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**Unformatted text preview: **ECE 2025 Summer 2010, Problem Set 3, Solutions 3.1 The desired signal has the form x ( t ) = cos( ( t )) ,- 2 < t < 2 , otherwise where ( t ) determines the instantaneous frequency over the time interval- 2 < t < 2. (a) To achieve the desired f i ( t ), we require d dt = 2 sin 2 2 t = (1- cos( t )) which implies ( t ) = t- sin( t ) + constant (b) To achieve the desired f i ( t ), we require d dt = 2 4 e- ( t +2) / 2 which implies ( t ) =- 16 e- ( t +2) / 2 + constant 3.2 For this problem, Fourier coefficients are obtained by using the inverse Euler identities cos = e j + e- j 2 sin = e j- e- j 2 j (a) The signal has radian frequencies of { , 8 , 18 } . Therefore, the signal is periodic with funda- mental frequency = 2 . The Fourier coefficients are a k = 2 2 e j 1- 3 8 , k =- 9 6 e- j 3 4 , k =- 4 7 e j , k = 0 6 e j 3 4 , k = 4 2 2 e- j 1- 3 8 , k = 9 , otherwise (b) The signal has radian frequencies of 9 , 9 . Since the ratio of these frequencies is an irrational number, 2 , the signal is not periodic. Thus, there is no fundamental frequency or Fourier coefficients to specify. (c) The signal can be reduced to a sum of complex exponential signals: x ( t ) =- 2cos(20 t + / 3)cos(- 12 t ) =- 1 2 e j/ 3 e j 20 t + e- j/ 3 e- j 20 t ( e- j 12 t + e j 12 t ) = 1 2 e- j 2 / 3 e j 20 t + e j 2 / 3 e- j 20 t ( e- j 12 t + e j 12 t ) = 1 2 e- j 2 / 3 e j 8 t + e- j 2 / 3 e j 32 t + e j 2 / 3 e- j 32 t + e j 2 / 3 e- j 8 t The signal has radian frequencies of { 8 , 32 } . Therefore, the signal is periodic with fundamental frequency = 8 . The Fourier coefficients are a k = 1 2 e j 2 / 3 , k =- 4 1 2 e j 2 / 3 , k =- 1 1 2 e- j 2 / 3 , k = 1 1 2 e- j 2 / 3 , k = 4 , otherwise 1 (d) The signal has radian frequencies of { 18 , 36 , 54 ,... } . Therefore, the signal is periodic with fundamental frequency = 18 . Since sin 18 kt + 3 = 1 2 j e j 3 e j 18 kt- 1 2 j e- j 3 e- j 18 kt = 1 2 e- j 6 e j 18 kt + 1 2 e j 6 e- j 18 kt it follows that a k =...

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