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MAE 4020/5020 – Numerical Methods with MATLAB
SECTION 10: FOURIER ANALYSIS
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Fourier Series – Trigonometric Form2
K. Webb MAE 4020/5020
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Periodic Functions
K. Webb MAE 4020/5020
A function is periodic if
where is the period of the function
The function repeats itself every seconds Here, we’re assuming a function of time, but could also be a spatial function, e.g. Elevation Pixel intensity along rows or columns of an image
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Frequency
K. Webb MAE 4020/5020
The frequency of a periodic function is the inverse of its period
1
We’ll refer to a function’s frequency as its fundamental frequency,
This is ordinary frequency, and has units of Hertz (Hz) (or cycles/sec)
Can also describe a function in terms of its angularfrequency, which has units of rad/sec
2 ⋅2
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Fourier Series
K. Webb MAE 4020/5020
In other words, any periodic signal of engineering interest
Then it can be represented as an infinite sum of harmonically‐related sinusoids, the Fourier series
Fourier discovered that if a periodic function satisfies the Dirichlet conditions:1) It is absolutely integrable over any period:
∞
2) It has a finite number of maxima and minima over any period
3) It has a finite number of discontinuities over any period
Joseph Fourier1768 – 1830
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Fourier Series
K. Webb MAE 4020/5020
The Fourier series
cos sin
where is the fundamental frequency,
and, the Fourier coefficients are given by
1
the average value of the function over a full period, and
2cos , 1,2,3…
and2
sin , 1,2,3…
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Sinusoids as Basis Functions
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Harmonically‐related sinusoids form a set of orthogonal basis functions for any periodic functions satisfying the Dirichlet conditions
Not unlike the unit vectors in space:
̂ 1,0 , ̂ 0,1
Any vector can be expressed as a linear combination of these basis vectors
̂ ̂
where each coefficient is given by an inner product
⋅ ̂⋅ ̂
These are the projections of onto the basis vectors
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Sinusoids as Basis Functions
K. Webb MAE 4020/5020
Similarly, any periodic function can be represented as a sum of projections onto the sinusoidal basis functions
Similar to vector dot products, these projections are also given by inner products:
2cos , 1,2,3…
and2
sin , 1,2,3…
These are projections of onto the sinusoidal basis functions
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Fourier Series – Example
K. Webb MAE 4020/5020
Consider a rectangular pulse train
2
0.5
⁄
Can determine the Fourier series by integrating over any full period, for example,
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Fourier Series – Example –
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10 0.500.5 1.511.5 2.0
First, calculate the average value
1 12
12 1
. 12 0
.
.
12 1.
12
. 12 .
0.25 0.25
0.5, as would be expected
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Fourier Series – Example –
K. Webb MAE 4020/5020
Next determine the cosine coefficients,
2cos
22 cos
. 22 cos.
1sin
. 1sin
.
1sin 2 0 0 sin 3 2
1sin 2 sin 3 2
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Fourier Series – Example –
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We know that
sin 3 2 sin 2 sin 2so
2sin 2 , 1,2,3…
The first few values of :
, 0 , , 0 ,
Zero for all even values of Only odd harmonics present in the Fourier Series
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Fourier Series – Example –
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Next, determine the sine coefficients,
2sin
22 sin
. 22 sin.
1cos
.cos
.
