SME
Signals & Systems · Problem type
A step into a first-order system
Use this for “Convolve u(t) with h(t) = e⁻²ᵗu(t). Find y(t) at t = 1 s.”
Chart
y(t) = 0.5(1 − exp(−t/τ)), τ = 0.5 s
After τ the response has made 63.2% of its change; after 5τ, 99.3%.
At t = 1 s: y = 0.4323 s.
This is an example. Type your own values to start a new problem.
τ (s)
=
1
÷
a (s⁻¹)
y∞ (s)
=
1
÷
a (s⁻¹)
y (s)
=
(1
÷
a (s⁻¹)
)
×
(1
−
e
−
s⁻¹
×
s
)
Formulas
τ
=
1
a
0.5
=
1
2
y∞
=
1
a
0.5
=
1
2
y
=
(
1
a
)
×
(1
−
e
−
a
×
t
)
0.4323
=
(
1
2
)
×
(1
−
e
−2 × 1
)
Assumptions
•
x(t) = u(t) and h(t) = e−atu(t) are pure numbers, so y, an area under h, is in seconds.
•
The area under e−aτ from τ = 0 to t gives y(t) = (1 ÷ a)(1 − e−at) for t ≥ 0.
•
y climbs toward 1 ÷ a with time constant τ = 1 ÷ a: 63.2% of the way after one τ.
•
y only creeps toward 1 ÷ a; after about 5τ it shows as 1 ÷ a.
Step-by-step
WE KNOW
a = 2 s⁻¹, t = 1 s
FIND
time constant (τ), final output (area under h) (y∞), output (area under h) (y)
Step 1 · Find time constant (τ)
Use τ = 1 ÷ a
The
exponential
falls
to
1
e
in
one
τ
=
1
a
.
τ
=
1
2
τ
=
0.5
s
Step 2 · Find final output (area under h): y∞
Use y∞ = 1 ÷ a
Long
after
the
step,
y
is
the
whole
area
under
e
−at
:
1
a
.
y∞
=
1
2
y∞ = 0.5 s
Step 3 · Find output (area under h): y
Use y = (1 ÷ a) × (1 − e−a × t)
Add
up
the
area
under
h
from
0
to
t
:
the
share
of
the
final
value
reached
so
far.
y
=
(
1
2
)
×
(1
−
e
−2 × 1
)
y
=
1
2
×
(1
−
0.1353)
y
=
1
2
×
0.8647
y
=
0.5
×
0.8647
y
=
0.4323
s
Check
Put the numbers back in. Both sides should match.
0.5
=
1
2
✓
0.5
=
1
2
✓
0.4323
=
(
1
2
)
×
(1
−
e
−2 × 1
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.