eces 390_week 1 lecture_summer 2015

ECES 390 Transforms IIWeek 1

Dr. Tom Chmielewski and Dr. Christopher Peters

ECE 390 - Transforms II - Chmielewski and Peters 1

Fourier/Inverse Fourier Transform

= =

1 = = 12

Fourier Transform:

Inverse Fourier Transform:

These are continuous functions

The Fourier transform provides the coefficients/weights (continuous function) that result by projecting f(t) onto the basis set. These coefficients provide a means to represent the signal in a different domain, while providing information of the strength of each basis function


Common Fourier Relationshipsf(t) F()

1

u(t) + 1

+ + sin

+ +

2 + + + 022

2

2

2Duality: () 2


Discretizing the Signal

Analog to Digital Converter

ComputerDigital to Analog Converter

signal

x(t) x(nTs) x(nTs) xp(t)

== + 1 What is the physicality of the above equation?


Zero Order Hold Model

Zero-Order Hold

signal

x(t) xs(t) xp(t)X

Impulse train

0

1

Ts

0

1

Ts

= = + 1

Single term input


Zero Order Hold

Square pulsePulse only contributes only during the pulse durationForms stairstep waveformReconstructed waveform is poor approximation to original waveform

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1-1

-0.5

0

0.5

1

Time (s)

Pulse Profile

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1-1

-0.5

0

0.5

1

Time (s)

Output Signal

How can we improve the reconstruction?


Zero Order Hold

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-1

-0.5

0

0.5

1

Time (s)

Pulse Profile - fs=200 samples/sec

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-1

-0.5

0

0.5

1

Time (s)

Output Signal

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045-1

-0.5

0

0.5

1

Time (s)

Pulse Profile - fs=479 samples/sec

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045-1

-0.5

0

0.5

1

Time (s)

Output Signal


Spectrum Representation of Sampled Signals

= =

Using superposition rule:

Any single point:

Fourier Identity: 0 0

Use Fourier transform property of superposition: the Fourier transform of total signal is equal to the sum of the individual Fourier transforms:

= =

This is called the Discrete Time Fourier Transform (DTFT)


Discrete Time Fourier Transform (DTFT)Let x(nTs)x[n], Ts

= =

=

=

Inverse DTFT: = 1211+2 = 1

22

DTFT

Discrete in time Continuous in frequency


DTFT Example = =

=

= =

= =0

= =0

Recall geometric series: =0

= 1 + + 2 + = 11 , < 1 =

=0

= 11 , < 1 = = 1 = ECE 390 - Transforms II - Chmielewski and Peters 10

MATLAB Exampleclose all;clear all;clc% Define the basic equationsyms wn=0:20;x=0.8.^n; % Data set x[n]X=sum(x.*exp(-j*w*n)); % DTFT of x[n]figuresubplot(2,1,1)h1=ezplot(abs(X),[-5*pi 5*pi]);set(h1,'color','b')xlabel('\Omega (rad/s)')ylim([0 5.8])title('Magnitude of DTFT')w1=-5*pi:0.01:5*pi;XX=zeros(size(w1));for K=1:length(w1)XX(K)=subs(X,w,w1(K));

endsubplot(2,1,2)plot(w1,angle(XX))xlabel('\Omega (rad/s)')title('Phase of DTFT')xlim([-5*pi 5*pi])

Is this fully correct?


-15 -10 -5 0 5 10 150

2

4

(rad/s)

Magnitude of DTFT

-15 -10 -5 0 5 10 15-1

-0.5

0

0.5

1

(rad/s)

Phase of DTFT

MATLAB Example


-15 -10 -5 0 5 10 150

1

2

(rad/s)

Magnitude of DTFT

-15 -10 -5 0 5 10 15-1

-0.5

0

0.5

1

(rad/s)

Phase (radians) of DTFT

close all;clear all;clcsyms w nx=0.6^n;X=symsum(x*exp(-j*w*n),n,0,Inf);w1=-5*pi:0.1:5*pi;X1=zeros(size(w1));for K=1:length(w1)

