# Signal and systems Linear Systems

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```					Signal and systems
Linear Systems
Luigi Palopoli

palopoli@dit.unitn.it

Signal and systems – p. 1/5
Wrap-Up

Signal and systems – p. 2/5
Fourier Series

•   We have see that is a signal is periodic, it can be conveniently expressed as a
Fourier Series:
X∞
s(t) =          sn ejnωt
n=−∞

or, if s(t) is real,
∞
X
s(t) =         2 |sn | cos(nωt + Φ(sn ))
n=1

•   This is essentially a representation of the signal in an appropriate base made of
exponential orthonormal functions
•   The convenience of this representation can easily be seen when we process a
signal through a linear systems

Signal and systems – p. 3/5
Fourier Series

•   Because exponential functions are eigenfunctions, the response to a sinusoidal
signal is:

st , H(s) = ∞ h(t)e−st dt →
˘ st ¯                    R
S e     = H(s)e             −∞
S {cos(ωt + φ0 )}                                    = |H(jω)| cos(ωt + φ0 + Φ(H(jω))
S {1} = H(0)

Where the last expression is true iff H(s) converges for s = jω.
•   Using the superimposition principle, for a generic periodic signal s(t), we get:

P∞
S {s(t)}   = H(0)s0 +       n=1   2 |sn | |H(jnω)| cos(ωnt + Φ(sn ) + Φ(H(jnω)))

•   The expression above reveals the power of our formalism. By simply knowing
H(jω), we are able to compute the response to any periodic signal without solving
differential equations. We would like to apply this to non-periodic signals as well.

Signal and systems – p. 4/5
Linear systems: Frequency Domain

Signal and systems – p. 5/5
Nonperiodic signals

Signal and systems – p. 6/5
Fourier series of a non periodic signal

•   A non-periodic signal can be obtained as the limit of a periodic signal:
s(t) = limT0 →∞ repT0 s(t) where repT0 s(t) = ∞
P
h=−∞ s(t + hT0 )
•   The coefﬁcients of the periodic signal repT0 s(t) are given by:

1
Z    T0 /2
sn =                   [repT0 s(t)]e−jnω0 t dt
T0       −T0 /2
T0 /2     ∞
1
Z              X
=                          s(t + hT0 )e−jnω0 t dt
T0       −T0 /2 h=−∞

Signal and systems – p. 7/5
Fourier series of a non periodic signal

•   Observing that e−j2πnh = 1, we can write

T0 /2    ∞
1
Z            X
sn =                             s(t + hT0 )e−jnω0 t e−jω0 nhT0 dt
T0         −T0 /2 h=−∞

T0 /2    ∞
1
Z            X
=                             s(t + hT0 )e−jnω0 (t+hT0 ) dt
T0         −T0 /2 h=−∞

∞        T0 /2
1          X Z
=                             s(t + hT0 )e−jnω0 (t+hT0 ) dt
T0     h=−∞        −T0 /2

∞        T0 (h+1/2)
1          X Z                                    ′
=                                     s(t′ )e−jnω0 t dt′
T0     h=−∞        T0 (h−1/2)
Z ∞
1                             ′
=                 s(t′ )e−jnω0 t dt′
T0       −∞

Signal and systems – p. 8/5
Fourier series of a nonperiodic signal

•   Deﬁne S(ω) =
R∞
−∞   s(t)e−jωt dt
•   We get: sn =    1
S(nω0 )
T0
•   Notice that two coefﬁcients are spaced out by ∆ω = ω0 . Drawing the spectrum in
the (ω − sn ) plan, the rows sn become closer and closer as T0 → ∞: the
spectrum tends to be continuous. Therefore, sn = ∆ω S(n∆ω)
2π
•   We can write: repT0 s(t) = n=−∞ ∆ω S(n∆ω)ejnω0 t
P∞
2π
•   As T → ∞ (∆ω → 0), we get the deﬁnition of the integral:

1
Z   ∞
s(t) = lim repT0 s(t) =              S(ω)ejω dω
T0 →∞             2π    −∞

Signal and systems – p. 9/5
Fourier Transform

•   Starting from the deﬁnition of Fourier series (that applies to periodic signal) we
have [informally] derived the fourier transform,
Z   ∞                 1
Z   ∞
S(ω) =    s(t)e−jωt dt s(t) =                     S(ω)ejωt dω
−∞                     2π             −∞

•   It can be seen that if the signal is absolutely integrable and bounded, (plus other
R∞
technical mathematical conditions) i.e., −∞ |s(t)| dt = L < ∞, then
◦   the fourier transform exists for almost all ω
◦   also the Fourier is squared -integrable: −∞ |S(ω)|2 dω < ∞
R∞

◦   its inverse transform converges to s(t) (in squared error sense)
•   Notice
◦   If a signal has ﬁnite energy, then it is absolutely integrable
◦   The above conditions are only sufﬁcient (we will ﬁnd the Fourier transform of
several signals that do not meet these conditions).

