Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
5. Fourier Transform and Spectrum
Analysis
1
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Spectrum of Non-periodic
Signals
• Fourier series help us to find the spectrum of
periodic signals
• Most signals are not periodic
• Speech, audio, etc.
• Need another tool to find the spectrum of nonperiodic (aperiodic) signals
⇒ Fourier Transform
2
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Fourier Transform of Discretetime Signals
• Let x(t) be an aperiodic continuous-time signal, x[n]
is the samples of x(t) such that:
x[n] = x(nTs)
Radian frequency
• The spectrum of x[n] is given by:
X p (ω ) =
3
∞
∑ x[n]e
n= −∞
− jωnTs
or X p (ωˆ ) =
ωˆ = ωTs
∞
∑ x[n]e
n= −∞
− jωˆ n
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Aperiodic Signals have Periodic
Spectrum
• It is interesting to note that Xp(ω) is periodic since
X p (ωˆ + 2πk ) =
=
∞
− j (ωˆ + 2πk ) n
x
[
n
]
e
∑
n= −∞
∞
∑ x[n]e
n= −∞
4
1
− jωˆ n − j 2πnk
e
= X p (ωˆ )
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
X p (ωˆ )
-2π
5
0
2π
ωˆ = ωTs
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
• If x(t) has a spectrum of X(ω) and x[n] = x(nTs) has
a spectrum of Xp(ω), it can be shown that
1
X p (ωˆ ) =
Ts
∞
∑ X (ωˆ + 2πk )
k = −∞
1
X (ωˆ − 2π )
= K+
Ts
1
+
X (ωˆ )
Ts
6
1
+
X (ωˆ + 2π )+ K
Ts
−∞ ≤ω ≤ ∞
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Time Domain
Frequency Domain
X(ω)
1
x(t)
0
t
7
Ideal low
pass filter
X p (ωˆ )
x[n]=x(nTs)
0 | |←Ts n
ω
-B 0 B
-2π
0
2π
ωˆ = ωTs
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Time Domain
x(t)
Frequency Domain
X(ω)
1
t
0
-B 0 B
X p (ωˆ )
x[n]
0
8
0
| |←Ts
x[n]
n
|←Ts
n
|
ω
-2π
0
-2π
0
2π
ωˆ = ωTs
2π
ωˆ = ωTs
X p (ωˆ )
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Time Domain
x(t)
Frequency Domain
X(ω)
1
0
t
-B 0 B
X p (ωˆ )
x[n]
0 | |←Ts
n
X’(ω)
x’(t)
9
0
ω
t
-B
0 B
ωˆ = ωTs
ω
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
• If the signal has frequency components beyond |π|,
after sampling, these frequency components will
affect the other replicas in the spectrum
• Even with an ideal low pass filter, the original
signal cannot be reconstructed. This is the socalled alias effect
• Restate the Shannon Sampling Theorem for
general aperiodic signals
ωˆ ≤ π ⇒ 2πf maxTs ≤ π
10
⇒ 2πf max ≤ πf s or f s ≥ 2 f max
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Shannon Sampling Theorem
A continuous-time aperiodic signal x(t) with
frequencies no higher than fmax can be
reconstructed exactly from its samples x[n] =
x(nTs) if the samples are taken at a rate fs =
1/Ts that is greater than 2fmax
Nyquist
NyquistFrequency
Frequency
11
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Real Examples
Time Domain
Frequency Domain
Resulted signal
Ideal low pass filter
12
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Time Domain
Frequency Domain
Resulted signal
Ideal low pass filter
13
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Time Domain
Frequency Domain
Resulted signal
Ideal low pass filter
14
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
How to Solve Aliasing
Problems?
1. Increase the sampling rate such that fs ≥ 2fmax
2. Use anti-aliasing filter first
Pre-filter
Pre-filterthe
theinput
inputsignal
signalsuch
suchthat
thatititwill
willnever
never
has
hasfrequency
frequencycomponents
componentsbeyond
beyond|π|
|π|
Prefilter
15
A/D
Digital Signal
Processor
D/A
Postfilter
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Time Domain
x(t)
Frequency Domain
X(ω)
1
t
0
-B 0 B
X(ω)
Anti-aliasing
filter
0
16
|
1
X p (ωˆ )
x[n]
|←Ts
n
ω
-2π 0 2π
ω
ωˆ = ωTs
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
With anti-aliasing filter
Without anti-aliasing filter
X p (ωˆ )
X p (ωˆ )
Ideal low
pass filter
Ideal low
pass filter
X’(ω)
ω
17
ωˆ = ωTs
ωˆ = ωTs
Look better
X’(ω)
ω
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Hear the effect!
Original
44.1kHz
sampling
18
8kHz
sampling
with
aliasing
8kHz
sampling
with antialiasing filter
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Discrete Fourier Transform
19
• Spectrum of aperiodic discrete-time signals is
periodic and continuous
• Difficult to be handled by computer
• Since the spectrum is periodic, there’s no point to
keep all periods – one period is enough
• Computer cannot handle continuous data, we can
only keep some samples of the spectrum
• Interesting enough, such requirements lead to a
very simple way to find the spectrum of signals
⇒ Discrete Fourier Transform
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
• Recall the Fourier transform of an aperiodic
discrete sequence
X p (ωˆ ) =
∞
− jωˆ n
x
[
n
]
e
∑
n= −∞
• Assume x[n] is an aperiodic sequence with N values,
i.e. {x[n] : n = 0, 1, ..., N-1}
X p (ωˆ ) =
20
N −1
− jωˆ n
x
[
n
]
e
∑
n=0
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
• If we are now interested only in N equally spaced
frequencies of 1 period of the Fourier spectrum, i.e.
 k .2π 
X [k ] = X p 

