Things to do
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Proposal / Human subject consent form
Solution to assignment 2 will be up at 1:05pm
Feedback on labs
Test 1 this Thursday (up to Chapter 2)
– Sampling, aliasing and quantization
– Transfer function, difference equation, zero-pole, signal flow
diagram
– Z-transform (system function / impulse response)
– Inverse Z transform (residue, partial fraction)
– General pre-filter stage design
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BIEN425 – Lecture 6
• By the end of the lecture, you should be able to:
– Compute the discrete-time Fourier transform (DTFT) of a signal
– Describe the difference between DTFT and discrete Fourier
transform (DFT)
– Compute the DFT of a discrete time signal
– Compute the power spectrum density of a signal
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• Recall that the Fourier transform (FT) converts a
continuous time-domain function to a continuous
frequency domain function.
• Now, given the Z-transform of a discrete-time signal X(z)
X ( f )  X ( z ) z  e  j 2 fT
• Also note that X(f) is complex
X ( f )  A( f )e
j ( f )
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Discrete-time Fourier transform
• Definition of DTFT:

X(f)
 x (k )e
 j 2  kfT
k 0
• In order for the series x(k) to converge, all poles of X(z)
must be within the unit circle
Remember partial fraction expansion? Infinite series and
convergence?
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To recover x(k) from X(f)
• Use inverse DTFT
fs / 2

X ( f )e
 fs / 2
fs / 2
ji 2  fT


 jk 2  fT
df     x ( k ) e
 fs / 2  k 0
fs / 2



e
x(k )
k 0
 x(k )
e


k 0
df
x(k )
j ( i  k )  f sT
e
 j ( i  k )  f sT
j 2 T ( i  k )
k 0

j ( i  k ) 2  fT
 fs / 2


 ji 2  fT
df
e

1 sin( i  k )
T
( i  k )


 x ( k ) f  (i  k )  x (i ) f
s
k 0
s
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Some characteristics
• X(f+fs) = X(f)
• X(-f) = X*(f)
A( f )  A( f )
 ( f )   ( f )
even
odd
• It is also common to represent DTFT in normalized
frequency:
fˆ  f s / 2
 0 . 5  fˆ  0 . 5
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Discrete Fourier transform (DFT)
• Limitation of DTFT in practice:
– Infinite number of arithmetic operations
– Infinite number of points in creating f-domain
• Solution: Evaluate at N distinct frequencies fi=fs/N,
where 0≤ i<N
N 1
X(f)

x (k )e
 j 2  fkT
k 0
• N determines
1) how many input values are needed
2) Resolution of freq domain results.
3) Processing time required for DFT.
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Introducing roots of unity
WN  e

 j 2
N

Where WNk = cos(2k/N)-jsin(2k/N)
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To get X(i) from x(k)
X (i )  X ( f )
f  f i  if s / N
N 1
X (i ) 
 x (k )e
0≤i<N
 j 2  ( if s / N ) kT
k 0
N 1
X (i ) 

x (k )e
 j ( 2 k / N ) i
k 0
N 1
X (i ) 
 x ( k )W
ik
N
k 0
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• DFT (Analysis equation)
If we write x(k) as a column matrix,
W as a square matrix, we can get
X(i) very easily.
X  Wx
• Inverse DFT (Synthesis equation)
x
1
N
*
W X
• Let’s recap our notations
x(t)
t: real
x(k) k: integer
x: continuous; time-domain; real function
x: discrete-time; time-domain; real function
X(z) z: complex X: Z-transform; z-domain; complex function
X(f)
f: real
X: DTFT; frequency-domain; complex function
X(i)
i: interger
X: DFT; frequency-domain ;complex function
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Example
 3 
  1
x  
 0 
 
 2 
(

(

W 
(

(
j)
0
( j)
0
( j)
0
j)
0
( j)
1
( j)
2
j)
0
( j)
2
( j)
4
j)
0
( j)
3
( j)
6
( j) 
3
( j)

6
( j) 
9
( j) 
0
 4 
3  j 3

X  Wx  
 2 


3

j
3


• Order of DFT = N2
Because is a N by N matrix multiplication
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Signal spectrum
• Recall from continuous domain

1
 j 2  if 0 t
ci   x a (t ) e
dt
 0
• Fourier coefficients ci can be obtained from the DFT of
the sampled xa(t)
ci 
1
N 1
x (k )e

N
 j 2  ik / N

X (i )
k 0
• Power spectrum density (PSD):
N
S N (i ) 
X (i )
N
2
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Full name
Abbrev.
Time
Freq
Fourier Transform
FT
C
C
Discrete-time Fourier Transform
DTFT
D
C
Discrete Fourier Transform
DFT
D
D
• As you can see DFT is not very affordable (N2), we will
introduce a faster and more effective way of computing
DFT, called Fast Fourier Transform (FFT)
• Let’s check out some examples of the FFT function using
Matlab (fft, fftshift) in more details
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