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Continuous Time Signals
All signals in nature are in continuous time
x(t )
t
From Discrete Time to Continuous Time
A continuous time signals can be viewed as the limit of a
discrete time signal with sampling interval TS 0
x(nTS )
TS 0
x(t )
From Discrete Time FT (DTFT) …
We saw the DTFT of a discrete time signal
X DTFT ( )
j n
x
(
nT
)
e
S
n
1
x(nTS )
2
jn
X
(
)
e
d
DTFT
F
Substitute 2 F 2 F TS and obtain:
S
TS X DTFT 2 F TS
FS / 2
j 2 F nTS
x
(
nT
)
e
TS
S
n
1
x(nTS )
2 TS X DTFT (2 F TS ) e j 2 F nTS dF
2
FS / 2
… to Continuous Time FT
Now take the limit
so that
TS 0
nTS t
discrete time -> cont. time
FS
sampling freq -> infinity
...T
S
...dt
sum -> integral
Then we obtain the Fourier Transform
X ( F ) FT x(t )
j 2Ft
x
(
t
)
e
dt
x(t ) IFTX ( F )
j 2Ft
X
(
F
)
e
dF
Fourier Transform
We want to represent a signal in terms of its frequency
components.
Define: Fourier Transform (FT)
X ( F ) FT x(t )
j 2Ft
x
(
t
)
e
dt
x(t ) IFTX ( F )
j 2Ft
X
(
F
)
e
dF
Example of a Fourier Transform
Take a Rectangular Pulse
t
x (t ) rect
T0
1
T0 / 2
T0 / 2
t
j 2FT0 / 2
j 2FT0 / 2
e
e
X ( F ) e j 2Ft dt
j 2F
T0 / 2
T0 / 2
F
sin FT0
T0sinc
F
F0
Example of a Fourier Transform
F
X ( F ) T0sinc
F0
t
x (t ) rect
T0
F0 1/ T0
0.1
0.08
T0 0.1sec
0.06
1
T0 / 2
T0 / 2
t
0.04
0.02
0
-0.02
-0.04
-50
-40
-30
-20
-10
0
10
20
30
40
50
F(Hz)
F0 10Hz
1
T0 1sec
0.8
1
0.6
T0 / 2
T0 / 2
t
0.4
0.2
0
-0.2
-0.4
-50
F0 1Hz
-40
-30
-20
-10
0
F(Hz)
10
20
30
40
50
Properties of the FT: 1. Symmetry
If the signal x(t )
is real, then its FT is symmetric as
X ( F ) X * ( F )
since
X ( F )
x(t )e
j 2Ft
*
j 2Ft
dt x(t )e
dt X * ( F )
Example: just verify the previous example
Symmetry of the FT
Magnitude has
“even”
symmetry
| X ( F ) || X ( F ) |
F
Phase has
“odd”
symmetry
X ( F ) X (F )
F
Properties of the FT: 2. Time Shift
FTx(t t0 ) e j 2Ft0 X ( F )
since
FTx(t t0 )
j 2Ft
x
(
t
t
)
e
dt
0
j 2Ft0
j 2Ft '
x(t ' )e
(let t ' t t0 ) e
dt
'
In other words a time shift affects the phase, not the magnitude
Bandwidth of a Baseband Signal
• A Baseband Signal has all frequency components at the low
frequencies, around F=0 Hz;
• Bandwidth: the frequency interval where most of the
frequency components are.
| X (F ) |
B
B
F
What does it mean?
If you take the signal at two different times
with t 1/ B
t
and
then x(t ) x(t t )
x(t t )
j 2Ft j 2Ft
X
(
F
)
e
e
dF
B
j 2Ft
j 2Ft
X
(
F
)
e
e
dF x(t )
B
1 since | Ft | Bt 1
t t
For Example:
x(t )
70
1
| X (F ) |
60
0.5
50
0
40
30
-0.5
20
-1
10
-1.5
0
10
20
30
40
50
60
t (msec)
70
80
90
zoom
100
0
0
0.5
1
1.5
2
2.5
3
3.5
4
F (kHz)
B 1kHz
1 / B 1m sec
0.4
0.2
0
0.1m sec
-0.2
-0.4
-0.6
31.2
31.4
31.6
31.8
t (msec)
32
32.2
samples spaced by less
than 0.1msec are fairly
close to each other
4.5
5
F (kHz)
Computation of the Fourier Transform
•
Whatever we do, physical signals are in continuous
time and, as we have seen, they are described by the
FT;
•
The FT can be computed in one of two ways:
1. Analytical: if we have an expression of the signal (like
in the example);
2. Numerical: by approximation using the Fast Fourier
Transform (FFT).