1cos 2 1 1 cos 2 0
0, 1,2,3…
All coefficients are zero Only cosine terms in the Fourier series
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Fourier Series – Example
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The Fourier series for the rectangular pulse train:
0.52sin 2 cos
Note that this is an equality as long as we include an infinite number of harmonics
Can approximate by truncating after a finite number of terms
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Fourier Series – Example
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Fourier Series – Example
K. Webb MAE 4020/5020
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Even and Odd Symmetry
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An even function is one for which
An odd function is one for which
Consider two functions, and If both are even (or odd), then
2
If one is even, and one is odd, then
0
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Even and Odd Symmetry
K. Webb MAE 4020/5020
Since is even, and is odd If is an even function, then
4cos
/, 1, 2, 3, …
0, 1, 2, 3, …
If is an odd function, then0, 1, 2, 3, …
4sin
/, 1, 2, 3, …
Recall the Fourier series for the pulse train, an even function, had only cosine terms
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Fourier Series – Cosine w/ Phase Form19
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Cosine‐with‐Phase Form
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Given the trigonometric identity
cos sin cos
where and tan
We can express the Fourier series in cosine‐with‐phase form:
cos
where
tan , 0
tan , 0
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Cosine‐with‐Phase Form – Example
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Consider, again, the rectangular pulse train
sin
0
So,
2sin 2
and
tan0
2 sin 2
0, 1, 5, 9, …, 3, 7, 11, …
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Line Spectra
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The cosine‐with‐phase form of the Fourier series is conducive to graphical display as amplitude and phase line spectra
Average value and amplitude of odd harmonics are clearly visible
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Fourier Series – Complex Exponential Form23
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Complex Exponential Fourier Series
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Recall Euler’s formula
cos sin
This allows us to express the Fourier series in a more compact, though equivalent form
where the complex coefficients are given by
1
Note that the series is now computed for both positive and negative harmonics of the fundamental
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Complex Exponential Fourier Series
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We can express the complex series coefficients in terms of the trigonometric series coefficients
12 , 1, 2, 3, …
12 , 1, 2, 3, …
Coefficients at are complex conjugates, so
and ∠ ∠
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Complex Exponential Fourier Series
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Similarly, the coefficients of the trigonometric series in terms of the complex coefficients are
2
2
Can also relate the complex coefficients to the cosine‐with‐phase series coefficients
12 , 1, 2, 3, …
∠ , 1, 2, 3, …, 1, 2, 3, …
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Even and Odd Symmetry
K. Webb MAE 4020/5020
For even functions, since 0, coefficients of the complex series are purely real:
12 , 1, 2, 3, …
For odd functions, since 0, coefficients of the complex series are purely imaginary (except ):
12 , 1, 2, 3, …
12 , 1, 2, 3, …
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Complex Series – Example
K. Webb MAE 4020/5020
10 0.500.5 1.511.5 2.0
The complex Fourier series for the rectangular pulse train:
The complex coefficients are given by1 /
/
12
12
.
.
12 .
.
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Complex Series – Example
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12 .
.
12
Rearranging into the form of a sinusoid
12
1sin 2
Given the even symmetry of , all coefficients are real, and also have even symmetry
1sin 2
1, 0,
13 , 0,
15 , 0, …
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Line Spectra
K. Webb MAE 4020/5020
The complex series coefficients can also be plotted as amplitude and phase line spectra Now, plot spectra over positive and negative frequencies
Note that the magnitude spectrum is an even function of frequency, and the phase spectrum is an odd function of frequency
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The Fourier series can also be understood by approaching it as a least‐squares curve‐fitting problem, where sinusoids are fit to a function or data set.
Sinusoidal Curve Fitting31
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Sinusoidal Curve Fitting
K. Webb MAE 4020/5020
In a previous section of the course we saw how we can fit different functions to data using linear least‐squares regression Can also fit sinusoids using this technique
The data we’re fitting could be:Measured data that we believe to be sinusoidal in nature
A periodic function, that, while not sinusoidal, we want to approximate as a sinusoid or sum of sinusoids
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Sinusoidal Curve Fitting
K. Webb MAE 4020/5020
Our fitting function is
cos
The fundamental frequency is
22
where is the period of the function or data we are fitting
The three fitting parameters are: , , and In order to be able to apply linear regression, we can’t have a fitting
parameter in the argument of a trigonometric function Apply a trig. Identity to recast the model as
cos sin
Assuming we know , this is a linear least‐squares model
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Sinusoidal Curve Fitting
K. Webb MAE 4020/5020
Assuming is known, the linear least‐squares model is
where1, cos , sin
and , , and
For a least‐squares fit, minimize the sum of the squares of the residuals
cos sin
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Normal Equations
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As we saw in the curve fitting section of the course, the matrix normal equations for this least‐squares fit are
where is the design matrix:
⋮ ⋮ ⋮
1 cos sin1 cos sin⋮ ⋮ ⋮1 cos sin
is the vector of fitting parameters
and is the vector of function or data values
…
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Normal Equations –
K. Webb MAE 4020/5020
The LHS of the normal equations is
1 1 ⋯ 1cos cos ⋯ cossin sin ⋯ sin
1 cos sin1 cos sin⋮ ⋮ ⋮1 cos sin
Σ cos Σ sinΣ cos Σ cos Σ cos sinΣ sin Σ sin cos Σ sin
