meeting 12 - electrical engineering and computer science · meeting 12 summer 2009 doing dsp...

Meeting 12

Summer 2009 Doing DSP Workshop

Today:

◮ Motivate Fourier Series to DFT.

◮ Properties of DFT.

◮ Using DFT to compute the IDFT.

◮ Dealing with discontinuities.

Many graphics from TI materials.

The essence of mathematics is not to make simple things complicated, but to

make complicated things simple. — S. Gudder

Doing DSP Workshop – Summer 2009 Meeting 12 – /54 Thursday – June 11, 2009

Projects

Audio waveform synthesizer –

sine, square wave, triangle, etc.

◮ Darin Rajabian

OFDM.

◮ Yu Wang

Motor speed control lab demon

stration.

◮ Zharori Cong

◮ B.K. Kim

Remote camera using ZigBee.

◮ James Kim

◮ Jordan Adams

Digital Filter Study.

◮ Vindhya Reddy

◮ Joanna Widjaja

Ultrasonic Vision Aide.

◮ Ronald Deang

Not cast in concrete.


Suggested Project Phases

◮ Start up.

◮ Basically define the task, locate useful resources, and verbalize

a possible plan of attack.

◮ Initial Start.

◮ Develop the initial proposal. If applicable, do MATLAB

simulation. Identify required parts and other resources needed

to be purchased. Should have a reasonably clear understanding

of what is to be done and how. Set up goals and time line.

◮ Work in earnest.

◮ Program, build, debug. Repeat.

◮ Completion. Sometime in August.

◮ Demonstration to the workshop.◮ Poster.

Feel free to use ChihWei and myself as resources.


Updated tentative schedule

Week of June 8: One Day Workshop continued.

Tuesday – discuss and decide on project topics.

Thursday – The DFT.

Week of June 15: Exercise 5, controlSTICK ADC, DAC, xfer meas..

Tuesday – FFTs using the controlSTICK.

Thursday – Xilinx 8bit PicoBlaze microcomputer (VHDL).

Week of June 22: Exercise 6, realtime FFT and waveform evaluation.

Tuesday – TBD. KM away.

Thursday – TBD. KM away.

Weeks following —

Lecture and lab complete, focus on projects.


Lab floor to be done next week

The floors in EECS 4341 are scheduled to be stripped and waxed next

week.

Except for the tables we will be getting everything up off of the floor on

Friday.

Hope to keep the lab functional. Will move the computers onto the

tables so as to retain access.


Today

◮ Will attempt to evolve continuous time Fourier Series into the

discrete time Fourier Transform (DFT). Some what hand wavy. Goal

is to develop some intuition. Can you do a development that is

intuitive, simple, direct and rigorous?

◮ Next define the DFT and it’s inverse, present some properties.

◮ Parseval’s theorem. Energy in the input time waveform is equal

energy in the frequency domain. (Not surprising.)

◮ Use of the DFT to compute the IDFT.

◮ Mapping DFT indices to realworld frequencies.

◮ Leakage and it’s mitigation.

We will consider fast algorithms for calculating the DFT next discussion

period.


Recall Fourier Series

Given a periodic function x(t) with period T one can write

x(t) = a0 +

∞∑

k=1

ak cos(2πkt/T)+

∞∑

k=1

bk sin(2πkt/T).

where

a0 =1

T

∫

Tx(t)dt,

ak =2

T

∫

Tx(t) cos(2πkt/T)dt,

bk =2

T

∫

Tx(t) sin(2πkt/T)dt.

Some restrictions apply.


Rewrite in complex form

Replace the sine and cosine with their complex exponential

representations and group terms to get

x(t) =

∞∑

k=−∞

cke−j2πkt/T .

Where

ck =1

T

∫

Tx(t)ej2πkt/Tdt.

Somewhat more concise.

The k/T can be interpreted as the complex exponential’s frequencies,

fk = k/T .


If waveform is bandlimited

Things get a bit hand wavy here. There is no such thing as a band limited

waveform, just models.

Assume that there is negligible energy outside of a band of frequencies

(B, B) Hz.

Then it is claimed that there are finite K such that K/T > B such all

cK = 0. Let K = ⌈B/T⌉. Then

x(t) =

K∑

k=−K

cke−j2πkt/T .

and we only need to calculate 2K values of ck.


