Signal is f t ht n n n

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Unformatted text preview: EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 24 The interpolated signal is a sum of shifted sincs, weighted by the samples f (nT ), which looks like: f (t ) f (0)sinc(2Bt ) −2T −T 0 T 2T t f (−T )sinc(2Bt + 1) f (0)sinc(2Bt ) f (t ) f (T )sinc(2Bt − 1) −2T −T 0 T 2T t The sinc shifted to nT is 1 at nT , and zeros at all other samples. The sum of the weighted shifted sincs will agree with all of the samples! EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 25 Sampled Sinusoids To see how remarkable this is, we need to look at some sampled sinusoids. A bandlimited signal with a bandwidth 2B can be reconstructed perfectly from its samples providing the sampling rate 1/T > 2B , Typically, we think of sampled sinusoids as looking like 1 Amplitude 0.5 0 ω0 = (2π )(200 Hz) !0.5 !1 ωs = (2π )(8192 Hz) 0 1 2 3 4 Time, ms 5 6 7 8 −ωs /2 −ω0 ω0 ωs /2 1 At this sampling rate, it is easy to believe that you can reconstruct the sinusoid from its samples. Amplitude 0.5 0 !0.5 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly !1 0 1 2 3 4 Time, ms 5 6 7 26 8 As the sampling rate decreases the reconstruction task looks harder, 1 0 !1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 0 !1 1 0 !1 1 0 !1 1 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 27 Amplitude 0.5 0 !0.5 !1 Most sampled0 sinusoids are not very recognizable! 1 2 3 4 5 6 7 8 6 7 8 Time, ms 1 Amplitude 0.5 0 !0.5 !1 0 1 2 3 4 Time, ms 5 ω0 = (2π )(3000 Hz) ωs = (2π )(8192 Hz) −ωs /2−ω0 ω0 ωs /2 The fact that we know the signal is bandlimited is a very powerful contraint on the reconstruction. EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 28 Aliased Frequencies If the sampling rate is insufficient, there will be frequencies that are indistinguishable from each other, cos((0.75)2π t) 1 0.5 0 !0.5 !1 0 1 2 3 0 1 2 4 3 5 6 7 8 9 10 4 5 6 7 8 9 10 cos((1.25)2π t) 1 0.5 0 !0.5 !1 Both cos((0.75)2π t) and cos((1.25)2π t) and have exactly the same samples, when sampled at 2 Hz. EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 29 A cosine sampled just below its Nyquist rate looks the same as a cosine just above its Nyquist rate 1 ω0 = 0.9 ωs /2 0.5 1/2 0 1/2 !0.5 !1 0 1 2 3 4 5 6 7 8 9 10 1 −ωs /2 0.5 ω0 = ωs /2 0 1/2 ωs /2 −j/2 1/2 !0.5 !1 0 1 2 3 4 5 6 7 8 9 10 1 −ωs /2 ωs /2 ω0 = 1.1 ωs /2 0.5 0 !0.5 !1 0 1 2 3 4 5 6 7 8 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 9 10 −ωs /2 ωs /2 30 A sine sampled just below its Nyquist rate looks the same as the negative of sine sampled just above its Nyquist rate. 1 ω0 = 0.9 ωs /2 0.5 j/2 0 !0.5 !1 0 1 2 3 4 5 6 7 8 9 −ωs /2 10 1 0.5 −j/2 ω0 = ωs /2 0 ωs /2 j/2 !0.5 !1 0 1 2 3 4 5 6 7 8 9 −j/2 10 1 ω0 = 1.1 ωs /2 0.5 j/2 0 j/2 !0.5 !1 0 1 2 3 4 5 6 7 8 9 10 −j/2...
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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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