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Material Type: Notes; Class: Classical Physics Lab; Subject: Physics; University: University of Illinois - Urbana-Champaign; Term: Fall 2004;
Typology: Study notes
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Alexander Graham Bell:
The human ear has logarithmic
response to power in a sound wave
Power
1m W
0.1mW
0.01mW
10m W
100m W
Hearing
0
- -
1
2
P (^) log P dB
20
10
0
reference
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The Same Signal in Frequency Domain
0005 sin( )
001 sin( )
( ) sin( )
0 3 3
0 2 2
0 1 1
V t
V t
V t V t
Example:
Small signal components
are not visible in the time
domain!
t
ω
V(t)
[Volts]
V(
ω
)
[db]
ω 1
ω 2
ω 3
0 dB
-60 dB
-67 dB
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10
2 V
10 V
1 V
10
3 V
10
4 V
V [Volts] (^) dB
80
60
40
20
( 1 Volt) 0
20 log
RMS
dB
10
- V
10
- V
10
- V - - -
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An Example (an odd function)
even k
vs odd k
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2
2
( ) ( )e
( ) ( )e
ift
ift
H f h t dt
h t H f df
ò
ò
( ) ( )e
( ) ( )e
i t
i t
H h t dt
h t H d
ò
ò
frequency f in Hz (^) angular frequency in rad / s
t
h(t)
ω
|H(ω)|
2
time domain frequency domain
Use Fourier Transform to Analyze
Continuous Functions
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If bandwidth in the signal is limited to frequencies below f c
i. e. H(f) = 0 for f > f c
then
sin 2
( )
c
n
n
f t n t
h t h
t n t t
é ù æ ö ê ú ç ÷ ê ú è ø ë û
æ ö
ç ÷ è ø
¥
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Continuous function represented by an infinite series!
But in experiments we collect a finite number of samples!
The Sampling Theorem shows how to
construct the continuous function from discrete values
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k
h h k t k N -
Discrete Fourier Transform
f
n
2
1
0
exp( 2 / )
N
n k
n
h H ikn N
å
1
0
exp(2 / )
N
n k
k
H h ikn N
å
(^)
t
h k
^
(^)
max
T N t
c
f
t
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Aliasing Exercise: 1.263 kHz sine wave with 20 kSa/s and f c
= 10 kHz
Measured Frequency from FFT: 1.27 kHz
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Aliasing Exercise: 1.263 kHz sine wave with 10 kSa/s and f c
= 5.0 kHz
Measured Frequency from FFT: 1.27 kHz
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Aliasing Exercise: 1.263 kHz sine wave with 2.0 kSa/s and f c
= 1.0 kHz
NB aliasing
Measured Frequency from FFT: 0.736 kHz
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Aliasing Exercise: 1.263 kHz sine wave with 1.0 kSa/s and f c
= 0.5 kHz
NB aliasing
Measured Frequency from FFT: 0.263 kHz