The Discrete Fourier Transform

Size: px

Start display at page:

Download "The Discrete Fourier Transform"

Cornelius Lucas
5 years ago
Views:

1 The Discrete Fourier Transform / 23

2 The Fourier Transform Any discrete time series exponentials: can be built up of a sum of complex = X m e i2πmn/, n = 0,, m=0 = [ ]. X 0 X e i2πn (/) X 2 e i2πn (2/) X i2πn (3/) 3 e So the m th term behaves as e i2πf m n = cos(2πf m n) + i sin(2πf m n), where m/. ote the sample interval is implicitly set to one! f m The quantity m/ is called the mth Fourier frequency. f m The period associated with f m m/ is / f m = /m. Thus m tells us the number of oscillations contained in the length time series. Δ 2 / 23

3 Continuous Time cos(2πfmt) and sin(2πfmt) fm = 0, /00, 2/00, 3/00 t = [0 00] 3 / 23

4 Discrete Time cos(2πfmn) and sin(2πfmn) fm = 0, /00, 2/00, 3/00 n = 0,, 2, 99 4 / 23

Musical Analogy We can think of the Fourier transform in musical terminology. Fundamental (First Harmonic) First Overtone (2nd Harmonic) 2nd Overtone (3rd Harmonic).

5 Musical Analogy We can think of the Fourier transform in musical terminology. Fundamental (First Harmonic) First Overtone (2nd Harmonic) 2nd Overtone (3rd Harmonic)... We can imagine that the sine terms in the Fourier transform describe the vibration of a string that is the twice the length of our time series. We must also include the cosine terms having antinodes at the endpoints. 5 / 23

6 The yquist Frequency The single most important frequency is the highest resolvable frequency, the yquist frequency. = f 2Δ ω 2π 2Δ π Δ The highest resolvable frequency is one cycle per two data points. e i2π f m nδ e i2π /(2Δ) nδ e iπn ) n = = = ( =,,,, ote that there is no sine component at yquist! 6 / 23

7 Aliasing Q: What if you try to observe a higher frequency than the yquist? 7 / 23

8 Aliasing Q: What if you try to observe a higher frequency than the yquist? A. You will think you see things that aren't really there. 7 / 23

9 The Wagon Wheel Effect Aliasing is a kind of optical illusion. In film, it's known as the wagon wheel effect. 0:00 Thanks to {Dora} for posting. 8 / 23

10 Bizarre Consequences 0:00 Thanks to {ick Moore} for posting. 9 / 23

11 Aliasing = Wagon Wheel Aliasing and the wagon-wheel effect are essentially the same thing. Unresolved frequencies are said to be aliased into resolved ones. 0 / 23

12 Aliasing = Wagon Wheel Aliasing and the wagon-wheel effect are essentially the same thing. Unresolved frequencies are said to be aliased into resolved ones. Q: For what δf does cos(2πf Δn) = cos(2πf 2 Δn) with f 2 = f + δf? Q: For what δf does e i2πf Δn i2π = e f 2 Δn with f 2 = f + δf? Q: If a one-spoked wheel filmed with a sample interval of one sample per second appears to turn clockwise at the rate of one revolution per three seconds, what are the possible true rotation rates? What about a two-spoked wheel? Three spokes? 0 / 23

13 The Rayleigh Frequency The second most important frequency is the lowest resolvable frequency, the Rayleigh frequency. f Δ ω 2π Δ The lowest resolvable frequency is one cycle over the entire record. Here the sample interval and the number of points is. Δ = = 0 If we think of the data as a vibrating string, the Rayleigh frequency is the first overtone. The fundamental does not appear in the DFT. / 23

14 Importance of Rayleigh The Rayleigh frequency is important because it gives the spacing between the Fourier frequencies: f 0 = 0, f =, f 2 Δ f 2 =, f n f f Δ Δ = n, = Thus, it controls the frequency domain resolution. For example, if you want to distiguish between two closely spaced peaks, you need the dataset duration to be sufficiently large so that the Rayleigh frequency is sufficiently small. The ratio of the Rayleigh to yquist frequencies tells you how many different frequencies you can resolve. f f Δ = = 2Δ 2 2 / 23

15 The Fourier Series Any discrete time series exponentials. can be built up of a sum of complex = X m e i2πmn/, n = 0,, m=0 If there are points in, then you also need different complex exponentials. Q: Is this obvious? 3 / 23

16 The Fourier Series Any discrete time series exponentials. can be built up of a sum of complex =, n = 0,, m=0 X m e i2πmn/ If there are points in, then you also need different complex exponentials. Q: Is this obvious? Subtlety (i): may be real-valued, but is complex. Subtlety (ii): There are different numbers in, but 2 different numbers in = R{ } + ii{ }. X m X m X m These two facts are resolved by noting that half of the information in the complex-valued series is redundant if is real-valued. X m X m 3 / 23

17 The Fourier Frequencies The first few Fourier frequencies while the last few are = m/ But notice that the last Fourier term will be are: because e i2πn for all integers n! Frequencies higher than then yquist cannot appear due to our sample rate. Therefore, these terms instead specify terms that have a frequency less than the yquist but that rotate in the negative direction. f m =, =, =, 0 f 0 f 2 f 2 2 2, f 2 = =, f = =. e i2π( )n/ = e = = i2πn( /) e i2πn e i2πn/ e i2πn/ = 4 / 23