Let’s sample

Assume that x(t) is sampled at rate fs . Write the sample times as n/fs .

x(n/fs) =

K∑

k=−K

cke−j2πkn/(Tfs ).

Restrict fs such that T = N/fs where N is integer. Then

x(n/fs) = x(n) =

K∑

k=−K

cke−j2πkn/N .

We now have a finite number of sample values and a finite number of

frequencies.

How to find the ck?


Guess the answer?

For a given k′ we certainly can write

N−1∑

n=0

x(n)ej2πk′n/N =

N−1∑

n=0

K∑

k=−K

ckej2πk′n/Ne−j2πkn/N .

Many proofs/demonstrations interchange operation order as soon as

possible.

N−1∑

n=0

x(n)ej2πk′n/N =

K∑

k=−K

ck

N−1∑

n=0

ej2π(k′−k)n/N .

The right most sum is the sum of geometric series. Leaving this as an

exercise, the value will be N when k = k′ and zero otherwise. Thus

ck =1

N

N−1∑

n=0

x(n)ej2πkn/N .


Almost there

We now have

x(n) =

K∑

k=−K

cke−j2πkn/N ,

ck =1

N

N−1∑

n=0

x(n)ej2πkn/N .

Recall also that T = N/fs .

If K/T > B to avoid aliasing then K > BT , K > BN/f2.

To avoid aliasing we need B < fs/2. So if B = fs/2 then K ≥ N/2. Let’s

choose K = N/2.


Finally

We now have

x(n) =

N/2∑

k=−N/2

cke−j2πkn/N ,

ck =1

N

N−1∑

n=0

x(n)ej2πkn/N .

Note that e−j2π(N−k)n/N = e−j2π(−k)n/N .

Without thinking too hard about one missing term we can write

x(n) =

N−1∑

k=0

cke−j2πkn/N ,

ck =1

N

N−1∑

n=0

x(n)ej2πkn/N .


This was meant to be motivational

◮ Assumes one has a good background in continuous time.

◮ That one believes “the” sampling theorem.

◮ Band limitness allows Fourier series terms to be dropped. If not

bandlimited then the energy is aliased when we sample.

◮ Note that the 1/N naturally shows up when computing the ck.

◮ Always it is true that T = N/fs .

◮ Discrete time reconstruction from coefficients is analogous with

continue time reconstruction.

◮ The assertion is made that the waveform is periodic outside of the

interval not included!

Rigor, and proper style, was somewhat lacking.


What happens if

the waveform is not band limited?

the waveform has a discontinuity?


The Discrete Fourier Transform (DFT)

The DFT is most commonly defined as

X[k] =

N−1∑

n=0

x[n]e−j2πkn/N , k = 0,1, . . . , N − 1.

The inverse DFT is

x[n] =1

N

N−1∑

k=0

X[k]ej2πkn/N , n = 0,1, . . . , N − 1.

In practice one often uses (my preference)

X[k] =1

N

N−1∑

n=0

x[n]e−j2πkn/N , k = 0,1, . . . , N − 1.

The associated inverse DFT is

x[n] =

N−1∑

k=0

X[k]ej2πkn/N , n = 0,1, . . . , N − 1.


A few DFT properties

The DFT is linear:

DFT(ax[n]+ by[n]) = aDFT(x[n])+ bDFT(y[n]) = aX[k]+ bY[k].

If x[n],n = 0,1,2, . . . , N − 1, is real then

X[−k] = X∗[k] = X[N − k], k = 0,1,2, . . . , N − 1.

This implies for real valued waveforms that:

For N even : the first N/2+ 1 values define the DFT.

For N odd : the first (N − 1)/2+ 1 values define the DFT.

For x[n] real: for N even let M = N/2 and for N odd let M = (N − 1)/2.

Writing X[k] = |X[k]ejφ[k] we have

|X[k]| = |X[N − k]| and φ[k] = −φ[N − k] for k = 1,2, . . . ,M − 1.


A few more DFT propertiesPeriodicity:

X[k] = X[k+N], ∀k and x[n] = x[n+N], ∀n.

