SME
Signals & Systems
Fourier series and transforms
Use this for “Find the amplitude of the third harmonic of a ±1 V, 1 kHz square wave.”
Chart
Square
wave
of
amplitude
1
(dashed)
and
its
sum
through
harmonic
3
(solid):
the
peak
1.2,
10%
of
the
jump
past
it
(Gibbs).
b₃
=
4A
kπ
=
4
×
1
3π
=
0.4244
V.
f₃
=
3
×
1
kHz
=
3
kHz.
Its
share
of
the
power:
b₃
2
÷ 2
A
2
=
9.006%.
This is an example. Type your own values to start a new problem.
fₖ (kHz)
=
k
×
f₀ (kHz)
bₖ (V)
=
4
×
A (V)
÷
(
k
×
π)
share
Share of the power
calculated · %
Formulas
fₖ
=
k
×
f₀
3 = 3 × 1
bₖ
=
4
×
A
k
× π
0.4244
=
4
×
1
3 × π
share
=
100
×
8
k
2
×
π
2
9.0063
=
100
×
8
3
2
×
π
2
Assumptions
•
An odd square wave between +A and −A, so its series has only odd sine terms, bₖ sin(2πkf₀t).
•
Parseval: the harmonics’ powers bₖ² ÷ 2 add to A², so each share is (bₖ² ÷ 2) ÷ A².
•
The fundamental holds 81.1% of the power; any partial sum still overshoots each jump by about 9% (Gibbs).
Step-by-step
WE KNOW
A = 1 V, f₀ = 1 kHz, k = 3
FIND
harmonic frequency (fₖ), harmonic amplitude (bₖ), share of the power (share)
Step 1 · Find harmonic frequency (fₖ)
Use fₖ = k × f₀
The
kth
harmonic
repeats
k
times
in
each
period
of
the
wave.
fₖ = 3 × 1
fₖ = 3 kHz
Step 2 · Find harmonic amplitude (bₖ)
Use bₖ = 4 × A ÷ (k × π)
Each
odd
sine
term
of
the
square
wave
has
amplitude
4
A
kπ
,
falling
as
1
k
.
bₖ
=
4
×
1
3 × π
bₖ = 0.4244 V
Step 3 · Find share of the power (share)
Use share = 100 × 8 ÷ (k² × π²)
Put
bₖ
=
4
A
kπ
into
bₖ² ÷ 2
A
2
:
A
cancels,
and
×
100
gives
a
percent.
share
=
100
×
8
3
2
×
π
2
share
=
100
×
8
9 × 9.87
share
=
100
×
8
88.83
share
=
800
88.83
share = 9.0063%
Check
Put the numbers back in. Both sides should match.
3 = 3 × 1 ✓
0.4244
=
4
×
1
3 × π
✓
9.0063
=
100
×
8
3
2
×
π
2
✓
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