EE 242 — Signals, Systems, and Data I
Assignment 1 Solutions
Total Points: 100
Autumn Quarter, 2025
Department of Electrical and Computer Engineering
University of Washington, Seattle, WA
Posted on October 2, 2025
Was due by 11:59 pm (PT) on October 1, 2025 via Canvas
Submission & Format
• This assignment is hand-computation only. You may submit either:
– Handwritten solutions (clearly written), scanned to PDF, or
– Typed solutions (Word or LATEX), exported to PDF.
• Submit a single PDF file to Canvas titled YourFirstName YourLastName EE242 A1.pdf.
• Show all steps for full credit. Box or clearly highlight final answers and include units where applicable.
• You may discuss concepts with classmates, but all work must be your own. List any collaborators
or resources you consulted.
• Figures may be neatly hand-drawn or generated and then embedded in your PDF. Label all axes and
key values.
Grading Notes
• Partial credit is awarded based on the clarity and correctness of intermediate steps.
• If a result depends on a parameter/assumption, state it explicitly and proceed.
• Unless otherwise stated, assume signals are real-valued and systems are LTI.
Academic Integrity
All submissions must comply with the UW Student Conduct Code and the course academic integrity policy
posted on Canvas.
1
Question 1 (10 × 2 = 20 points): Time-Axis Transformations (CT)
A continuous-time signal x(t) is shown in Fig. Q1 of the handout. A convenient piecewise description
consistent with the figure is


0,
t < −2,






t + 2, −2 ≤ t ≤ −1,
x(t) =
−t,




t,




1,
−1 ≤ t ≤ 0,
with x(t) = 0 for t > 2 on the right-hand segment (1, 2] via x(t) = 2 − t.
0 ≤ t ≤ 1,
t ≥ 1,
(a) y(t) = x(2t + 1).
Let u = 2t + 1. Features of x(u) occurring at u = t0 appear in y(t) at
t=
t0 − 1
2
(time shift left by 0.5 and compression by factor 2).
The support of x on [−2, 2] maps to
2t + 1 ∈ [−2, 2] ⇐⇒ t ∈
h −3 1 i
, .
2 2
Within this interval, substitute u = 2t + 1 into the pieces of x(u):


t < − 23 ,
0,



(2t + 1) + 2 = 2t + 3, − 3 ≤ t ≤ −1,

2



−(2t + 1) = −2t − 1, −1 ≤ t ≤ − 1 ,
2
y(t) = x(2t + 1) =
1

≤
t
≤
0,
2t
+
1,
−

2





1,
0
≤
t ≤ 12 ,



0,
t > 1.
2
Key checkpoints:
t = − 32 7→ y = 0,
t = −1 7→ y = 1,
t = − 12 7→ y = 0,
t = 0 7→ y = 1,
The waveform is a compressed/shifted version of x(t) confined to − 32 , 12 .
t = 21 7→ y = 1.
t
(b) y(t) = x 4 −
.
2
t
Let u = 4 − 2 . This is a time-reversal and expansion by 2 with a shift:
u=4−
t
2
⇒
t = 8 − 2u.
Support mapping:
u ∈ [−2, 2] ⇐⇒ t ∈ [4, 12].
As t increases from 4 to 12, u decreases from 2 to −2 (time-reversal). Using u = 4 − 2t in each piece:
2
y(t) = x 4 − 2t =


0,





1,




2 − t ,
t < 4,
4≤t≤8
(comes from u ≥ 1),
8 ≤ t ≤ 10 (u ∈ [0, 1]),
2
t

10 ≤ t ≤ 11 (u ∈ [−1, 0]),

2 − 9,



t


− 2 + 11, 11 ≤ t ≤ 12 (u ∈ [−2, −1]),



0,
t > 12.
Checkpoints:
t = 4 7→ y = 1,
t = 8 7→ y = 1,
t = 10 7→ y = 1,
t = 11 7→ y = 0,
t = 12 7→ y = 0.
Question 2 (10 × 2 = 20 points): Index Transformations (DT)
The given sequence (Fig. Q2 in the handout) has the nonzero samples
x[−4] = −1, x[−3] = 2, x[−2] = 2, x[−1] = 1, x[0] = 1, x[1] = 2, x[2] = 1, x[3] = −1,
and x[n] = 0 elsewhere.
(a) y[n] = x[3n + 1].
We need indices k = 3n+1 that hit nonzero locations of x[k]. Solve 3n+1 ∈ {−4, −3, −2, −1, 0, 1, 2, 3}:
k
n = (k − 1)/3
−4
−5/3
−3
−4/3
−2
−1
−1
−2/3
0
−1/3
1
0
2
1/3
3
2/3
Only k = −2 and k = 1 yield integer n values (n = −1, 0), hence
y[−1] = x[−2] = 2,
y[0] = x[1] = 2,
y[n] = 0 otherwise.
Figure 1: Solution Sketch for Question 2(a)
(b) y[n] = x[(n − 1)2 ].
Let m = (n−1)2 . Since m is a perfect square, within the nonzero support we only need m ∈ {0, 1, 4}:
(n − 1)2 = 0 ⇒ n = 1
⇒
y[1] = x[0] = 1,
2
(n − 1) = 1 ⇒ n = 0 or 2 ⇒ y[0] = x[1] = 2, y[2] = x[1] = 2,
(n − 1)2 = 4 ⇒ n = −1 or 3 ⇒ y[−1] = x[4] = 0, y[3] = x[4] = 0.
3
Therefore,
y[0] = 2,
y[1] = 1,
y[2] = 2,
and y[n] = 0 for all other n.
Figure 2: Solution Sketch for Question 2(b)
Question 3 (10 × 2 = 20 points): Even/Odd Decomposition (CT)
Given the x(t) in Fig. Q3 of the handout, a consistent analytic model is