X1(K)=eval(subs(X,w,w1(K)));endfiguresubplot(2,1,1)plot(w1,abs(X1))xlabel('\Omega (rad/s)')title('Magnitude of DTFT')xlim([-5*pi 5*pi])ylim([0 2.8])subplot(2,1,2)plot(w1,angle(X1))xlim([-5*pi 5*pi])xlabel('\Omega (rad/s)')title('Phase (radians) of DTFT')

DTFT Linearity Example


-3 -2 -1 0 1 2 35

10

15

20

25

(rad/s)

Magnitude of DTFT

-3 -2 -1 0 1 2 3-1

-0.5

0

0.5

1

(rad/s)

Phase of DTFT

close all;clear all;clcsyms wn=0:10;x1=0.8.^n;x2=0.7.^n;a=3;b=4;f=a*x1+b*x2;X1=sum(a*x1.*exp(-j*w*n));X2=sum(b*x2.*exp(-j*w*n));F=sum(f.*exp(-j*w*n));w1=-pi:0.01:pi;

figuresubplot(2,1,1)h1=ezplot(abs(X1+X2),[-pi pi]);hold onh2=ezplot(abs(F),[-pi pi]);set(h2,'color','r')xlabel('\Omega (rad/s)')title('Magnitude of DTFT')subplot(2,1,2)x1h=zeros(size(w1));x2h=zeros(size(w1));Fh=zeros(size(w1));for K=1:length(w1)

x1h(K)=subs(X1,w,w1(K));x2h(K)=subs(X2,w,w1(K));Fh(K)=subs(F,w,w1(K));

endplot(w1,angle(x1h+x2h));hold onplot(w1,angle(Fh),'r');xlabel('\Omega (rad/s)')title('Phase of DTFT')xlim([-pi pi])

1 + 2 1 + 2

DTFT Time Shift Example


-3 -2 -1 0 1 2 3

5

10

15

20

Magnitude of DTFT

UnshiftedShifted

-3 -2 -1 0 1 2 3-4

-2

0

2

4

Phase (radians) of DTFT

UnshiftedShifted

close all;clear all;clcsyms wn1=0:5;n2=0:7;x1=[1 2 3 4 5 6];x2=[0 0 1 2 3 4 5 6];DTFT_unshifted=sum(x1.*exp(-j*w*n1));DTFT_shifted=sum(x2.*exp(-j*w*n2));figuresubplot(2,1,1)h1=ezplot(abs(DTFT_unshifted),[-pi pi]);hold onh2=ezplot(abs(DTFT_shifted),[-pi pi]);set(h2,'color','r')legend('Unshifted','Shifted',0)xlabel('\Omega')title('Magnitude of DTFT')subplot(2,1,2)w1=-pi:.01:pi;phase_unshifted=zeros(size(w1));phase_shifted=zeros(size(w1));for K=1:length(w1)

phase_unshifted(K)=angle(eval((subs(DTFT_unshifted,w,w1(K)))));phase_shifted(K)=angle(eval((subs(DTFT_shifted,w,w1(K)))));

endplot(w1,phase_unshifted);hold onplot(w1,phase_shifted,'r');xlim([-pi pi])legend('Unshifted','Shifted',0)xlabel('\Omega')title('Phase (radians) of DTFT')

0 0

DTFT Frequency Shifting


close all;clear all;clcw0=1; % Frequency shiftsyms wn=-5:0;x1=sin(0.1*n);x2=sin(0.1*n).*exp(j*w0*n);unshifted=sum(x1.*exp(j*w*n));shifted=sum(x2.*exp(-j*w*n));figuresubplot(2,1,1)h1=ezplot(abs(unshifted),[-pi pi]);set(h1,'color','b')hold onh2=ezplot(abs(shifted),[-pi pi]);set(h2,'color','r')xlabel('\Omega')title ('Magnitude of DTFT')legend('Unshifted','Shifted',0)subplot(2,1,2)w1=-pi:.01:pi;phase_unshifted=zeros(size(w1));phase_shifted=zeros(size(w1));for K=1:length(w1)

phase_unshifted(K)=angle(eval(subs(unshifted,w,w1(K))));phase_shifted(K)=angle(eval(subs(shifted,w,w1(K))));

endplot(w1,phase_unshifted)hold onplot(w1,phase_shifted,'r')xlabel('\Omega')ylabel('Phase (radians) of DTFT')legend('Unshifted','Shifted',0)xlim([-pi pi])

