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Digital Signal Processing – Chapter 10 Fourier Analysis of Discrete-Time Signals and Systems Dr. Ahmed Samir Fahmy Associate Professor Systems and Biomedical Eng. Dept. Cairo University

Digital Signal Processing – Chapter 10

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Digital Signal Processing – Chapter 10. Fourier Analysis of Discrete-Time Signals and Systems Dr. Ahmed Samir Fahmy Associate Professor Systems and Biomedical Eng. Dept. Cairo University. Review . Complex Variables. Complex Variables. - PowerPoint PPT Presentation

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Page 1: Digital Signal Processing – Chapter 10

Digital Signal Processing – Chapter 10Fourier Analysis of Discrete-Time Signals and Systems

Dr. Ahmed Samir FahmyAssociate ProfessorSystems and Biomedical Eng. Dept.Cairo University

Page 2: Digital Signal Processing – Chapter 10

Review Complex Variables

Page 3: Digital Signal Processing – Chapter 10

Complex Variables• Most of the theory of signals and systems is

based on functions of a complex variable • However, practical signals are functions of a real

variable corresponding to time or space• Complex variables represent mathematical tools

that allow characteristics of signals to be defined in an easier to manipulate form▫Example: phase of a sinusoidal signal

Page 4: Digital Signal Processing – Chapter 10

Complex Numbers and Vectors• A complex number z represents any point (x, y):

z = x + j y, ▫x =Re[z] (real part of z) ▫y =Im[z] (imaginary part of z)▫ j =Sqrt(-1)

• Vector representation ▫Rectangular or polar form▫Magnitude and Phase

Page 5: Digital Signal Processing – Chapter 10

Complex Numbers and Vectors• Identical: use either depending on operation

▫Rectangular form for addition or subtraction▫Polar form for multiplication or division

• Example: let

More Difficult

Page 6: Digital Signal Processing – Chapter 10

Complex Numbers and Vectors• Powers of complex numbers: polar form

• Conjugate

Page 7: Digital Signal Processing – Chapter 10

Functions of a Complex Variable• Just like real-valued functions

▫Example: Logarithm

• Euler’s identity

▫Proof: compute polar representation of R.H.S.

▫Example:

Page 8: Digital Signal Processing – Chapter 10

Functions of a Complex Variable• Starting from Euler’s Identity, one can show:

Page 9: Digital Signal Processing – Chapter 10

Phasors and Sinusoidal Steady State• A sinusoid is a periodic signal represented by,

• If one knows the frequency, cosine is characterized by its amplitude and phase.

• Define Phasor as complex number characterized by amplitude and the phase of a cosine signal

▫Such that

Page 10: Digital Signal Processing – Chapter 10

example• Example: Steady state solution of RC circuit with

input

▫Assume that the steady-state response of this circuit is also a sinusoid

▫ Then, we can let▫Substitute in

Page 11: Digital Signal Processing – Chapter 10

Discrete Time Fourier Transform

Page 12: Digital Signal Processing – Chapter 10

The Fourier TransformOn Board Illustration

this slide is intentionally left blank

Page 13: Digital Signal Processing – Chapter 10

Discrete-Time Fourier Transform• Sampled time domain signal periodic Fourier

transform• Periodic time domain signal sampled Fourier

transform• Periodic and sampled time domain signal

Fourier transform that is both periodic and sampled

Forward DTFT

Inverse DTFT

Page 14: Digital Signal Processing – Chapter 10

Notes on DTFT

• DTFT is periodic, with period 2π:

• For convergence:

DTFT

Page 15: Digital Signal Processing – Chapter 10

Notes on DTFT Eigen functions:• Let the input to LTI be: • Using Convolution:

DTFT of h[k] calculated at input frequency

The O/P is similar to I/P (multiplied by a factor)

Page 16: Digital Signal Processing – Chapter 10

DTFT vs. Z-Transform

• Compare to z-transform, it is clear that:

• DTFT is the values on the unit circle in z-Transform

• Unit circle must be included in the ROC

DTFT

Page 17: Digital Signal Processing – Chapter 10

DTFT vs. Z-Transform

• Consider the noncausal signal , with ▫Determine its DTFT. ▫Use the obtained DTFT to find

• Solution:▫Z-trans:

▫The ROC is: 1st term: 2nd term: Overall ROC:

includes unit circle!

Example

Page 18: Digital Signal Processing – Chapter 10

DTFT vs. Z-Transform

• Thus the DTFT is given by

• Also, the DC component is given by:

• Another way to calculate is:

Example

Page 19: Digital Signal Processing – Chapter 10

Notes on DTFT • Problem: the sampled signal is still infinite in

length▫Impossible to obtain using computers▫Consider cases when the signal has compact

support▫The frequency variable is continuous!

Page 20: Digital Signal Processing – Chapter 10

Discrete Fourier Transform (DFT)

• Defined as the Fourier transform of a signal that is both discrete and periodic▫Fourier transform will also be discrete and

periodic▫Can assume periodicity if we have a finite

duration of a signal: better than zero assumption since it allows reducing the frequency domain to a sampled function

Page 21: Digital Signal Processing – Chapter 10

Discrete Fourier Transform (DFT)

Page 22: Digital Signal Processing – Chapter 10

Discrete Fourier Transform (DFT)

Page 23: Digital Signal Processing – Chapter 10

Discrete Fourier Transform (DFT)

Page 24: Digital Signal Processing – Chapter 10

Discrete Fourier Transform (DFT)

21

0

2

2

( )

k=0, 2N, ...10 o.w.

