VII. Bandwidth Limited Time Series

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1 VII. Bandwidth Limited Time Series

2 To summarize the discussion up to this point: (1) In the general case of the aperiodic time series, which is infinite in time and frequency, both the time series and the Fourier series are continuous functions. (2) When we consider a time series of finite length T, the Fourier series becomes discrete at integer multiples of the fundamental frequency, which is inversely related to T. (3) However, at this point the time series is still continuous and the frequency ranges from negative to positive infinity (or from 0 to positive infinity using real notation). The frequencies that can be resolved are the fundamental frequency and all integer multiples of it.

3 Consider now the case where, in addition to being of finite length, the time series is also discrete rather than continuous. This is the situation that results when a continuous analog signal is digitized at a fixed sampling interval Δt. The sampled time series is now referred to as X(kΔt), where k indexes the interval. Discrete temporal sampling has the effect of limiting the highest frequency that can be resolved.

4 In this example, we see the same periodic time series (300 Hz signal) sampled at 3 different sampling rates (10000, 625, and Hz). Sampling at Hz is fully adequate to represent the time series; sampling at 625 Hz is barely adequate; and sampling at Hz is not adequate. This example shows that for any frequency component in a time series, the sampling rate must be high enough to represent that component. Alternatively, we can say that for any sampling rate, frequency components in the time series must be low enough to be resolved.

5 What is the highest frequency that can be resolved for a given sampling rate? First, note that the sampling rate (S), which is the number of samples per unit of time, equals 1/Δt. ow, consider that we need at least 2 time points to define a periodic wave at any frequency. The 2 points specify the maximum and minimum of the wave. Thus, for an arbitrary sampling interval Δt: (a) The maximum and minimum are separated by Δt (1/2 cycle occurs in Δt). (b) 0.5 cycle / Δt sec (or 1 cycle / 2 Δt sec) is the highest frequency that is uniquely determined (i.e., resolved) by sampling interval Δt.

6 yquist Frequency The maximum frequency that can be resolved given a particular sampling interval Δt is called the yquist frequency. It represents 1 cycle in 2Δt. 1 F = cycles /sec( Hz) 2Δt 2π = radians / sec 2 Δ t There being a maximum resolvable frequency means that the resolvable spectrum of a sampled time series is bandwidth (or band) limited between 0 and F.

7 The yquist frequency (F) is half of the sampling rate (S): F 1 1 S = = = 2Δt S

8 We now define as the number of sampled intervals within the observation period T. =T/Δt (and Δt=T/ and T=Δt). Remember that the index n represents the integer multiplier of the fundamental frequency in the Fourier series representation. The yquist frequency is itself part of that representation, so there must be a value of n corresponding to the yquist frequency. That value of n is /2: 2π 2π 2π F = = = ( )( ) 2Δt 2T 2 T 2π Sinceω 0 = ; F = ( 2 T ) ω 0

9 We can now write the Fourier series representation of the discrete time series X(kΔt) (in real notation): A(0) X(k t) = + [ A(n) (n k t) + B(n) (n k t)] /2 Δ 2 n=1 cos ω0 Δ sin ω0 Δ We see that n ranges from 1 to /2. ote that there are /2 Fourier components in addition to A(0). Since each component has 2 coefficients, the total number of coefficients needed to represent X(kΔt) would appear to be +1. However, it turns out that B(/2)=B(0)=0. Therefore, a total of coefficients is needed to represent X(kΔt).

10 Bandwidth-limited DFT In complex notation, the bandwidth-limited versions of the DFT and the inverse DFT are: Z(n) = 1 X(kΔt) = k= 1 2 n= _ 2 X(kΔt)e and Z(n)e -jnω kδt 0 0 jnω kδt

11 yquist-shannon sampling theorem: if the coefficients of the Fourier series for a continuous time series are all 0 for frequencies higher than the yquist frequency, then sampling at Δt is guaranteed to represent all the information in the time series. In other words, if all the power of the time series is contained in frequency components at the yquist frequency or lower, then sampling at Δt will be adequate to evaluate the Fourier coefficients uniquely. That is, a continuous time series can be completely reconstructed by the sampled time values, provided it is band limited by the yquist frequency. In short, all is well if the time series bandwidth is less than the yquist frequency.

