Signal and systems - DISI, University of Trentodisi.unitn.it/~palopoli/courses/SS/SSlect6.pdf ·...

Signal and systems

Linear Systems

Luigi Palopoli

[email protected]

Signal and systems – p. 1/52

Wrap-Up


Fourier Series

• We have see that is a signal is periodic, it can be conveniently expressed as aFourier Series:

s(t) =

∞X

n=−∞

snejnωt

or, if s(t) is real,

s(t) =∞

X

n=1

2 |sn| cos(nωt + Φ(sn))

• This is essentially a representation of the signal in an appropriate base made ofexponential orthonormal functions

• The convenience of this representation can easily be seen when we process a

signal through a linear systems


Fourier Series

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

S˘

est¯

= H(s)est, H(s) =R

∞

−∞h(t)e−stdt →

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:

S {s(t)} = H(0)s0 +P

∞

n=1 2 |sn| |H(jnω)| cos(ωnt + Φ(sn) + Φ(H(jnω)))

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


Linear systems: Frequency Domain


Nonperiodic signals


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 repT0s(t) =

P

∞

h=−∞s(t + hT0)

• The coefficients of the periodic signal repT0s(t) are given by:

sn =1

T0

Z T0/2

−T0/2[repT0

s(t)]e−jnω0tdt

=1

T0

Z T0/2

−T0/2

∞X

h=−∞

s(t + hT0)e−jnω0tdt


Fourier series of a non periodic signal

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

sn =1

T0

Z T0/2

−T0/2

∞X

h=−∞

s(t + hT0)e−jnω0te−jω0nhT0dt

=1

T0

Z T0/2

−T0/2

∞X

h=−∞

s(t + hT0)e−jnω0(t+hT0)dt

=1

T0

∞X

h=−∞

Z T0/2

−T0/2s(t + hT0)e−jnω0(t+hT0)dt

=1

T0

∞X

h=−∞

Z T0(h+1/2)

T0(h−1/2)s(t′)e−jnω0t′dt′

=1

T0

Z

∞

−∞

s(t′)e−jnω0t′dt′


Fourier series of a nonperiodic signal

• Define S(ω) =R

∞

−∞s(t)e−jωtdt

• We get: sn = 1T0

S(nω0)

• Notice that two coefficients are spaced out by ∆ω = ω0. Drawing the spectrum inthe (ω − sn) plan, the rows sn become closer and closer as T0 → ∞: thespectrum tends to be continuous. Therefore, sn = ∆ω

2πS(n∆ω)

• We can write: repT0s(t) =

P

∞

n=−∞

∆ω2π

S(n∆ω)ejnω0t

• As T → ∞ (∆ω → 0), we get the definition of the integral:

s(t) = limT0→∞

repT0s(t) =

1

2π

Z

∞

−∞

S(ω)ejωdω


Fourier Transform

• Starting from the definition of Fourier series (that applies to periodic signal) wehave [informally] derived the fourier transform,

S(ω) =

Z

∞

−∞

s(t)e−jωtdt s(t) =1

2π

Z

∞

−∞

S(ω)ejωtdω

• It can be seen that if the signal is absolutely integrable and bounded, (plus othertechnical mathematical conditions) i.e.,

R

∞

−∞|s(t)| dt = L < ∞, then

◦ the fourier transform exists for almost all ω

◦ also the Fourier is squared -integrable:R

∞

−∞|S(ω)|2 dω < ∞

◦ its inverse transform converges to s(t) (in squared error sense)

• Notice◦ If a signal has finite is absolutely integrable, then it has finite energy:

R

∞

−∞|s(t)|2dt is finite.

◦ The above conditions are only sufficient (we will find the Fourier transform ofseveral signals that do not meet these conditions).


