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102_1_lecture8_student - UCLA Fall 2011 Systems and Signals...

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UCLA Fall 2011 Systems and Signals Lecture 8: Continuous Time Fourier Series II October 19, 2011 EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 1
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Introduction Today’s topics: Review: Fourier series Properties of continuous time Fourier series Introduction to filtering EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 2
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Review . Last time we’ve seen that complex exponentials are useful for studying LTI systems. If a system has a transfer function H ( s ) and an input can be written as: x ( t ) = X k a k e s k t The system output will be: y ( t ) = X k a k H ( s k ) e s k t LTI system scales complex exponentials by its transfer function H ( s ) This allows us to avoid calculating the convolution integral! EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 3
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We’ve also seen how to decompose continuous, periodic signals into sums of complex exponentials. Continuous Time Fourier Series x ( t ) = X k = -∞ a k e jkω 0 t (Synthesis) a k = 1 T Z T x ( t ) e - jkω 0 t dt (Analysis) The two equations allow us to alternate between time domain and frequency domain. The transformation is invertible: you can first compute a k from x ( t ) (go from time to frequency domain), and then compute x ( t ) from a k (return to time domain). Both representations uniquely specify the signal. EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 4
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We will sometimes use a shorthand notation for this transformation (your book also uses this): x ( t ) F S --→ a k a k F S - 1 ---→ x ( t ) Notice some facts about the representations. Signals in time domain are continuous and periodic (for now!..). Frequency domain is discrete (by construction, signals we deal with only contain frequencies at ω = 0 , where ω 0 = 2 π T ). Fourier coefficients are, in general, complex (even if the signal is real). We can gain insight into frequency content of a signal by considering magnitude and phase of Fourier coefficients ( | a k | and 6 a k ). EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 5
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Example . Given Fourier coefficients a k , find associated signal x ( t ) that’s periodic in T = 8 . a k = 1 if k = 0 2 j if k = 1 - 2 j if k = - 1 0 otherwise . In this case, it’s enough to look at the synthesis equation: x ( t ) = X k = -∞ a k e jkω 0 t (Synthesis) x ( t ) = 1 e j 0 ω 0 t + 2 j e j 1 ω 0 t - 2 j e j ( - 1) ω 0 t EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 6
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x ( t ) = 1 + 4 e 0 t - e - 0 t 2 j You immediately see this as a sin function. Since ω 0 = 2 π/ 8 , this can be rewritten as: x ( t ) = 1 + 4 sin( π 4 t ) Example. Find the Fourier series coefficients of the periodic (period T , ω 0 = 2 π T ) square wave defined as: x ( t ) = 1 if | k | < T 1 0 if T 1 < | t | < T . t T 1 T x(t) -T 1 -T EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 7
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Use the analysis equation. Integrate from - T/ 2 to T/ 2 . a k = 1 T Z T/ 2 - T/ 2 x ( t ) e - jkω 0 t dt a k = 1 T Z T 1 - T 1 e - jkω 0 t dt a k = - 1 jkω 0 T e - jkω 0 t T 1 - T 1 a k = - 1 jkω 0 T e - jkω 0 T 1 - e jkω 0 T 1 Difference of complex exponentials is can be rewritten a sin function EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 8
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a k = 2 0 T e jkω 0 T 1 - e - jkω 0 T 1 2 j a k = 2 0 T sin( 0 T 1 ) Recalling that ω 0 = 2 π T , 2 0 T = 2 T k 2 πT = 1 .
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