SME
Signals & Systems · Problem type
Two rectangular pulses
Use this for “Convolve a 2 V pulse lasting 3 s with h(t) = 0.5 s⁻¹ for 1 s. Find the length and peak of y(t).”
Chart
This is an example. Type your own values to start a new problem.
L (s)
=
w₁ (s)
+
w₂ (s)
F (s)
=
w₁ (s)
−
w₂ (s)
yₘₐₓ (V)
=
A (V)
×
B (s⁻¹)
×
w₂ (s)
x(t) on the chart is a pulse A tall lasting w₁; h is B tall lasting w₂. y climbs until t = w₂, stays flat for F, then falls to 0 at t = L.
Formulas
L
=
w₁
+
w₂
4 = 3 + 1
F
=
w₁
−
w₂
2 = 3 − 1
yₘₐₓ
=
A
×
B
×
w₂
1 = 2 × 0.5 × 1
Assumptions
•
x(t) = A from 0 to w₁ and h(t) = B from 0 to w₂, with w₂ ≤ w₁; both are 0 elsewhere.
•
As h slides across x, y ramps up while the overlap grows, stays flat while h lies inside x, then ramps down.
•
h is in s⁻¹, so the area under x(τ)h(t − τ), which is y(t), is in volts like x.
Step-by-step
WE KNOW
A = 2 V, w₁ = 3 s, B = 0.5 s⁻¹, w₂ = 1 s
FIND
length of y (L), flat top of y (F), peak of y (yₘₐₓ)
Step 1 · Find length of y (L)
Use L = w₁ + w₂
The overlap begins at t = 0 and ends when the start of h passes the end of x.
L
=
3
+
1
L
=
4
s
Step 2 · Find flat top of y (F)
Use F = w₁ − w₂
h
lies
wholly
inside
x
from
t
=
w₂
to
t
=
w₁
,
so
the
top
stays
flat
that
long.
F
=
3
−
1
F
=
2
s
Step 3 · Find peak of y (yₘₐₓ)
Use yₘₐₓ = A × B × w₂
While
h
lies
inside
x,
the
overlap
is
all
of
h:
height
A
×
B
over
the
width
w₂
.
yₘₐₓ = 2 × 0.5 × 1
yₘₐₓ = 1 V
Check
Put the numbers back in. Both sides should match.
4 = 3 + 1 ✓
2 = 3 − 1 ✓
1 = 2 × 0.5 × 1 ✓
Topic check
See what you know: 5 questions on this topic
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