Periodic-Function and Harmonic-Building Exercises

(Example)

FS01E1: Periodic-Function and Harmonic-Building Exercises

This companion is designed to make the ideas in FS01 automatic before any Fourier coefficient formula is introduced.

The problems focus on:

recognizing periodicity,
distinguishing a period from the fundamental period,
converting among T, f, and ω,
interpreting phase,
combining sine and cosine at one frequency,
identifying harmonic number,
finding common fundamental frequencies,
distinguishing commensurate from incommensurate frequencies,
proving finite harmonic sums remain periodic,
evaluating early square-wave partial sums.

The guiding rule is simple:

do not memorize a period or harmonic result until you can reproduce it from the definitions.

Figure 1. Periodicity means exact repetition after a horizontal shift. A function may have many periods, but the fundamental period is the smallest positive one.

1 Problem set

All problems appear before the complete solutions.

Problem 1: test periodicity directly

Show directly from the definition that

f(t) = sin (5t)

(1)

is periodic. Determine its fundamental period.

Then show that twice the fundamental period is also a period.

Problem 2: from frequency to period and angular frequency

A periodic electrical signal has frequency

f0 = 60 Hz.

(2)

Find:

the fundamental period T₀,
the angular frequency ω₀,
the number of cycles completed in 0.50 s.

Problem 3: from angular frequency to period

A cosine has angular frequency

ω = 20π rad∕s.

(3)

Find its ordinary frequency and fundamental period.

Then verify the period by checking the phase advance over one cycle.

Problem 4: phase as a horizontal shift

Consider

y(t) = 3 cos(4t − π ∕3).

(4)

Determine:

amplitude,
angular frequency,
ordinary frequency,
period,
phase angle,
time shift relative to 3 cos(4t).

State whether the waveform is shifted to the right or left.

Problem 5: combine sine and cosine at one frequency

Rewrite

y(t) = 5cos(ωt) + 12 sin(ωt)

(5)

in the form

y (t) = C cos(ωt − δ).

(6)

Derive C and δ rather than quoting them.

Problem 6: identify harmonic content

For

f (t) = 1 − 2sin(3ω0t) + 4cos(7ω0t ),

(7)

identify:

the DC component,
every harmonic number present,
the sine and cosine coefficients of each present harmonic,
the highest harmonic number.

Problem 7: find a common fundamental from two frequencies

Consider

f (t) = cos(8t) + sin(12t).

(8)

Find the smallest positive angular frequency ω₀ for which both terms are integer harmonics.

Then find the fundamental period.

Problem 8: common period by direct component periods

For

f(t) = cos(2t) + cos(3t),

(9)

find the period of each component separately.

Then determine the smallest common positive period of the sum.

Verify your answer by direct substitution into f(t + T).

Problem 9: a hidden smaller fundamental period

Suppose

g(t) = cos(4ω t) + 1-sin(8ω t). 0 2 0

(10)

The number

T0 = 2π- ω0

(11)

is certainly a period.

Show that it is not the fundamental period and find the actual fundamental period.

Problem 10: commensurate frequencies written in an awkward way

Consider

h(t) = cos(15t) + 2 sin(21t) − cos(27t).

(12)

Determine a fundamental angular frequency and the fundamental period.

State the harmonic numbers of all three terms.

Problem 11: prove that an incommensurate pair has no common period

Consider

√ -- f(t) = sint + sin( 3 t).

(13)

Assume a common period T exists.

Use the individual sinusoidal periodicity conditions to show that this would force √--
3 to be rational.

Conclude that no finite common period exists.

Problem 12: prove that a finite harmonic sum is periodic

Let

∑N S (t) = A + [A cos(n ω t) + B sin(nω t)]. N 0 n 0 n 0 n=1

(14)

Prove directly that

SN (t + T0) = SN (t)

(15)

when

T = 2π. 0 ω0

(16)

Do not simply say that every term is periodic. Carry the phase shift through the sine and cosine arguments.

Problem 13: build and sample a finite harmonic waveform

Consider

y(t)	= 2 + 3 cos(ω₀t) − sin(2ω₀t)
	+ 0.5 cos(3ω₀t).	(17)

Evaluate the waveform at

ω t = 0, π-, π, 3π, 2π. 0 2 2

(18)

Explain why the first and last values must agree before doing any arithmetic.

Problem 14: square-wave partial sums

Use

S (t) = -4sint, 1 π

(19)

( ) 4- 1- S3(t) = π sint + 3 sin 3t ,

(20)

and

( ) 4- 1- 1- S5(t) = π sin t + 3 sin3t + 5 sin 5t .

(21)

Evaluate all three at

π- t = 2

(22)

and at

π- t = 6.

(23)

Describe how adding the third and fifth harmonics changes the result.