Circular convolution:

DFT−1(H[k]X[k]) = h[n]⊗

x[n]

=

N−1∑

m=0

x[m]h[〈n−m〉N]

=

N−1∑

m=0

h[m]x[〈n−m〉N]

Shifting:

x[〈n−n0〉N] ⇔ Wkn0N X[k]

W−k0nN x[n] ⇔ X[〈k− k0〉N]

where WN = e−j2π/N .


Yet a few more properties

For an arbitrary x[n] of length N :

x[n] ⇔ X[k]

x∗[n] ⇔ X∗[〈−k〉N]

x∗[〈−n〉N] ⇔ X∗[k]

Re{x[n]} ⇔12 {X[〈k〉N]+X

∗[〈−k〉N]}

Im{x[n]} ⇔12 {X[〈k〉N]−X

∗[〈−k〉N]}


Parseval’s theorem for the DFT

Parseval’s theorem relates the energy in a finite duration time

waveform to the energy in its frequency domain representation.

For the standard definition of the DFT we have

N−1∑

n=0

|x[n]|2 =1

N

N−1∑

k=0

|X(k)|2 .

For the alternate normalization (my preferred) Parseval’s

theorem becomes

1

N

N−1∑

n=0

|x[n]|2 =

N−1∑

k=0

|X(k)|2 .


DFT: 2 N-pt real using 1 N-pt complex

The standard definition of the DFT/IDFT assumes the use of

complex values. Consider the DFT of a set of complex values

z[n] = x[n]+ jy[n] where x[n] and y[n] are real:

DFT(z[n]) = DFT(x[n])+ jDFT(y[n]).

DFT(z∗[n]) = Z∗[N − k].

Recall that x[n] and y[n] are real and that

z[n] = x[n]+ jy[n].

DFT(x[n]) =DFT(z[n]+ z∗[n])

2=

Z[k]+ Z∗[N − k]

2,

DFT(y[n]) =DFT(z[n]− z∗[n])

j2=

Z[k]− Z∗[N − k]

j2.


N/2 complex transform for N real values

For {x[n]}, 0 ≤ n < N , real valued and N even the DFT of the {x[n]} is

XN[k] =

N−1∑

n=0

x[n]e−j2πkn/N ,

=

N/2−1∑

n=0

x[2n]e−j2π2kn/N +

N/2−1∑

n=0

x[2n+ 1]e−j2π(2n+1)k/N .

Write the even numbered sample values as xE[m] = x[2m] and the odd

numbered sample values as xO[m] = x[2m+ 1]. Then

XN[k] =

N/2−1∑

m=0

xE[m]e−j2πkm/(N/2)

+ e−j2πk/NN/2−1∑

m=0

xO[m]e−j2πmk/(N/2),

= XE,N/2[k]+ e−j2πk/NXO,N/2[k].

Because x[n] is real we only need to consider k = 0,1, . . . , N/2.


N real in N/2 complex continued

The positive frequency portion of the DFT of N real values can be

written as a weighted sum of two k = N/2 value transforms. A few

slides back it was shown how an N/2 value complex DFT can

accomplish this task.

Combining the previous slide’s result with the procedure for using an

N value DFT to compute the individual DFTs of two real N valued data

sets gives

XN[k] =XN/2[k]+X

∗N/2[N/2− k]

2+ e−j2πk/N

XN/2[k]−X∗N/2[N/2− k]

j2

for k = 0,1, . . . , N/2. The k = N/2 value is calculated as the difference

of the k = 0 values of the two N/2 value DFTs.


Yet another detail

Assume we wish to use a N/2 element complex array to hold

the frequencies and yet wish to be able to reconstruct the

complete N value transform. There is a problem with where to

put the k = N/2 value.

The input array has N real values and produces N/2 complex

values, k = 0,1, . . . , N − 1. Fortunately the k = N/2 value is real

as is the k = 0 value. Simply use the imaginary part of the k = 0

value to hold the k = N/2 value.

This procedure, using an N/2 value transform to compute the

DFT of a N real valued data set, has an overhead associated

with it. The speed up is not going to be 2 times, but slightly

less. However, it still is generally very much worth doing.


There are many ways to slice bread

I have seen an article developing an algorithm for use with real valued

waveforms that uses an N complex DFT to simultaneously form the

direct transform of N samples of a real valued waveform while also

computing an N value inverse transform of a real valued waveform.

