SME
Signals & Systems · Problem type
Final and initial value
Use this for “Find the final value of x(t) when X(s) = 10 ÷ (s(s + 5)).”
Chart
x(t) = 2(1 − exp(−t/τ)), τ = 0.2 s
After τ the response has made 63.2% of its change; after 5τ, 99.3%.
At t = 0.2 s: x = 1.264.
This is an example. Type your own values to start a new problem.
X(s)
=
K
s(s
+
)
x(∞)
=
K
÷
a
=
τ
Time constant
worked out (not typed)
t
Time
example
x
Signal at t
calculated
Formulas
x
(∞)
=
K
a
2
=
10
5
τ
=
1
a
0.2
=
1
5
x
=
x
(∞)
×
(1
−
e
−
t
÷
τ
)
1.2642
=
2
×
(1
−
e
−0.2 ÷ 0.2
)
Assumptions
•
X(s) = K ÷ (s(s + a)), with a > 0: one pole at s = 0 and one at −a, left of the jω axis.
•
Final value theorem: x(∞) = lim sX(s) as s → 0. It holds only when every other pole is left of the jω axis.
•
Initial value theorem: x(0⁺) = lim sX(s) as s → ∞, which is K ÷ (s + a) → 0 here, so x starts at 0.
•
Partial fractions give x(t) = (K ÷ a)(1 − e−at), with time constant τ = 1 ÷ a.
Step-by-step
WE KNOW
K = 10, a = 5 s⁻¹, t = 0.2 s
FIND
final value (x(∞)), time constant (τ), signal at t (x)
Step 1 · Find final value (x(∞))
Use x(∞) = K ÷ a
sX(s)
=
K
s +
a
;
put
s
=
0,
since
the
theorem
takes
the
limit
as
s
→
0.
x
(∞)
=
10
5
x
(∞)
=
2
Step 2 · Find time constant (τ)
Use τ = 1 ÷ a
The
pole
at
−a
decays
as
e
−at
=
e
−
t
τ
,
so
τ
=
1
a
.
τ
=
1
5
τ
=
0.2
s
Step 3 · Find signal at t (x)
Use x = x(∞) × (1 − e−t ÷ τ)
The
residues
are
K
a
at
s
=
0
and
−
K
a
at
−
a
:
a
step
that
climbs
to
x
(∞).
x
=
2
×
(1
−
e
−0.2 ÷ 0.2
)
x
=
2
×
(1
−
e
-1
)
x
=
2
×
(1
−
0.3679)
x
=
2
×
0.6321
x
=
1.2642
Check
Put the numbers back in. Both sides should match.
2
=
10
5
✓
0.2
=
1
5
✓
1.2642
=
2
×
(1
−
e
−0.2 ÷ 0.2
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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