Problem 15: distinguish common period from fundamental period

A learner says:

“If every term repeats after T₀, then T₀ must be the fundamental period of the sum.”

Give a counterexample and explain exactly what is wrong with the statement.

Problem 16: physical harmonic interpretation

A vibrating system contains frequencies

120 Hz, 240 Hz, 360 Hz, 600 Hz.

(24)

Assume these are all harmonics of one fundamental.

Determine:

the fundamental frequency,
the harmonic number of each listed frequency,
the fundamental period,
whether a 480 Hz component would fit the same harmonic family.

2 Hints

A sinusoid repeats when its phase changes by an integer multiple of 2π.
To find a common fundamental angular frequency among integer-valued frequencies, look for their greatest common divisor.
A common period need not be the smallest common period.
For amplitude-phase conversion, expand C cos(ωt − δ) using the cosine subtraction identity.
For an incommensurate pair, write one integer phase condition for each sinusoid and divide them.
In the square-wave problem, use the exact values of sin(π∕2), sin(3π∕2), sin(5π∕2), sin(π∕6), sin(π∕2), and sin(5π∕6).

3 Complete worked solutions

Solution 1: test periodicity directly

Start with

f(t) = sin(5t).

(25)

A period T must satisfy

f(t + T) = f (t)

(26)

for every t.

Substitute:

f(t + T)	= sin[5(t + T)]	(27)
	= sin(5t + 5T).	(28)

Sine repeats when its phase increases by 2πm, where m is any integer:

5T = 2πm.

(29)

Therefore

T = 2-πm-. 5

(30)

The smallest positive value occurs at

m = 1.

(31)

Hence the fundamental period is

2π- T0 = 5 .

(32)

Now test twice that value:

4π- 2T0 = 5 .

(33)

Then

f(t + 2T₀)	= sin	(34)
	= sin(5t + 4π)	(35)
	= sin(5t).	(36)

Thus 2T₀ is also a period, but it is not the fundamental period.

Solution 2: frequency, period, and angular frequency

Given

f0 = 60 Hz,

(37)

use

T0 = 1-. f0

(38)

Therefore

T₀	= s	(39)
	≈ 0.0166667 s.	(40)

In milliseconds,

T0 ≈ 16.67 ms.

(41)

Angular frequency is

ω0 = 2πf0.

(42)

Therefore

ω₀	= 2π(60)	(43)
	= 120π rad∕s	(44)
	≈ 376.99 rad∕s.	(45)

The number of cycles in a time interval Δt is

N = f0Δt.

(46)

For

Δt = 0.50 s,

(47)

N	= (60 cycles∕s)(0.50 s)	(48)
	= 30 cycles.	(49)

Solution 3: angular frequency to ordinary frequency

Given

ω = 20π rad∕s,

(50)

use

f = -ω-. 2π

(51)

Thus

f	=	(52)
	= 10 Hz.	(53)

The period is

T₀	=	(54)
	= 0.10 s.	(55)

Now verify the phase advance:

Δϕ	= ωT₀	(56)
	= (20π)(0.10)	(57)
	= 2π.	(58)

A 2π phase advance is one complete cycle, so the period is correct.

Solution 4: phase as a horizontal shift

The waveform is

y(t) = 3 cos(4t − π ∕3).

(59)

Compare with

y (t) = A cos(ωt − ϕ).

(60)

Therefore

A = 3, ω = 4 rad∕s, ϕ = π-. 3

(61)

The ordinary frequency is

f	=	(62)
	=	(63)
	= Hz	(64)
	≈ 0.63662 Hz.	(65)

The period is

T₀	=	(66)
	=	(67)
	= s.	(68)

To convert phase angle into time shift, factor the argument:

4t −

= 4

(69)

Hence

[ ( )] y(t) = 3cos 4 t −-π- . 12

(70)

The waveform is shifted to the right by

Δt = -π-s. 12

(71)

Figure 2. The same-frequency sine and cosine coefficients can be viewed as perpendicular components. Their resultant magnitude and angle give the amplitude-phase form.

Solution 5: amplitude-phase conversion

Start with

y(t) = 5 cos(ωt) + 12sin(ωt).

(72)

Assume

y (t) = C cos(ωt − δ).

(73)

Use the cosine subtraction identity:

cos(A − B ) = cosA cosB + sin A sinB.

(74)

Therefore

C cos(ωt − δ)	= C cos δ cos(ωt)	(75)
	+ C sin δ sin(ωt).	(76)

Match the cosine coefficient:

C cosδ = 5.

(77)

Match the sine coefficient:

C sin δ = 12.

(78)

Square both equations and add:

C² cos ²δ + C² sin ²δ	= 5² + 12²	(79)
C²(cos ²δ + sin ²δ)	= 25 + 144	(80)
C²	= 169.	(81)

Thus

C = 13.