This would be very useful in an FPGA. One could overlap flowing data

through the steps of DFT, processing in the frequency domain and IDFT.


Using the DFT to compute the IDFT

The conventional form of the forward transform is

X[k] =

N−1∑

n=0

x[n]e−j2πkn/N , k = 0,1, . . . , N − 1.

The conventional form of the inverse transform is

x[n] =1

N

N−1∑

k=0

X[k]ej2πkn/N , n = 0,1, . . . , N − 1.

Let n = N −n′, n′ = 0,1, . . . , N − 1.

x[N −n′] =1

N

N−1∑

k=0

X[k]ej2πk(N−n′)/N =

1

N

N−1∑

k=0

X[k]e−j2πkn′/Nej2πN/N

Thus

DFT(X[k]) is Nx[N −n′] where n′ = 0,1, . . . , N − 1.


Second method

Conjugate X[k] and form the DFT.

N−1∑

k=0

X∗[k]e−j2πkn/N =

N−1∑

k=0

N−1∑

m=0

x∗[m]ej2πkm/N

e−j2πkn/N

=

N−1∑

m=0

x∗[m]

N−1∑

k=0

ej2πk(m−n)/N .

N−1∑

k=0

ej2πk(m−n)/N = N for m = n and 0 otherwise.

DFT{

X∗[k]}

= Nx∗[n], n = 0,1, . . . , N − 1.


Third method

Consider the inverse DFT of X[k] = Xr [k]+ jXi[k]

IDFT {X[k]} =1

N

N−1∑

k=0

X[k]ej2πkn/N , n = 0,1, . . . , N − 1

=1

N

N−1∑

k=0

{Xr [k] cos(2πkn/N)−Xi[k] sin(2πkn/N)}+

j {Xr [m] sin(2πkn/N)+Xi[k] cos(2πkn/N)} .

Consider forming the DFT{

Xi[k]+ jXr [k]}

=

N−1∑

k=0

{

Xi[k]+ jXr [k]}

e−j2πkn/N

=

N−1∑

k=0

{Xr [k] sin(2πkn/N)+Xi[k] cos(2πkn/N)}+

j {Xr [k] cos(2πkn/N)−Xi[k] sin(2πkn/N)} .


Third method continued

Comparing the two expressions we see that

x[n] =1

Nswap{FFT[swap(X[k])]}.

Where swap interchanges the real and imaginary parts of a complex

value.(Swap notation from the Wikepedia.)


Short summary

All three methods:

can easily be accomplished when moving data and results,

require dividing the (inverse) DFT values by N .


Making light

Bill Hoest

“Question: How many signal processing engineers does it take to

change a light bulb?

Answer: Three.

One to Fourier transform the light bulb, one to apply a complex

exponential rotational shifting operator, and one to inverse transform

the removed light bulb. ”

Drawing from the web via Google, joke from the November 2006 IEEE Signal Processing Magazine.


How do we program complex numbers?

As pairs of real numbers. Even indices are the real part and the

odd indices are the imaginary part. For example to access the

kth complex element in the y array,

yr = y[2*k]; yi = y[2*k+1];

How do we conjugate a complex number? Change the sign of

the imaginary part.


Consider a cosine waveformGiven a cosine waveform:

cos(2πft) =ej2πft + e−j2πft

2.

The DFT is linear · · · superposition applies.

Consider only the positive frequency term.

Acquire N samples using a sample rate of fs Hz.

Data set duration is T = N/fs .

Sample times are

tn = 0/fs ,1/fs ,2/fs , . . . , (N − 1)/fs .

The DFT is

X[k] =1

2

N−1∑

n=0

ej2πnf/fs e−j2πkn/N =1

2

N−1∑

n=0

ej2πn(f/fs−k/N) .

When f/fs − k/N = 0 it is clear that X[k] = N/2.


DFT frequency step size

This implies that there is an integer k′ such as f = k′fs/N .

For this frequency it can be shown that X[k] = 0 for all other k

in the given range.

Frequencies that are integer multiples of fs/N are sometimes

termed the eigenfrequencies or, alternatively, analysis

frequencies of the DFT.

What happens if f 6= k′fs/N?