12 Aliasing When the time series bandwidth extends beyond the yquist frequency, it can cause a problem called aliasing. Even though the Fourier coefficients may be computed in the usual way, they may suffer from errors due to aliasing. The size of the errors depends on the amount by which the range of power in the time series exceeds the yquist frequency.

13 The more power there is in the time series above the yquist frequency, the greater are the errors in the Fourier coefficients and the less adequate are the coefficients for representing the time series. Aliasing may result in serious misinterpretation of spectra if unrecognized. The onus is upon the analyst to ensure that sampling is adequate for the bandwidth of the time series being analyzed.

14 How to Avoid Aliasing One common way to avoid aliasing is to use low-pass analog filtering prior to A-D conversion. The bandwidth of the data must be limited to the yquist frequency PRIOR to sampling in order to prevent these errors. Example: If one decides to sample at 200 samples/sec, then Δt is 5 msec and the yquist frequency is 100 c/sec. Then, to prevent errors the analog time series must be lowpass filtered to remove any appreciable power above 100 c/sec.

15 We now want to examine how spectral analysis can be falsified by sampling when the time series contains power at frequencies above the yquist frequency. That is, how does aliasing arise? To answer this, we must consider the case where we sample a continuous time series X(t), assuming that it is bandwidth limited, when in fact it is not. The sampled time series is X(kΔt).

16 We apply the bandwidth-limited version of the DFT to the sampled time series to get the apparent Fourier coefficients: 1 -jn 0k t Z(n) = ω Δ X(kΔt)e k = 1 where T = Δt. The integer n represents all the frequency components in the Fourier series representation of X(kΔt).

17 Aliasing arises because sampling causes the loss of time points containing the fine structure of the time series. Consequently, in the frequency domain, components above the yquist frequency can no longer be represented properly. What happens to these higher frequency components? To answer this question, we must consider the true Fourier series representation, given by: jm 0t X(t) = Z(m)e ω m= - The integer m represents all the frequency components in the Fourier series representation of X(t). We want to see how these frequency components distribute when we naively perform the bandwidth-limited DFT (i.e. incorrectly assuming that there is no power above the yquist frequency).

18 In essence, we are using the bandwidth-limited Fourier series representation (only containing frequency components n, where /2 <= n <= /2): jn 0 X(kΔt) = Z(n)e ω Δ n= _ 2 to represent a time series (containing frequency components m that extend above the yquist frequency), whose true Fourier series representation is: 2 k t jm 0t X(t) = Z(m)e ω m= -

19 Aliasing results when the frequency components in X(t) above the yquist frequency cannot properly be represented. To understand the errors caused by aliasing, we must understand the relation between n (frequencies within the yquist range) and m (all frequencies in X(t)).

20 To derive the relation between m and n, we first substitute the true Fourier series representation into the equation for the bandwidth-limited DFT: 1 jmω0kδt jnω 1 0kΔt j( m n) ω0kδt Z( n) = Z( m) e e Z( m) e = k= 1 m= k= 1 m= Rearranging, we get: 1 j( m n) ω0kδt 1 j( m n) ω0kδt Z( n) = Z( m) e Z( m) e = k= 1 m= m= k= 1

21 We next need to resolve the expression in curly brackets. (a) First note that the frequencies involved are (m-n)ω 0, where m ranges from - to. (b) ext consider that for any frequency of the form (m-n)ω 0, an integer number of periods exists in the time interval from k = 1 to. (c) Since the sum of a cosine or sine wave over an integer number of periods equals 0, the sum of the exponential term exp[jω 0 (m-n)kδt] over the time interval equals 0. Then, the expression in curly brackets also equals 0. (d) However, there is an exception to this general conclusion. In the special case when (m-n) = 0, e 0 =1, and the sum over the time interval equals. In this case, the expression in curly brackets equals 1.

22 (e) In fact, there is also an exception to the general conclusion if (m-n) is a non-zero integer multiple of, i.e. (m-n)=i, where i is a non-zero integer multiplier of. (f) If (m-n) = i, with i 0, then the exponential term exp[jω 0 ikδt] reduces to exp(j2πik). [Here we use the facts that ω 0 =2π/T and T=Δt.] (g) We now use Euler s relation: exp(j2πik) = cos(2πik)+jsin(2πik) = = 1 for all i. Again, the sum over the time interval equals, and the expression in curly brackets equals 1. (h) Thus, the expression in curly brackets equals 1 or 0 depending on whether (m-n) is equal or not equal to i.