1

kN j nN

n

j k

kjN

X k e

Ne

e

1 1sin( ( 1/ 2))( )sin( / 2)

jwNjw w NX e ew

Page 25: Digital Signal Processing – Chapter 10

DFT Formula• Given a periodic signal x[n] of period N, its DFT

is given by,

• Its inverse is given by,

• Both X[k] and x[n] are periodic of the same period N.

Page 26: Digital Signal Processing – Chapter 10

Computation of DFT Using FFT• A very efficient computation of the DFT is done

by means of the FFT algorithm, which takes advantage of some special characteristics of the DFT▫DFT requires O(N2) computations▫FFT requires O(N log2N) computations▫Both compute the same thing

• It should be understood that the FFT is NOT another transformation but an algorithm to efficiently compute DFTs.

• Always remember that the signal and its DFT are both periodic

Page 27: Digital Signal Processing – Chapter 10

Improving Resolution in DFT• When the signal x[n] is periodic of period N, the DFT

values are normalized Fourier series coefficients of x[n] that only exist for the N harmonic frequencies {2k/N}, as no frequency components exist for any other frequencies▫The number of possible frequencies depend on the

length L chosen to compute its DFT. ▫Frequencies at which we compute the DFT can be seen

as frequencies around the unit circle in the z-plane• We can improve the frequency resolution of its DFT by

increasing the number of samples in the signal without distorting the signal by padding the signal with zeros▫Increase samples of the units circle

Page 28: Digital Signal Processing – Chapter 10

Zero-Padding ExampleN= 32;N2= 128;n=0:N-1;n2= 0:N2-1;x= zeros(N,1);x(1:16)= 1;X= fftshift(fft(x));X2= fftshift(fft(x,N2));subplot(3,1,1)stem(n,x)subplot(3,1,2)stem(n,abs(X))subplot(3,1,3)stem(n2,abs(X2))

Matlab Code

OriginalSignal

32-Point FFT

128-Point FFT

No zero-padding

With zero-padding

Page 29: Digital Signal Processing – Chapter 10

Phase in DFT• Output of DFT is complex numbers• Phase is just as important as magnitude in DFT

(in fact, it is more important!)• Time shift amounts to linear phase

▫Phase wrapping occurs for large shifts• Linear phase can be unwrapped to its correct

form using the “unwrap” function of MatlabN= 32; x= zeros(N,1); x(11:15)= [0.2 0.5 1 0.5 0.2];X= fftshift(fft(x));subplot(4,1,1); stem(1:N,x); title('Signal')subplot(4,1,2); plot(abs(X)); title('DFT Magnitude')subplot(4,1,3); plot(angle(X)); title('Computed Phase')subplot(4,1,4); plot(unwrap(angle(X))); title('Unwrapped Phase')

Page 30: Digital Signal Processing – Chapter 10

More Time ShiftLess Time Shift

Steeper linear phaseSlower linear phase

Phase in DFT: Examples

Page 31: Digital Signal Processing – Chapter 10

Properties of DFT• Linearity

• Duality

kbXkaXnbxnaxkXnxkXnx

21DFT

21

2DFT

2

1DFT

1

N

DFT

DFT

kNxnXkXnx

Page 32: Digital Signal Processing – Chapter 10

Properties of DFT• Duality: example (n=0:9)

Page 33: Digital Signal Processing – Chapter 10

Properties of DFT• Circular Shift of a Sequence

mNkjDFT

N

DFT

ekXmnxkXnx

/21-Nn0;

Page 34: Digital Signal Processing – Chapter 10

Linear vs. Circular Convolution• Convolution in continuous and discrete time

aperiodic signals is called linear convolution• Convolution can be calculated as a multiplication

in the frequency domain▫Convolution property of the Fourier transform

• Problem: when using DFT to compute convolution of aperiodic signals, the periodicity assumption results in errors in computing the convolution▫Signal outside the N points used is not zero:

periodic extension of the N points disturb the computation

▫What we compute is called circular convolution

Page 35: Digital Signal Processing – Chapter 10

Linear vs. Circular Convolution• Given x[n] and h[n] of lengths M and K, the

linear convolution sum y[n] of length N=(M+K-1) can be found by following these three steps:▫Compute DFTs X[k] and H[k] of length LN for

x[n] and h[n] using zero-padding▫Multiply them to get Y[k]= X[k]H[k].▫Find the inverse DFT of Y[k] of length L to

obtain y[n]

Page 36: Digital Signal Processing – Chapter 10

Linear vs. Circular Convolution: Example

N1= 16;N2= 16;N= 32; % N>N1+N2-1x1= zeros(N1,1);x2= zeros(N2,1);x1(1:4)= 1;x2(:)= 1; X1= fft(x1,N1);X2= fft(x2,N1);X_circ= X1.*X2;x_circ= ifft(X_circ); X1= fft(x1,N);X2= fft(x2,N);X_lin= X1.*X2;x_lin= ifft(X_lin); figure(1); subplot(4,1,1); stem(1:N1,x1)subplot(4,1,2); stem(1:N2,x2)subplot(4,1,3); stem(1:N1, abs(x_circ))title('Circular Convolution');subplot(4,1,4); stem(1:N,abs(x_lin))title('Circular Convolution');

Page 37: Digital Signal Processing – Chapter 10

Fourier Transformations Chart

SignalClassification

Frequency

Continuous Discrete

Time

Continuous Continuous Fourier Transform (CFT)

Discrete Fourier Series (DFS)

Discrete Discrete-Time Fourier Transform

(DTFT)

Discrete Fourier Transform (DFT)

Only one that can be done on a computer

Most General Form: others are special cases

Page 38: Digital Signal Processing – Chapter 10

Problem Assignments• Problems: 10.12, 10.26, 10.27, 10.29• Partial Solutions available from the student

section of the textbook web site