Signals & Systems · Problem type

Energy and power of a signal

Use this for “Find the energy of x(t) = 2e⁻⁴ᵗu(t) and the average power of 2 cos(10t + 30°).”

Chart

x²(t) = 22·e−2·4·t00.10.20.30.40.50.60.700.511.522.533.544.5Time t (s)Power x² (W)∫ = 0.4988
Shaded: ∫ from 0 to 0.75 of x²(t) dt = 0.4988
This is an example. Type your own values to start a new problem.
E (J)
=
(
A (V)
)
2
÷
(2
×
a (s⁻¹)
)
P (W)
=
(
A (V)
)
2
÷
2

Formulas

E
=
A
2
2 ×
a
0.5
=
2
2
2 × 4
P
=
A
2
2
2
=
2
2
2

Assumptions

•
Energy and power are normalized: the signal is a voltage across 1 Ω, so the instantaneous power is x(t)².
•
For Ae−atu(t) with a > 0, E = ∫ A²e−2at dt from 0 to ∞ = A² ÷ (2a); its average power is 0.
•
For A cos(ωt + φ), cos² averages ½ over a period, so P = A² ÷ 2 for any ω and φ; its energy is infinite.

Step-by-step

WE KNOW
A = 2 V, a = 4 s⁻¹
FIND
pulse energy (E), sinusoid power (P)
Step 1 · Find pulse energy (E)
Use E = A² ÷ (2 × a)
x
2
=
A
2
e
−2at
starts
at
A
2
and
falls
with
rate
2
a
,
so
the
area
under
it
is
A
2
2
a
.
E
=
2
2
2 × 4
E
=
2
2
8
E
=
4
8
E
=
0.5
J
Step 2 · Find sinusoid power (P)
Use P = A² ÷ 2
x
2
=
A
2
cos²(ωt
+
φ)
swings
between
0
and
A
2
,
averaging
half
of
A
2
.
P
=
2
2
2
P
=
4
2
P
=
2
W
Check
Put the numbers back in. Both sides should match.
0.5
=
2
2
2 × 4
✓
2
=
2
2
2
✓
Topic check
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