The Discrete Fourier Transform. Signal Processing PSYCH 711/712 Lecture 3

Transcription

1 The Discrete Fourier Transform Signal Processing PSYCH 711/712 Lecture 3

2 DFT Properties symmetry linearity shifting scaling

3 Symmetry x(n) X(m) n m X(m) = complex conjugate{ X(N-m) }

4 DFT of even-symmetric for continuous waves: - even-symmetry: f(t) = f(-t) for discrete, sampled waves - even symmetry: x(n) = x(n-n) Fourier transform of even-symmetric waveforms is Real (i.e., Imaginary part is zero)

5 Even to Even if x(n) is a real, even-symmetric function, then X(m) is a real, even-symmetric function x(n) n X(m) Re{X(m)} m m

6 Odd-symmetry Fourier transform of a real, odd-symmetric function is an imaginary, odd-symmetric function x(n) X(m) n m f(t) = -f(-t), and f(0)=0 or x(n) = -x(n-n), and x(0)=0

7 DFT Linearity X sum (m) = X 1 (m) + X 2 (m) spectral components (m 19) = cos(2π3t π/8) + sin(2π10t) t mag{x(m)} / N phase{x(m)} m magnitude phase m

8 Shifting Property if we start to sample x(n) at n=k, rather than n=0, the DFT of the shifted samples will be related to the DFT of the unshifted samples according to the equation: X shifted (m) = e j2πkm/n X(m)

9 Linear Phase Shifts the shifting property implies that the effect of a shift, or delay, is to add a linear phase term to each spectral component: X shifted (m) = e j2πkm/n A(m)e jφ(m) = A(m)e j(φ(m)+2πkm/n)

10 φ(m) = k2π N m k = 3; N = 20 2*pi*k*m/N m

11 Scaling Property if the width of a function is increased, its Fourier Transform becomes narrower and taller; if its width is decreased, its transform becomes wider and shorter: [ ( t )] F x b = b F (bf)

12 Scaling Example mag[x(m)] t m

13 DFT Leakage

14 Frequency Representations of Sinusoids DFT of sinusoidal waves that are harmonics of 1/T, where T is the width of the sample, has non-zero magnitude only at spectral components corresponding to ± frequency of pattern

15 DFT Leakage Frequency representations of sine waves that are not harmonics of 1/T are represented as the sum of many non-zero spectral components

16 Why does leakage occur?

17 Review Origins of DFT = 1 + cos(2*pi*t) continuous time continuous transform Mag time (sec) f(hz) f s = 10 Hz sampled time continuous transform Mag time (sec) f (Hz)

18 finite set of sampled points continuous transform samples Mag time (sec) f (Hz) frequency representation of finite set of samples is a smeared version of the original

19 1.0 continuous transform Mag time (sec) f (Hz) Quartz (4) - Inactive Mag samples finite set of sampled points f (Hz) frequency representation of finite set of samples is a smeared version of the original

20 where does this smoothing come from?

21 digression: convolution

22 Convolution A convolution is an integral that expresses the amount of overlap of one function g(t) as it is shifted over another function. u(t) = f(t) g(t) = + f(τ)g(t τ)dτ

23 Origins of Temporal Smearing truncation = multiplication by rect rect & sinc functions are related multiplication in time domain is equivalent to convolution in the frequency domain convolution causes frequency smearing

24 Convolution u(t) = f(t) g(t) = + f(τ)g(t τ)dτ g(t-τ) is a flipped version of g(t) (i.e., reversed in time) that is centered on t the 2 functions are multiplied point-by-point the product is integrated, and the result is u(t) g(t-τ) is centered on new time t, and the process is repeated

25 u(t) = f(t) g(t) = + f(τ)g(t τ)dτ

26 covolution & smoothing f(x) * h(x) = g(x) convolution with a delta function g(x) = f(x) increased smoothing

27 convolution is linear g(t) = f(t) * h(t) replications of h(t) sum of replications

28 convolution & multiplication convolution in the time domain is equivalent to multiplication in the frequency domain - Fourier Transform{x(n) * y(n)} = X(m) Y(m) convolution in the frequency domain is equivalent to multiplication in the time domain - Fourier Transform{x(n) y(n)} = X(m) * Y(m)

29 What does this have to do with spectral smoothing?

30 Temporal Truncation time (sec) X = time (sec) time (sec) Using a finite set of discrete values is equivalent to multiplying the original infinite set of samples by a rect function

31 rect(t) rect(t) width = t

32 rect(t-1) rect(t-1) t rect(t/2) rect(t/2) t rect((t-1)/2) rect((t-1)/2) t

33 Multiplication & Convolution was multiplied by rect(t) equivalent to X(f) * Y(f) - * is convolution - Y(f) is Fourier Transform{rect(t)} Fourier Transform{rect(t)} = sinc(m) in this case, X(f) consists of sets of 3 delta functions centered on kf s so, X(f) * sinc(f) creates a series of sinc functions centered on each delta function

