A digital signal processing primer with applications to by Ken Steiglitz

By Ken Steiglitz

This article is directed on the marketplace of DSP clients led to by way of the advance of strong and cheap software program instruments to research signs. those instruments permit refined manipulation of indications yet don't supply an figuring out of the speculation or the basis for the suggestions. This paintings develops an method of the advance of the maths of DSP and makes use of examples from components of the spectrum everyday to newcomers, including questions and instructed experiments

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6. Fourier transform of periodic functions Let us consider a periodic function f (t ) with a period of T and Cn the coefficient of the exponential Fourier series. 7. Energy density The energy contained in a non-periodic signal appears as follows: E= +∞ ∫ −∞ f 2 (t ) dt Fourier Transform +∞ ∫ We demonstrate that f 2 (t ) dt = −∞ 55 +∞ 1 ∫ F ( jω )F ( − jω ) d ω if f (t ) is 2π −∞ real. 8. 15). 2. Exercises The exercises presented in this chapter come from [LAT 66]. 1.

N = +∞ ⎧ ( t ) v Cn e jnω0t = ⎪ ∑ n = −∞ ⎪ 16π 2 ⎪⎪ ( j ) Z jn 0 . 01 ω ω = + + ⎨ 0 jnω0 ⎪ n = +∞ n = +∞ ⎪ Cn e jnω0t = ∑ Cn' e jnω0t ⎪i (t ) = ∑ ⎪⎩ n = −∞ Z ( jnω0 ) n = −∞n Both voltages have the same period T = 1 meaning ω0 = 2π . 01 + j 2π n2 − 4 n For n=0 which is the same as saying ω = 0 , the impedance becomes infinite. The continuous component, or mean value of the current will be null which is explained by the presence of a capacitor in the circuit. 001 for n = ±2 or ω = ±4π = ±2ω0 which is the resonant frequency of the RLC series circuit.

As the spectrum is real, we will trace an = 2Cn . 2. 2 f (t ) is orthogonal to cos nt on Let us begin by proving that interval [0,2π ] : I= 2π ∫ 0 π 2π 0 π f ( t ) cos nt dt = ∫ cos nt dt + ∫ − cos nt dt The integral is null, f (t ) is hence orthogonal to cos nt . This tells us that the Fourier decomposition will not include any cosine values. Let us now show that the error function f e ( t ) is orthogonal to sin t on interval [0,2π ] . I' = 2π ∫ 0 π 2π 4 4 ⎞ ⎞ ⎛ ⎛ f e ( t ) sin t dt = ∫ ⎜ 1 − sin t ⎟ dt + ∫ ⎜ − 1 − sin t ⎟ dt π π ⎠ ⎠ ⎝ ⎝ π 0 The integral is null and f e (t ) is hence orthogonal to sin t .

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