Signal and systems – p. 10/5
Example

•   Compute the FT of a rectangular impulse:
8
<A if τ /2 ≤ t ≤ τ /2
s(t) = AGτ (t) =
:0 Otherwise

•   By applying the deﬁnition:
Z   ∞
S(ω) =              s(t)e−jωt dt
−∞
Z    τ /2
=A                e−jωt dt
−τ /2
1     −jωt ˛τ /2
˛
=A      e       −τ /2
−jω
1                τ
=A     (−2j sin(ω )
−jω               2
1        τ           τ
= Aτ τ sin(ω ) = Aτ Sinc(ω )
ω2        2           2

Signal and systems – p. 11/5
Example

•   Compute the FT of a rectangular impulse:

s(t) = δ(t)

•   By applying the deﬁnition:
Z   ∞
S(ω) =           δ(t)e−jωt dt
−∞
=1

Signal and systems – p. 12/5
Properties of the Fourier transform

•   Linearity:
F(u1 (t)) = U1 (ω),
F(u2 (t)) = U2 (ω)
↔
F(a1 u1 (t) + a2 u2 (t)) = a1 U1 (ω) + a2 U2 (ω)

This is a fairly obvious consequence of the linearity of the integral operator.
•   Time shifting:

F(u1 (t)) = U1 (ω)
↔
Z   ∞
F(u1 (t − t0 )) =           u1 (t − t0 )e−jωt dt =
−∞
Z   ∞
=           u1 (t − t0 )e−jω(t−t0 ) e−jωt0 dt =
−∞

= e−jωt0 U1 (ω)

Signal and systems – p. 13/5
Properties of the Fourier transform

•   Shiﬁting in the frequency domain

F(u1 (t)) = U1 (ω),
↔
Z   ∞
jω0 t
F(e           u1 (t)) =           ejω0 t u1 (t)e−jωt dt =
−∞
Z∞
=            u1 (t)e−j(ω−ω0 )t dt =
−∞
= U1 (ω − ω0 )

Signal and systems – p. 14/5
Properties of the Fourier transform

•   Time scaling. Assume a positive

F(u1 (t)) = U1 (ω)
↔
Z   ∞
F(u1 (at)) =             u1 (at)e−jωt dt =
−∞
Z∞1             ω
=       u1 (t′ )e−j a t dt′ =
−∞ a
1    ω
= U1 ( )
a    a

•   For general a we can easily see

F(u1 (t)) = U1 (ω)
↔
1      ω
F(u1 (at)) =        U1 ( )
|a|     a

Signal and systems – p. 15/5
Properties of the Fourier transform

•   Differentiation in the time domain:

F(u1 (t)) = U1 (ω),
↔
Z   ∞
u1 (t) =               S(ω)ejωt dt   =
−inf ty
↔
Z ∞
d            d
u1 (t) =            S(ω)ejωt dt =
dt          dt −inf ty
Z ∞
d
=          S(ω) ejωt dt      =
−inf ty     dt
= jωU1 (ω)

•   More generally
F(u1 (t)) = U1 (ω),
↔
dn u1
F( n ) = (jω)n U1 (ω)
dt                                       Signal and systems – p. 16/5
Properties of the Fourier transform

•   Duality
F(f (t)) = F (ω)
↔
F(F (t)) = 2πf (−ω)
•   Proof:
F(f (t)) = F (ω)
↔
1
Z   ∞
f (t) =              F (ω)ejωt dt
2π   ∞

If we simply swap the two variables t and ω, we ﬁnd:
Z ∞
1
f (ω) =         F (t)ejωt dt
2π ∞
1
=    F(F (t))|−ω
2π

Signal and systems – p. 17/5
An example

•   Consider the signal s(t) = 1. This is not absolutely integrable (it does not
converge to 0).
•   However, if we apply duality we get:

F(δ(t)) = 1
↔
F(1) = 2πδ(ω)

•   This is extremely important because it shows that if we consider generalised
functions (δ(.)) we can ﬁnd the Fourier Transform of function that are not
absolutely integrable

Signal and systems – p. 18/5
Another (important) example

•   Let us consider the signal (non absolutely integrable)
8
<1     t≥0
1(t) =
:0     Otherwise

•   We can see that:
1
1(t) =     (1 + sgn(t))
2
8
<1     t≥0
sgn(t) =
:−1 Otherwise

•   Function sgn(t) is not an absolutely integrable function, but we can manage it with
some trick.....

