35
Linear systems: Frequency Domain
Signal and systems – p. 2/35
Properties of the Fourier Series
Signal and systems – p. 3/35
Example
• Compute the spectrum of the sawtooth signal:
s(t) = repT sc(t)
sc(t) =
tT
if t ∈ [0, T ]
0 otherwise
• Let us start from the power:
P =1
T
∫ T
0
t2
T 2dt
=1
T 3
t3
3|T0
=1
3
• The signal has a finite power. Hence, it is possible to compute the spectrum.
Signal and systems – p. 4/35
Example
• The DC component:
s0 =1
T
∫ T
0
t
Tdt
=1
T 2
t2
2|T0
=1
2
• For the other coefficients it is very useful to bear in mind the following result(integration by parts):
∫ b
ate−jωntdt =
t
−jωne−jωnt|ba−
1
−jωn
∫ n
ae−jωntdt
=b
−jωne−jωnb −
a
−jωne−jωna −
1
−jωn
1
−jωne−jωnt|ba
=j(
be−jωnb − ae−jωna)
ωn+
1
ω2n2
(
e−jωnb − e−jωna)
Signal and systems – p. 5/35
Example
• The DC component:
s0 =1
T
∫ T
0
t
Tdt
=1
T 2
t2
2|T0
=1
2
• For the other coefficients it is very useful to bear in mind the following result(integration by parts):
∫ b
ate−jωntdt =
t
−jωne−jωnt|ba−
1
−jωn
∫ n
ae−jωntdt
=b
−jωne−jωnb −
a
−jωne−jωna −
1
−jωn
1
−jωne−jωnt|ba
=j(
be−jωnb − ae−jωna)
ωn+
1
ω2n2
(
e−jωnb − e−jωna)
Signal and systems – p. 6/35
Example
• Now, we compute the nth coeeficient
sn =1
T
∫ T
0
t
Te−jωntdt
=1
T 2
∫ T
0te−jωntdt
=1
T 2
[
jTe−jωnT − 0e−jωn0
ωn+
1
ω2n2
(
e−jωnT − e−jωn0)
]
=1
T 2
jTe−jωnT
ωn+
1
T 2
1
ω2n2
(
e−jωnT − 1)
• Observing that e−jnωT = cos(−jnωT ) + j sin(−jnωT ) =
cos(nωT )− j sin(nωT ) = cos(n2π)− j sin(n2π) = 1, we get:
sn =1
T 2
jT
ωn
=j
2πn
Signal and systems – p. 7/35
Example
• How much of the power is in the first 4 harmonics?
• The power is P = 1/3 ad the power is (Parseval equality):∑3
k=−3 |sk|2 = 0.319 ≈ 95.68%P
Signal and systems – p. 8/35
Example - 2
• Now let us consider the triangular wave:
s(t) = repT sc(t)
sc(t) =
2tT
if t ∈ [0, T/2]
2− 2tT
if t ∈ [T/2, T ]
0 otherwise
• Compute the Power and say how much of the power is in the first 7 elements of theFourier Series (4 harmonics).
• The power and the DC component can be computed as in the previous exampleand we get, once again, P = 1/3, s0 = 1/2.