(82)

Now divide the sine equation by the cosine equation:

tan δ

(83)

Hence

( 12) δ = tan−1 --- ≈ 67.38∘. 5

(84)

The result is

y(t) = 13 cos(ωt − 67.38∘).

(85)

Solution 6: identify harmonic content

The function is

f (t) = 1 − 2sin(3ω0t) + 4cos(7ω0t ).

(86)

The constant term is the DC component:

A0 = 1.

(87)

The term

− 2 sin (3ω0t)

(88)

is the third harmonic because its angular frequency is

3ω0.

(89)

Its sine coefficient is

B3 = − 2,

(90)

and its cosine coefficient is zero.

The term

4 cos(7ω0t)

(91)

is the seventh harmonic.

Its cosine coefficient is

A7 = 4,

(92)

and its sine coefficient is zero.

Therefore the present harmonic numbers are

3 and 7,

(93)

and the highest harmonic number is

7.

(94)

Solution 7: greatest common angular frequency

The angular frequencies are

8 rad∕s and 12 rad∕s.

(95)

We seek the largest angular frequency that divides both into integers.

The greatest common divisor is

gcd(8,12) = 4.

(96)

Thus

ω0 = 4 rad ∕s.

(97)

Then

8 = 2ω0

(98)

and

12 = 3ω . 0

(99)

So the waveform contains the second and third harmonics.

The fundamental period is

T₀	=	(100)
	=	(101)
	= s.	(102)

Figure 3. Finding a common fundamental frequency is analogous to finding a greatest common divisor when all angular frequencies are integer multiples of one underlying frequency.

Solution 8: common period from the component periods

For

cos(2t),

(103)

the period is

2 π T1 = --- = π. 2

(104)

For

cos(3t),

(105)

the period is

2π- T2 = 3 .

(106)

We need the smallest positive T satisfying

T = mT1 = nT2

(107)

for integers m and n.

Set

2π- m π = n 3 .

(108)

Cancel π:

2n m = --. 3

(109)

The smallest positive integers are

m = 2, n = 3.

(110)

Therefore

T = 2 π.

(111)

Now verify directly:

f(t + 2π)	= cos[2(t + 2π)] + cos[3(t + 2π)]	(112)
	= cos(2t + 4π) + cos(3t + 6π)	(113)
	= cos(2t) + cos(3t)	(114)
	= f(t).	(115)

Thus the fundamental period is

2π.

(116)

Solution 9: hidden smaller fundamental period

The waveform is

g(t) = cos(4ω0t ) + 1-sin(8ω0t). 2

(117)

The first term has period

T₁	=	(118)
	= .	(119)

The second term has period

T₂	=	(120)
	= .	(121)

The smallest common positive period of T₀∕4 and T₀∕8 is

T0-. 4

(122)

Indeed, over T₀∕4, the first harmonic term completes one cycle and the second completes two cycles.

Therefore the actual fundamental period is

T Tfund = -0. 4

(123)

The originally chosen T₀ is a period, but it is four times larger than necessary.

Figure 4. A chosen common period can hide a smaller fundamental period. Here both harmonics repeat after only one quarter of the original period T₀.

Solution 10: commensurate frequencies in awkward form

The angular frequencies are

15, 21, 27 rad∕s.

(124)

Find their greatest common divisor:

gcd (15,21,27) = 3.

(125)

Therefore

ω0 = 3 rad ∕s.

(126)

The harmonic numbers are

15 = 5ω0,

(127)

21 = 7ω0,

(128)

and

27 = 9ω0.

(129)

Thus the fifth, seventh, and ninth harmonics are present.

The fundamental period is

T₀

(130)

Solution 11: incommensurate frequencies

Assume

√ -- f (t) = sint + sin( 3t)

(131)

has a common period T.

For the first sine to repeat,

T = 2πm

(132)

for some integer m.

For the second sine to repeat,

√ -- 3T = 2πn

(133)

for some integer n.

Substitute the first condition into the second:

(2πm)

= 2πn.

(134)

Cancel 2π:

√ -- 3 m = n.

(135)

Thus

√ -- n 3 = -- . m

(136)

The right side is rational because n and m are integers.

But √ --
3 is irrational.

This is a contradiction.

Therefore no finite common period exists.

Figure 5. Integer-related frequencies repeatedly realign after a common period. Incommensurate frequencies such as 1 and √ --
3 do not possess an exact finite realignment time.

Solution 12: prove a finite harmonic sum is periodic

Start from Equation (14):

S_N(t)

= A₀ + ∑ _n=1^N [An cos(nω0t ) + Bn sin(nω0t)]

(137)

Evaluate at t + T₀:

S_N(t + T₀)	= A₀ + ∑ _n=1^NA_n cos(nω₀(t + T₀))
	+ B_n sin(nω₀(t + T₀)).	(138)

Distribute the argument:

nω0(t + T0) = nω0t + nω0T0.

