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hw03su10_soln

# hw03su10_soln - ECE 2025 Summer 2010 Problem Set 3...

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ECE 2025 Summer 2010, Problem Set 3, Solutions 3.1 The desired signal has the form x ( t ) = cos( ψ ( t )) , - 2 < t < 2 0 , otherwise where ψ ( t ) determines the instantaneous frequency over the time interval - 2 < t < 2. (a) To achieve the desired f i ( t ), we require dt = 2 π sin 2 π 2 t = π (1 - cos( πt )) which implies ψ ( t ) = πt - sin( πt ) + constant (b) To achieve the desired f i ( t ), we require dt = 2 π 4 e - ( t +2) / 2 which implies ψ ( t ) = - 16 πe - ( t +2) / 2 + constant 3.2 For this problem, Fourier coefficients are obtained by using the inverse Euler identities cos θ = e + e - 2 sin θ = e - e - 2 j (a) The signal has radian frequencies of { 0 , 8 π, 18 π } . Therefore, the signal is periodic with funda- mental frequency ω 0 = 2 π . The Fourier coefficients are a k = 2 2 e j 1 - 3 8 π , k = - 9 6 e - j 3 4 π , k = - 4 7 e , k = 0 6 e j 3 4 π , k = 4 2 2 e - j 1 - 3 8 π , k = 9 0 , otherwise (b) The signal has radian frequencies of 9 π, 9 π . Since the ratio of these frequencies is an irrational number, π 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 ) = - 2 cos(20 πt + π/ 3) cos( - 12 πt ) = - 1 2 e jπ/ 3 e j 20 πt + e - jπ/ 3 e - j 20 πt ( e - j 12 πt + e j 12 πt ) = 1 2 e - j 2 π/ 3 e j 20 πt + e j 2 π/ 3 e - j 20 πt ( e - j 12 πt + e j 12 πt ) = 1 2 e - j 2 π/ 3 e j 8 πt + e - j 2 π/ 3 e j 32 πt + e j 2 π/ 3 e - j 32 πt + e j 2 π/ 3 e - j 8 πt The signal has radian frequencies of { 8 π, 32 π } . Therefore, the signal is periodic with fundamental frequency ω 0 = 8 π . The Fourier coefficients are a k = 1 2 e j 2 π/ 3 , k = - 4 1 2 e j 2 π/ 3 , k = - 1 1 2 e - j 2 π/ 3 , k = 1 1 2 e - j 2 π/ 3 , k = 4 0 , otherwise 1

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(d) The signal has radian frequencies of { 18 π, 36 π, 54 π, . . . } . Therefore, the signal is periodic with fundamental frequency ω 0 = 18 π . Since sin 18 πkt + π 3 = 1 2 j e j π 3 e j 18 πkt - 1 2 j e - j π 3 e - j 18 πkt = 1 2 e - j π 6 e j 18 πkt + 1 2 e j π 6 e - j 18 πkt it follows that a k = 13 , k = 0 1 k 2 +1 1 2 e - j π 6 , k > 0 1 k 2 +1 1 2 e j π 6 , k < 0 3.3 One period of a periodic signal with fundamental period T 0 is specified by x ( t ) = cos π T 0 t , for - T 0 2 < t < T 0 2
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hw03su10_soln - ECE 2025 Summer 2010 Problem Set 3...

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