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

jn
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 2Ft
x
(
t
)
e
dt


x(t )  IFTX ( F ) 

j 2Ft
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 2Ft
x
(
t
)
e
dt


x(t )  IFTX ( F ) 

j 2Ft
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 2FT0 / 2
j 2FT0 / 2
e

e
X ( F )   e  j 2Ft dt 
 j 2F
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 2Ft
*


 j 2Ft
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
FTx(t  t0 )  e j 2Ft0 X ( F )
since
FTx(t  t0 ) 

 j 2Ft
x
(
t

t
)
e
dt
0





 j 2Ft0
 j 2Ft '
  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 2Ft j 2Ft
X
(
F
)
e
e
dF


B

j 2Ft
j 2Ft
X
(
F
)
e
e
dF  x(t )

B
 1 since | Ft | Bt  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 )  FTx(t )   x(t )e  j 2Ft dt   x(nTS )e  j 2FnTS 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 ]  FFTx(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(2FC t   )
2
2
1.8
cos(2FC 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 2FC 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 2FC t

X ( F )  FTx (t )   e s(t )e
j
j 2FC t  j 2Ft
e

x(t )  Rex (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 2Ft
*


j 2Ft

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 2FC t
F
X (F ) 
Usually B  FC
 FC
1 j
1
e S ( F  FC )  e  j S * ( F  FC )
2
2
FC
F