SME
Communication Systems · Problem type
Modulation index from the envelope
Use this for “An AM envelope peaks at 1.5 V and dips to 0.5 V. Find the modulation index.”
Diagram
μ
=
Aₘₐₓ − Aₘᵢₙ
Aₘₐₓ + Aₘᵢₙ
=
1.5 − 0.5
1.5 + 0.5
μ = 0.5
Ac
=
Aₘₐₓ + Aₘᵢₙ
2
=
1
V,
the
middle
of
the
envelope.
Each
sideband
is
μAc
2
=
0.25
V;
the
band
is
2
×
fₘ
wide
whatever
μ
is.
The carrier is drawn 9 cycles to each message cycle, far slower than it is.
This is an example. Type your own values to start a new problem.
μ
=
Aₘₐₓ (V)
−
Aₘᵢₙ (V)
Aₘₐₓ (V)
+
Aₘᵢₙ (V)
Ac
Carrier amplitude
worked out (not typed)
Aₘ
Message amplitude
worked out (not typed)
Formulas
μ
=
Aₘₐₓ − Aₘᵢₙ
Aₘₐₓ + Aₘᵢₙ
0.5
=
1.5 − 0.5
1.5 + 0.5
Ac
=
Aₘₐₓ + Aₘᵢₙ
2
1
=
1.5 + 0.5
2
Aₘ
=
Aₘₐₓ − Aₘᵢₙ
2
0.5
=
1.5 − 0.5
2
Assumptions
•
Tone modulation: the envelope swings between Aₘₐₓ and Aₘᵢₙ as Ac(1 + μ cos 2πfₘₜ).
•
The envelope never reaches zero below 100% modulation, so μ ≤ 1 and Aₘᵢₙ ≥ 0.
•
Both amplitudes are peak values of the envelope, read as heights above zero.
Step-by-step
WE KNOW
Aₘₐₓ = 1.5 V, Aₘᵢₙ = 0.5 V
FIND
modulation index (μ), carrier amplitude (Ac), message amplitude (Aₘ)
Step 1 · Find modulation index (μ)
Use μ = (Aₘₐₓ − Aₘᵢₙ) ÷ (Aₘₐₓ + Aₘᵢₙ)
The envelope swings μAc either side of Ac: half the swing over half the sum.
μ
=
1.5 − 0.5
1.5 + 0.5
μ
=
1
2
μ
=
0.5
Step 2 · Find carrier amplitude (Ac)
Use Ac = (Aₘₐₓ + Aₘᵢₙ) ÷ 2
The carrier’s amplitude is the middle of the envelope.
Ac
=
1.5 + 0.5
2
Ac
=
2
2
Ac = 1 V
Step 3 · Find message amplitude (Aₘ)
Use Aₘ = (Aₘₐₓ − Aₘᵢₙ) ÷ 2
The
message
moves
the
envelope
this
far
either
side
of
the
middle;
μ
=
Aₘ
Ac
.
Aₘ
=
1.5 − 0.5
2
Aₘ
=
1
2
Aₘ = 0.5 V
Check
Put the numbers back in. Both sides should match.
0.5
=
1.5 − 0.5
1.5 + 0.5
✓
1
=
1.5 + 0.5
2
✓
0.5
=
1.5 − 0.5
2
✓
Topic check
See what you know: 5 questions on this topic
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