Signal and systems – p. 9/35
Example - 2
• Now let us move to the computation of the coefficients:
sk =1
T
∫ T
0sc(t)e
−jnωtdt
=2
T
[
∫ T/2
0
t
Te−jnωtdt+
∫ T
T/2(1−
t
T)e−jnωtdt
]
• Let
A =
∫ T/2
0
t
Te−jnωtdt
B =
∫ T
T/2(1−
t
T)e−jnωtdt
Signal and systems – p. 10/35
Example - 2
• We can compute A as follows
A =1
T
∫ T/2
0te−jnωtdt
=1
T
j T2e−jnω T
2 − 0e−jnω0
ωn+
1
ω2n2
(
e−jnωT/2 − 1)
=j
2
e−jnω T2
ωn+
1
T
1
ω2n2
(
e−jnωT/2 − 1)
Signal and systems – p. 11/35
Example - 2
• Likewise we can compute B:
B =1
T
∫ T
T/2
(
1−t
T
)
e−jnωtdt
=
∫ T
T/2e−jnωtdt−
1
T
∫ T
T/2te−jnωtdt
=e−jnωT − e−jnωT/2
−jnω−
1
T
[
jTe−jnωT − T/2e−jnωT/2
ωn+
1
ω2n2
(
e−jnωT − e−jnωT/2)
= je−jnωT − e−jnωT/2
nω− j
e−jnωT − 1/2e−jnωT/2
ωn−
1
ω2n2
(
1− e−jnωT/2)
=j
nω−
je−jnωT/2
nω−
j
nω+
je−jnωT/2
2nω+
1
Tω2n2
(
e−jnωT/2 − 1)
Signal and systems – p. 12/35
Example - 2
• Therefore:
sn =2
T(A+B)
=2
T(A+B)
=4
T 2ω2n2
(
e−jnωT/2 − 1)
=1
π2n2
(
e−jnωT/2 − 1)
=1
π2n2((−1)n − 1)
• Summarising s0 = 12
, Sk = 0 for even k and sk = −2(πk)2
for odd k
• The power is P = 1/3 ad the power is (Parseval equality):∑3
k=−3 |sk|2 = 0.33314 ≈ 99.95%P
Signal and systems – p. 13/35
Amplitude Spectrum
0
0.1
0.2
0.3
0.4
0.5
0 2 4 6 8 10
sawtoothtriangle
Signal and systems – p. 14/35
Consideration
• In the examples above we have seen the sawtooth signalhas a smaller percentage of its power concentrated in thelow order harmonics than the triangle
• This is because the triangle changes more smoothly thanthe sawtooth
• To have quick changes we need high frequency terms in theFourier Series
Signal and systems – p. 15/35
Properties of the Fourier Series
• consider two signals x(t) ∈ SET and y(t) ∈ SE
T . Suppose that Fs {x(t)} = xn andFs {y(t)} = yn are respectively their Fourier series.
• Linearity:
αx(t) + βy(t) ↔ αxn + βyn
This is a straightforward implication of the linearity of the integral operator.
Signal and systems – p. 16/35
Properties of the Fourier Series
• Product
x(t)y(t) ↔ αcn =
∞∑
k=−∞
xkyn−k
• Proof:
z(t) = x(t)y(t) =
=(
∑
∞
k=−∞xke
jkωt)(
∑
∞
m=−∞ymejmωt
)
=
=∑
∞
m=−∞
∑
∞
k=−∞xkymej(k+m)ωt
(change of variable n = m+ k)
=∑
∞
n=−∞
∑
∞
k=−∞xkyn−ke
jnωt
=∑
∞
n=−∞
(
∑
∞
k=−∞xkyn−k
)
ejnωt
Hence the thesis.
• The sum c′n =∑
∞
k=−∞xkyn−k is called convolution.
Signal and systems – p. 17/35
Properties of the Fourier Series
• Time shifting:
Fs {x(t)} = xn → Fs {x(t− τ)} = xne−jnωτ
• Proof:
x(t) =∑
∞
n=−∞xnejnωt →
x(t− τ) =∑
∞
n=−∞xnejnω(t−τ)
=∑
∞
n=−∞xne−jnωτ ejnωt
• When we delay the signal, we do not change the amplitude spectrum but we
simply introduce a shift in the phase.