0 0

-3 -2 -1 0 1 2 3

0.5

1

1.5

Magnitude of DTFT

UnshiftedShifted

-3 -2 -1 0 1 2 3-4

-2

0

2

4

Pha

se (r

adia

ns) o

f DTF

T

UnshiftedShifted

Discrete Fourier Transform Continuous spectrum

Fourier transform Discrete Time Fourier Transform

Concern with modern technology Cant store an infinite set of frequencies on electronic media. HOW DO WE APPROXIMATE THE FREQUENCY CONTENT OF A DISCRETE SIGNAL?

Practical Issues Finite set of discrete values N = number of samples chosen to represent the discrete time series N must be large enough to represent the signal

Our objective: We seek an approximation of Fourier transforms that can be computed from a finite set of discrete samples of an analog signal which produces a finite set of discrete frequency spectrum values

Solution: Compute discrete-frequency samples of continuous frequency function X()ECE 390 - Transforms II - Chmielewski and Peters 16

Approach to a Solution

Define a rectangular window

= 1, = 0, 1, 2, 10, Multiply infinite set x[n] by wR[n]

= = , = 0, 1, 2, 10, Intermediate step yields discrete time Fourier transform of windowed signal

= = =

=

=0

1

Still continuous We can drop the subscript N if desiredECE 390 - Transforms II - Chmielewski and Peters 17

Approach to a Solution Select N equal-distance samples of XN() to represent the frequency spectrum

Need at least N discrete frequency values (could use more) to avoid errors if the inverse DFT is desired Recall the DTFT is periodic with a periodicity of 2 = + 2

= =0

1

2

=

=0

1

2

2

Value of frequency spectrum at each desired, discrete frequency Impulse in frequency domain

Set of discrete values 2

is information we select to represent the frequency spectrum of discrete

time sequence x[n} and xN[n]


Discrete Fourier Transform Continued

Convention: = 2 , = 0,1,2, 1Since x[n] is finite, it is a nonperiodic discrete time sequence, and therefore we only need N samples

= 1=0

1

2 , = 0, 1, 2, 1

N discrete samples from a set of N samples of discrete time signal

X = = =01 2 , = 0, 1, 2, 1x = 1 = 1

=01

2 , = 0, 1, 2, 1


DFT Example

-0.5 0 0.5 1 1.5 2 2.5 3 3.5

0

2

4

6

k

|Xk|

-0.5 0 0.5 1 1.5 2 2.5 3 3.5

-2

0

2

k

X k

(rad

ians

)

close all;clear all;clcx=[1 2 2 1];N=length(x);X=zeros(1,length(x));Xk=zeros(1,length(x));for k=0:N-1

for n=0:N-1X(n+1)=x(n+1)*exp(-j*2*pi*k*n/N);

endXk(k+1)=sum(X);

endfiguresubplot(2,1,1)stem(0:N-1,abs(Xk),'filled')xlim([-.5 3.5])ylim([-0.5 6.5])xlabel('k')ylabel('|X_k|')subplot(2,1,2)stem(0:N-1,angle(Xk),'filled')xlim([-.5 3.5])ylim([-3.5 3])xlabel('k')ylabel('\angleX_k (radians)')