If we assume our data points span exactly one period, then we know the following mean values
Σ cos Σ sin Σ cos sin0
andΣ cos Σ sin 1
2
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Normal Equations –
K. Webb MAE 4020/5020
Using these known mean values, the normal equations simplify to
0 00 /2 00 0 /2
ΣΣ cosΣ sin
We can solve for the vector of fitting parameters,
The inverse of the diagonal matrix is a diagonal matrix, where the diagonal elements are inverted, so
1/ 0 00 2/ 00 0 2/
ΣΣ cosΣ sin
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Sinusoidal Least‐Squares Fit
K. Webb MAE 4020/5020
The fitting parameters areΣ
2Σ cos
2Σ sin
Note the similarity to the Fourier series coefficients
The least‐squares, best‐fit sinusoid is given byΣ 2
Σ cos cos2Σ sin sin
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Sinusoidal Least‐Squares Fit – Example
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As expected, 0 due to the even symmetry of the function being fit
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Least‐Squares Fit of Two Harmonics
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Now, consider extending the fitting model to include the first two harmonics
cos sin cos 2 sin 2
We’ve added two more basis functions to the linear least‐squares model
The design matrix is now1 cos sin cos 2 sin 21 cos sin cos 2 sin 2⋮ ⋮ ⋮ ⋮ ⋮1 cos sin cos 2 sin 2
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Least‐Squares Fit of Two Harmonics
K. Webb MAE 4020/5020
If we again assume samples spanning exactly one period, the off‐diagonal terms on the LHS of the normal equations go to zero, leaving
0 0 0 00 /2 0 0 00 0 /2 0 00 0 0 /2 00 0 0 0 /2
ΣΣycosΣysinΣycos 2Σysin 2
Solve for as
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Least‐Squares Fit of Two Harmonics
K. Webb MAE 4020/5020
Solving for gives the following fitting parametersΣ
2Σ cos
2Σ sin
2Σ cos 2
2Σ sin 2
This model could obviously be extended to include an arbitrary number of harmonics
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Least‐Squares Fit – Example
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Sawtooth wave has odd symmetry, so B 0, and only sine terms are present
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The Fourier transform extends the frequency‐domain analysis capability provided by the Fourier series to aperiodic signals.
Fourier Transform44
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Fourier Transform
K. Webb MAE 4020/5020
The Fourier Series is a tool that provides insight into the frequency content of periodic signals
where the complex coefficients are given by /
/
These values provide a measure of the energy present in a signal at discrete values of frequency , integer multiples (harmonics) of the fundamental
Frequency‐domain representation is discrete, because the time‐domain signal is periodic
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Fourier Transform
K. Webb MAE 4020/5020
Many signals of interest are aperiodic They never repeat Equivalent to an infinite period, → ∞
As , the mapping from the time domain to the frequency domain is given by the Fourier transform
where is a complex, continuous function of frequency
The continuous frequency‐domain representation corresponds to the aperiodic time‐domain signal
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Inverse Fourier Transform
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We can also map frequency‐domain functions back to the time domain using the inverse Fourier transform
12
The forward ( or transform) and the inverse ( or transform) provide the mapping between Fourier transform pairs
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Fourier Transform – Rectangular Pulse
K. Webb MAE 4020/5020
Consider a pulse of duration,
Calculate the Fourier transform
/
/
1 1
22
2sin 2
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Fourier Transform – Rectangular Pulse
K. Webb MAE 4020/5020
Here, we can introduce the sinc function
sin
Letting , we have
2sin 2
sin 2
2
2
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Fourier Transform – Triangular Pulse
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Next, consider a triangular pulse of duration,
Λ
Λ
2
1, 2 02
1, 0 20, otherwise
The Fourier transform is
Λ2
1/
21
/
Integrating by parts, or symbolically in MATLAB, gives
8sin 4
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Fourier Transform – Triangular Pulse
K. Webb MAE 4020/5020
This, too, can be recast into the form of a sincfunction
Letting , we have
8sin 4
2sin 4
4
2 4
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Rectangular vs. Triangular Pulse
K. Webb MAE 4020/5020
Average value in time domain translates to 0 value in frequency domain
More abrupt transitions in time domain correspond to more high‐frequency content
Multiplication in one domain corresponds to convolution in the other
Convolution of two rectangular pulses is a triangular pulse
becomes in the frequency domain
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Fourier Transform – Impulse Function
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The impulse function is defined as
0, 0
1
Its Fourier transform is
Since 0 for 0, and since 1 for 0
1
The Fourier transform of the time‐domain impulse function is one for all frequencies Equal energy at all frequencies
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Fourier Transform – Decaying Exponential
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Consider a decaying exponential
⋅ 1
where 1 is the unit step function
The Fourier transform is:
1 10 1
1
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Fourier Transform – Decaying Exponential
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Fourier transform of this exponential signal is complex
Plot magnitude and phase separately
Note the even symmetry of magnitude, and odd symmetry of the phase of
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Fourier Transform – Decaying Exponential
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On logarithmic scales, this Fourier transform should look familiar
could be the impulse response of a first‐order system Convolution of an impulse
with the system’s impulse response
looks like the frequency response of a first‐order system Multiplication of the F.T. of an
impulse ( 1) with the system’s frequency response
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Even and Odd Symmetry
K. Webb MAE 4020/5020
We are mostly concerned with real time‐domain signals Not true for all engineering disciplines, e.g. communications, signal processing, etc.
