1


0≤t<1
t
z (t) = 2 − t 1 ≤ t ≤ 2


0
else
Find Z (ω).
Answer:
Lets dene the following signal:
(
x (t) =
1 − |t| |t| ≤ 1
, tri (t)
0
else
Notice that:
x (t) = (w ∗ w)t
where:
(
1 |t| ≤ 0.5
0 else
ω
W (ω) = sinc
2
w (t) =
⇒ X (ω) = W 2 (ω) = sinc2
ω
2
z (t) = x (t − 1)
⇒ Z (ω) = X (ω) e−jω = sinc2
ω
2
e−jω
Correct Answer: 2
2
Given the following signal:
x (t) =
∞
X
k=−∞
sinc
πk
2
πk
·δ t−
2
and the following lter:
(
1 |ω| ≤ 2
H (ω) =
0 else
We dene the output as:
y (t) = (x ∗ h)t
Find the energy of y (t), namley:
D
E
energy [y (t)] = ky (t)k22 = y (t) , y (t)
Answer:
∞
X
πk
·δ t−
x (t) =
sinc
2
k=−∞
∞
X
πk
=
sinc (t) · δ t −
2
k=−∞
∞
X
1
πk
= π · sinc (t) ·
δ t−
π
2
k=−∞
=
πk
2
(
1 |ω| ≤ 1
1
F
sinc (t) , wk (ω) =
π
0 else
⇒ X (ω)
= π·
∞
X
1
2πk
2π
δ
ω
−
w (ω) ∗
π
π
2π
2
2
k=−∞
1
w (ω) ∗
2
=
=
2
∞
X
∞
X
4δ (ω − 4k)
k=−∞
w (ω − 4k)
k=−∞
y (t) = (x ∗ h)t
(
⇒ Y (ω) =
2 |ω| ≤ 1
0 else
Parseval:
2
kyk
=
=
=
1
2
· kY k
2π
1
hY, Y i
2π
ˆ∞
1
Y (ω) Y (ω)dω
2π
−∞
=
1
2π
=
4
π
ˆ1
4 · dω
−1
Correct Answer: 4
3
Find the impulse response for the following initially at rest system:
y [n] − y [n − 2] = x [n] + 2x [n − 1]
Answer:
The impulse response h [n] can be derived when the input signal is δ [n]:
h [n] − h [n − 2] = δ [n] + 2δ [n − 1]
since the system is casual (initially at rest):
h [n] = 0
∀n < 0
thus:
h [0] − h [0 − 2] = δ [0] + 2δ [0 − 1] = 1
| {z } |{z} | {z }
=0
=1
=0
h [1] − h [1 − 2] = δ [1] +2 δ [1 − 1] = 2
| {z } |{z}
| {z }
=0
=0
=1
For n > M (in this case M = 1), we get homogeneous equation:
h [n] − h [n − 2] = δ [n] +2 δ [n − 1] = 0
| {z }
|{z}
0
0
h [n] − h [n − 2] = 0
Characteristic polynomial:
⇒ z n − z n−2
z n−2 z 2 − 1
=
0
=
0
⇒ z1,2 = ±1
⇒ h [n] = A · z1n + B · z2n = A + B (−1)
n
using n = 0, 1 as initial conditions:
h [0]
=
1=A+B
= 2=A−B
(
A = 32
⇒
B = − 12
3 1
n
⇒ h [n] =
− (−1) u [n]
2 2
h [1]
Correct Answer: 5
4
Given the following system, h1 , h2 are an LTI systems.
x [n] −→ h1 [n] −→ h2 [n] −→ y [n]
where:


x [n] = δ [n] − aδ [n − 1]
h1 [n] = cos (6n)