DFT Function

function Xk=dft(x)N=length(x);Xk=zeros(1,length(x));X=zeros(1,length(x));for k=0:(N-1)

for n=0:(N-1)X(n+1)=x(n+1)*exp(-j*2*pi*k*n/N);

endXk(k+1)=sum(X);

endend


DTFT Property - Periodicity

0 2 4 6 8 10 12 140

10

20

30

k

|Xk|

0 2 4 6 8 10 12 14-4

-2

0

2

4

k

X k

close all;clear all;clcn=0:4;x=n.^2;N=length(x);for k=0:(3*N-1);a


endXk(k+1)=sum(X);

end

figuresubplot(2,1,1)stem(0:(3*N-1),abs(Xk),'filled')xlabel('k')ylabel('|X_k|')subplot(2,1,2)stem(0:(3*N-1),angle(Xk),'filled')xlabel('k')ylabel('\angleX_k')


DFT Property - Linearity

0 0.5 1 1.5 2 2.5 3 3.5 40

20

40

60

k

X k

X1+X2X3

0 0.5 1 1.5 2 2.5 3 3.5 4-2

-1

0

1

2

k

X k

X1+X2X3

close all;clear all;clcx1=[5 2 1 9 6];x2=[9 1 7 8 6];x3=x1+x2;k=0:(length(x1)-1);X1=dft(x1);X2=dft(x2);X3=dft(x3);figuresubplot(2,1,1)stem(k,abs(X1+X2),'filled')hold onstem(k,abs(X3),'filled','r')xlabel('k')ylabel('\angleX_k')legend('X1+X2','X3',0)subplot(2,1,2)stem(k,angle(X1+X2),'filled')hold onstem(k,angle(X3),'filled','r')xlabel('k')ylabel('\angleX_k')legend('X1+X2','X3',0)


DFT Property Parsevals

close all;clear all;clcx=[9 1 2 9 3 4]Ex=sum(x.^2)X=dft(x)EX=1/length(x)*sum(abs(X.^2))


Inverse DFT

function x=idft(X)N=length(X);x=zeros(1,length(X));K=zeros(1,length(X));for n=0:(N-1)

for k=0:(N-1)K(k+1)=X(k+1)*exp(j*2*pi*k*n/N);

endx(n+1)=sum(K)/N;

endend


Compare and Contrast DFT and IDFT

function Xk=dft(x)N=length(x);Xk=zeros(1,length(x));X=zeros(1,length(x));for k=0:(N-1)


endXk(k+1)=sum(X);

endend

function x=idft(X)N=length(X);x=zeros(1,length(X));K=zeros(1,length(X));for n=0:(N-1)

for k=0:(N-1)K(k+1)=X(k+1)*exp(j*2*pi*k*n/N);

endx(n+1)=sum(K)/N;

endend

X = =01 2/ x = 1=01 2/


Comparison of DFT to DTFT

= = =0

1

X = = =01 2 , = 0, 1, 2, 1


= =

2

, = 0, 1, 2, 1

Example DFT - 3 Data Points (N=3)

= =0

2

2

0 = 0 2 0 03 + 1 2 0 13 + 2 2 0 23 = 0 + 1 + 2 1 = 0 2 1 03 + 1 2 1 13 + 2 2 1 23 = 0 + 1 23 + 2 43 2 = 0 2 2 03 + 1 2 2 13 + 2 2 2 23 = 0 + 1 43 + 2 23


Example IDFT-3 Data Points (N=3)

= 13=0

2

2

0 = 13 0 2 0 03 + X 1 2 1 03 + X 2 2 2 03 = 13 0 + 1 + 2 1 = 13 0 2 0 13 + X 1 2 1 13 + X 2 2 2 13 = 13 0 + 1 23 + 2 43 2 = 13 0 2 0 23 + X 1 2 1 23 + X 2 2 2 23 = 13 0 + 1 43 + 2 23The symmetry of two sets of equations (DFT and IDFT) allows computation using one set of computer codes


Periodic signals

-3 -2 -1 0 1 2 3 4 5 6

x[-3] x[-2] x[-1] x[0] x[1] x[2] x[3] x[4] x[5] x[0] x[1] x[2] x[0] x[1] x[2] x[0] x[1] x[2]