Example

• Compute the FT of a rectangular impulse:

s(t) = AGτ (t) =

8

<

:

A if τ/2 ≤ t ≤ τ/2

0 Otherwise

• By applying the definition:

S(ω) =

Z

∞

−∞

s(t)e−jωtdt

= A

Z τ/2

−τ/2e−jωtdt

= A1

−jωe−jωt

˛

˛

τ/2

−τ/2

= A1

−jω(−2j sin(ω

τ

2)

= Aτ1

ω τ2

sin(ωτ

2) = AτSinc(ω

τ

2)


Example

• Compute the FT of a dirac impulse:

s(t) = δ(t)

• By applying the definition:

S(ω) =

Z

∞

−∞

δ(t)e−jωtdt

= 1


Properties of the Fourier transform

• Linearity:F(u1(t)) = U1(ω),

F(u2(t)) = U2(ω)

↔

F(a1u1(t) + a2u2(t)) = a1U1(ω) + a2U2(ω)

This is a fairly obvious consequence of the linearity of the integral operator.

• Time shifting:

F(u1(t)) = U1(ω)

↔

F(u1(t − t0)) =

Z

∞

−∞

u1(t − t0)e−jωtdt =

=

Z

∞

−∞

u1(t − t0)e−jω(t−t0)e−jωt0dt =

= e−jωt0U1(ω)



• Shifiting in the frequency domain

F(u1(t)) = U1(ω),

↔

F(ejω0tu1(t)) =

Z

∞

−∞

ejω0tu1(t)e−jωtdt =

=

Z

∞

−∞

u1(t)e−j(ω−ω0)tdt =

= U1(ω − ω0)



• Time scaling. Assume a positive

F(u1(t)) = U1(ω)

↔

F(u1(at)) =

Z

∞

−∞

u1(at)e−jωtdt =

=

Z

∞

−∞

1

au1(t′)e−j ω

atdt′ =

=1

aU1(

ω

a)

• For general a we can easily see

F(u1(t)) = U1(ω)

↔

F(u1(at)) =1

|a|U1(

ω

a)



• Differentiation in the time domain:

F(u1(t)) = U1(ω),

↔

u1(t) =1

2π

Z

∞

−∞

U1(ω)ejωtdt =

↔

F(d

dtu1(t)) = F(

d

dt

Z

∞

−∞

U1(ω)ejωtdt) =

= F(1

2π

Z

∞

−∞

jωU1(ω)ejωtdt) =

= jωU1(ω)

• More generallyF(u1(t)) = U1(ω),

↔

F(dnu1

dtn) = (jω)nU1(ω)



• DualityF(f(t)) = F (ω)

↔

F(F (t)) = 2πf(−ω)

• Proof:F(f(t)) = F (ω)

↔

f(t) =1

2π

Z

∞

∞

F (ω)ejωtdt

If we simply swap the two variables t and ω, we find:

f(ω) =1

2π

Z

∞

∞

F (t)ejωtdt

=1

2πF(F (t))|−ω


An example

• Consider the signal s(t) = 1. This is not absolutely integrable (it does notconverge 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


Another (important) example

• Let us consider the signal (non absolutely integrable)

1(t) =

8

<

:

1 t ≥ 0

0 Otherwise

• We can see that:

1(t) =1

2(1 + sgn(t))

sgn(t) =

8

<

:

1 t ≥ 0

−1 Otherwise

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

some trick.....



• We can write:sgn(t) = lim

α→0Sα(t)

Sα(t) =

8

<

:

e−αt t ≥ 0

−eαt t < 0

• Sα(t) is absolutely integrable, hence we can deal with it:

F(Sα(t)) =

Z

∞

−∞

Sα(t)e−jωtdt

= −

Z 0

−∞

eαte−jωtdt +

Z

∞

0e−αte−jωtdt

= −1

α − jωe(α−jω)t|0

−∞−

1

α + jωe−(α+jω)t|∞0

= −1

α − jω+

1

α + jω



• Hence:

F(sgn(t)) = limα→0

−1

α − jω+

1

α + jω

=2

jω

• We can conclude:

