, # 3 12 Harmonic Fourier Pairs Delta function in frequency domain: jΩ0t e ↔ 2π δ(Ω − Ω0), cos(Ω0t) ↔ π [δ(Ω − Ω0) + δ(Ω + Ω0)], π sin(Ω t) ↔ [δ(Ω − Ω ) − δ(Ω + Ω )]. 0 j 0 0 EE 524, Fall 2004, # 3 13 Parseval’s Theorem for CTFT Z ∞ 1 Z ∞ |x(t)|2dt = |X(Ω)|2dΩ −∞ 2π −∞ 1 Z ∞ |X(Ω)|2dΩ 2π −∞ 1 Z ∞ Z ∞ Z ∞ = x(t)x∗(τ)e−jΩ(t−τ)dtdτ dΩ 2π −∞ −∞ −∞ Z ∞ Z ∞ 1 Z ∞ = x(t)x∗(τ) ejΩ(τ−t)dΩ dtdτ −∞ −∞ 2π −∞ | {z } δ(τ−t) Z ∞ = |x(t)|2dt. −∞ EE 524, Fall 2004, # 3 14 Discrete-time Fourier Transform (DTFT) Represent continuous signal x(t) via discrete sequence x(n): ∞ X x(t) = x(n)δ(t − nT ). n=−∞ Substituting this equation into the CTFT formula, we obtain: ∞ Z ∞ X X(Ω) = x(n)δ(t − nT )e−jΩtdt −∞ n=−∞ ∞ X Z ∞ = x(n) δ(t − nT )e−jΩtdt n=−∞ −∞ ∞ X = x(n)e−jΩnT . n=−∞ Switch to the discrete-time frequency, i.e. use ω = ΩT : ∞ X X(ω) = x(n)e−jωn. n=−∞ X(ω) is periodic with period 2π: ∞ X X(ω) = x(n)e−jωn e−j2πn | {z } n=−∞ 1 EE 524, Fall 2004, # 3 15 ∞ X = x(n)e−j(ω+2π)n = X(ω + 2π). n=−∞ Trick: in computing inverse DTFT, use only one period of X(ω): ∞ X X(ω) = x(n)e−jωn DTFT, n=−∞ 1 Z π x(n) = X(ω)ejωndω Inverse DTFT. 2π −π Inverse DTFT: Proof. 1 Z π x(n) = X(ω)ejωndω 2π −π ∞ 1 Z π X = x(m)ejω(n−m)dω 2π −π m=−∞ ∞ X 1 Z π = x(m) ejω(n−m)dω = x(n). 2π −π m=−∞ | {z } δ(n−m) 2 EE 524, Fall 2004, # 3 16 Fourier Series vs. DTFT ∞ Z T/2 X j 2πnt 1 −j 2πnt x(t) = X e T ,X = x(t)e T dt FS n n T n=−∞ −T/2 ∞ X 1 Z π X(ω) = x(n)e−jωn, x(n) = X(ω)ejωndω DTFT 2π n=−∞ −π Observation: Replacing, in Fourier Series x(t) → X(ω), Xn → x(n), t → −ω, T → 2π, we obtain DTFT! An important conclusion: DTFT is equivalent to Fourier series but applied to the “opposite” domain. In Fourier series, a periodic continuous signal is represented as a sum of exponentials weighted by discrete Fourier (spectral) coefficients. In DTFT, a periodic continuous spectrum is represented as a sum of exponentials, weighted by discrete signal values. EE 524, Fall 2004, # 3 17 Remarks: • DTFT can be derived directly from the Fourier series, • All Fourier series results can be applied to DTFT • Duality between time and frequency domains. EE 524, Fall 2004, # 3 18 Parseval’s Theorem for DTFT ∞ X 1 Z π |x(n)|2 = |X(ω)|2dω 2π n=−∞ −π 1 Z π |X(ω)|2dω 2π −π ∞ ∞ 1 Z π X X = x(n)x∗(m)e−jω(n−m)dω 2π −π n=−∞ m=−∞ ∞ ∞ X X 1 Z π x(n)x∗(m) e−jω(n−m)dω 2π −π n=−∞ m=−∞ | {z } δ(n−m) ∞ X = |x(n)|2. n=−∞ EE 524, Fall 2004, # 3 19 When Does DTFT Exist (i.e. |X(ω)| < ∞)? Sufficient condition: ∞ X |x(n)| < ∞. n=−∞ ∞ X −jωn |X(ω)| = x(n)e n=−∞ ∞ X ≤ |x(n)| |e−jωn| n=−∞ | {z1 } ∞ X = |x(n)| < ∞. n=−∞ Example: Finite-energy rectangular signal: N/2 N/2 X X |X(ω)| = Ae−jωn = A e−jωn n=−N/2 n=−N/2 N+1 sin( 2 ω) = A(N + 1) ω (N + 1) sin( 2 ) EE 524, Fall 2004, # 3 20 N+1 sin( 2 ω) ≈ A(N + 1) ω for ω π (N + 1) 2 | {z } well-known function Both functions look very similar in their “mainlobe” domain. EE 524, Fall 2004, # 3 21 DTFT — Convolution Theorem If X(ω) = F{x(n)} and H(ω) = F{h(n)} and ∞ X y(n) = x(k)h(n − k) = {x(n)} ? {h(n)} k=−∞ then Y (ω) = F{y(n)} = X(ω)H(ω). ∞ ∞ X X Y (ω) = F{y(n)} = x(k)h(n − k) e−jωn | {z } n=−∞ k=−∞ m ∞ ∞ X X = x(k)h(m)e−jω(m+k) m=−∞ k=−∞ ∞ ( ∞ ) X X = x(k)e−jωk h(m)e−jωm k=−∞ m=−∞ = X(ω)H(ω). Windowing theorem: If X(ω) = F{x(n)}, W (ω) = F{w(n)}, and y(n) = x(n)w(n), then 1 Z π Y (ω) = F{y(n)} = X1(λ)X2(ω − λ)dλ.