If we compute k>(N-1) and soon

X 3 = =0

2

2 3

3 = 0 + 1 + 2 = 0ECE 390 - Transforms II - Chmielewski and Peters 30

Shorthand Notation of the DFT

Let = 2 = =

=0

1

, = 0,1, 1

= = =0

1

, = 0,1, 1


Frequency Resolution of DFT

N frequency samples, with a period of 2

Separation between samples: = 2

Given N1 samples in the discrete time domain, is fixed. To increase the resolution, there needs to be more sampling in the frequency domain, which means increasing N, the number of discrete time domain points

In certain circumstances, zero padding (adding zeros at the end) can be done, but the resolution DOES NOT get better (we will discuss this later, but this comes from information theory)


Understanding WN = 2Let N=8: k=0, 1, 2, 7

8 = 28 = 4

= 0: 80 = 1 = 1: 81 = 4

= 2: 82 = 2 = 3: 83 = 34 = 4: 84 = = 1

= 5: 85 = 54 = 6: 86 = 23 = = 7: 87 = 78 Adjacent vector: Angle of 2

radians

Each vector has an opposite of equal magnitude=1. This leads to the sum of all vectors=0. =07

= 0 Each vector has an associated conjugate pair which

is in


Understanding WN = 2Expanding Notation: = 2

80 = 1

81 = 14

82 = 12 87 = 174

Observations: n=0: All vectors=1n=1: Previous vector field shown previouslyn>1: Vectors repeat pattern

82 5 = 125 = 1 12+42 = 12

87 5 = 1354 = 1 34+324 = 134

Summarizing: =0

=1

0 =

=0

=1

= 0, = 1, 2, 1


Validity of DFT = 2

= 1=0

1

= 1

=0

1

=0

1

= 1=0

1

=0

1

= 1

=0

1

=0

1

Note: =0

1

= , = 0,

= 1 =


MATLAB Implementationclose all;clear all;clc % Initialize scriptx=[1 2 3 4]; % Define discrete time series dataN=length(x); % Gets the number of points in sequenceW=exp(-j*2*pi/N)% Display magnitude and phase of Wdisp(['Magnitude of W = ' num2str(abs(W))])disp(['Angle of W = ' num2str(angle(W)) ' radians'])% Calculate the N point DFTXk=zeros(1,N); % Create and initialize frequency arrayfor k=0:1:(N-1)

for n=0:1:(N-1)Xk(k+1)=Xk(k+1)+x(n+1)*W^(k*n);

endend% Plot the magnitude and phase of Xkfiguresubplot(2,1,1)stem(0:1:(N-1),abs(Xk),'filled','r','linewidth',3)title('Magnitude of X[k], the DFT of x[n]')axis([-1,N,0,1.1*max(abs(Xk))])grid onxlabel('k - frequency sample')subplot(2,1,2)stem(0:1:(N-1),angle(Xk),'filled','r','linewidth',3)title('angle of X[k], the DFT of x[n]')axis([-1,N,-1.1*max(abs(angle(Xk))),1.1*abs(max(angle(Xk)))])grid onxlabel('k - frequency sample')ylabel('angle of x[k], radians')

-1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 40

5

10

Magnitude of X[k], the DFT of x[n]

k - frequency sample

-1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4

-2

0

2

angle of X[k], the DFT of x[n]

k - frequency sample

angl

e of

x[k

], ra

dian

s


Calculating the IDFT using the DFT

= 1=0

1

=

=0

1

DFT: IDFT:

Conjugate IDFT

= 1=0

1

Conjugate result to get x[n]

DFT of X*[k]

= = 1

=0

1

2

1) Change X[k] to X*[k]2) Conduct DFT on X*[k] to get Nx*[n]3) Conjugate Nx*[n] to get Nx[n]4) Divide by N to get x[n]

Process


Using FFT in MATLAB to get the DFT and IDFT

FFT is efficient when N=2m, a power of 2

FFTx[n] X[k]

FFT x[n]X[k]

conjugateX*[k] Nx*[n]

conjugateNx[n]

