# part4 - Part 4 Fourier Series Representation of Periodic...

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Part 4: Fourier Series Representation of Periodic Signal Periodic Signals z Different types of periodic signals z Difficult to analyze

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Questions z Is it possible to represent any periodic waveform by trigonometric series? z If the answer is positive, how to find the representation? z Note: With such representation, the output of the system can be found more easily. Example: Linear System x(t, ω ) Linear System s 1 (t, ω ) y(t, ω ) s 3 (t, ω ) s 2 (t, ω ) : : y 1 (t, ω ) y 3 (t, ω ) y 2 (t, ω ) : : x(t, ω )=s 1 (t, ω )+s 2 (t, ω )+s 3 (t, ω ) … y(t, ω )=y 1 (t, ω )+y 2 (t, ω )+y 3 (t, ω ) … Trigonometric Fourier Series () = = + + = 1 0 1 0 0 sin cos n n n n t n b t n a a t x ω where = 0 0 0 0 1 T dt t x T a () ( ) = 0 0 0 0 cos 2 T m dt t m t x T a for 0 m = 0 0 0 0 sin 2 T m dt t m t x T b for 0 m (1) Note: The range of the integral can be started from any time. Eg. 2 / 2 / 0 0 T T
EE3118 Linear System & Signal Analysis (KST) Fourier Series for periodic function Given a periodic function x(t) with period T 0 , let ( ) () = = + + = 1 0 1 0 0 sin cos n n n n t n b t n a a t x ω (1) where 0 0 2 2 T f π = = . To obtain the coefficient, 0 a , we take the integral over one period (any period can be done. In the following example, t from 0 to 0 T ): = = + + = 0 0 0 0 0 0 1 1 0 0 0 0 0 sin cos T n n n T n T T dt t n b dt t n a dt a dt t x Since 0 cos 0 0 0 = T dt t n and ( ) 0 sin 0 0 0 = T dt t n over the period, = 0 0 0 0 1 T dt t x T a To obtain the coefficient, i a , multiply Eqn.(1) with ( ) t m 0 cos and then take the integral over one period: ( ) ( ) ( ) = = + + = 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 cos sin cos cos cos cos T n n n T n T T dt t m t n b dt t m t n a dt t m a dt t m t x Given that = = 0 2 0 cos cos 0 0 0 0 0 n m T n m dt t m t n T Hence, = 0 0 0 0 cos 2 T m dt t m t x T a for 0 m . To obtain the coefficient, i b , multiply Eqn.(1) with ( ) t m 0 sin and then take the integral over one period: ( ) ( ) ( ) = = + + = 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 sin sin sin cos sin sin T n n n T n T T dt t m t n b dt t m t n a dt t m a dt t m t x 0 0 0 0

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EE3118 Linear System & Signal Analysis (KST) Similar, due to the fact that () ( ) = = 0 2 0 sin sin 0 0 0 0 0 n m T n m dt t m t n T ω Hence, = 0 0 0 0 sin 2 T m dt t m t x T b for 0 m . In summary, = = ω + ω + = 1 0 1 0 0 n n n n t n b t n a a t x sin cos where = 0 0 0 0 1 T dt t x T a = 0 0 0 0 cos 2 T m dt t m t x T a for 0 m . () ( ) = 0 0 0 0 sin 2 T m dt t m t x T b for 0 m .
EE3118 Linear System & Signal Analysis (KST) Example: Show that the Fourier series of a square wave is () = = 1 0 1 2 sin 1 2 4 k t k k t x ω π Referring to (1), we have () ( ) = = + + = 1 0 1 0 0 sin cos n n n n t n b t n a a t x where 0 1 1 1 1 2 2 0 0 0 = + = = T T T T dt dt T dt t x T a 0 sin sin 2 cos cos 2 cos 2 2 0 0 2 0 0 0 2 0 2 0 0 0 0 = = = = T T T T T T T m m t m m t m T dt t m dt

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part4 - Part 4 Fourier Series Representation of Periodic...

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