Consider the FM signal mentioned UFM (t) = Ac cos(2n fet + 0(t)), wherein the message...

  

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Consider the FM signal mentioned UFM (t) = Ac cos(2n fet + 0(t)), wherein the message signal m(t) = Am cos(2 fmt) is a sinusoid. The phase deviation 0(t) is also defined in Problem I. Assume that Ac = 1 for simplicity. Refer to page 114 in the textbook. Clearly the complex envelope, u(t), of the FM signal UAM (t), is ej(t), since UFM(t) = Re{u(t)ej² fet}. Since (t) is a periodic signal with period 1/fm, so is e(t) and hence the complex envelope u(t) can be written as a Fourier series (Refer to Chapter 2, page 40 in the textbook) as follows. u(n)j²fmt u(t) = 11=-x where u(n) are the complex Fourier coefficents. IT (Bain(2)-ns) dr (refer to the expression in the middle of page 114 in the textbook). Jn (3) refers to the Bessel function of the first kind of order n. ß is defined in page 110. 1. Show that Jn (3) can be rewritten as u(n) = Jn (B) = 6 2π Jn (3) = = √ √ ² cos (Bsin(x) - nr)dx. π (Hint: Use the fact that sin (Bsin(x) - nx) is an odd function.) 2. Show that Jn (B) = (-1)" Jn (B), that is, the odd numbered Fourier or Bessel coefficients with suffix +n and -n have opposite sign. For example, with n = 5, J5(B) = -J-5(3). (Hint: In the integral in part 1) make use of the change of variable x = - y.) 3. Show that Σ (3) = 1. 1==∞ 4. Since UFM (t) = Re{u(t)e2fet}, and u(t) = M=Ix for the Fourier Transform of uFM (t), namely UFM (f). n(B)e2nfmt, obtain an expression Consider the FM signal mentioned UFM (t) = Ac cos(2n fet + 0(t)), wherein the message signal m(t) = Am cos(2 fmt) is a sinusoid. The phase deviation 0(t) is also defined in Problem I. Assume that Ac = 1 for simplicity. Refer to page 114 in the textbook. Clearly the complex envelope, u(t), of the FM signal UAM (t), is ej(t), since UFM(t) = Re{u(t)ej² fet}. Since (t) is a periodic signal with period 1/fm, so is e(t) and hence the complex envelope u(t) can be written as a Fourier series (Refer to Chapter 2, page 40 in the textbook) as follows. u(n)j²fmt u(t) = 11=-x where u(n) are the complex Fourier coefficents. IT (Bain(2)-ns) dr (refer to the expression in the middle of page 114 in the textbook). Jn (3) refers to the Bessel function of the first kind of order n. ß is defined in page 110. 1. Show that Jn (3) can be rewritten as u(n) = Jn (B) = 6 2π Jn (3) = = √ √ ² cos (Bsin(x) - nr)dx. π (Hint: Use the fact that sin (Bsin(x) - nx) is an odd function.) 2. Show that Jn (B) = (-1)" Jn (B), that is, the odd numbered Fourier or Bessel coefficients with suffix +n and -n have opposite sign. For example, with n = 5, J5(B) = -J-5(3). (Hint: In the integral in part 1) make use of the change of variable x = - y.) 3. Show that Σ (3) = 1. 1==∞ 4. Since UFM (t) = Re{u(t)e2fet}, and u(t) = M=Ix for the Fourier Transform of uFM (t), namely UFM (f). n(B)e2nfmt, obtain an expression

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The problem asks us to find the Fourier transform of the positive part of an FM signal ... View the full answer