Signal and systems – p. 18/35
Properties of the Fourier Series
• Time reversal:
Fs {x(t)} = xn → Fs {x(−t)} = x−n
• Conjugation:
Fs {x(t)} = xn → Fs {x⋆(t)} = x⋆
−n
• Proof:
x(t) =∑
∞
n=−∞xnejnωt →
x⋆(t) =(
∑
∞
n=−∞xnejnωt
)⋆
=∑
∞
n=−∞x⋆ne
−jnωτ
=∑
∞
m=−∞x⋆−mejmωτ
Where the last step was obtained replacing n = −m
Signal and systems – p. 19/35
Properties of the Fourier Series
• Derivation
Fs {x(t)} = xn → Fs
{
dx(t)
dt
}
= jωnxn
• Proof:
x(t) =∑
∞
n=−∞xnejnωt →
dx(t)dt
=d∑
∞
n=−∞xnejnωt
dt→
=∑
∞
n=−∞xn
dejnωt
dt→
=∑
∞
n=−∞xn (jωn) ejnωt
• The derivation emphasises the high frequency
Signal and systems – p. 20/35
Properties of the Fourier Series
• Example....
• Consider the function cos (ωt) + 2ω cos 2ωt
• The spectrum is
Xn =
12
n = 1, n = −1
1 n = 2, n = −2
0 otherwise
• If we compute the derivation we get
X′
n =
ω2
n = 1, n = −1
ω n = 2, n = −2
0 otherwise
• The derivation generates a signal that changes more quickly. Hence the shift
toward higher frequency
Signal and systems – p. 21/35
Comparison of a signal with its derivative
-6
-4
-2
0
2
4
6
0 2 4 6 8 10
y(t)=cos (wt) + 2cos(2 w t)d(y(t))/dt
Signal and systems – p. 22/35
Example - 3
• Now let us consider the signal;
s(t) = repT sc(t)
sc(t) =
2T
if t ∈ [0, T/2]
− 2tT
if t ∈ [T/2, T ]
0 otherwise
• Compute the spectrum
• We can observe that this is the derivative of the triangular wave introduced above(Example 2).
• Thereby
sn = (jωn)1
π2n2((−1)n − 1)
= j2
Tπn((−1)n − 1)
Signal and systems – p. 23/35
Properties of the Fourier Series
• We have seen how the derivation “reshapes” the spectrum of a periodic signal
• The idea is that each component of the spectrum is changed according to acertain rule
• If we consider the derivation as a “transformation”, on the signal (i.e., a system), itmeans that we can characterise this transformation in terms of how it changes thedifferent components of the spectrum.
• This is what happens with any linear transformation operating between vectorspaces: we can describe the application by saying how each component of thebasis is transformed.
• This is generally true....but we have to see how to do it for a general linear system.
Signal and systems – p. 24/35
Some more examples
Signal and systems – p. 25/35
Example - 4
• Compute the spectrum of the signal:
s(t) = repT sc(t)
sc(t) =
sin(ω(t− T/8) if t ∈ [−T/8, 3T/8]
0 otherwise
and say how many components we need to reconstruct 85% of the power
• Let us start from the power:
P =1
T
∫ T/2
−T/2sc(t)dt =
1
T
∫ 3T/8
−T/8s2c(t)dt
=1
T
∫ T/4
−T/4s2c(t+ T/8)dt =
1
T
∫ T/4
−T/4sin2(ωt)dt
=1
T
∫ T/4
−T/4
1− cos(2ωt)
2dt =
1
T
(
T/2−1
4ωsin(2ωt)|
T/4−T/4
)
=1
4−
1
4ωT2 sin(2ωT/4) =
1
2 Signal and systems – p. 26/35
Example - 4
• To compute the spectrum, we take advantage of the property
s(t) ↔ Sn implies
s(t− t0) ↔ e−jnωt0Sn
• Therefore we can first translate the signal by T/8 to the right, compute thespectrum and then multiply by e−jnωT/8 = e−jnπ/4
• The DC component is clearly 0. We can then move to the computation of the othercomponents:
sn =e−jnπ/4
T
∫ T/4
−T/4sin(ωt)e−jnωtdt
∫ T/4
−T/4sin(ωt)e−jnωtdt =
1
2j
∫ T/4
−T/4
(
ejωt − e−jωt)
e−jnωtdt
=1
2j
∫ T/4
−T/4
(
ejω(1−n)t − e−jω(n+1)t)
dt
Signal and systems – p. 27/35
Example - 4
• As far as∫ T/4−T/4
ejω(1−n)tdt is concerned....