 N 
k = 0,1, K N − 1
IfIfNN==13
13
X p (ωˆ )
k=4
k=0
-2π
21
0
2π/N
2π
ωˆ = ω T s
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
• Now if we want to compute the value of these N
frequencies,
X [k ] =
=
 k .2π

−
j
N −1
Ts N

x[n]e
∑

 nTs

W Nnk = e − j 2πnk / N
n=0
N −1
N −1
n=0
n=0
− jk .2πn / N
x
[
n
]
e
=
∑
nk
x
[
n
]
W
∑
N
for k = 0,1,…, N-1
Discrete Fourier Transform
22
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
• Discrete Fourier Transform (DFT) is exactly the
output of the Fourier Transform of an aperiodic
sequence at some particular frequencies
• Inherently
periodic since
X[k+N] = X[k],
although we
always only
consider one
period of X[k]
23
X [k ] =
N −1
− j 2πnk / N
x
[
n
]
e
∑
n=0
X [k + N ] =
N −1
− j 2πn( k + N ) / N
x
[
n
]
e
∑
n=0
=
N −1
1
− j 2πnk / N − j 2πn
x
[
n
]
e
e
= X [k ]
∑
n=0
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
• If we know X[k], we can reconstruct back the signal
x[n] via the inverse discrete Fourier transform
1
x[n] =
N
1
=
N
N −1
j 2πnk / N
X
[
k
]
e
∑
n=0
N −1
− nk
X
[
k
]
W
∑
N
for n = 0,1,…, N-1
n=0
Inverse Discrete Fourier Transform
24
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
• It can be proven as follows:
1
N
N −1
jk 2πn / N
[
]
X
k
e
∑
k =0
1
=
N
1
=
N
0

N
25
if n ≠ m
otherwise
N −1 N −1
∑
− jk 2πm / N jk 2πn / N
[
]
x
m
e
e
∑
∑
jk .2π ( n − m ) / N
[
]
x
m
e
∑
k =0 m =0
N −1 N −1
k =0 m =0
 N −1 jk .2π ( n− m ) / N 

∑ x[m ] ∑ e

m =0
 k =0
= x[n]
for n = 0,1,..., N − 1
1
=
N
N −1
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
• Although DFT gives exact frequency response of a
signal, sometimes it may not give the desired
spectrum
• Example
Fourier
Transform
One period of
X p (ωˆ )
x[n]
n
0
9
NN==10
10
26
DFT
10
X[k] if N = 10
So different from
X p (ωˆ )
k
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
• Need improved resolution
• Achieve by padding zero to the end of x[n] to make
N bigger
x[n]
Pad
Pad40
40zeros
zeros
n
0
9
50
x[n] = {1 1 1 1 1 1 1 1 1 1 0 0 0 0 … 0}
40 zeros
27
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
N
N == 50
50
28
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
N
N == 100
100
29
Signal Processing Fundamentals – Part I
Spectrum Analysis and Filtering
5. Fourier Transform and Spectrum Analysis
Exercise
Given that x[n] is defined in the following figure,
determine its spectrum using DFT with N = 4
2
1
0
30
x[n]
2
1
3
n