Signal and systems – p. 19/5
Another (important) example

•   We can write:
sgn(t) = lim Sα (t)
α→0
8
<e−αt t ≥ 0
Sα (t) =
:−eαt t < 0

•   Sα (t) is absolutely integrable, hence we can deal with it:
Z   ∞
F(Sα (t)) =           Sα (t)e−jωt dt
−∞
Z 0                        Z   ∞
αt −jωt
=−            e    e        dt +           e−αt e−jωt dt
−∞                        0
1                    1
=−        e(α−jω)t |0 −
−∞         e−(α+jω)t |∞
0
α − jω               α + jω
1         1
=−        +
α − jω    α + jω

Signal and systems – p. 20/5
Another (important) example

•   Hence:
1        1
F(sgn(t)) = lim −          +
α→0    α − jω   α + jω
2
=
jω
•   We can conclude:                   „            «
1    2
F(1(t)) =        + 2πδ(ω)
2 jω
1
=     + πδ(ω)
jω

Signal and systems – p. 21/5
Properties of the Fourier transform

•   Convolution
F(u1 (t)) = U1 (ω),
F(u2 (t)) = U2 (ω),
↔
F(u1 (t) ∗ u2 (t)) = U1 (ω)U2 (ω)
•   Proof                            Z   ∞   Z   ∞
F(u1 (t) ∗ u2 (t)) =                    u1 (τ )u2 (t − τ )dτ e−jωt dt
−∞          −∞
Z∞      Z    ∞
=                     u1 (τ )u2 (t − τ )dτ e−jωt dt
−∞          −∞
Z∞                »Z   ∞                    –
=            u1 (τ )            u2 (t − τ )e−jωt dt dτ
−∞                  −∞
Z∞
=            u1 (τ )U1 (ω)e−jωτ dτ
−∞
= U1 (ω)U2 (ω)

Signal and systems – p. 22/5
Properties of the Fourier transform

•   Product
F(u1 (t)) = U1 (ω),
F(u2 (t)) = U2 (ω),
↔
1
F(u1 (t)u2 (t)) =      U1 (ω) ∗ U2 (ω)
2π
•   Proof: It descends from duality + convolution

Signal and systems – p. 23/5
Properties of the Fourier transform

•   Integration

F(u1 (t)) = U1 (ω),
↔
Z   t                                1
F(                       u1 (τ )dτ ) =      U1 (ω) + πU1 (0)δ(ω)
τ =−∞                               jω

•   Proof                      Z       t
u1 (τ )dτ = u1 (t) ∗ 1(t)
τ =−∞
↔
Z   t
F(                      u1 (τ )dτ ) = F (u1 (t) ∗ 1(t))
τ =−∞
1
=(     + πδ(ω))U1 (ω)
jω
U1 (ω)
=        + πδ(ω))U1 (0)
jω

Signal and systems – p. 24/5
Example

•   Consider the f (t) = B cos ω0 t

B `` jω0 t
+ e−jω0 t
´´
F(f (t)) =     F e
2

•   We have seen that F(δ(t)) = 1; applying the duality property:

F (1) = 2πδ (−ω)

•   Now, we apply frequency shifting property:
` jω t ´
F e 0 = 2πδ (ω − ω0 )

•   Therefore, we get:

F (B cos ω0 t) = Bπ (δ(ω − ω0 ) + δ(ω + ω0 ))

Signal and systems – p. 25/5
Example

• We have seen that

τ
F (AGτ (t)) = Aτ Sinc ω
2
• Let us ﬁnd: F AGτ /2

• We can apply time scaling rule:

Aτ      ωτ
F AGτ /2   =    Sinc
2      22

Signal and systems – p. 26/5
Spectrum

•   Also for non periodic signals g(t) we can associate a frequency domain spectrum
G(ω)
•   It is typically depicted by giving its norm |G(ω)| and its phase ∠(G(ω))
•   For real signals the following hold:
◦   |G(ω)| = |G(−ω)|, ∠(G(ω)) = −∠(G(−ω))
◦   If the signal is even, then G(ω) is real
◦   If the signal is odd, then G(ω) is imaginary
•   An interesting example is the following ideal ﬁlter.
•   Its spectrum is given by: |H(ω)| = 1GateB (ω) and ∠(H(ω) = −ωt0

Signal and systems – p. 27/5
Why is an ideal ﬁlter ideal?