Divide by N

Important MATLAB Commands to use: FFT, conj


Convolution Review

LTI System

= =

=

Convolution in one domain yields multiplication in the other domain

Let 1 1 , 2 2 1 2 1 2

1 2 12

1 2

=


Linear Convolution of Discrete Signals

Discrete, linear convolution = =

Let N be the length of one sequence, and M be the length of the other sequence. The convolution yields a sequence of length N+M-1

Let = 1 2 3 , = 2 4 6 = [2 8 20 24 18] Graphically (fold and shift) Use equation

Synthetic multiplication Table Computer program (e.g.,MATLAB)

Methods to conduct linear convolution


Example: Using equation

= =

Hint: only need to use indices that add up to n

Let = 1 2 3 , = 2 4 6 0 = 0 0 = 1 2 = 2 1 = 0 1 + 1 0 = 1 4 + 2 2 = 8 2 = 0 2 + 1 1 + 2 0 = 1 6 + 2 4 + 3 2 = 20 3 = 1 2 + 2 1 = 2 6 + 3 4 = 24 4 = 2 2 = 3 6 = 18


Example: Using Synthetic Multiplication

= =

Hint: only need to use indices that add up to n

Let = 1 2 3 , = 2 4 6 2 4 6 1 2 3

6 12 18

4 8 12

2 4 6

2 8 20 24 18


Convolving with sequences of nonzero starting sequences

= =

Consider the sequences 1 = 5 1 9 2 4 2 = 1 4 2 3

Recall: only need positions where m+n=new index for y

Most negative index: min(-2-1,-2-2)=-4

1 2 = =

1 2 = [5 21 23 55 33 47 14 12]

Most positive index: min(1-(-2),2-(-2))=-3


Can We Use the DFT to Convolve Linearly?Can we find y[n] if Y[k]=X[k]H[k}?

= = =0

1

2

=0

1

2

DFT(x) DFT(h)

= 1=0

1

=0

1

2

=0

1

2

2

Use math rules to pull out x[l] and h[m], and combine exponents

= 1=0

1

=0

1

=0

1

2

=0

1

2

= , = 00, ECE 390 - Transforms II - Chmielewski and Peters 44

Can We Use the DFT to Convolve Linearly?

= 1=0

1

=0

1

Thus,

= 1 = Therefore,

= 1=0

1

This looks like linear convolution, but the summation is over one period rather than all time. This is called circular convolution, with the operation denoted by

= = =0

1

DFT In circular convolution, both

data sets must have the same number of elements


Circular Convolution Example1 = 1 2 3 42 = 0 1 2 3 = 1 2

MATLAB code = 1 . 2 = 10 2 + 2 2 2 2 . 6 2 + 2 2 2 2 = 60 4 = 16 18 16 10


Circular Convolution-General Case 0

1

2

3 4

5

6

7 0

1 2

3 4 5

6 7

Outer ring is in counter-clockwise (CCW) order, inner ring is in clockwise (CW) order

Only rotate inner ring CCW. Outer ring is stationary.

Multiply, add, and shift inner ring CCW


Rotating Method-Previous Example1 = 1 2 3 4 2 = 0 1 2 3

1

2

3

4

0

1

2

3

0 = 1 0 + 2 3 + 3 2 + 4 1 = 16

1

2

3

4

1

2

3

0

1 = 1 1 + 2 0 + 3 3 + 4 2 = 18ECE 390 - Transforms II - Chmielewski and Peters 48

Rotating Method-Previous Example1 = 1 2 3 4 2 = 0 1 2 3

1

2

3

4

2

3

0

1

2 = 1 2 + 2 1 + 3 0 + 4 3 = 16

1

2

3

4

3

0

1

2

3 = 1 3 + 2 2 + 3 1 + 4 0 = 10ECE 390 - Transforms II - Chmielewski and Peters 49

Can We Use the FFT To Conduct Linear Convolution?1 2 3 4 0 1 2 3 = 0 1 4 10 16 17 12 (length N=7)Because the convolution is 7 points long, we can add enough zeroes to each sequence such that the length of each modified sequence is 7 points long1 2 3 4 0 0 0 0 1 2 3 0 0 0