If we define the duration of a data set as T = N/fs then the

frequency step size ∆f between DFT values is

∆f = 1/T = fs/N .

ALWAYS!


Negative frequencies

Consider k′ = N − k.

X[N − k′] =1

2

N−1∑

n=0

ej2πn[f/fs−(N−k′)/N]

=1

2

N−1∑

n=0

e−j2πnej2πn(f/fs+k′/N)

X[−k′] =1

2

N−1∑

n=0

ej2πn(f/fs+k′/N)

Negative frequencies of the form f = −k′fs/N are also

eigenfrequencies of the DFT.


Naming

We could name DFT analysis frequencies for k = 0,1, . . . , N − 1

as

0, fs/N,2fs/N, . . . , (N − 2)fs/N, (N − 1)fs/N

or

0,−(N − 1)fs/N,−(N − 2)fs/N, . . . ,−2fs/N,−fs/N

or

0, fs/N,2fs/N, . . . ,−(N − 2)fs/N,−(N − 1)fs/N

or

0, fs/N,2fs/N, . . . ,−2fs/N,−fs/N

The last of these namings is the most common.


fftshift — even and odd

Two cases, N even, and N odd.

Let’s see what MATLAB does.

For the N = 8 input values [1 0 0 0 0 0 0 0]

MATLAB’s fftshift gives [0 0 0 0 1 0 0 0]

For the N = 9 input values [1 0 0 0 0 0 0 0 0]

MATLAB’s fftshift gives [0 0 0 0 1 0 0 0 0]

After doing a fftshift the analysis frequencies are:

(-floor(N/2)+[0:N-1])*fs/N.

If you aren’t sure, try it out.


End values

For N even, the range includes

−fs/2 through (fs/2)(N − 2)/N .

For N odd, the range includes

−(fs/2)(N − 1)/N through (fs/2)(N − 1)/N .


Be careful when using linspace

Want 10 samples of one period of a sine wave.

What is wrong with using

sin(linspace(0, 2*pi, 10))?

The last value returned is actually the first value in the

immediately following period!

Use

sin(linspace(0, 2*pi*(N-1)/N, N))

It’s so easy to misuse linspace that you might as well do the

required calculations yourself. If you don’t know how, you

shouldn’t be using linspace anyway.


Summary

Analysis frequencies are integer multiples of 1/T = fs/N .

Only two of T , fs , and N can be chosen independently.

Analysis range is nominally from −fs/2 to fs/2 Hz.

For N even:

there are N/2 nonnegative analysis frequencies.

there are N/2 negative analysis frequencies.

For N odd:

there are (N + 1)/2 nonnegative analysis frequencies.

There are (N − 1)/2 negative analysis frequencies.

After using MATLAB’s fftshift analysis frequencies will be:

(-floor(N/2)+[0:N-1])*fs/N .


Leakage, no leakage


Recall

Analysis frequencies are integer multiples of 1/T = fs/N .

Only two of T , fs , and N can be chosen independently.

Analysis range is nominally from −fs/2 to fs/2 Hz.

For N even:

there are N/2 nonnegative analysis frequencies.

there are N/2 negative analysis frequencies.

For N odd:

there are (N + 1)/2 nonnegative analysis frequencies.

There are (N − 1)/2 negative analysis frequencies.

After using MATLAB’s fftshift analysis frequencies will be:

(-floor(N/2)+[0:N-1])*fs/N .


Sampling a cosine waveform

Given a cosine waveform:

cos(2πft) =ej2πft + e−j2πft

2.

The DFT is linear · · · superposition applies. Consider only the

positive frequency term. Acquire N samples using a sample rate

of fs Hz. The data set duration is T = N/fs .

Sample times are

tn = 0/fs ,1/fs ,2/fs , . . . , (N − 1)/fs .

The DFT is

X[k] =1

2

N−1∑

n=0

ej2πnf/fse−j2πkn/N =1

2

N−1∑

n=0

ej2πn(f/fs−k/N) .

When f/fs −k/N = 0 it is clear that X[k] = N/2. Otherwise · · · .


Arbitrary frequency sinusoidal input to DFT

◮ Fix the index value being looked at.

◮ Slowly tune an oscillator from 0 Hz to fs/2.