•   Consider, for simplicity, t0 = 0
“ ”
•   Applying duality, we get: h(t) = 2πBSinc t B
2
•   As we can see this ﬁlter is not a causal system....
•   Therefore the system is not pysically implementable

Signal and systems – p. 28/5
Fourier Transform of periodic signals

•   We have seen that the fourier transform of a cosine is the sum of two δ
•   The same applies also to other periodic signals
•   For a periodic signal, we have seen that it is possible to write them in terms of
Fouries series:
X∞
s(t) =          sn ejnω0 t
n=−∞

•   we have seen that F ejnω0 t = 2πδ(ω − nω0 )
`       ´

•   Therefore we get:
∞
X
F (s(t)) =          sn δ(ω − nω0 )
n=−∞

Signal and systems – p. 29/5
Fourier Transform of periodic signals

•   What if we construct a periodic repeating a non periodic signal?

∞
X
s(t) =            sc (t + iT )
i=−∞

•   The signal can be expressed as

∞
X
s(t) = sc (t) ∗          δ(t + iT )
i=−∞

•   We can compute the Fourier series of signal sr (t) =
P∞
i=−∞   δ(t + iT ):

∞
X
sr (t) =          sn ejnω0 t
n=−∞

1
Z   T0 /2                       1
sn =                  δ(t)e−jnω0 t dt =
T0    −T0 /2                         T0

Signal and systems – p. 30/5
Fourier Transform of periodic signals

•   Therefore:
∞
1    X
s(t) = sc (t) ∗             ejnω0 t
T0   n=−∞

Which, corresponds, in the frequency domain, to:

∞
1    X
S(ω) = Sc (ω)             δ(ω − nω0 )
T0   n=−∞

Signal and systems – p. 31/5
Fourier Transform of periodic signals

Example:

Signal and systems – p. 32/5
Mathematical Complements

Signal and systems – p. 33/5
Discussion

•   We have seen that for signals compying with the following conditions:
◦   s(t) limited
◦   Finite number of minima and maxima and of singularities
◦   Absolutely integrable
The Fourier transform exists and the inverxe transform converges to s(t) .
•   For these signals we have the Parseval equality:
Z    ∞               1
Z   ∞
|s(t)|2 dt =              |S(ω)|2 dω
−∞              2π   −∞

•   We can compute the integral in the easier domain (for instance for a low pass ﬁlter
it is much easier in the frequency domain)

Signal and systems – p. 34/5
Discussion

•   If we consider signals with ﬁnite power (e.g. periodic signals) we can still compute
the fourier transform using generalized functions (δ)
•   We have derived the Fourier transform from the Fourier series, but we also have
seen that the Fourier series is a special case of the Fourier transform.

Signal and systems – p. 35/5
Two interesting applications

Signal and systems – p. 36/5
Amplitude Modulation

• We want to use the same medium (e.g., the air), to transmit
multiple signals (e.g., different channels)
• Assume that each transmission can have a limited
bandwidth
• One of the oldest ways for doing this was to modulate the
amplitude of the signal by multiplying it by a sinusoidal
oscillation:
xAM (t) = x(t) cos(5t)

Signal and systems – p. 37/5
Amplitude Modulation

x(t)

1.0

0.7

0.5

0.3

0.1

K             K
10            5               0
t
5   10

*
cos(5t)

1.0

0.5

K             K              0

K
10            5                          5   10

=
0.5

K
1.0

x(t)cos(5t)

0.8

0.6

0.4

0.2

K
10            K
5
K
0.2
0              5   10

K
t
0.4

K
0.6

Signal and systems – p. 38/5
Amplitude Modulation - Frequency domain

• It is important to see what happens in the frequency
domain.
• Remember cos ω0 t =   ejω0 t +e−jω0 t
2
• Therefore

ejω0 t + e−jω0 t
xAM (t) = x(t) cos(ω0 t) = x(t)                  ↔
2
X(ω − ω0 ) + X(ω + ω0 )
XAM (ω) =
2

Signal and systems – p. 39/5
Fourier Transform of periodic signals

Example:

Signal and systems – p. 40/5
FDM

• The idea outlined above can be used to do a Frequency
Division Demultiplexing
• In practice, the spectrum of each singnal is translated to a
different frequency range: Xi (ω) → Xi (ω − ωi )
• In order for the idea to work, the frequency used to translate
the signal must be sufﬁciently spaced out so as to avoid
interference: ωi+1 − ω > B/2, where B is the bandwidth
• To demodulate the signal, we ﬁrst isoltate the part of the
spectrum we are interested in, translate the spectrum by ωi
and then elimnate spurious component by a low pass ﬁlter.

