# Example f1 j 120 80 40 0f1 j0 0 40 80 ee102asignal

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Unformatted text preview: in the time domain? If f (t) is a signal, then f (t)δT (t) is the sampled signal, ¯ f (t) = f (t)δT (t) and it Fourier transform is ¯ F (j ω ) = F [f (t)δT (t)] 1 = F [f (t)] ∗ F [δT (t)] 2π 1 = F (j ω ) ∗ (ω0δω0 (ω )) 2π 1 = F (j ω ) ∗ δω0 (ω ) T EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 18 If we expand δω0 (ω ), ¯ F (j ω ) = = ∞ ￿ 1 F (j ω ) ∗ δ (ω − nω0) T n=−∞ ∞ 1￿ F (j (ω − nω0)) T n=−∞ The spectrum of the sampled signal consists of shifted replicas of the original spectrum, scaled by 1/T . What this looks like exactly depends on the sampling frequency ω0. We’ll deﬁne the bandwidth of f (t) to be ±B Hz. The following plot shows the case where 2π B ￿ ω0/2 This is the case where the signal bandwidth is much less than the sampling rate. EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 19 F ( j!) −!0 −2!B 0 ∗ −!0 1 F ( j(! + !0)) T −!0 2!B ! 1 !" (") T0 0 ! = 1 F ( j!) T −2!B 0 lowpass ﬁlter 2!B ¯ F (!) 1 F ( j(! − !0)) T ! ¯ If we lowpass ﬁlter F (j ω ), then we can perfectly recover F (j ω ), and f (t)! EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 20 As the sampling frequency ω0 decreases (sampling period T increases) the spectral replicas get closer: 2!B −2!B −!0 0 −!0 −2!0 −3!0 −2!0 −!0 !0 0 2!0 !0 0 !0 ! ! 2!0 3!0 ! Eventually the replicas overlap, and F (j ω ) cannot be recovered. EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 21 The overlap is called aliasing because the low frequencies of one band appear (alias) as high frequencies of the next band. High frequencies from one band also alias as low frequencies of the next band. 2!B −2!B −!0 −!0/2 !0/2 0 !0 ! No aliasing occurs only if 2π B < ω0/2, or 2B < ω0/2π = 2π 1 1 = Hz T 2π T The signal can be recovered exactly only if the signal bandwidth 2B Hz is less than or equal to the sampling rate 1/T Hz. The sampling rate 2B is the Nyquist rate for f (t), the lowest rate that allows f (t) to be perfectly recovered. T is the Nyquist interval. EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 22 Signal Reconstruction: the Interpolation Formula The ideal lowpass ﬁlter for 2B = 1/T , sampling at the Nyquist rate, is T rect ￿!￿ 4"B 1 F (!) T −!0 −!0/2 0 −2!B !0 ! !0/2 2!B The lowpass ﬁlter has the Fourier transform ￿ω￿ H (j ω ) = T rect 4π B Using the fact that sinc(t) ⇔ rect(ω /2π ) and the scaling theorem, h(t) = 2BT sinc(2Bt) = sinc(2Bt) EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 23 since we are assuming 2B = 1/T for sampling at the Nyquist rate. The reconstructed signal is ¯ f (t) ∗ h(t) = = ￿ ∞ ￿ n=−∞ ∞ ￿ n=−∞ = ∞ ￿ n=−∞ = ∞ ￿ n=−∞ ￿ f (nT )δ (t − nT ) ∗ h(t) f (nT )h(t − nT ) f (nT ) sinc(2B (t − nT )) f (nT ) sinc(2Bt − n) This is the Whittaker-Kotelnikov-Shannon sampling theorem....
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