F(1(t)) =1

2

„

2

jω+ 2πδ(ω)

«

=1

jω+ πδ(ω)



• ConvolutionF(u1(t)) = U1(ω),

F(u2(t)) = U2(ω),

↔

F(u1(t) ∗ u2(t)) = U1(ω)U2(ω)

• Proof

F(u1(t) ∗ u2(t)) =

Z

∞

−∞

Z

∞

−∞

u1(τ)u2(t − τ)dτe−jωtdt

=

Z

∞

−∞

Z

∞

−∞

u1(τ)u2(t − τ)dτe−jωtdt

=

Z

∞

−∞

u1(τ)

»

Z

∞

−∞

u2(t − τ)e−jωtdt

–

dτ

=

Z

∞

−∞

u1(τ)U1(ω)e−jωτdτ

= U1(ω)U2(ω)



• ProductF(u1(t)) = U1(ω),

F(u2(t)) = U2(ω),

↔

F(u1(t)u2(t)) =1

2πU1(ω) ∗ U2(ω)

• Proof: It descends from duality + convolution



• Integration

F(u1(t)) = U1(ω),

↔

F(

Z t

τ=−∞

u1(τ)dτ) =1

jωU1(ω) + πU1(0)δ(ω)

• ProofZ t

τ=−∞

u1(τ)dτ = u1(t) ∗ 1(t)

↔

F(

Z t

τ=−∞

u1(τ)dτ) = F (u1(t) ∗ 1(t))

= (1

jω+ πδ(ω))U1(ω)

=U1(ω)

jω+ πδ(ω))U1(0)


Example

• Consider the f(t) = B cos ω0t

F(f(t)) =B

2F

``

ejω0t + e−jω0t´´

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

F (1) = 2πδ (−ω)

• Now, we apply frequency shifting property:

F`

ejω0t´

= 2πδ (ω − ω0)

• Therefore, we get:

F (B cos ω0t) = Bπ (δ(ω − ω0) + δ(ω + ω0))


Example

• We have seen that

F (AGτ (t)) = AτSinc

(

ωτ

2

)

• Let us find: F(

AGτ/2

)

• We can apply time scaling rule:

F(

AGτ/2

)

=Aτ

2Sinc

(

ω

2

τ

2

)


Spectrum

• Also for non periodic signals g(t) we can associate a frequency domain spectrumG(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


Why is an ideal filter ideal?

• Consider, for simplicity, t0 = 0

• Applying duality, we get: h(t) = 2πBSinc“

t B2

”

• As we can see this filter is not a causal system....

• Therefore the system is not pysically implementable


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 ofFouries series:

s(t) =

∞X

n=−∞

snejnω0t

• we have seen that F`

ejnω0t´

= 2πδ(ω − nω0)

• Therefore we get:

F (s(t)) =∞

X

n=−∞

2πsnδ(ω − nω0)



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

s(t) =∞

X

i=−∞

sc(t + iT0)

• The signal can be expressed as

s(t) = sc(t) ∗∞

X

i=−∞

δ(t + iT0)

• We can compute the Fourier series of signal sr(t) =P

∞

i=−∞δ(t + iT0):

sr(t) =∞

X

n=−∞

snejnω0t

sn =1

T0

Z T0/2

−T0/2δ(t)e−jnω0tdt =

1

T0



• Therefore:

s(t) = sc(t) ∗1

T0

∞X

n=−∞

ejnω0t

Which, corresponds, in the frequency domain, to:

S(ω) = Sc(ω)2π

T0

∞X

n=−∞

δ(ω − nω0)



Example:


Mathematical Complements


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

∞

−∞

|s(t)|2dt =1

2π

Z

∞

−∞

|S(ω)|2dω

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


Discussion

• If we consider signals with finite power (e.g. periodic signals) we can still computethe fourier transform using generalized functions (δ)

• We have derived the Fourier transform from the Fourier series, but we also haveseen that the Fourier series is a special case of the Fourier transform.