23 This conclusion is stated formally as: 1 j m n ω 1 0kΔt e = k = 1 ( ), for m n= i 0, for m n i Essentially, the frequencies m-n=i survive because their exponential terms equal 1. All other frequencies are eliminated because their exponential terms equal 0. Thus, the frequencies m-n=i have a special meaning.

24 Conclusion about the expression: 1 j( m n) ω0kδt 1 j( m n) ω0kδt Z( n) = Z( m) e Z( m) e = k= 1 m= m= k= 1 1. most values of m, i.e. those for which the sum of the exponential term equals 0, make only zero contributions to Z(n). 2. only the values of m for which (m-n) = i, i.e. those for which the sum of the exponential term equals 1, make a non-zero contribution to Z(n).

25 Thus, we can use: to express: 1 j m n ω 1 0kΔt e = k = 1 ( ), for m n= i 0, for m n i 1 j( m n) ω0kδt 1 j( m n) ω0kδt Z( n) = Z( m) e Z( m) e = k= 1 m= m= k= 1 as: Z(n)= i= - Z( n +i )

26 This relation says that each term in the DFT of X(kΔt) is the sum of a possibly infinite set of Fourier coefficients associated with the higher frequency components in X(t). The higher frequency components are those corresponding to frequencies that are greater than n by an amount i. If X(t) has no Fourier series components greater than /2 (corresponding to the yquist frequency), then the apparent Fourier coefficients are correct. In other words, the DFT of X(kΔt) yields correct results only if X(t) is bandwidth-limited to frequencies below the yquist frequency (the yquist-shannon sampling theorem).

27 Errors arise when X(t) has frequency components higher than the yquist frequency. These higher frequency components add to (i.e. mix with) components below the yquist frequency. This mixing is what causes aliasing.

28 Aliasing Cannot Be Reversed Once aliasing occurs, there is no way to separate out the true coefficient from the aliased one. This is why the lowpass cutoff frequency of the analog filter must be matched to the sampling rate such that the highest frequency with any appreciable power must be half of the sampling frequency, i.e. the yquist frequency. This is essential for analysis of sampled continuous data.

29 Example: Consider a cosine at the yquist frequency (F) sampled at the negative and positive peaks. If the frequency of the wave increases or decreases by a small increment a, sampling appears to be indistinguishable. In other words, it appears that a wave of frequency (F + a) will, after sampling, be confused with a wave of frequency (F a).

30 Generalization of the Example To determine all the frequencies that will be aliased onto F a, start with: Z(n)= i= - Z( n +i ) with n=(/2) k, where k is an integer multiplier of ω 0 corresponding to some arbitrary frequency component below the yquist frequency. ote about this equation: 1. Index i extends from negative to positive infinity. 2. Since Z is complex, the representation of frequency n is both at Z(n) and Z(-n).

31 Let n in this equation have the value of [(/2) k]: Z(n) becomes Z[(/2) k] and Z( n) becomes Z[( /2)+k]. Then: n+i =(/2)- k +i = (i +1/2) - k and - n+i = (-/2)+k +i =(i - 1/2) +k

32 The two sequences are as follows: i (i+1/2)-k (i-1/2)+k (-3/2)-k (-5/2)+k -1 (-/2)-k (-3/2)+k ** 0 (/2)-k (-/2)+k 1 (3/2)-k (/2)+k 2 (5/2)-k (3/2)+k... ** -- this is the term that shows aliasing

33 In real frequency terms: A real term at (/2)+k comes from the complex pair at (/2)+k and (-/2)-k. A real term at (3/2)-k comes from the complex pair at (3/2)-k and (-3/2)+k. A real term at (3/2)+k comes from the complex pair at (3/2)+k and (-3/2)-k. A real term at (5/2)-k comes from the complex pair at (5/2)-k and (-5/2)+k. And so on. In effect, the original Fourier representation of x(t) has been folded in accordion fashion about frequencies that are multiples of F, and collapsed into the frequency region extending from 0 to F (which is also called the folding frequency). In other words, the frequency axis is folded back onto the range 0-F in pieces of length F.

34 Graphical Representation of Aliasing

IB Paper 6: Signal and Data Analysis

IB Paper 6: Signal and Data Analysis Handout 5: Sampling Theory S Godsill Signal Processing and Communications Group, Engineering Department, Cambridge, UK Lent 2015 1 / 85 Sampling and Aliasing All of