By Lehmer D. N.

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**Example text**

K=1 Since t varies from 0 to T , θ = 2πt/T varies from 0 to 2π, we have g(θ + 2π) = g(θ). Note that g(θ) completes one cycle over any interval of length 2π, including the commonly used [−π, π]. 5. 49) Xk ejkθ . g(θ) = k=−∞ 6. In Chapter 5, we will learn that the frequency contents of a nonperiodic function x(t) are de ned by a continuous-frequency function X(f ), and we will also encounter the Fourier series representation of the periodically extended X(f ), which appears in the two forms given below.

5. 5 An example: the sum of 11 cosine and 11 sine components. 4 Using Complex Exponential Modes By using complex arithmetics, Euler s formulaejθ = cos θ + j sin θ, where j ≡ resulting identities cos θ = ejθ − e−jθ ejθ + e−jθ , sin θ = , 2 2j √ −1, and the CHAPTER 1. REPRESENTATION OF FUNCTION CONTENTS 8 we can express y(t) in terms of complex exponential modes as shown below. 2) n y(t) = Ak cos(2πfk t) + Bk sin(2πfk t) k=1 n ej2πfk t + e−j2πfk t 2 Ak = k=1 n Ak − jBk 2 = k=1 n ej2πfk t + ej2πfk t − e−j2πfk t 2j + Bk Ak + jBk 2 Note: X±k ≡ Xk ej2πfk t + X−k ej2πf−k t , = e−j2πfk t k=1 Ak ∓ jBk , f−k ≡ −fk 2 n (Note: the term X0 ≡ 0 is added) Xk ej2πfk t + X−k ej2πf−k t , = X0 + k=1 n Xk ej2πfk t .

XN −1 } to the sequence of coef cients {X0 , X1 , . . , XN −1 } without solving a system of equations. 1. 7) N −1 −r x ωN , for r = 0, 1, . . , N − 1. 5). Since X±k are the coef cients of the complex exponential modes e±j2πkt/T , the corresponding frequencies ±fk = ±k/T are marked on the frequency grid. 2 Equally-spaced samples and computed DFT coef cients. 3 t 2 Frequency Grid (∆f = 1/T) Y ±k 1 −f5 0 −f5 0 f5 0 f5 = 5/T We defer the matrix formulation of the DFT until Chapter 4. It turns out that because of the special properties of the DFT matrix, the DFT coef cient Xr can be computed more ef ciently using various fast Fourier transform algorithms (commonly known as the FFT).