MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.161 Signal Processing - Continuous and Discrete 1 Sampling and the Discrete Fourier Transform 1 Sampling Consider a continuous function f(t) that is limited in extent, T1 · t < T2. In order to process this function in the computer it must be sampled and represented by a ¯nite set of numbers. The most common sampling scheme is to use a ¯xed sampling interval ¢T and to form a sequence of length N: ffng (n = 0 : : : N ¡ 1), where fn = f(T1 + n¢T ): In subsequent processing the function f(t) is represented by the ¯nite sequence ffng and the sampling interval ¢T . In practice, sampling occurs in the time domain by the use of an analog-digital (A/D) converter. The mathematical operation of sampling (not to be confused with the physics of sampling) is most commonly described as a multiplicative operation, in which f(t) is multiplied by a Dirac comb sampling function s(t; ¢T ), consisting of a set of delayed Dirac delta functions: X1 s(t; ¢T ) = ±(t ¡ n¢T ): (1) n=¡1 ? We denote the sampled waveform f (t) as X1 ? f (t) = s(t; ¢T )f(t) = f(t)±(t ¡ n¢T ) (2) n=¡1 ? as shown in Fig. 1. Note that f (t) is a set of delayed weighted delta functions, and that the waveform must be interpreted in the distribution sense by the strength (or area) of each component impulse. The implied process to produce the discrete sample sequence ffng is by integration across each impulse, that is Z + Z + n¢T n¢T X1 ? fn = f (t)dt = f(t)±(t ¡ n¢T )dt (3) ¡ ¡ n¢T n¢T n=¡1 or fn = f(n¢T ) (4) by the sifting property of ±(t). ? 1.0.1 Spectrum of the Sampled Waveform f (t): Notice that sampling comb function s(t; ¢T ) is periodic and is therefore described by a Fourier series: X1 1 jn­ t s(t; ¢T ) = e 0 ¢T n=¡1 1D. Rowell October 2, 2008 1 f ( t ) t s ( t ; DT ) t f * (t) = f(x)s(t; DT ) t Figure 1: Sampling by a comb impulse function. The function f(t) is multiplied by the sampling ? function s(t; ¢T ) to produce the sampled waveform f (t). where all the Fourier coe±cients are equal to (1=¢T ). Using this form, the spectrum of the sampled ? waveform f (t) may be written Z 1 ? ? ¡j­t F (j­) = f (t)e dt ¡1 Z X1 1 1 jn­ t ¡j­t = f(t)e 0 e dt ¢T n=¡1 ¡1 1 1 X = F (j (­ ¡ n­0)) ¢T n=¡1 1 � � ¶¶ 1 X 2¼n = F j ­ ¡ (5) ¢T n=¡1 ¢T ? where ­0 = 2¼=¢T . This tells us that the Fourier transform of a sampled function f (t) is periodic in the frequency domain with period ­0, and is a superposition of an in¯nite number of shifted Fourier transforms F (j­) of the original function, scaled by a factor of 1=¢T . 1.1 The Nyquist Sampling Theorem Given a set of samples ffng and its generating function f(t), an important question to ask is whether the sample set uniquely de¯nes the function that generated it? In other words, given ffng can we unambiguously reconstruct f(t)? The answer is clearly no, as shown in Fig. 2 where there are obviously many functions that will generate the given set of samples. In fact there are an in¯nity of candidate functions that will generate the same sample set. 2 f ( t ) t Figure 2: Demonstration that a set of samples does not uniquely de¯ne a continuous function. There are clearly many functions that would generate the same set of samples. The Nyquist sampling theorem places restrictions on the candidate functions and, if satis¯ed, will uniquely de¯ne the function that generated a given set of samples. The theorem may be stated in many equivalent ways, we present three of them here to illustrate di®erent aspects of the theorem: ² A function f(t), sampled at equal intervals ¢T , can not be unambiguously reconstructed from its sample set ffng unless it is known a-priori that f(t) contains no spectral energy at or above a frequency of ¼=¢T radians/s. ² In order to uniquely represent a function f(t) by a set of samples, the sampling interval ¢T must be su±ciently small to capture more than two samples per cycle of the highest frequency component present in f(t). ² There is only one function f(t) that is band-limited to below ¼=¢T radians/s that is satis¯ed by a given set of samples ffng. Note that the sampling rate, 1=¢T , must be greater than twice the highest cyclic frequency fmax in f(t). Thus if the frequency