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Question fullscreenExpand Transcribed Image Text We have seen that the Hilbert transform introduces a 90° phase shift in the compo- nents of a signal and the transfer function of a quadrature filter can be written as e H(f) = {0 f >0 f =0. We can generalize this concept to a new transform that introduces a phase shift of 0 in the frequency components of a signal, by introducing e-je f > 0 f = 0 eje Hạ(f) = {0 f < 0 and denoting the result of this transform by xa(1), i.e., X4(f) = X(f)H9(f), where Xe(f) denotes the Fourier transform of xe (t). Throughout this problem, assume that the signal x (t) does not contain any DC components. 1. Find hø (1), the impulse response of the filter representing this transform. 2. Show that xe(t) is a linear combination of x(t) and its Hilbert transform. 3. Show that if x(t) is an energy-type signal, xe(t) will also be an energy-type signal and its energy content will be equal to the energy content of x(t). check_circle Expert Answer Hilbert transform analysis According to the Hilbert transform, the transfer function of the quadrature filter is written as:   In terms of the frequency, the given transform is generalized as:       part 1 The impulse response of the filter representing the given transform is to be calculated. The function  can be written as:   The inverse Fourier transform of the above function is given as: result The value of the impulse response of the filter  comes out to be:     part 2 to show that the  is the linear combination of the x(t) and its Hilbert transform. if the given function is a linear combination of the x(t) , then it will be written in the form as: the Fourier transform of the above equation gives.   from equations 1 and 2. the value of the constant b will be calculated as shown below.         calculation of the constants the value of the constant b is calculated above, the value of the constant a is calculated by plugging the values of the constant b into any of the equation. The expression is given as: The given function is a linear combination. part 3 the above can be solved using some results as shown below.   Final conclusion the signals  and the  are the orthogonal signals.  therefore the equation of the energy reduces to the form as shown below:     result the signal comes out to be an energy signal from the above calculation. also, the energy content of the signal comes out to be equal to x(t). 谢谢亲的回惠顾，期待您的下次光临！ 添加客服微信:bbwxnly，购买沟通交流so easy 欢迎咨询以下我们的一站式服务内容：课后习题解答查题｜文件文档解锁下载｜会员帐号包月月卡 Coursehero｜Scribd｜Studymode｜Oneclass｜Chegg｜Studyblue｜Termpaperwarehouse｜SolutionInn｜Bartleby｜eNotes｜Bookrags｜SaveMyEaxms｜Study.com｜Studypool｜Numerade ｜QuillBot｜and more… 谢谢亲的支持，祝您学习愉快:)