ECEN3513 Signal Analysis
Lecture #23
16 October 2006
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Read 6.1, 6.2; Review 6.3
Problems 4.7-2, 6.2-1, 6.3-5
Test #2 on 27 October
Quiz 5 results: Hi = 7.8, Lo = 2.9, Ave = 5.35
Standard Deviation = 1.35
Read 6.4, 6.5 (1st section), 6.7 (1st section)
Problems: 6.4-3, 6.4-4, 6.5-1
Test #2 on 27 October
Quiz 6 Results: Hi =8.0, Lo = 2.0, Ave. = 5.00
Standard Deviation = 2.04
yp(t): Base Periodic Function
i
i  0  200
t  5  10
j  0  500
f  5.000001  10
i
j
sin 2   f 
time varies from -5 to +5 seconds
200
j
500
1 
500
i
Ts  50
  1 
 sin 2   f  
  4     4 
1
1
2   f  
2   f  
 4
 4
envelope ( f) 
x 
Frequencies vary from -5 to +5 Hz
in 0.02 Hertz increments
500
 Reenvelopefjcos 2fjti   Imenvelopefjsin2fjti
j0
x
x 
i
j0
1.5
1
xi
0.5
0
0.5
6
4
2
0
ti
2
4
6
i
Ts
Fourier Transform of yp(t)
2
envelope f j
1
0
6
4
2
0
2
4
6
2
4
6
fj
4
atan2 Re envelope f j   Im envelope f j  
2
0
6
4
2
0
fj
y(t): Periodic Time Domain Waveform
i
i  0  200
t  5  10
j  0  20
f  5.00000  10
i
j
sin 2   f 
envelope ( f) 
20
x 
i
time varies from -5 to +5 seconds
200
 
j
Frequencies vary from -5 to +5 Hz
in 0.5 Hertz increments
20
1 
1
sin 2   f  


  4     4 
1
1
2   f  
2   f  
 4
 4
20
 j 
Re envelope f
 cos 2  f  t
   
j i
j0

 j 
Im envelope f
 sin 2  f  t
x

x 
j i
i
j0
1
xi
0.5
0
6
Ts  2
4
2
0
ti
2
4
6
i
Ts
Fourier Series of y(t)
2
envelope f j
1
0
6
4
2
0
2
4
6
2
4
6
fj
4
atan2 Re envelope f j   Im envelope f j  
2
0
6
4
2
0
fj
Linear Time Invariant? yes
δ(t) → h(t)... an impulse response exists
 convolution works

 y(t)

transfer function exists
 Y(f)

= x(t) * h(t)
= X(f) H(f)
Energy & Power transfer functions exist
 |Y(f)|2
= |X(f)|2 |H(f)|2 J/Hz
 SYY(f) = SXX(f) |H(f)|2 W/Hz
Linear Time Invariant? no
δ(t) → h(t)... an impulse response exists
 convolution gives you the wrong answer

 y(t)

transfer function does not exist
 Y(f)

≠ x(t) * h(t)
≠ X(f) H(f)
Energy & Power transfer functions don't
exist
≠ |X(f)|2 |H(f)|2 J/Hz
 SYY(f) ≠ SXX(f) |H(f)|2 W/Hz
 |Y(f)|2
Autocorrelation (Power Signal)
RXX(τ) = lim 1
T→∞ T
x(t)x(t+τ)dt
T

When in doubt, evaluate this one first
 If
= 0, evaluate energy signal autocorrelation
SXX(f) = lim |X(f)|2 non-zero at same
T→∞ T
freqs as X(f)!
 RXX(τ) ↔ SXX(f) W/Hz

 Autocorrelation
& Power Spectrum form a
Fourier Transform pair
Autocorrelation (Power Signal)
RXX(τ) = lim 1
T→∞ T
x(t)x(t+τ)dt =
T




+∞
SXX(f) e -j2πfτ df
-∞
units
are
W/Hz
RXX(0) = ??
 Average (normalized) power of x(t)
How can you find RXX(0) from SXX(f)?
 Evaluate at τ = 0
 The area under power spectrum =
average (normalized) power of x(t)
What's SXX(f) for x(t) = 3cos2π100t?
What's SXX(f) for x(t) = 3sin2π100t?
Autocorrelation (Energy Signal)
RXX(τ) = lim
T→∞
x(t)x(t+τ)dt
T

RXX(τ) ↔ |X(f)|2 J/Hz
 Autocorrelation
& Energy Spectrum form a
Fourier Transform pair

GXX(f) = |X(f)|2 non-zero at
same freqs as X(f)!
Autocorrelation (Energy Signal)
RXX(τ) = lim
T→∞
x(t)x(t+τ)dt =
T




+∞
|X(f)|2 e -j2πfτ df
-∞
units
are
J/Hz
RXX(0) = ??
 Average (normalized) energy of x(t)
How can you find RXX(0) from |X(f)|2?
 Evaluate at τ = 0
 The area under energy spectrum =
average (normalized) energy of x(t)
What's |X(f)|2 for x(t) = rect[(t-0.5)/1]?
What's |X(f)|2 for x(t) = rect[t/1]?
Autocorrelation (Real World)
^
RXX(τ) = 1
T
x(t)x(t+τ)dt
T

observation
interval
^
^
RXX(τ) ↔ SXX(f) W/Hz
 FT
of autocorrelation yields estimate of
Power Spectrum

^
SXX(f) =
^
|X(f)|2,
T
^
where X(f) =
x(t) e-jωt dt
T
Autocorrelation (Real World)
^
RXX(τ) =
x(t)x(t+τ)dt
T

observation
interval
^
^
RXX(τ) ↔ GXX(f) W/Hz
 FT
of autocorrelation yields estimate of
Energy Spectrum

^
GXX(f) =
^
|X(f)|2,
^
where X(f) =
x(t) ejωt dt
T
sinusoid + d.c.
i
i  0  200
t  50  100
j  0  400
f  4.00000  8
i
time varies from -50 to +50 seconds
200
j
j
Ts  50
Frequencies vary from -4 to +4 Hz
in 0.02 Hertz increments
400
envelope ( f)   0.5j  ( f  0.099)   ( f  .101)  0.5j  ( f  .101)   ( f  .099)   ( f  .001)   ( f  .001)   50  .000001
δ(f-0.1)
400
x 
i
 
 j 
Re envelope f
δ(f+0.1)
400
 cos 2  f  t
   
j i
j0

δ(f)
 j 
Im envelope f
 sin 2  f  t
x

x 
j i
i
j0
2
xi
1
0
60
40
20
0
ti
20
max( x)  2
40
60
i
Ts
Magnitude & Phase Plots
60
envelope f j
40
20
0
4
3
2
1
0
1
2
3
4
fj
 
 j  Imenvelopefj
phase  atan2 Re envelope f
j
f
205
 0.1
phase

180
205 
 90
2
phase j
0
2
4
3
2
1
0
fj
1
2
3
4