For a real time‐domain signal, , If is even will be real and even If is odd, will be imaginary and odd If has neither even nor odd symmetry, will be complex with an even real part and an odd imaginary part.
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For discrete‐time signals, mapping from the time domain to the frequency domain is accomplished with the discrete Fourier transform (DFT).
Discrete Fourier Transform58
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Discrete‐Time Fourier Transform (DTFT)
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The Fourier transform maps a continuous‐time signal, defined for ∞ ∞, to a continuous frequency‐domain function defined
for ∞ ∞ In practice we have to deal with discrete‐time, i.e. sampled, signals
Only defined at discrete sampling instants
→
Now, mapping to the frequency domain is the discrete‐time Fourier transform (DTFT)
DTFT maps a discrete, aperiodic, time‐domain signal to a continuous, periodic function of frequency
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Aliasing
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Aliasing is a phenomena that results in a signal appearing as a lower‐frequency signal as a result of sampling
In order to avoid aliasing, the sample rate must be at least the Nyquist rate
2
where is the highest frequency component present in the signal
For a given sample rate, the Nyquist frequency is the highest frequency signal that will not result in aliasing
2
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Aliasing – Examples
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Discrete‐Time Fourier Transform (DTFT)
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Discrete‐time generated from by sampling at a sample rate of , with a sample period of
Sampled signals can only accurately represent frequencies up to the Nyquist frequency
2
Higher frequency components of are aliased down to lower frequencies in the range of
2 2
The DTFT is a periodic function of frequency, with a period Due to aliasing, sampling in the time domain corresponds to periodicity in the
frequency domain
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The Discrete Fourier Transform (DFT)
K. Webb MAE 4020/5020
The DTFT
utilizes discrete‐time, sampled, data, but still requires and infinite amount of data
In practice, our time‐domain data sets are both discrete and finite The discrete Fourier transform, DFT, maps discrete and finite
(periodic) time‐domain signals to periodic and discrete frequency‐domain signals
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The Discrete Fourier Transform (DFT)
K. Webb MAE 4020/5020
Consider samples of a time‐domain signal, Sampled with sampling period and sampling frequency Total time span of the sampled data is ⋅
The DFT of is
/
A discrete function of the integer value, The DFT consists of complex values: , , … , Each value of represents a discrete value of frequency from
0 to
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The Inverse Discrete Fourier Transform
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A discrete, finite set of frequency‐domain data can be transformed back to the time domain
The inverse discrete Fourier Transform (IDFT)
1 /
Note the scaling factor In practice, this is often applied when computing the DFT Must exist in either the DFT or IDFT, not both
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DFT Frequencies
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/
A dot product of with a complex exponential
⋅
The frequency of the exponential is Ω, integer multiples of the normalized frequency, Ω
Ω 2 /
which has units of / Normalized frequency is related to the ordinary frequency by the
sample rate,
Ω2
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DFT Frequencies
K. Webb MAE 4020/5020
/
# of samples: , sample rate: , sample period: Maximum detectable frequency
/2 Nyquist frequency Corresponds to /2, Ω
Frequency increment (bin width, resolution)
Δ1⋅
Last ⁄ 1 points of , ⁄ … correspond to negative frequency
2 Δ … Δ
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DFT Frequencies
K. Webb MAE 4020/5020
For example, consider 10 samples of a signal sampled at 100 , 10
Δ⋅ .
