ENGR E-250
Systems, Signals and Control
Fall 2026
Periodic signals and Fourier series
We have already seen that a signal 𝑥(𝑡) is periodic if there exists a 𝑇 such that 𝑥(𝑡) =
𝑥(𝑡 + 𝑘𝑇) for all 𝑡 and for all integers 𝑘.
The smallest 𝑇, i.e., the smallest repetition period, is called the fundamental period T₀.
The fundamental frequency is
𝑓₀ = 1/𝑇₀
The fundamental frequency f₀ is the largest frequency such that
𝑓ₖ = 𝑘 𝑓₀
where 𝑓ₖ is the 𝑘𝑡ℎ harmonic of 𝑓₀.
Examples: Find 𝒇₀ and 𝑻₀
Example 1: Consider 𝑐𝑜𝑠(40𝜋𝑡)
𝑓₀ = 20 𝐻𝑧 and 𝑇₀ = 1/𝑓₀ = 0.05 𝑠
Example 2: Consider 𝑠𝑖𝑛(𝜋𝑡 + 𝜋/4)
𝑓₀ = 0.5 𝐻𝑧 and 𝑇₀ = 2 𝑠
Example 3: Consider 𝑐𝑜𝑠²(4𝜋𝑡)
For 𝑐𝑜𝑠(4𝜋𝑡), the fundamental period is 𝑇0 = 0.5 𝑠.
Using the trigonometric identity: 𝑐𝑜𝑠²(𝜃) =
we obtain 𝑐𝑜𝑠²(4𝜋𝑡) =
1 + 𝑐𝑜𝑠(2𝜃)
2
,
1 + 𝑐𝑜𝑠(8𝜋𝑡)
2
Thus, the fundamental frequency is 𝑓₀ = 4 𝐻𝑧 and the
fundamental period 𝑇₀ = 0.25 𝑠.
Example 4: Given the sum of sinusoids 𝑐𝑜𝑠(24𝜋𝑡) + 𝑐𝑜𝑠(40𝜋𝑡) + 𝑐𝑜𝑠(120𝜋𝑡)
The corresponding frequencies are 12 𝐻𝑧, 20 𝐻𝑧, and 60 𝐻𝑧, respectively.
If
𝑓ₖ = 𝑘 𝑓₀,
Silvina L. Ferradal
1
ENGR E-250
Systems, Signals and Control
Fall 2026
then 𝑓₀ is the greatest common divisor (gcd) of the harmonic frequencies 𝑓ₖ.
In this case,
𝑓₀ = 4 𝐻𝑧
because
𝑓₃ = 3𝑓₀ = 12 𝐻𝑧
𝑓₅ = 5𝑓₀ = 20 𝐻𝑧
𝑓₁₅ = 15𝑓₀ = 60 𝐻𝑧.
Therefore, the fundamental period is 𝑇₀ = 1/𝑓₀ = 1/4 𝑠
A sum of harmonics of a common frequency, given by the fundamental frequency
𝒇₀, is a periodic signal.
In mathematical terms,
𝑁
𝑥(𝑡) = Σ𝑘=−𝑁
𝑎ₖ 𝑒 𝑗2𝜋𝑘 𝑓0 𝑡
where 𝑓ₖ = 𝑘𝑓₀ is the 𝑘𝑡ℎ harmonic.
Fourier series
Any periodic signal can be constructed from a (possibly infinite) sum of harmonic
sinusoids.
When 𝑁 → ∞, the previous expression becomes the Fourier series of 𝑥(𝑡), i.e.,
∞
𝑥(𝑡) = Σ𝑘=−∞
𝑎ₖ 𝑒 𝑗2𝜋𝑘 𝑓0 𝑡
or
𝑗
2𝜋
∞
𝑥(𝑡) = Σ𝑘=−∞
𝑎ₖ 𝑒 𝑇0
𝑘𝑡
which is also known as the Fourier Synthesis Summation (see DSP First, page 74).
The Fourier series coefficients are obtained from
𝑎ₖ =
Silvina L. Ferradal
2𝜋
1 𝑇0
−𝑗 𝑘 𝑡
∫ 𝑥(𝑡)𝑒 𝑇0 𝑑𝑡
𝑇0 0
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ENGR E-250
Systems, Signals and Control
Fall 2026
Which is also known as the Fourier Analysis Integral (see DSP First, page 75).
Example: A rectified sine wave
Consider the following signal
𝑥(𝑡) = |𝑠𝑖𝑛 (
2𝜋𝑡
)|
𝑇1
The original sine wave has
period 𝑇₁, but taking its absolute
value causes the waveform to
repeat twice during each original
period.
Therefore,
𝑇₀ =
𝑇1
2
The Fourier coefficients of 𝑥(𝑡) are given by
𝑎ₖ =
2𝜋
1 𝑇0
2𝜋𝑡
−𝑗 𝑘 𝑡
∫ |𝑠𝑖𝑛 (
)| 𝑒 𝑇0 𝑑𝑡
𝑇0 0
𝑇1
Since
|𝑠𝑖𝑛 (
2𝜋𝑡
)| ≥ 0
𝑇1
we can remove the absolute value brackets and rewrite 𝑎𝑘 as
1 𝑇0
2𝜋𝑡 −𝑗2𝜋
𝑘𝑡
𝑎ₖ =
∫ 𝑠𝑖𝑛 (
) 𝑒 𝑇0 𝑑𝑡
𝑇0 0
𝑇1
Using Euler’s inverse formula 𝑠𝑖𝑛(𝜃) =
𝑒 𝑗𝜃 − 𝑒 −𝑗𝜃
2𝑗
𝑗2𝜋𝑡
1 𝑇0 𝑒 𝑇1 − 𝑒
𝑎ₖ =
∫ {
𝑇0 0
2𝑗
𝑗2𝜋𝑡
− 𝑇
1
}𝑒
𝑗2𝜋𝑘𝑡
− 𝑇
0 𝑑𝑡
Since 𝑇₁ = 2𝑇₀
Silvina L. Ferradal
3
ENGR E-250
Systems, Signals and Control
Fall 2026
the expression becomes
𝑎ₖ =
𝑇0 𝑗(1−2𝑘)𝜋𝑡
𝑇0
𝑗(1+2𝑘)𝜋𝑡
1
1
−
𝑇0
∫ 𝑒 𝑇0 𝑑𝑡 −
∫ 𝑒
𝑑𝑡
2𝑗𝑇0 0
2𝑗𝑇0 0
Solving the integrals and because (1 − 2𝑘) and (1 + 2𝑘) are always odd for all 𝑘, the
evaluated exponential terms give −1.
Then, the Fourier coefficients are given by
𝑎ₖ =
2
𝜋(1 − 4𝑘 2 )
and the Fourier series of the rectified sine wave 𝑥(𝑡) is given by
∞
𝑥(𝑡) = Σ𝑘=−∞
2𝜋
2
𝑗 𝑘𝑡
𝑇0
𝑒
𝜋(1 − 4𝑘 2 )
Note that the first few coefficients are
𝑎₀ = 2/𝜋
𝑎±1 = −2/(3𝜋)
𝑎±2 = −2/(15𝜋)
𝑎±3 = −2/(35𝜋)
The spectrum of 𝑥(𝑡), therefore, contains components at
… , −3𝑓₀, −2𝑓₀, −𝑓₀, 0, 𝑓₀, 2𝑓₀, 3𝑓₀, …
with amplitudes given by the Fourier coefficients 𝑎𝑘 .
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