34 sinc(x) = sin(xπ) xπ sinc function sinc(t) t sinc(t)=0 for all integer values of t except t=0, where sinc(0)=1

35 Rect & Sinc are Fourier Transform Pairs rectangle wave sinc function abs(sinc) function Amplitude Mag time (sec) f (Hz) f (Hz) Fourier Transform {rect(x)} = sinc(f)

36 Fourier rect(t) Transform sinc(f) Pairs Amplitude time (sec) f (Hz) sinc(t) rect(f) Amplitude time (sec) f (Hz) Fourier Transform {rect(x)} = sinc(f) Fourier Transform {sinc(x)} = rect(f)

37 Multiplication & Convolution rect(t) = X(f) * sinc(f) but, X(f) consists of sets of 3 delta functions centered on kf s so, X(f) * sinc(f) creates a series of sinc functions centered on each delta function

38 Spectral Smoothing Mag X(f) is a series of delta functions rect(t) = X(f) * sinc(f) 0.0 so, X(f) * sinc(f) is a series of sinc functions f (Hz) Mag sum f (Hz) close up of smoothed spectrum that shows 3 individual sinc functions, one for each delta function, as well as the sum of the sinc functions.

39 sampling creates periodic X(m) truncating creates smoothed X(m) but, we are using sampled (i.e., discrete) versions of X(m) consequences of sampling X(m)?

40 periodic sampling creates a periodic X(m) - if sampling interval is t s, then replicates of X(m) are separated by f s = 1/t s and sampling X(m) creates a periodic - sample frequency spectrum at sampling interval f s =1, then replicates of separated by 1/f s = 1 s after sampling frequency spectrum t t

41 what does this have to do with DFT leakage?

42 non-harmonic frequencies leakage occurs when f is not an integer multiple of the number of samples when we concatenate the samples to create a periodic waveform, do we get a sine wave?

43 1.5 Hz sine wave time (sec) sampled & truncated tmp time (sec) concatenated waveforms are periodic, but not sinusoidal t seq(0, 49)

44 non-harmonic frequencies leakage occurs when f is not an integer multiple of the number of samples when we concatenate the samples to create a periodic waveform, do we get a sine wave? no, the periodic waveform is not sinusoidal and therefore the spectrum of that waveform will not be the same as a spectrum of a sine wave this is one explanation for why the spectrum of a non-harmonic sine wave looks strange another explanation can be found by inspecting the sampled spectrum itself

45 sampled spectra X(f) signal f = 4 Hz f X(f) signal f = 4.5 Hz f leakage occurs as a result of sampling a continuous sinc(f) function that is centered on non-integer values of f

46 sampling frequency spectra cos(2 pi f 0 t) rect(t) has a spectrum that consists of sinc functions centered on ±f 0 sinc(f 0 ) is zero whenever f = f 0 ± n - n = 1, 2, 3,... etc when f 0 is an integer, X(m) will be exactly zero at all integer frequencies, m ±f 0 - X(m) = 0 for all integer m ±f 0 but, when f 0 is not an integer, the sampled spectrum X(m) will not be zero at m f 0

47 coarse frequency sampling Continuous frequency spectra of truncated data always contain the fingerprints of sinc functions we sometimes fail to see them because of the coarse sampling of the frequency spectrum (f s =1) but we can get a better look by increasing the rate at which the frequency spectrum is sampled...

48 increasing frequency resolution Suppose we record 1s worth of data sampled at 10 Hz: - T = 1s; t s = 0.1 s; n=(0...9); N = 10 - The spectral components m=1 and m=2 are complex exponentials with frequencies of 1 & 2 cycles per second - or 1 and 2 Hz Let s increase T to 2s: - T = 2s; t s = 0.1 s; n=(0...19); N = 20 - The spectral components m=1 and m=2 have frequencies of 1 & 2 cycles per two seconds -... or 0.5 and 1 Hz doubling N doubled our frequency resolution

49 f = 1.5 Hz mag[x(m)] sampling rate = f s = 10 Hz N = 10 t f = m f s / N = m m f = 1.5 Hz t mag[x(m)] sampling rate = f s = 10 Hz N = 20 f = m f s / N = m/2 m

50 f = 1.25 Hz t f = 1.25 Hz mag[x(m)] mag[x(m)] m t m

51 zero stuffing (also improves frequency resolution) f = 1.25 Hz mag[x(m)] f = 1.25 Hz continuous abs(sinc(m)) function t m mag[x(m)] t m N.B. Amplitude of X(m) not altered by zero stuffing.

52 Application of Windows Hamming Window Hanning Window t t Triangle Window t t

53 Windows Reduce Frequency Spread x(n) mag[x(m)] n m Hamming Window x(n) mag[x(m)] n m

54 the end