Signal and systems – p. 41/5
FDM

Signal and systems – p. 42/5
FDM - demodulation

Signal and systems – p. 43/5
Sampling

Signal and systems – p. 44/5
Ideal sampler

• Ideal sampling can intuitively be seen as generated by
multiplying a signal by a sequence of dirac’s δ

Signal and systems – p. 45/5
Properties of δ

•    +∞
−∞ f (t)δ(t   − a)dt = f (a)
•    t
−∞ δ(τ )dτ    = 1 → F (δ(t)) = 1
• f (t) ∗ δ(t − t0 ) = f (t − t0 )

Signal and systems – p. 46/5
F -trasform of r∗

•                                             to
Using the above properties it is possible ” write:
“    P∞
F (r∗ (t)) = F r(t) n=−∞ δ(t − nT )
•   We can express the sampling signal using the Fourier series:

∞                    ∞
X                1   X            2π t
δ(t − nT ) =            ejh   T

n=−∞
T   h=−∞

“                       ”
•                                 1 P∞      jh 2π t
Hence F   (r∗ (t))   = F r(t) T  h=−∞ e
T

•   Applying the frequency shifting property we get:
1 P∞                 2π
F (r∗ (t)) = T    h=−∞ R(ω − h T )

Signal and systems – p. 47/5
Example

1                                                                            1

0.9                                                                          0.9

0.8                                                                          0.8

0.7                                                                          0.7

0.6                                                                          0.6
|R(j ω)|

|R (jω)|
0.5                                                                          0.5

*
0.4                                                                          0.4

0.3                                                                          0.3

2 π /T
0.2                                                                          0.2

0.1                                                                          0.1

0                                                                            0
−1   −0.8   −0.6   −0.4   −0.2   0   0.2   0.4   0.6   0.8   1               −2   −1.5   −1   −0.5   0   0.5        1   1.5         2
ω                                                                   ω

Signal and systems – p. 48/5
Aliasing

The spectrum might be altered (i.e., signal not attainable from
samples!)
2

sin(2 π t/3)
1.5
samples collected with T = 3/2

1

0.5

0

−0.5

−1

sin(2 π t)
−1.5

−2
0        0.5      1             1.5             2                 2.5   3

Signal and systems – p. 49/5
Shannon theorem

• If the signal has a ﬁnite badwidth then the signal can be reconstructed
1
from samples collected with a period such that T   ≥ 2B
• Band-limited signals have inﬁnite duration; many signals of
interest have inﬁnite bandwidth
• Typically a low-pass ﬁlter is used to de-emphasize higher
frequencies

Signal and systems – p. 50/5
Data Extrapolation

•   If the following hypotheses hold
◦   the signal has limited bandwidth B
◦   the signal is sampled at frequency fs =        1
≥ 2B
T
•   then the signal can be reconstructed using an ideal lowpass ﬁlter L(s) with
8
<T              π      π
if ω ∈ [− T ,    T
]
|L(ω)| =
:0    elsewhere.

•   Signal l(t) is given by:

1
Z   pi/T                 sin(πt/T )
l(t) =                  T ejωT dω =              = sinc(πt/T )
2π    −pi/T                       πt/T

Signal and systems – p. 51/5
Data Extrapolation I

•   The reconstructed signal is computed as follows:

R +∞    P              π(t−τ
r(t) = r∗ (t) ∗ l(t) = −∞ r(τ ) δ(τ − kT )sinc T dτ =
π(t−kT )
= +∞ r(kT )sinc
P
−∞                  T

•   The function sinc is not causal and has inﬁnite duration
•   In communication applications
◦   The duration problem can be solved truncating the signal
◦   The causality problem can be solved introducing a delay and collecting some
sample before the reconstruction
•   Not viable in control applications since large delays jeopardise stability

Signal and systems – p. 52/5

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