0,
t ≤ −2,





t + 2, −2 ≤ t ≤ −1,
x(t) = −t,
−1 ≤ t ≤ 0,


t,
0 ≤ t ≤ 1,



1,
t ≥ 1.
x(t) + x(−t)
.
2
Compute by regions (use symmetry; results are even in t):

1

,
|t| ≥ 2,



2

xe (t) = 3 − |t| , 1 ≤ |t| ≤ 2,


2



|t|,
|t| ≤ 1.
(a) xe (t) =
Explanation/sketch guide:
• For |t| ≤ 1: x(t) = |t| and x(−t) = |t| ⇒ xe = |t|.
• For 1 ≤ |t| ≤ 2: one side equals 1 (the flat segment), the other is linear to 0; averaging gives
(3 − |t|)/2.
• For |t| ≥ 2: one side is 1 and the other is 0, hence xe = 21 .
4
Figure 3: Solution Sketch for Question 3(a)
x(t) − x(−t)
.
2
Results (odd in t):
(b) xo (t) =

1


,
t ≥ 2,


2



t
−
1


, 1 ≤ t ≤ 2,


2

−1 ≤ t ≤ 1,
xo (t) = 0,



t
+
1


,
−2 ≤ t ≤ −1,


2




− 1 ,
t ≤ −2.
2
(You can verify x(t) = xe (t) + xo (t) and xe even, xo odd.)
Figure 4: Solution Sketch for Question 3(b)
Question 4 (10 × 2 = 20 points): Even/Odd Decomposition (DT)
For the sequence in Fig. Q4, the nonzero samples are
x[−4] = −1, x[−3] = 2, x[−2] = 2, x[−1] = 1, x[0] = 1, x[1] = 2, x[2] = 1, x[3] = −1.
(a) xe [n] =
x[n] + x[−n]
(even).
2
5
Pair samples at ±n:
n x[n]
0
1
1
2
2
1
3 −1
4
0
x[−n]
1
1
2
2
−1
xe [n]
1
3
2
3
2
1
2
− 12
Symmetry gives xe [−n] = xe [n], and xe [n] = 0 for |n| ≥ 5.
xe [n]
2
xo [n]
3
2
3
2
1
1
1
2
n
0
-4
-3
-2
-1
0
1
2
3
n
0
4
-4
-3
(a) xe [n]
−1
-2
-1− 1 0
2
1
2
3
4
−1
3
−
(b)
2 xo [n]
Solution sketches for Q4.
(b) xo [n] =
n
0
1
2
3
4
x[n] − x[−n]
(odd).
2
x[n]
1
2
1
−1
0
x[−n]
1
1
2
2
−1
xo [n]
0
1
2
− 12
− 32
1
2
Odd symmetry: xo [−n] = −xo [n], and xo [n] = 0 for |n| ≥ 5.
Finally, confirm x[n] = xe [n] + xo [n] for all n.
Question 5 (2.5 × 4 = 10 points): Periodicity and Time-Scaling (CT)
Let y1 (t) = x(2t) and y2 (t) = x 2t . Suppose x has fundamental period Tx (if periodic).
(1) True. If x is periodic with Tx , then
y1 (t + Tx /2) = x 2(t + Tx /2) = x(2t + Tx ) = x(2t) = y1 (t).
Thus Ty1 = Tx /2.
(2) True. If y1 is periodic with Ty1 , then
x(u + 2Ty1 ) = x(u)
so x is periodic with Tx = 2Ty1 .
6
(set u = 2t),
(3) True. If x is periodic with Tx , then
t
t
= y2 (t),
y2 (t + 2Tx ) = x + Tx = x
2
2
so Ty2 = 2Tx .
(4) True. If y2 is periodic with Ty2 , then
x(u + Ty2 /2) = x(u)
(set u = t/2),
so x is periodic with Tx = Ty2 /2.
Question 6 (5 × 2 = 10 points): Periodicity of DT Signals
(a) x[n] = cos
π
n2 .
8
Check if there exists a nonzero integer N such that x[n + N ] = x[n] for all n:
x[n + N ] = cos π8 (n + N )2 = cos π8 n2 + π8 (2nN + N 2 ) .
We need π8 (2nN + N 2 ) = 2πkn for all integers n, i.e.,
πN 2
πN
n+
= 2πkn .
4
8
For the term linear in n to be an integer multiple of 2π for all n, we must have πN
4 = 2πL for some
integer L, i.e. N = 8L. With N = 8, the phase increment becomes
π
(16n + 64) = 2πn + 8π,
8
which is an integer multiple of 2π. Thus x[n + 8] = x[n] for all n. Smaller N fail (e.g. N = 4 leaves a
residual πn term). Hence the fundamental period is N0 = 8.
π π (b) x[n] = cos n cos n .
2
4
Use the product-to-sum identity:
h
i
π
x[n] = 12 cos 3π
n
+
cos
.
4
4n
A DT cosine cos(ω0 n) is periodic iff ω0 /2π is rational, with period N the smallest positive integer
satisfying ω0 N = 2πm. For ω0 = π/4, the least N is 8. For ω0 = 3π/4, the least N satisfying
(3π/4)N = 2πm is also N = 8 (take m = 3). Therefore the fundamental period is N0 = lcm(8, 8) = 8.
7