Sampling Signals Overview: ¯ We use the Fourier transform to understand the discrete sampling and re-sampling of signals. ¯ One key question is when does sampling or re-sampling provide an adequate representation of the original signal? Terminology: ¯ sampling – creating a discrete signal from a continuous process. ¯ downsampling (decimation) – subsampling a discrete signal ¯ upsampling – introducing zeros between samples to create a longer signal ¯ aliasing – when sampling or downsampling, two signals have same sampled representation but differ between sample locations. Matlab Tutorials: samplingTutorial.m, upSample.m 320: Sampling Signals ­c A.D. Jepson and D.J. Fleet, 2005 Page: 1 Sampling Signals I(x) continuous signal x I[n] ● ● ● ● ● sampled ● ● ● ● ● ● ● ● ● signal ● ● ● ● n g[m] downsampling ● ● ● ● ● ● ● I[n] 2 g[m] ● ● m f[n] ● ● upsampling ● ● ● ● ● ● ● g[m] 2 f[n] ● ● ● ● ● ● ● ● ● n 320: Sampling Signals Page: 2 Up-Sampling [Ò] Æ Consider up-sampling a signal Á of length : ¯ Æ Ò Increase number of samples by a factor of × . ¯ Ò ½ Á [Ò] f [Ò] ¼ Step 1. Place × zeros after every sample of , to form Ò Æ of length × , namely ´ Á [Ò=Ò ] Ò ÑÓ d Ò = ¼; × × for f [Ò] = ¼ (1) ¼ otherwise. Signal Raw Upsampled Signal 1.5 2 1.5 1 1 0.5 0.5 0 0 f 0 I[n] −0.5 −0.5 −1 −1.5 −1 −2 −1.5 0 5 10 15 0 5 10 15 20 25 30 n n 320: Sampling Signals Page: 3 Up-Sampling (Cont.) ¯ f Step 2. Interpolate signal ¼ : f = Ë £ f ; Ë [Ò]: ¼ for some smoothing kernel (2) f [Ò] f [Ò] Ë [¼] = ½ Here, in order for to interpolate ¼ , we require that Ë [jÒ ] = ¼ j and × for nonzero integers . Sinc Filtered Upsampled Signal (b) 1.5 1 0.5 0 f[n] −0.5 −1 −1.5 0 5 10 15 20 25 30 n ¯ Many smoothing kernels can be used (see upSample.m). 320: Sampling Signals Page: 4 Frequency Analysis of Up-Sampling f [Ò] Step 1. Fourier transform of the raw up-sampled signal ¼ : ÆÒ ½ Æ ½ × X X × × Ò i! jÒ i! × k k F ´f [Ò]µ[k ] = f [Ò]e f [jÒ ]e ¼ ¼ ¼ × Ò=¼ j =¼ Æ ½ X × i! jÒ × k = Á [j ]e j =¼ ¾ × k = Here ! , so the last line above becomes k ÆÒ × Æ ½ X ¾ ÆÒ ÆÒ × × kj i Æ F ´Á µ[k ]; F ´f µ[k ] = Á [j ]e k < : ¼ for (3) ¾ ¾ j =¼ F ´Á µ Æ F ´f µ Since is -periodic, equation (3) implies that ¼ consists of Ò F ´Á µ × copies of concatenated together. Amplitude Spectrum of I[n] Amplitude Spectrum of Raw Upsampled Signal 6 6 5 5 4 4 3 3 |FT| |FT| 2 2 1 1 0 0 −5 0 5 −15 −10 −5 0 5 10 15 k k 320: Sampling Signals Page: 5 Fourier Analysis of Up-Sampling Step 2 f [Ò] = Ë £ f Ë Recall Step 2 is to form ¼ , for some interpolation filter . However, notice from the inverse Fourier transform that (for Æ even) Æ=¾ ½ X ¾ ½ kj i ^ Æ Á [j ] = Á [k ]e Æ k = Æ=¾ Æ=¾ ½ X ¾ Ò × i kj Ò × ^ ÆÒ × f [k ]e = f [jÒ ]: = ¼ × (4) ÆÒ × k = Æ=¾ f ´jÒ µ = Á ´j µ Here we used × in the last line. Notice the left term in the ^ Ò B [k ]f [k ] ¼ last line above is × times the inverse Fourier transform of B [k ] = ½ Æ=¾ k < Æ=¾ where B is the box function, for and [k ] = ¼ B otherwise. We can therefore evaluate this inverse Fourier Ò jÒ transform at every pixel , and not just at the interpolation values × , [Ò] to construct a possible interpolating function f ½ ^ f [Ò] = Ò F ´B ´k µf [k ]µ: ¼ × (5) Sinc Filtered Upsampled Signal (b) Amp. Spec. (Upsampled Sig.