content of f(t) is limited to ­max radians/s (or fmax cycles/s) the sampling interval ¢T must be chosen so that ¼ ¢T < ­max or equivalently 1 ¢T < 2fmax The minimum sampling rate to satisfy the sampling theorem fmin = ­max=¼ samples/s is known as the Nyquist rate. 1.2 Aliasing Consider a sinusoid f(t) = A sin(at + Á) sampled at intervals ¢T , so that the sample set is ffng = fA sin(an¢T + Á)g ; and noting that sin(t) = sin(t + 2k¼) for any integer k, �� ¶ ¶ 2¼m f = A sin(an¢T + Á) = A sin a + n¢T + Á (6) n ¢T where m is an integer, giving the following important result: 3 Given a sampling interval of ¢T , sinusoidal components with an angular frequency a and a + 2¼m=¢T , for any integer m, will generate the same sample set. Figure 3 shows a sinusoid sampled at three di®erent rates. In Fig. 3(a) the waveform is sampled at a rate above the Nyquist rate, and the function is uniquely de¯ed by the samples. In (b) the sampling interval ¢T is at the Nyquist rate (two samples/cycle) and the sample set is ambiguous; note that the function f(t) = 0 generates the same samples. In (c) The sinusoid is undersampled and a lower frequency sinusoid, shown as a dashed line, also satis¯es the sample set. The phenomenon demonstrated in Fig. 3 is known as aliasing. After sampling any spectral component in F (j­) above the Nyquist frequency ¼=¢T will \masquerade" as a lower frequency component within the reconstruction bandwidth, thus creating an erroneous reconstructed function. The phenomenon is also known as frequency folding since the high frequency components will be \folded" down into the assumed system bandwidth. One-half of the sampling frequency (i.e. 1=(2¢T ) cycles/second, or ¼=¢T radians/second) is known as the aliasing frequency, or folding frequency for these reasons. Equation (6) de¯nes how frequency components above the folding frequency will be aliased into the range below the folding frequency. This is illustrated in Fig. 4. Figure 5 shows the e®ect of folding in another way. In Fig. 5(a) a function f(t) with Fourier transform F (j­) has two disjoint spectral regions. The sampling interval ¢T is chosen so that the folding frequency ¼=¢T falls between the two regions. The spectrum of the sampled system between the limits ¡¼=¢T < ­ · ¼=¢T is shown in Fig. 5(b). The higher frequency components have been folded down into the region ¡¼=¢T < ­ · ¼=¢T . 1.2.1 Anti-Aliasing Filtering: Once a sample set ffng has been taken, there is nothing that can be done to eliminate the e®ects of aliased frequency components. The only way to guarantee that the sample set unambiguously represents the generating function is to ensure that the sampling theorem criteria have been met, either by 1. Selecting a sampling interval ¢T su±ciently small to capture all spectral components, or 2. Processing the function f(t) to \eliminate" all components at or above the Nyquist rate. The second method involves the use of a continuous-domain processor before sampling f(t). A low-pass aanti-aliasing ¯lter is used to eliminate (or at least attenuate) spectral components at or above the Nyquist frequency. Ideally the anti-aliasing ¯lter would have a transfer function H(j­) = 1 for j­ j < ¼=¢T = 0 otherwise,: In practice it is not possible to design a ¯lter with such characteristics, and a more realistic goal is to reduce the o®ending spectral components to insigni¯cant levels, while maintaining the ¯delity of components below the folding frequency. Figure 6 illustrates the use of an anti-aliasing ¯lter. 2 Reconstruction of a Function from its Sample Set ? ? Equation (5) demonstrates that the spectrum F (j­) of a sampled function f (t) is in¯nite in extent and consists of a periodic extension of F (j­) with a period of 2¼=¢T , i.e. � � ¶¶ X1 ? 1 2¼n F (j­) = F j ­ ¡ : ¢T n=¡1 ¢T 4 f(t) o o o o o t o 0 Δ T o o o o o o o 0 (a) f(t) t o o o o o o o o 0 Δ T (b) f(t) o o Δ t o T 0 o o o o 0 (c) Figure 3: Demonstration that a set of samples does not necessarily uniquely define a continuous function. In (a) the sampling rate satisfies the sampling theorem and there is no ambiguity, in (b) and (c) the sampling theorem is not satisfied and lower frequency waveforms also generate the same sample set.