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华科奥本海姆讲义4
CHAPTER 4THE CONTINUOUS-TIME FOURIER TRANSFORM4.0 INTRODUCTION Represent continuous-time aperiodic signals as linear combinations of complex exponentials. Fourier transform and inverse Fourier transform. (傅立叶变换) (傅立叶逆变换) Use Fourier methods to analyze and understand signals and LTI systems. 4.1 REPRESENTATION OF APERIODIC SIGNALS: THE CONTINUOUS-TIME FOURIER TRANSFORM Using the convolution property, we have to the input signal Example 4.15 Consider the response of an LTI system with impulse response the Fourier transforms of x(t) and h(t) are Therefore, Expanding Y(jω) in a partial-fraction expansion(部分分式展开) where A and B are constants to be determined. When b ≠ a Therefore, The inverse transform for each of the terms can be recognized by inspection, then we have When b = a Recognizing this as We can use the differentiation in the frequency-domain property. Thus, Consequently, Example 4.16 Determine the response of an ideal low-pass filter to an input signal x(t) that has the form of a sinc function. That is, The impulse response of the ideal low-pass filter is of a similar form: Therefore, where ω0 = min (ωi , ωc) . Finally, the inverse Fourier transform of Y(jω) is given by That is, depending on which of ωi andωc is smaller, the output is equal to either x(t) or h(t). 4.5 THE MULTIPLICATION PROPERTY modulation property (调制定理) Example 4.17 Let s(t) be a signal whose spectrum S(jω) is depicted in Figure (a). Also, consider the signal (a) Then -ω1 ω1 S(jω) A ω -ω0 ω0 P(jω) π π ω (b) the spectrum R(jω) of r(t) = s(t)p(t) is obtained by an application of the multiplication property: The spectrum of r(t) consists of the sum of two shifted and scaled versions of S(jω). -ω0 ω0 (-ω0-ω1) (-ω0+ω1) (ω0-ω1) (ω0+ω1) R(jω)= [S(jω)* P(jω)]/2π A/2 ω Example 4.18 Let us consider r(t) as obtained in Example 4.17, and let g(t) = r(
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