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PHYS 352

Intro Signal Processing: Fourier Transforms and Sampling

  any† periodic function with period T (and f0 =1/T) can be represented as an infinite sum:

  re-writing gives:

  create a purely mathematical artifact → negative frequencies:

Fourier Series

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  exponential form is useful since math with exponential functions is simpler than with trig functions and you take care of sine and cosine together (i.e. magnitude and phase)

  magnitude of frequency component nf0 is:   phase of frequency component nf0 is:

  which is just the phase angle of cn

Fourier Coefficients

an = cn+c-n = 2 Re cn a0 = c0 bn = j(cn-c-n) = 2 Im cn

Example: Fourier Coefficients

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Example cont'd

magnitude phase

Example cont'd   let's plug in some real numbers, let's pick

x = 0, τ = T/5 and A = 1; w = 2π/T   coefficients are:

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  let n/T→k and let T→∞ (periodic pulse train becomes a single pulse); then k becomes continuous; cnT→ F(k)

  with:

  note: ω = 2πk and

F(ω) is the frequency spectrum easy to transform back and forth from time domain to frequency domain

Fourier Transform

electrical engineering convention: most physics texts use: for both

  symmetry between time and frequency   square wave to sinc function   sinc(x) =

time frequency square sinc sinc square delta constant constant delta impulse train frequency domain periodic spikes cosine/sine two, real/imaginary, even/odd, d functions f(t-t0)

Fourier Transform Pairs

sinc(x)

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Sampling: Key Fourier Transforms   impulse train: infinite periodic pattern of delta functions, also called a

Dirac comb

  convolution theorem: Fourier transform of the convolution of two functions is the product of their separate Fourier transforms and vice versa

sums to zero unless ωT=2π

F

  consider a signal which has limited bandwidth (all frequency content f<fmax)

  sampling is like applying an “impulse train” function

What Happens When Sampled?

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  previous page mathematically – boils down to the shifting property of the delta function under convolution

  if fs < 2 fmax, aliasing occurs

Nyquist criterion: must sample at over twice fmax to preserve signal content

Nyquist-Shannon Sampling Theorem

Frequency Limited Signals   can use samples to determine Fourier

coefficients if the frequency content is limited   if Fourier series has nmax then f0*nmax is

maximum frequency contained in the signal contents

  need to determine 2 nmax+1 coefficients (nmax complex values and 1 for c0) and hence need the same number of samples to generate that many linear equations

  of course, that's consistent with the sampling theorem

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Sampling Example   f0 = 1 Hz, fmax = 2 Hz thus nmax = 2; need 5 samples   sampling frequency fs = 5 Hz, Δt = 0.2 s   5 samples are: [0 1 0 −1 0] t=0: 0 = a0 + a1 cos(0) + b1 sin(0) + a2 cos(0) + b2

sin(0) t=0.2: 1 = a0 + a1 cos(0.4π) + b1 sin(0.4π) + a2 cos(0.8π) + b2

sin(0.8π) t=0.4: 0 = a0 + a1 cos(0.8π) + b1 sin(0.8π) + a2 cos(1.6π) + b2

sin(1.6π) t=0.6: −1 = a0 + a1 cos(1.2π) + b1 sin(1.2π) + a2 cos(2.4π) + b2

sin(2.4π) t=0.8: 0 = a0 + a1 cos(1.6π) + b1 sin(1.6π) + a2 cos(3.2π) + b2

sin(3.2π)   solve linear system (e.g. re-write as a matrix equation,

invert the matrix using MATLAB to get a0, a1, b1, a2, b2)   a0= 0, a1=0.4472, b1=0.6155, a2=−0.4472, b2=−0.1453

  solving K equations for K samples is not the way one normally determines Fourier coefficients

  integrating over one period times 1/T → averaging over the samples for one period, when discrete

becomes

discrete Fourier transform

Coefficients: Sampling

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Reprise: Sampling Example

  sample K times in one period T; T = K Δt

  previous example: K = 5 samples, (K−1)/2 coefficients of cn, plus c0

  when you use MATLAB [c0 c1 c2 c3 c4] is returned when samples [f1 f2 f3 f4 f5] and one can show that   c3 = c2* and c4 = c1*

  note also: MATLAB does not include factor 1/K (# of samples)

Discrete Fourier Transform

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  FFT's are algorithms that decrease the number of computations to perform a discrete Fourier transform   # computations: reduced from O(N2) to O(N log N)

  generally based upon factorization of N, where N is the number of samples; however, note that even when N is a prime number, there are FFT algorithms that achieve the above computational reduction

  how do the algorithms work?   too much detail for this course

Fast Fourier Transform

  when you know the Fourier coefficients and want the value of the function

→

  FFT algorithms work on the inverse DFT also   note: the upper half of the cm coefficients are the

complex conjugates of the lower half (negative frequencies)

  note also: MATLAB includes factor 1/K (# of samples)

Inverse DFT

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Recap: for Math Wonks   Fourier transform

  continuous in time and frequency domain

  Fourier series   continuous and periodic in time, discrete in

frequency

 Discrete-Time Fourier Transform   discrete in time, continuous and periodic in

frequency   we did not discuss

 Discrete Fourier Transform   both are discrete (time and frequency)   both are finite, but are periodic if extended

DFT Example of Use   signal waveform is sampled

  501 samples every 1 µs, starting at t=0 s, until t=500 µs and that’s T=500 µs

  get the DFT using an FFT algorithm   each frequency step is: 1/T = 2000 Hz   there will be 250 frequency components plus c0

  each of those c1 to c250 is complex, has real and imaginary values, thus has 500 values

  or think 250 positive frequencies and 250 negative

  maximum frequency is: 250(2000) = 500 kHz   note: sampling frequency was 1 MHz, twice the

maximum frequency, so Mr. Nyquist is happy   summary: 501 time domain samples generated 501

frequency domain values

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DFT Even Number of Samples   say you have and even number of samples, what are

the frequency components?   20 samples, say every 1 ms; T = 20 ms

  f0 = 1/T = 50 Hz   fs = 1 kHz

  you expect 20 “complex” numbers from the DFT   #1 is the DC coefficient (c0, real-valued)   #2 to #10 are the c1 to c9 complex-valued coefficients   #11 is the c10 coefficient and it will be real   #12 to #20 are the c9 to c1 complex conjugates

  TOTAL: 9 complex values + 2 real values = 20 real quantities from the DFT

  fmax = 10*f0 = 500 Hz   that’s ½ of fs