(r), Sinc Filter(g), Filtered Sig.(b) 1.5 12 1 10 0.5 8 0 6 f[n] |FT| 4 −0.5 2 −1 0 −1.5 −15 −10 −5 0 5 10 15 0 5 10 15 20 25 30 k n 320: Sampling Signals Page: 6 Down-Sampling [Ò] Æ Consider down-sampling a signal Á of length : ¯ Æ Ò Ò × Reduce number of samples by a factor of × , where is a divi- sor of Æ . ¯ Define the comb function: ´Æ=Ò µ ½ × X Æ C ´Ò; Ò µ = Ò;ÑÒ × × Ñ=¼ I [n] ns ns = 4 ●●●●●●●●●●●●●●●●● n Á [Ò] ¯ Step 1. Introduce zeros in at unwanted samples. g [Ò] = C [Ò; Ò ]Á [Ò]: × ¼ (6) ¯ g Step 2. Downsample signal ¼ : g [Ñ] = g [ÑÒ ]; ¼ Ñ < Æ=Ò : × × ¼ for (7) 320: Sampling Signals Page: 7 Frequency Domain Analysis of Down-Sampling Proposition 1. The Fourier transform of the comb function is another comb function: Æ F ´C [Ò; Ò ]µ = C [k ; Æ=Ò ]: × × (8) Ò × ∧ ω C( ) 2π n = 4 ns s −π 0 π ω ¾ k ! = k Recall the frequency ! and wave number are related by .So Æ ¾ Æ ¾ ! = = the spacing in the plot above is × . Æ Ò Ò × × Proposition 2. Pointwise product and convolution of Fourier trans- [Ò] g [Ò] Æ forms. Suppose f and are two signals of length (extended to be Æ -periodic). Then ½ F ´f µ £F´g µ F ´f [Ò]g [Ò]µ = Æ Æ=¾ ½ X ½ ^ f [j ] g^[k j ]: (9) Æ j = Æ=¾ ^ g^ f g where f and denote the Fourier transforms of and , respectively. The proofs of these two propositions are straight forward applications of the definition of the Fourier transform given in the preceeding notes, and are left as exercises. 320: Sampling Signals Page: 8 Fourier Analysis of Down-Sampling Step 1 g [Ò] = C [Ò; Ò ]Á [Ò] × Recall Step 1 is to form ¼ . By Prop. 1 and 2 above, we have F ´g µ = F ´C [Ò; Ò ]Á [Ò]µ ¼ × ½ Æ ^ C [¡; Æ=Ò ] £ Á [¡] = × Ò Æ × Æ=¾ X ½ ^ C [j ; Æ=Ò ]Á [k j ] × Ò × j = Æ=¾·½ Ò Ö × ¼ X ½ Æ ^ = Á [k Ö ; ] (10) Ò Ò × × Ö = Ö ·½ ¼ ^ Á [k ] Ö where, due to the periodicity of , we can use any integer ¼ (eg. Ö = Ò =¾ Ò × × ¼ for even ). ∧ ∧ I(ω) g (ω) 2π 0 n s ns = 4 −π 0 π ω −π 0 π ω ^ g^ [k ] Á [k ÖÆ=Ò ] ¼ Therefore consists of the sum of replicas × of the Á Æ=Ò Fourier transform of the original signal , spaced by wavenumber × ¾ ! = or, equivalently, by frequency × . Ò × g^ ¾ Note the Fourier transform ¼ has period , so the contribution sketched ¾ above for ! can be shifted to the left. 320: Sampling Signals Page: 9 Fourier Analysis of Down-Sampling Step 2 g [Ò] In Step 2 we simply drop the samples from ¼ which were set to zero C [Ò; Ò ] by the comb function × . That is g [Ñ] = g [ÑÒ ]; ¼ Ñ < Æ=Ò : × × ¼ for In terms of the Fourier transform, it is easy to show F ´g µ[k ] = g^[k ] = g^ [k ]; Æ=´¾Ò µ k < Æ=´¾Ò µ: × × ¼ for (11) ¾ ! = k g [Ñ] Rewriting this in terms of the frequency ×;k (note is ´Æ=Ò µ × ¾ Æ=Ò ! = k k a signal of length × ), and the corresponding frequency Æ g [Ò] = ! Ò ¼ × ×;k of the longer signal ,wehave g^[! ] = g^ [! ] = g^ [! =Ò ]; ! < : ¼ k ¼ ×;k × ×;k ×;k for (12) ∧ ω f( ) ns = 4 original signal −π ω π ω 0 s ∧ g(ω) downsampled signal −π 0 π 4π ω ∧ ω f( ) upsampled and filtered signal −π 0 π ω 320: Sampling Signals Page: 10 Nyquist Sampling Theorem [Ò] Sampling Theorem: Let f be a band-limited signal such that ^ f [! ] = ¼ j! j > ! for all ¼ ! f [Ò] g [Ñ] = for some ¼ . Then is uniquely determined by its samples f [ÑÒ ] × when ¼ > ! ! =¾ = Ò < ¼ × × or equivalently Ò ¾ × = ¾=! ¼ where ¼ . In words, the distance between samples must be smaller than half a wavelength of the highest frequency in the signal. ! In terms of the previous figure, note that the maximum frequency ¼ ! must be smaller than one-half of the spacing, × , between the replicas introduced by the sampling. This ensures the replicas do not overlap. [Ñ] When the replicas do not overlap, we can up-sample the signal g [Ò] and interpolate it to recover the signal f , as discussed above. g [k ] Otherwise, when the replicas overlap, the Fourier transform ^ con- ^ [k ] tains contributions from more than one replica of f . Due to these [Ò] aliased contributions, we cannot then recover the original signal f . 320: Sampling Signals Page: 11 Sampling Continuous Signals ´Üµ Ü ¾ [¼;ĵ A similar theorem holds for sampling signals f for .We can represent f as the Fourier series ½ X i! Ü ^ k f [k ]e ; f ´Üµ = a:e: k = ½ ¾ ! = k = a:e: where k and denotes equals almost everywhere. Suppose Ä f ´Üµ ! > ¼ is band-limited so that for some ¼ ^ j! j > ! : f [k ] = ¼ ¼ for all k ´Üµ Á [Ñ] = f ´Ñµ Then f is uniquely determined by its samples when ¼ < (13) ¾ = ¾=! ¼ where ¼ . In words, the distance between samples must be smaller than half the wavelength of the highest frequency in the signal. ´Üµ The link with the preceeding analysis is that sampling f withasam- ple spacing of causes replicas in the Fourier transform to appear with ! = ¾= spacing × . As before, the condition that these replicas do not ! < ! =¾ × overlap is ¼ , which is equivalent to condition (13). 320: Sampling Signals Page: 12 Aliasing Aliasing occurs when replicas overlap: Consider a perspective image of an infinite checkerboard. The signal is dominated by high frequencies in the image near the horizon. Properly designed cameras blur the signal before sampling, using ¯ the point spread function due to diffraction, ¯ imperfect focus, ¯ averaging the signal over each CCD element. These operations attenuate high frequency components in the signal.