SME
Communication Systems · Problem type
FM bandwidth by Carson’s rule
Use this for “Broadcast FM swings 75 kHz with a 15 kHz tone. Find β and the bandwidth.”
Diagram
β
=
Δf
fₘ
=
75
15
=
5.
Carson’s rule: B = 2(Δf + fₘ) = 2 × (75 + 15) = 180 kHz, the lines out to ±(β + 1)fₘ.
The carrier’s line is |J₀(5)| = 0.178 of the unmodulated carrier; the rest of the power moved into the sidebands.
The carrier is drawn 11 cycles to each message cycle, far slower than it is.
This is an example. Type your own values to start a new problem.
β
=
Δf (kHz)
fₘ (kHz)
B (kHz)
=
2(
Δf (kHz)
+
fₘ (kHz)
)
Formulas
β
=
Δf
fₘ
5
=
75
15
B
=
2
×
(Δf
+
fₘ)
180 = 2 × (75 + 15)
Assumptions
•
Tone modulation: one message frequency fₘ, a cosine.
•
Carson’s rule: about 98% of the power lies within (β + 1)fₘ of the carrier.
•
The lines sit every fₘ, each |Jₙ(β)| of the unmodulated carrier’s amplitude.
Step-by-step
WE KNOW
Δf = 75 kHz, fₘ = 15 kHz
FIND
modulation index (β), Carson bandwidth (B)
Step 1 · Find modulation index (β)
Use β = Δf ÷ fₘ
The swing of the frequency in units of the message frequency.
β
=
75
15
β
=
5
Step 2 · Find Carson bandwidth (B)
Use B = 2 × (Δf + fₘ)
Carson’s rule: the deviation plus one more message frequency, on each side of the carrier.
B
=
2
×
(75
+
15)
B
=
2
×
90
B
=
180
kHz
Check
Put the numbers back in. Both sides should match.
5
=
75
15
✓
180 = 2 × (75 + 15) ✓
Topic check
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