MM3FC Mathematical Modeling 3 LECTURE 10

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MM3FC Mathematical Modeling 3LECTURE 10

• Times
• Weeks 7,8 9.
• Lectures Mon,Tues,Wed 10-11am, Rm.1439
• Tutorials Thurs, 10am, Rm. ULT.
• Clinics Fri, 8am, Rm.4.503

Dr. Charles Unsworth, Department of Engineering
Science, Rm. 4.611 Tel [phone] ext.
82461 Email c.unsworth_at_auckland.ac.nz
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This LectureWhat are we going to cover Why ?

• Spectral Analyser.
• (Understand the DSP behind a spectrum
  analyser)
• Discrete Fourier Transform (DFT).
• (From understanding the spectrum analyser
• we will learn a DFT can be represented as a
  filter bank)
• Calculating a DFT.
• Determine the Magnitude spectrum of a
  complicated digitised sequence.

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Spectral Analysis

• The basic idea of spectral analysis is to create
  a system that will compute the values of the
  spectral components from the digital sequence of
  numbers xn
• The general form for representing a real signal
  xn as a sum of complex exponentials is
• For the case where N3, the spectrum would be

(10.1)
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• However, an equivalent plot could be used to
  translate the ve frequencies to their ve alias
  frequencies, using
• e-jwn ej2?n e-jwn ej(2? jw)n
• We will find this form of the spectrum useful
  when we discuss the DFT.

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A Spectrum Analyser

• The job of a Spectrum Analyser is to extract
  the frequency, magnitude and phase of each
  spectral component in xn.
• The analyser will only have a finite number of
  electronic channels (N1) to separate the
  spectral components into.
• However, there is an infinite amount of
  frequencies that could exist in the range 0lt w lt
  2? of the signal xn.
• Thus, we must choose the frequencies of the
  spectral components that are going to be
  dedicated to each channel first.
• So we will have to pick our frequencies ahead of
  time, independent of the signal.
• Since there is a finite number of channels (N1),
  one obvious is to cover the total 2? frequency
  range with equally spaced frequencies
• wk 2?k/N
• Thus, the spectrum is only an approximate
  representation.

(10.2)
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Spectrum Analysis by Filtering

• Each channel of the analyser must do two jobs
• - Isolate the frequency of interest.
• - Compute the complex amplitude (Xk) for that
  frequency.
• Isolating the frequency of interest
• Is suited to a BPF with a narrow passband.
• Once xn is passed through the BPF we can
  measure the ouput for its amplitude and phase.
• Rather than design individual BPFs for each
  channel, we shall introduce the
  frequency-shifting property of the spectrum.
• (In this way, all channels can share the same
  filter).

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Frequency-Shifting

• The frequency of a complex exponential can be
  shifted if it is multiplied by another complex
  exponential.
• Let the signal xn be multiplied by exp(-jwsn).
• The entire spectrum of xn if shifted in the ve
  direction by (ws).
• The idea is to shift the desired frequency to the
  zero frequency position.

(10.3)
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• Thus, if we wanted to move the frequency (w1) to
  the zero frequency position. We would multiply
  xn by exp(-jw1n).
• Thus, the power spectrum of xn below
• Would become the power spectrum of xn
  exp(-jw1n).

X0
X2
X2
X1
X1
X3
X3
w
w3-w1
-2w1
-w2-w1
-w1
-w3-w1
w2-w1
0
?
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Channel Filters

• Why have we translated the desired frequency to
  the w0 position ?
• Measurement at w0 is equivalent to finding the
  average value of the signal with a lowpass
  filter.
• When a lowpass filter extracts the DC component
  of the shifted spectrum it is actually measuring
  the complex amplitude of the shifted spectral
  components in xn.
• A system for doing this must not only measure the
  average value but null out all other competing
  sinusoidal components.

Preserves whats at w0
Places over
spectrum
SHIFTS spectrum By w to the left
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Spectrum Analysis of Periodic Signals

• A spectrum analyser will work perfectly for
  periodic signals.
• The same approach can be used for non-periodic
  signals but will have approximation errors.
• A periodic signal can be defined by the shifting
  property
• xn N xn
• Where, N is the period. (i.e if we delay xn by
  N we will get xn again)
• If a complex exponential is periodic with period
  N,
• For the last equation to be true,
• Equating exponents we get wN 2?k, Thus any
  periodic exponential can be expressed as

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• Once the period is specified we can create
  different complex exponential signals by changing
  the value of (k). However, there are actually
  only (N) different frequencies because
• So k has the range k 0,1,2, , N-1
• Spectrum of a Periodic Discrete-time Signal
• If xn has period (N). Then its spectrum can
  contain only complex exponentials with the same
  period. Hence, the general representation of a
  periodic discrete-time signal xn is
• The range is from 0 to (N-1) as there are only N
  different frequencies.
• The frequencies are all multiples of a common
  fundamental frequency
• w 2?/N

(10.4)
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• Now we will change the notation in anticipation
  for the DFT.
• We will replace Xk by X k/N. Thus,
• Later on, we will see that this is the inverse
  DFT.
• Filtering with a running Sum
• Earlier, we showed that FIR filters can
  calculate a running sum or running average.
• Lets consider the L-point running sum with the
  difference equation
• As we can see all the co-efficients are equal to
  one.