1 2

This gives the correct results when using the FFT in approach

1 2 = 1 2

1 2

1 . 2 = ECE 390 - Transforms II - Chmielewski and Peters 50

Convolution Example Equal Lengths

close all;clear all;clcx=[5 1 9 2]h=[1 9 2 4]L=length(x)+length(h)-1;y1=conv(x,h) % MATLAB linear convolution command% Conduct linear convolution using DFTy2=ifft(fft(x,L).*fft(h,L))% Conduct circular convolution using DFTy3=ifft(fft(x).*fft(h))


Convolution Example-Unequal Lengths

close all;clear all;clcx=[5 1 9 2 9 9]h=[1 9 2 4]L=length(x)+length(h)-1;y1=conv(x,h)y2=ifft(fft(x,L).*fft(h,L))


Signal EnergyRecall Parsevals Equation

=

2 = 12

2 = 10

2

Energy spectral density function

= 1 2 = 1 Energy in frequency band of interest

1

2


Example of Parsevals Equation

% Define data set x[n]x=[1 2 -3 4 5]% Calculate DFT of x[n]Xk=fft(x)% Calculate energy in time domainEx=sum(abs(x).^2)% Calculate energy in frequency domainEXk=sum(abs(Xk.^2))/length(x)


Parsevals Equation Continued

General case of Parsevals equation: =0

1

= 1=0

1

If y[n]=x[n], =0

1

2 = 1=0

1

2

What Parsevals equation is stating is the total energy time domain is equal to the total energy in the frequency domain


Discrete Time Fourier Series (DTFS)x[n] is periodic with period N: = + , Because there are N points in the period, there are N harmonically related exponential functions

2 , = 0, 1, . . 1

Synthesis equation: = =0

1

2

Analysis equation: = 1=0

1

2

Periodic condition: + =

Because the Fourier Series is for periodic signals, we only need to have N points (one period) to provide a complete description of the signal


Power Spectral Density of Periodic Signals

= 1=0

1

2 Power in the discrete time domain

= =0

1

2 Average power in the discrete frequency domain

If x[n] is a real sequence, =


ECES 390 Transforms IIWeek 1Fourier/Inverse Fourier TransformCommon Fourier RelationshipsDiscretizing the SignalZero Order Hold ModelZero Order HoldZero Order HoldSpectrum Representation of Sampled SignalsDiscrete Time Fourier Transform (DTFT)DTFT ExampleMATLAB ExampleMATLAB ExampleDTFT Linearity ExampleDTFT Time Shift ExampleDTFT Frequency ShiftingDiscrete Fourier TransformApproach to a SolutionApproach to a SolutionDiscrete Fourier Transform ContinuedDFT ExampleDFT Function DTFT Property - PeriodicityDFT Property - LinearityDFT Property ParsevalsInverse DFTCompare and Contrast DFT and IDFTComparison of DFT to DTFTExample DFT - 3 Data Points (N=3)Example IDFT-3 Data Points (N=3)Periodic signalsShorthand Notation of the DFTFrequency Resolution of DFTUnderstanding WNUnderstanding WNValidity of DFTMATLAB ImplementationCalculating the IDFT using the DFTUsing FFT in MATLAB to get the DFT and IDFTConvolution ReviewLinear Convolution of Discrete SignalsExample: Using equationExample: Using Synthetic MultiplicationConvolving with sequences of nonzero starting sequencesCan We Use the DFT to Convolve Linearly?Can We Use the DFT to Convolve Linearly?Circular Convolution ExampleCircular Convolution-General CaseRotating Method-Previous ExampleRotating Method-Previous ExampleCan We Use the FFT To Conduct Linear Convolution?Convolution Example Equal LengthsConvolution Example-Unequal LengthsSignal EnergyExample of Parsevals EquationParsevals Equation ContinuedDiscrete Time Fourier Series (DTFS)Power Spectral Density of Periodic Signals

eces 390_week 1 lecture_summer 2015

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individual fourier transforms

discrete time fourier

transforms iiweek

tom chmielewski

use fourier

fourier identity

time continuous

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