◮ Plot. xaxis is the frequency of the oscillator and the yaxis

is the magnitude of the selected DFT “bin”.

On the following sheet the complex exponential input was

sampled at 48000 Hz.

The frequency of the top goes from −fs/2 to fs/2.

The magnitude of the value at index k equal to 21 was plotted.

The lower plot expands the plot.


DFT as a filter

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

x 104

−80

−60

−40

−20

0

dB

N = 1024, k = 21

DFT as a Filter

0 500 1000 1500 2000−80

−60

−40

−20

0N = 1024, k = 21

dB

Hz

Magnitude of bin 21 output as ej2πft sweeps through

frequencies from −fs/2 to fs/2.


Windowing

Select portion of waveform to analyze.

DFT enforces periodicity. . . what happens at the ends?

Weight or shade the data to minimize end effects.

Multiplication in time corresponds to convolution in frequency.

X(k) =

N−1∑

n=0

w[n]x[n]e−j2πkn/N , k = 0,1, . . . , N − 1.

Multiplication in the time domain corresponds to convolution

(filtering) in the frequency domain.


Window functions (shades) used in lab

0 200 400 600 800 10000

0.2

0.4

0.6

0.8

1rectangle

Hamming

Chebyshev

sample index

ampl

itude

Window Functions


Rectangle (no) window

Plot of samples and the

window function.

Weighted samples shown

recentered at end point

splice.

dB plot of the spectrum

of the windowed samples.

200 400 600 800 1000−1

−0.5

0

0.5

1

Rectangle (No) Window

−500 0 500−1

−0.5

0

0.5

1

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

x 104

−80

−60

−40

−20

0

dB

Hz


Hamming window


window function.



splice.



200 400 600 800 1000−1

−0.5

0

0.5

1

Hamming Window

−500 0 500−1

−0.5

0

0.5

1

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

x 104

−80

−60

−40

−20

0

dB

Hz


Chebyshev 72 dB window


window function.



splice.



200 400 600 800 1000−1

−0.5

0

0.5

1

Chebyshev 72 dB Window

−500 0 500−1

−0.5

0

0.5

1

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

x 104

−80

−60

−40

−20

0

dB

Hz


Windowed DFT as a filter

0 500 1000 1500 2000−100

−80

−60

−40

−20

0

dB

Unwindowed

N = 1024, k = 21

0 500 1000 1500 2000−100

−80

−60

−40

−20

0

dB

HzDFT as a Filter

Chebyshev 72 dB Window

N = 1024, k = 21

Magnitude of bin 21 output as ej2πft sweeps through

frequencies from −fs/2 to fs/2.


Unmasking a low level sinewave

0 0.1 0.2 0.3 0.4 0.5−100

−90

−80

−70

−60

−50

−40

−30

−20

−10

0Raw DFT magnitude (N=256)

normalized frequency

dB

0 0.1 0.2 0.3 0.4 0.5−100

−90

−80

−70

−60

−50

−40

−30

−20

−10

0Hamming windowed DFT magnitude (N=256)


dB

0 0.1 0.2 0.3 0.4 0.5−100

−90

−80

−70

−60

−50

−40

−30

−20

−10

0Chebyshev windowed DFT magnitude (N=256)


dB

0 0.1 0.2 0.3 0.4 0.5−100

−90

−80

−70

−60

−50

−40

−30

−20

−10

0Chebyshev windowed DFT magnitude (N=512)


dB


Leakage and windowing examples

jçîáÉ jçîáÉ

jçîáÉ jçîáÉ

jçîáÉ jçîáÉ

0 200 400 600 800 10000

0.5

1

ampl

itude Rectangle window

0 200 400 600 800 10000

0.5

1

ampl

itude Hamming window

0 200 400 600 800 10000

0.5

1

ampl

itude Chebyshev window

0 200 400 600 800 10000

0.5

1

bin index

ampl

itude R, H and C windows


What are the Windowing Costs?

Main lobe width Spreading of energy near a line. . . loss of nearby

resolution.

Line amplitudes are reduced.

Scalloping loss . . . apparent attenuation when line falls midway

between adjacent lines.

May need increased numeric precision. Especially for the more

effective windows.


meeting 12 - electrical engineering and computer science · meeting 12 summer 2009 doing dsp...

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