h2 [n] = an u [n]
, 0 < |a| < 1
Find y [n].
Answer:
y [n]
(x1 ∗ h2 )
=
=
(x1 ∗ h1 ∗ h2 )n
=
((x1 ∗ h2 ) ∗ h1 )n
((δ [k] − aδ [k − 1]) ∗ h2 [k])n
= h2 [n] − ah2 [n − 1]
= an u [n] − a · an−1 u [n − 1]
= an (u [n] − u [n − 1])
= an · δ [n]
= δ [n]
⇒ y [n]
=
((x1 ∗ h2 ) ∗ h1 )n
=
(δ ∗ h1 )n
= h1 [n]
=
cos (6n)
Correct Answer: 3
5
For some signal x (t) we dene the following:
g (t) = x (t) cos (t)
g (t) Fourier transform is given by:
(
1 |ω| ≤ 2
G (ω) =
0 else
Which from the following can be x (t):
1. x (t) = π2 sinc (2t)
2. x (t) = π1 sinc (2t)
3. x (t) = π2 sinc (t)
4. x (t) =
1
2π sinc (2t)
5. x (t) = π1 sinc (t)
Answer:
Notice that for x (t) = π2 sinc (t):
(
F {x (t)} =
2 |ω| ≤ 1
0 else
Thus:
F {x (t) cos (t)}
1
[X (ω − 1) + X (ω + 1)]
2
(
1 |ω| ≤ 2
=
0 else
=
Correct Answer: 3
6
For some IOM (input output map) we know that for the input signal x (t) = ej 8 t the output is y (t) = cos
Which of the following statements in correct:
π
π
8t
.
1. The system is not LTI.
2. This is an LTI system with frequency response H (ω) =
1 + δ ω + 2 · π8 .
3. This is an LTI system with frequency response H (ω) = 12 δ (ω) + δ ω + 2 · π8 .
4. This is an LTI system with frequency response H (ω) = 12 1 + δ ω − 2 · π8 .
1
2
5. If H (ω) is an odd function H (ω) = −H (ω) then the system is LTI.
Answer:
The output of an LTI system is Y F (ω) = H F (ω) · X F (ω),
which means that X F (ω) = 0 ⇒ Y F (ω) = 0.
Thus, no new frequency can be added when a signal pass through the system.
π
π
Since x (t) comprise only from the frequency ej 8 t , and y (t) contains also the frequency e−j 8 t the system is not an LTI.
Correct Answer: 1
7
Given the real and periodic signal x (t) with T0 = 4, and its Fourier series coecients ak .
The following fact are also given:
1. ak = 0 for all |k| > 1
2. The signal y (t) =
∞
P
ak e−j 2 k ejω0 kt is odd (y (t) = −y (−t)).
π
k=−∞
3.
2́
2
|x (t)| dt = 4
−6
Find x (t) up to its sign.
Answer:
ω0 =
x (t) =
∞
X
2π
π
=
4
2
ak ejω0 kt = a−1 e−jω0 t + a0 + a1 ejω0 t
k=−∞
π
π
π
= a−1 · ej 2 · e−j 2 t + a0 + a1 e−j 2 ejω0 t
y (t)
= ja−1 · e−j
π
2t
+ a0 − ja1 ejω0 t
y (t) = −y (−t)
π
π
⇒ ja−1 · e−j 2 t + a0 − ja1 ejω0 t = −ja−1 · ej 2 t − a0 + ja1 e−jω0 t
(
⇒
a0 = −a0 = 0
a1 = a−1
π
π
⇒ x (t) = a1 ej 2 t + e−j 2 t
Since x (t + 4) = x (t)
ˆ2
ˆ6
2
2
|x (t)| dt =
−2
|x (t)| dt
2
ˆ2
2
⇒
|x (t)| dt =
2
−2
Parseval:
1
T0
ˆ2
2
|x (t)| dt =
1
2
2
2
2
= |a1 | + |a0
| + |a−1 | = 2 |a1 |
2
−2
2
⇒ |a1 | =
1
4
Since x (t) is both real and even (a1 = a−1 ), a1 is real as well.
⇒ a1 = ±
⇒ x (t) = ±
Correct Answer: 3
1
2
π π 1 −j π t
e 2 + ej 2 t = ± cos
t
2
2
8
x [n] = h [n] = αn u [n]
y [n] = (x ∗ h)n
Find y [n].
Answer:
y [n]
∞
X
=
αm u [m] αn−m u [n − m]
m=−∞
∞
X
n
= α
u [n − m]
m=0
n
= u [n] α (n + 1)
Correct Answer: 3
9
Given the signal x (t) with its Fourier transform X (ω).
Find the Fourier transform of:
1
1
y (t) = x (t) + x t −
∗
2
jt
Answer:
Duality:
F
1
jt
=
1
F
2
2
jt
=
1
· 2π sign (−ω) = π sign (−ω)
2
Translation:
1
1
= X (ω) e−j 2 ω
x t−
2
1
1
1
⇒F x t−
∗
= π sign (−ω) X (ω) e−j 2 ω
2
jt
F
F {y (t)} = X (ω) + π sign (−ω) X (ω) e−j 2 ω
1
Correct Answer: 4
10
Given the periodic signal x (t), x (t + T0 ) = x (t).
The Fourier series coecients are cl .
We dene the following signal:
y (t) = 2x (t − 1) − 1
Find the Fourier series coecients of y (t).
Answer:
y (t)
=
=
2x (t − 1) − 1
∞
X
2π
2
cl ej T l(t−1) − 1
l=−∞
=
∞
X
l=−∞
2π
2π
2cl e−j T l · ej T
lt
−1
so the Fourier series coecients bl of y (t) are:
(
2π
2cl e−j T l − 1 l = 0
2π
⇒ bl =
= 2cl e−j T l − δ [l]
l
−j 2π
T
l=
6 0
2cl e
Correct Answer: 1