(10.5)
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• Procedure is Take a typical signal xn with
  power spectrum below

• Shift desired spectral component to zero
  position.
• Put through running sum filter.

Gain 10
Null
Frequency response of a Running sum filter
Null

• Only the spectral component at zero position is
  preserved.

5x10 50
Spectral component of interest Is magnified by
the gain of 10. All other components are nulled.
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• As we can see, the running-sum FIR has nulls
  in its frequency response curve at w 2?k/L.
• And this works very well for nulling out a
  signal if its constituient frequencies are also
  equally spaced.
• If xn is periodic and
• Running sum filter length period of xn
• yn is
• A constant signal
• Thus, for a periodic signal, each channel is
  composed of the filter combination below that
  produces one spectral component.
• The structure is called a filter bank.

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The Discrete-Fourier Transform (D.F.T.)

• The previous discussion is a roundabout way to
  describe the DFT.
• The Forward DFT (Xk) or Analysis Equation
  is given below
• By substituting in xn we can calculate the
  spectral components Xk can be found.
• Similarly, the Inverse DFT (xn) or Synthesis
  Equation is given below
• By subsituting, in the spectral components Xk
  we can synthesise the input signal xn.

(10.6)
(10.7)
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• The filter bank of the DFT to compute Xk for
  k0,1,2,,(N-1)

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• Thus,
• We will use the DFT formulas for computation.
• And the filter-bank for interpretation and
  insight.
• The DFT transforms an n-domain(or time-domain)
  signal
• into its frequency-domain representation.

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Computing the Forward DFT

• So how do we go about computing the DFT and
  determining the Magnitude Spectrum from an input
  sequence xn ?
• Consider a simple continuous sinusoid with a
  frequency f 1Hz
• x(t) cos(2?(1)t)
• Thus, 1 period of the signal will occur in 1
  second.
• And the one sided Magnitude spectrum will give a
  single peak at 1Hz.
• Lets see if the forward DFT can predict this !
• Okay lets sample the signal 8 times in 1 sec,
  fs 8Hz and Ts1/8 secs
• The digitised signal is therefore
• xn x(nTs) cos(2?nTs) cos(2?n(1/8))
  cos((?/4)n)
• We have 8 samples, N8 occurring when n
  0,1,2,3, , (N-1 7)

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• Using the forward DFT formula we will determine
  the spectral components.

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For k0,
For k1,
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• The best way to calculate what xn is
  multiplied is to identify the smallest increment
  that the exponential goes up in. Consider X1
  again

• The exponential power is always a multiple of
  ?/4.
• In general the exponential power is a multiple
  of 2?/no.samples
• Next create a unit circle which is split up into
  these partitions.
• Then at each of the multiples write in the
  values the exponent has at these positions.

Values of xn at each position
Unit circle in partitions
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For k 2,

• We can read off the values from the unit circle
  to determine what the exponent value is.

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For k3,
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k1 ?/4 increments
k2 ?/2 increments
k3 3?/4 increments

• So as (k) increases, the sample frequency
  decreases.
• Also we are autocorrelating the sequence xn
  with differently sampled versions of itself.

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• We can write all of this in a matrix notation

Power of the exponent (e-j?/4)
n increases
xn
k increases
Goes up in p/4 s
Goes up in p/2 s
Goes up in 3p/4 s
Goes up in p s
Goes up in 5p/4 s
Goes up in 3p/2 s
Goes up in 7p/4 s
So only need to remember this bit really.

• Now we can read off the unit circle the values
  at these frequencies.

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Reflection in the line of 1s and 1s
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• Thus, we get

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• So discounting the aliassed frequency, due to
  finite data length, we have located the frequency
  of the sinusoid.
• This is a special case, as the sinusoid had an
  integer number of cycles in the sequence.
• What DFT do we get if a sinusoid had a
  non-integer number of cycles in the sequence ?
• Thus, for a non-integer number of cycles in the
  sequence, leakage at the outputs occurs.
  However, the main component still has the most
  power in it.

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So endeth the course ! Good luck in the exams,
folks !!!