10
50
ΔΩ ⁄ 0.2 ⁄
Units
Ω . . . . . . . . /
/ . . . . . . . . .
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DFT ‐ Example
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Consider the following signal0.3 0.5 cos 2 ⋅ 50 ⋅ cos 2 ⋅ 120 ⋅ 0.8 cos 2 ⋅ 320 ⋅
Sample rate: 1 Record length: 100 Bin width: Δ 10
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DFT ‐ Example
K. Webb MAE 4020/5020
Plotting magnitude of (real)
Components at 0, 50, 120, and 310 are clearly visible
Plot spectrum as a function of Index value, Normalized frequency Ordinary frequency
values divided by so that is the average value of
Amplitude of other components given by the sum of and magnitudes
0.3 0.5 cos 2 ⋅ 50 ⋅ cos 2 ⋅ 120 ⋅ 0.8 cos 2 ⋅ 320 ⋅
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Spectral Leakage
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0.3 0.5 cos 2 ⋅ 50 ⋅ cos 2 ⋅ 120 ⋅ 0.8 cos 2 ⋅ 320 ⋅
For 1 and 100, Δ 10 , and all signal components fall at integer multiples of Δ All components lie in exactly one frequency bin
Now, increase the number of samples to 105 Bin width decreases to Δ 9.52 Each non‐zero signal component now falls between frequency bins – Spectral
Leakage
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Spectral Leakage
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Signal components now fall between two bins
Why non‐zero over more than two bins? Truncation (windowing)
Finite record length is equivalent to multiplication of by a rectangular pulse (window) F.T. of pulse is a sinc Multiplication in the time domain →
convolution in frequency domain
Truncated signal is assumed periodic True only if windowing function
captures an integer number of periods of all signal components
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Summary of Fourier Analysis Tools
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Time Domain Frequency Domain
Fourier series continuousperiodic (or truncated)
aperiodicdiscrete
Fourier transform
continuousaperiodic
aperiodiccontinuous
DTFT discreteaperiodic
periodiccontinuous
DFT discreteperiodic (or truncated)
periodicdiscrete
In general:Frequency DomainDiscreteContinuousAperiodicPeriodic
Time DomainPeriodic
AperiodicContinuous
Discrete
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DFT Algorithm74
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Implementing the DFT in MATLAB
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/
A dot product of complex ‐vectors for each of the values of
⋅ /
Simple to code multiplications for each
value – operations Inefficient, particularly for large
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Fast Fourier Transform – FFT
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The fast Fourier transform (FFT) is a very efficient algorithm for computing the DFT The Cooley‐Tukey algorithm
Requires on the order of operations Significantly fewer than
For example, for : DFT: operations FFT: operations – ( faster)
Requires be a power of two If not, data record is padded with zeros
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It is very simple to implement a straight DFT algorithm in MATLAB, but the FFT algorithm is, by far, more efficient .
FFT in MATLAB77
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Fast Fourier Transform in MATLAB – fft.m
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Xk = fft(x,n)
x: vector of points for DFT computation n: optional length of the DFT to compute Xk: complex vector of DFT values – size(x) or an n‐vector
If n is not specified, x will either be truncated or zero‐padded so that its length is n
If x is a matrix, the fft for each column of x is returned
fft.m uses the Cooley‐Tukey algorithm
Fastest for length(x) or n that are powers of two
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Inverse FFT in MATLAB – ifft.m
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x = ifft(Xk,n)
Xk: vector of points for inverse DFT computation n: optional length of the inverse DFT to compute x: complex vector of time‐domain values – size(x) or an n‐vector
If n is not specified, x will either be truncated or zero‐padded so that its length is n
If Xk is a matrix, the inverse fft for each column of Xk is returned
ifft.m uses the Cooley‐Tukey algorithm Fastest for length(Xk) or n that are powers of two
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Shifting Negative Frequency Values – fftshift.m
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Xshift = fft(Xk)
Xk: vector of FFT values with zero frequency point at Xk(1) Xshift: FFT vector with the zero‐frequency point moved to the middle of
the vector
If N = length(Xk) is even, first and second halves of Xk are swapped Xshift = [Xk(N/2+1:N),Xk(1:N/2)]
Frequency points are: … Δ
If N = length(Xk) is odd, zero frequency point moved to the Xshift((N+1)/2) position Xshift = [Xk((N+3)/2):N),Xk(1:(N-1)/2)]
Frequency points are: …
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