18 The Fourier Frequencies In the vicinity of m = /2 for even we have f /2 =, f /2 =, f /2+ = +, 2 but actually the first frequency higher than the yquist is the highest negative frequency 2 2 f /2 =, =, = ( ),. 2 f /2 2 f /2+ 2 Thus the positive frequencies and negative frequencies are both increasing toward the middle of the Fourier transform array. For this reason Matlab provides fftshift, to shifts the zero frequency, not the yquist, to be in the middle of the array. ote! If you give fftshift an array of multiple time series, it will shift both dimensions, assuming you had carried out a 2-D FFT! 5 / 23

19 Conjugate Pairs Let's examine individual terms in the discrete-time Fourier series = X m e i2πmn/. m=0 The second ( m = ) term is X e i2πn/. The final ( m = ) term is X e i2πn( )/ = X e i2πn e i2πn/ = X e i2πn/ which rotates at the same rate as the second ( m = ) term, but in the opposite direction! Assuming is even-valued, we have { +[ ] [ ] = X X e i2πn/ X e i2πn/ X 2 e i2πn(2/) + X 2 e i2πn(2/) +[ + ] } X 3e i2πn(3/) X +. 3 e i2πn(3/) X /2 e iπn For to be real-valued, we must have =, =, etc. X X X 2 X 2 6 / 23

20 One Sided Representation The discrete Fourier transform represents a time series as a sum of positively rotating and negatively rotating contributions. When is real-valued, these contributions cancel. In the usual form of the DFT, the negatively-rotating contributions are represented by aliasing frequencies higher than the yquist! We can write the DFT in the one sided form, for even, 2 = + cos(2π n + ) + ( X 0 /2 A m f m Φ m X /2 ) m m= where A n and Φ n are an amplitude and phase, with = A n e iφ n. The first and last contributions are the mean and the yquist. Their coefficients must be real-valued, as there is nothing to cancel them. X n 7 / 23

21 One Sided vs. Two Sided The two-sided and one-sided representations of the discrete Fourier transform are entirely equivalent. = = m=0 X m e i2πmn/ X0 /2 + 2 cos(2πmn/ + ) + ( m= A m Φ m X /2 ) m The two-sided representation is more compact mathematically. For real-valued, the one-sided representation is more intuitive as it expresses as a sum of phase shifted cosinusoids. 8 / 23

22 One Sided vs. Two Sided The two-sided and one-sided representations of the discrete Fourier transform are entirely equivalent. = = m=0 X m e i2πmn/ X0 /2 + 2 cos(2πmn/ + ) + ( m= A m Φ m X /2 ) m The two-sided representation is more compact mathematically. For real-valued, the one-sided representation is more intuitive as it expresses as a sum of phase shifted cosinusoids. A price of the one-sided form is that even and odd different! This expression is for even-valued. are somewhat 8 / 23

23 Review: Orthogonality If you have complex exponentials at two different frequencies, and q/, and you multiply one by the conjugate of the other and sum over all n, you have p/ n=0 e i2πpn/ e i2πqn/ = e i2π(p q)n/ = n=0 δ pq where δ pq is called the Kronecker delta function: = δ pq { 0 p = q p q Thus, the sum over the product these two cosines is frequencies are the same, and zero otherwise. if their This occurs because complete periods unless e i2π(p q)n/ p = q executes an integral number of, thus summing to zero. 9 / 23

24 The Fourier Coefficients How do we know the values of the Fourier coefficients? Multiply by another complex exponential, e i2πpn/ with frequency p/, and sum over all. X n e i2πpn/ = X m e i2πmn/ [ ] m=0 e i2πpn/ n=0 e i2πpn/ = m=0 X m e i2π(m p)n/ n=0 n=0 e i2πpn/ = X m δ pm = X p m=0 Because of orthogonality, all Fourier components vanishes except the coefficient of e i2πpn/. 20 / 23

25 The Forward & Inverse DFT The previous slide shows X p = n=0 e i2πpn/. However, the p is just a label, so we can use the letter m instead. Thus we obtain: X m, = X m e i2πmn/. n=0 e i2πmn/ m=0 The first of these is called the discrete Fourier transform of. It transforms from the time domain to the frequency domain. The DFT defines a sequence of complex-valued numbers, X n, for, which are termed the Fourier coefficients. n = 0,, 2, The second expression is the inverse discrete Fourier transform. It expresses how may be constructed using the Fourier coefficients multiplying complex exponentials or, as we saw earlier, phase-shifted sinusoids. 2 / 23

26 Proof of Orthogonality Using cos A cos B = 2 [cos(a + B) + cos(a B)], we find n=0 cos(2πpn/) cos(2πqn/) = [cos(2π(p + q)n/) + cos(2π(p q)n/)]. 2 n=0 Both terms are of the form cos(2πkn/) where is an integer. If k 0, both terms execute an integer number of complete periods as n varies from 0 to, so that n=0 cos(2πkn/) = 0. If p = q however, then the second term is always cos(0) =. If is even and k =, cos(2π(n + /2)/) = cos(2πn/ + π), the first /2 terms cancel the last /2 terms. Thus we have if and otherwise, or /2 p = q 0. 2 δ pq 22 / 23

27 Homework. Please review the notes up through page Take the DFT of your data using fft, then invert this using ifft. Compare the result with the original. 3. Remove the last point from your dataset to make it even length, if necessary. Repeat the previous exercise, but after first setting the (i) yquist component and (ii) zero-frequency component of the DFT of your data to zero. Does the result match your expectation? What happens if the data length is odd? 23 / 23

FOURIER ANALYSIS. (a) Fourier Series

FOURIER ANALYSIS. (a) Fourier Series (a) Fourier Series FOURIER ANAYSIS (b) Fourier Transforms Useful books: 1. Advanced Mathematics for Engineers and Scientists, Schaum s Outline Series, M. R. Spiegel - The course text. We follow their notation