Dt t 2 1 t t 2 t 2 tej 2ntt dt t

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Unformatted text preview: Fourier series is then ∞ ￿1 f (t) = ej 2πnt/T T n=−∞ = ∞ 1 ￿ jnω0t e . T n=−∞ EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 9 The Fourier transform of δT (t) is then F [δT (t)] = ∞ 1￿ 2πδ (ω − nω0) T n=−∞ = ω0 ∞ ￿ n=−∞ δ (ω − nω0) = ω0δω0 (ω ) since ω0 = 2π /T . We then have the transform pair δT (t) ⇔ ω0δω0 (ω ) The Fourier transform of an array evenly spaced δ ’s is another array of evenly spaced δ ’s! EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 10 The delta train δT (t) is !T (t ) 1 -3T -2T -T 0 T 2T 3T t and its Fourier transform ω0δω0 (t) is !0"!0 (!) !0 −3!0 −2!0 −!0 !0 0 !0 = 2"/T 2!0 3!0 ! These are very useful functions! EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 11 Ideal Sampling One of the most important uses of δT (t) is to represent sampling. If f (t) is a signal, then f (t)δT (t) = f (t) ∞ ￿ n=−∞ = ∞ ￿ n=−∞ = ∞ ￿ n=−∞ δ (t − nT ) f (t)δ (t − nT ) f (nT )δ (t − nT ) where we have used the fact that f (t)δ (t − T ) = f (T )δ (t − T ) for the last step. EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 12 Original Signal -3T f (t ) -2T -T 0 T 2T 3T t 2T 3T t !T (t ) 1 -2T 3T t × Sampling Function -3T 2T -T 0 T = Sampled Signal -3T -2T f (t )!T (t ) -T 0 T EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 13 Example: Square Wave, Revisited We can write the square wave function from earlier today as a convolution: f (t) = rect(t) ∗ δ2(t) which is illustrated below rect(t ) −3 −3 −2 −2 −1 −1 0 ∗ 0 = 1 2 3t 2 3t 2 3t !2(t ) 1 f (t ) −3 −2 −1 0 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 1 14 Using the convolution theorem, the Fourier transform of f (t) is F [f (t)] = F [rect(t)] F [δ2(t)] = sinc(ω /2π )ω0δω0 (ω ) = π sinc(ω /2π )δπ (ω ) where ω0 = 2π /T = 2π /2 = π . To get this into the form we found earlier, expand δπ (ω ) F [f (t)] = π sinc(ω /2π ) =π ∞ ￿ n=−∞ =π ∞ ￿ n=−∞ ∞ ￿ n=−∞ δ (ω − π n) sinc(ω /2π )δ (ω − π n) sinc(n/2)δ (ω − π n) EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 15 which is the same thing we obtained before. In general, if f1(t) is one cycle of a periodic function with period T , then f (t) is f (t) = f1(t) ∗ δT (t) and the Fourier transform is F (j ω ) = ω0F1(j ω )δω0 (ω ) Recall that multiplying by δω0 (ω ) samples F1(j ω ) at multiples of ω0. The Fourier transform of the periodic signal is the sampled Fourier transform of one period EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 16 This is what we saw the periodic rect signal example: F1( j!) −12!0 −8!0 −4!0 !0F1( j!)"!0 (!) 0 4!0 8!0 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 12!0 ! 17 Sampling Theorem What is the Fourier transform of a signal that has been sampled...
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This note was uploaded on 09/28/2013 for the course EE 108b taught by Professor Mukamel during the Fall '13 term at Singapore Stanford Partnership.

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