Two interesting applications


Amplitude Modulation

• We want to use the same medium (e.g., the air), to transmitmultiple signals (e.g., different channels)

• Assume that each transmission can have a limitedbandwidth

• One of the oldest ways for doing this was to modulate theamplitude of the signal by multiplying it by a sinusoidaloscillation:

xAM (t) = x(t) cos(5t)


Amplitude Modulation

x(t)

tK10 K5 0 5 10

0.1

0.3

0.5

0.7

1.0

*cos(5t)

K10 K5 0 5 10

K1.0

K0.5

0.5

1.0

=x(t)cos(5t)

tK10 K5 0 5 10

K0.6

K0.4

K0.2

0.2

0.4

0.6

0.8


Amplitude Modulation - Frequency domain

• It is important to see what happens in the frequencydomain.

• Remember cos ω0t = ejω0t+e−jω0t

2

• Therefore

xAM (t) = x(t) cos(ω0t) = x(t)ejω0t + e−jω0t

2↔

XAM (ω) =X(ω − ω0) + X(ω + ω0)

2



Example:


FDM

• The idea outlined above can be used to do a FrequencyDivision Demultiplexing

• In practice, the spectrum of each singnal is translated to adifferent frequency range: Xi(ω) → Xi(ω − ωi)

• In order for the idea to work, the frequency used to translatethe signal must be sufficiently spaced out so as to avoidinterference: ωi+1 − ω > B/2, where B is the bandwidth

• To demodulate the signal, we first isoltate the part of thespectrum we are interested in, translate the spectrum by ωi

and then elimnate spurious component by a low pass filter.


FDM


FDM - demodulation


Sampling


Ideal sampler

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


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)


F -trasform of r∗

• Using the above properties it is possible to write:

F (r∗(t)) = F“

r(t)P

∞

n=−∞δ(t − nT )

”

• We can express the sampling signal using the Fourier series:

∞X

n=−∞

δ(t − nT ) =1

T

∞X

h=−∞

ejh 2πT

t

• Hence F (r∗(t)) = F“

r(t) 1T

P

∞

h=−∞ejh 2π

Tt”

• Applying the frequency shifting property we get:F (r∗(t)) = 1

T

P

∞

h=−∞R(ω − h 2π

T)


Example

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ω

|R(j

ω)|

−2 −1.5 −1 −0.5 0 0.5 1 1.5 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ω|R

* (jω)|

2 π /T


Aliasing

The spectrum might be altered (i.e., signal not attainable fromsamples!)

0 0.5 1 1.5 2 2.5 3−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

sin(2 π t)

sin(2 π t/3)

samples collected with T = 3/2


Shannon theorem

• If the signal has a finite badwidth then the signal can be reconstructed

from samples collected with a period such that 1T ≥ 2B

• Band-limited signals have infinite duration; many signals ofinterest have infinite bandwidth

• Typically a low-pass filter is used to de-emphasize higherfrequencies


Data Extrapolation

• If the following hypotheses hold◦ the signal has limited bandwidth B (i.e., the spectrum is not null in the range

[−B, B].◦ the signal is sampled at frequency fs = 1

T≥ 2B

• then the signal can be reconstructed using an ideal lowpass filter L(s) with

|L(ω)| =

8

<

:

T if ω ∈ [− πT

, πT

]

0 elsewhere.

• Signal l(t) is given by:

l(t) =1

2π

Z pi/T

−pi/TTejωT dω =

sin(πt/T )

πt/T= sinc(πt/T )


Data Extrapolation I

• The reconstructed signal is computed as follows:

r(t) = r∗(t) ∗ l(t) =R +∞

−∞r(τ)

P

δ(τ − kT )sinc π(t−τT

dτ =

=P+∞

−∞r(kT )sinc π(t−kT )

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 - DISI, University of Trentodisi.unitn.it/~palopoli/courses/SS/SSlect6.pdf ·...

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