Fourier Transform and FFT
Consider a signal of a finite duration
x(t ), 0 t T0
with Bandwidth B . Then we can approximate, by
simple arguments,
T0
M 1
0
n 0
X ( F ) FTx(t ) x(t )e j 2Ft dt x(nTS )e j 2FnTS TS
1
(say at least an order of magnitude smaller)
B
M round(T0 / TS )
where TS
Fourier Transform and FFT
Using the FFT:
•
Take an even integer N M . Then compute the N
point FFT of the sampled data, padded with zeros:
X [k ] FFTx(0),...,x(M 1),0,0,...,0, k 0,..., N 1
•
Assign the frequencies:
X k
X k
N
T
X
[
N
k
],
k
,...,1
S
2
FS
N
k 0,..., 1
TS X [k ],
N
2
FS
N
negative frequencies
positive frequencies
Example
Take a sinusoid with frequency F0 10kHz
and length T0 5m sec
Let the sampling frequency be FS 200kHz
Example
X=fft(x, N);
F=(-N/2:N/2 -1)*Fs/N;
plot(F,fftshift(20*log10(abs(X))))
50
| X ( F ) |dB
dB
0
-50
-100
-80
-60
-40
-20
0
kHz
20
40
60
80
100
F (kHz)
Example (Zoom in at the Peak)
Max at F=10kHz
| X ( F ) |dB
0
dB
-20
-40
-60
Sidelobes due to
finite data length
-80
7
8
9
10
11
kHz
12
13
14
F (kHz)
Complex Signals
All signals in nature are real. There is not such as a thing as
“complex” signal.
However in many cases we are interested in processing and
transmitting “pairs” of signals. We can analyze them “as if”
they were just one complex signal:
x(t ) a(t ) jb(t )
a(t )
x(t )
Complex Signal
b(t )
Real Signals
j
Amplitude Modulation: Real Signal
You want to transmit a signal over a medium (air, water,
space, cable…). You need to “modulate it” by a carrier
frequency:
s (t )
xAM (t ) s(t ) cos(2FC t )
2
2
1.8
cos(2FC t )
1.6
1.4
1.2
1
1.5
1
0.5
0.8
0
0.6
0.4
-0.5
0.2
0
0
50
100
150
200
250
300
350
400
450
-1
-1.5
0
50
100
150
200
250
300
350
400
450
500
Amplitude Modulation: Complex Signal
However most of the times the signal we modulate is
Complex
x(t ) a(t ) cos(2 FC t )
s(t ) a(t ) jb(t )
b(t ) sin(2 FC t )
Re{.}
e j e j 2FC t
Notice now that the modulated signal is real and it contains
both signals a(t) and b(t).
FT of Modulated Signal
x (t )
s (t )
See the different steps:
x(t )
Re{.}
e j e j 2FC t
X ( F ) FTx (t ) e s(t )e
j
j 2FC t j 2Ft
e
x(t ) Rex (t )
x (t )e
FT x (t )
*
dt e j s(t )e j 2 ( F FC )t dt e j S ( F FC )
1
1
x (t ) x * (t )
2
2
*
j 2Ft
*
j 2Ft
dt x (t )e
dt X * ( F ) e j S * ( F FC )
FT of Modulated Signal
Put things together:
x (t )
s (t )
| X (F ) |
Re{.}
| S (F ) |
B
x(t )
e j e j 2FC t
F
X (F )
Usually B FC
FC
1 j
1
e S ( F FC ) e j S * ( F FC )
2
2
FC
F