Exam I: Solutions
BME 171-02, Signals and Systems
Exam I: Solutions
100 points total
1. (15 pts.) Fill in the following table (for each column, give a “yes/no” answer and briefly
justify it):
System
Linear
Time-invariant
Causal
yes
no
yes
x2 (t)
no
yes
yes
+1, x[n] ≥ 0
−1, x[n] < 0
no
yes
yes
x(λ)dλ
yes
yes
no
y[n] = 2(x[n + 1]u[n] − x[n]) + 1
no
no
no
y(t) = 3x(t) cos(t)
y(t) =
p
y[n] =
(
y(t) =
Z
t+1
t
Exam I: Solutions
2. (15 pts.) Consider the following system:
x[n]
z[n]
+
2
_
y[n]
F
Δ
Here, the system F is defined by the input-output relationship
n
o
F z[n] = z[n] − z[n − 1],
and ∆ is the unit delay
n
o
∆ w[n] = w[n − 1].
Write down the linear difference equation describing this system.
Solution.
Let z[n] be the output of the summer, as shown above. Then
n
o
y[n] = F z[n] = z[n] − z[n − 1].
Now,
n
o
z[n] = 2x[n] − ∆ y[n] = 2x[n] − y[n − 1].
Therefore, substituting the expression for z[n] into the first equation, we can write
y[n] = z[n] − z[n − 1]
=
2x[n] − y[n − 1] − 2x[n − 1] − y[n − 2]
|
{z
}
=z[n]
|
{z
=z[n−1]
= 2x[n] − y[n − 1] − 2x[n − 1] + y[n − 2].
Simplify to get
y[n] + y[n − 1] − y[n − 2] = 2x[n] − 2x[n − 1]
2
}
Exam I: Solutions
3. (20 pts.) Consider the following circuit:
R1
L
+
i2(t)
i1(t)
i0(t)
x(t)
y(t)
R2
C
_
+
_
Write down the input-output differential equation for this circuit in terms of the input voltage
x(t) and the output voltage y(t).
Solution.
Let
i0 (t) =
current through R1
i1 (t) =
current through C
i2 (t) =
current through R2
Then, using Kirchhoff’s voltage law, we write
−x(t) + R1 i0 (t) + L
di0 (t)
+ y(t) = 0.
dt
On the other hand, using Kirchhoff’s current law, we have
i0 (t) = i1 (t) + i2 (t).
Moreover,
i1 (t) = C
dy(t)
dt
and
i2 (t) =
y(t)
.
R2
Therefore,
i0 (t) = C
dy(t) y(t)
+
dt
R2
and
di0 (t)
d2 y(t)
1 dy(t)
=C
+
.
2
dt
dt
R2 dt
Substituting this into the KVL equation, we get
−x(t) + R1
dy(t) y(t)
+
C
dt
R2
1 dy(t)
d2 y(t)
+
+L C
2
dt
R2 dt
!
+ y(t) = 0.
Simplify to get the final answer:
2
y(t)
LC d dt
+ R1 C +
L
R2
3
dy(t)
dt
+
R1
R2
+ 1 y(t) = x(t).
Exam I: Solutions
4. (25 pts.) Compute the convolution of the following two signals and plot the result on the set
of axes provided. Show all your work!
x[n]
ν[n]
1
1
n
0
n
0
-1
x[n] * ν[n]
1
n
0
Solution. First, rename the time variable n into k. Next, flip and shift one of the signals.
We will flip ν[k] to get ν[−k]:
ν[-k]
x[k]
1
1
k
0
k
0
-1
Now, shift by n to get ν[n − k]. Note that there is no overlap between x[k] and ν[n − k] as
long as n ≤ −2 or n ≥ 4:
ν[-2-k]
x[k]
1
1
k
0
k
0
-1
ν[4-k]
x[k]
1
1
k
0
-1
4
0
k
Name: ..................................................
So, y[n] = x[n] ⋆ ν[n] = 0 for n ≤ −2 and for n ≥ 4
Now, shift and compute the overlap:
y[−1] = x[−1]ν[0] + x[0]ν[−1] + x[1]ν[−2]
= (−1) · 1 + 0 · 0 + 1 · 0
= −1
y[0] = x[−1]ν[1] + x[0]ν[0] + x[1]ν[−1]
= (−1) · 1 + 0 · 1 + 1 · 0
= −1
y[1] = x[−1]ν[2] + x[0]ν[1] + x[1]ν[0]
= (−1) · 1 + 0 · 1 + 1 · 1
= 0
y[2] = x[−1]ν[3] + x[0]ν[2] + x[1]ν[1]
= (−1) · 0 + 0 · 1 + 1 · 1
= 1
y[3] = x[−1]ν[4] + x[0]ν[3] + x[1]ν[2]
= (−1) · 0 + 0 · 0 + 1 · 1
= 1
Overall,
y[n] = x[n] ⋆ ν[n] =

0,





 −1,
0,



1,



0,
5
n ≤ −2
n = −1, 0
n=1
n = 2, 3
n≥4
Exam I: Solutions
5. (25 pts.) Compute the convolution of the following two signals. Write down its analytical
form and sketch its plot in the set of axes provided. Show all your work!
ν(t)
x(t)
2
1
t
1
0
1
0
t
-2
x(t)* ν(t)
1
-1
0 1
-1
2
t
Solution. First, rename the time variable t into λ, say. Next, flip and shift one of the
signals. We will flip x(λ) to get x(−λ):
ν(λ)
x(-λ)
2
1
1
0
λ
0
1
λ
-2
Now, shift by t to get x(t − λ). Note that there is no overlap between x(t − λ) and ν(λ) as
long as t ≤ −1 or t ≥ 2:
ν(λ)
x(-1-λ)
2
1
0
1
λ
0
1
λ
1
λ
-2
ν(λ)
x(2-λ)
2
1
0
1
λ
-2
6
0
Exam I: Solutions
So, y(t) = x(t) ⋆ ν(t) = 0 when t ≤ −1 or when t ≥ 2.
Now, we consider the remaining cases:
−1 ≤ t ≤ 0:
2
1
t-1 t 0 t+1
λ
y(t) =
Z
t+1
h
it+1
it
− λ2
(−2) · λdλ = − λ2
0
0
= −(t + 1)2
0 ≤ t ≤ 1:
2
t-1 0 t 1 t+1
y(t) =
Z
t
2 · λdλ +
0
λ
Z
1
h
(−2) · λdλ = λ2
t
0
h
i1
t
= t2 − 1 + t2 = 2t2 − 1
1 ≤ t ≤ 2:
2
1
0 t-1
y(t) =
t
Z
t+1
1
λ
h
2 · λdλ = λ2
t−1
i1
t−1
= 1 − (t − 1)2 = −t2 + 2t
Overall,
y(t) = x(t) ⋆ ν(t) =

0,




2

 −(t + 1) ,
2t2 − 1,



−t2 + 2t,



0,
t ≤ −1
−1 ≤ t ≤ 0
0≤t≤1
1≤t≤2
t≥2
7