• for n = 1 we get:∫ T/4
−T/4ejω(1−n)tdt =
∫ T/4
−T/4dt =
T
2
• For n > 1
∫ T/4
−T/4ejω(1−n)tdt =
1
jω(1− n)ejω(1−n)t|
T/4−T/4
=1
jω(1− n)
(
ejω(1−n)T/4 − e−jω(1−n)T/4)
=1
jω(1− n)2j sin(ω(1− n)T/4)
Signal and systems – p. 28/35
Example
• As for∫ T/4−T/4
e−jω(1+n)tdt:
∫ T/4
−T/4e−jω(1+n)tdt =
1
−jω(1 + n)e−jω(1+n)t|
T/4−T/4
=1
−jω(1 + n)
(
e−jω(1+n)T/4 − ejω(1+n)T/4)
=1
jω(1 + n)2j sin(ω(1 + n)T/4)
Signal and systems – p. 29/35
Example -4
• Overall for n = 1 we have sn = e−jπ/4 14
and for n > 1
sn =e−jnπ/4
T
∫ T/4
−T/4sin(ωt)e−jnωtdt
=e−jnπ/4
2jT
(
1
jω(1− n)2j sin(ω(1− n)T/4)−
1
jω(1 + n)2j sin(ω(1 + n)T/4)
)
=e−jnπ/4
2π
(
−j
(n− 1)sin(ω(n− 1)T/4) +
j
(1 + n)sin(ω(1 + n)T/4)
)
=e−jnπ/4
2πjs′n
s′n =
(
−1
(n− 1)sin(ω(n− 1)T/4) +
1
(1 + n)sin(ω(1 + n)T/4)
)
Signal and systems – p. 30/35
Example - 4
• The power is only by the squared amplitude of the coefficients. Therefore if wetruncate to N , we have got PN =
∑N−N |sn|2 =
∑N−N |s′n|
2. Hence,
P1 = 2 ∗ s′12= 0.125 = 50%P
P2 = 2
2∑
n=1
s′n2= 0.215 = 86%P
• It is sufficient to truncate to the first two components.
• Trigonometric Form:
s(t) =
∞∑
n=1
2(rn cos(nωt)− in sin(nωt)
rn =sin(nπ/4)
2πs′n
in =cos(nπ/4)
2πs′n
Signal and systems – p. 31/35
Example - 4
• Harmonic Form:
s(t) =
∞∑
n=1
2||sn|| cos(nωt+Φ(sn))
||sn|| =1
2π|s′n|
Φ(sn) =
π/2− nπ/4 ifs′n > 0
−π/2− nπ/4 ifs′n > 0
Signal and systems – p. 32/35
Example - 5
• Compute the spectrum of the signal:
s(t) = repT sc(t)
sc(t) =
1 + ta
t ∈ [−a, −b]
1− ba
t ∈ [−b, b]
1− ta
t ∈ [b, a]
0 otherwise
with T/2 > a > b.
• It is useful to start from another signal.
sh(t) = repT shc (t)
sh(t) =
1 + th
t ∈ [−h, 0]
1− th
t ∈ [0, h]
0 otherwise
with T/2 > h. Signal and systems – p. 33/35
Example - 5
• Proceeding as for Example 2 we get:
shn =
aT
if n = 0
Ta
12π2n2
(1− cosnωa) OW
with T/2 > h.
• Now, we can observe that our signal is given by:
s(t) = sa(t)−
b
asb(t)
Signal and systems – p. 34/35
Example - 5
• Hence
sn = san −b
asbn
sn =
a/T − ba
bT
n = 1
Ta
12π2n2
(1− cosnωa)− ba
Tb
12π2n2
(1− cosnωb) OW
sn =
1/T(
a− b2/a)
n = 1
Ta
12π2n2
(cosnωb− cosnωa) OW
Signal and systems – p. 35/35