(139)

Since

2π- T0 = ω0,

(140)

we have

nω₀T₀	= nω₀	(141)
	= 2πn.	(142)

Therefore

cos[nω (t + T )] = cos(n ω t + 2πn ). 0 0 0

(143)

Because n is an integer,

cos(n ω0t + 2 πn) = cos(nω0t).

(144)

Similarly,

sin(n ω0t + 2 πn) = sin(nω0t).

(145)

Hence every term in the sum returns to its original value, and

S (t + T ) = S (t). N 0 N

(146)

This proves that any finite sum of integer harmonics has T₀ as a common period.

It does not prove that T₀ is always the smallest period.

Solution 13: sample a finite harmonic waveform

The waveform is

y(t)	= 2 + 3 cos(ω₀t) − sin(2ω₀t)
	+ 0.5 cos(3ω₀t).	(147)

Let

𝜃 = ω0t.

(148)

Then

y(𝜃) = 2 + 3cos 𝜃 − sin 2𝜃 + 0.5cos3 𝜃.

(149)

At 𝜃 = 0:

y(0)	= 2 + 3(1) − 0 + 0.5(1)	(150)
	= 5.5.	(151)

At 𝜃 = π∕2:

y(π∕2)	= 2 + 3(0) − sin π + 0.5 cos(3π∕2)	(152)
	= 2.	(153)

At 𝜃 = π:

y(π)	= 2 + 3(−1) − sin 2π + 0.5 cos 3π	(154)
	= 2 − 3 − 0.5	(155)
	= −1.5.	(156)

At 𝜃 = 3π∕2:

y(3π∕2)	= 2 + 3(0) − sin 3π + 0.5 cos(9π∕2)	(157)
	= 2.	(158)

At 𝜃 = 2π:

y(2π)	= 2 + 3(1) − sin 4π + 0.5 cos 6π	(159)
	= 5.5.	(160)

The first and last values had to agree because 2π is one complete fundamental phase advance.

ω₀t	y(t)

0	5.5
π∕2	2
π	−1.5
3π∕2	2
2π	5.5

Figure 6. The sampled waveform is built from a DC offset plus first, second, and third harmonics. The separate components repeat together after one common fundamental period.

Solution 14: square-wave partial sums

π- t = 2,

(161)

we use

π 3π 5π sin --= 1, sin ---= − 1, sin ---= 1. 2 2 2

(162)

Thus

S₁	=	(163)
	≈ 1.27324.	(164)

For S₃,

S₃	=	(165)
	=	(166)
	≈ 0.84883.	(167)

For S₅,

S₅	=	(168)
	=	(169)
	≈ 1.10347.	(170)

Now evaluate at

π t = -. 6

(171)

Use

π 1 sin 6-= 2,

(172)

π- sin 2 = 1,

(173)

and

sin 5π- = 1-. 6 2

(174)

Therefore

S₁	=	(175)
	=	(176)
	≈ 0.63662.	(177)

For S₃,

S₃	=	(178)
	=	(179)
	≈ 1.06103.	(180)

For S₅,

S₅	=	(181)
	=	(182)
	≈ 1.18836.	(183)

Adding harmonics does not simply increase or decrease the value everywhere. The new harmonics reshape the waveform differently at different points.

Figure 7. The first three odd-harmonic partial sums at two sample phases. Each added harmonic can raise the approximation at one point and lower it at another.

Solution 15: common period versus fundamental period

The learner’s claim is false because a common period may be larger than necessary.

Use the counterexample

f(t) = cos(2 ω0t) + sin(4ω0t).

(184)

Both terms repeat after

T = 2π, 0 ω0

(185)

so T₀ is a period.

But the first term has period

T0 ---, 2

(186)

and the second term has period

T0-. 4

(187)

Both therefore repeat after

T0- 2 .

(188)

Hence the fundamental period is at most T₀∕2. Indeed it is exactly T₀∕2.

The corrected statement is:

if every term repeats after T₀, then T₀ is a common period of the sum;
an additional check is required to determine whether a smaller positive common period exists.

Solution 16: physical harmonic interpretation

The listed frequencies are

120, 240, 360, 600 Hz.

(189)

Their greatest common divisor is

120 Hz.

(190)

Therefore the fundamental frequency is

f0 = 120 Hz.

(191)

The harmonic numbers are

120 = 1f0,

(192)

240 = 2f0,

(193)

360 = 3f0,

(194)

and

600 = 5f0.

(195)

Thus the first, second, third, and fifth harmonics are present.

The fundamental period is

T₀	=	(196)
	≈ 8.333 × 10⁻³ s	(197)
	= 8.333 ms.	(198)

For 480 Hz,

480 = 4(120 ),

(199)

so it would be the fourth harmonic and fits the same harmonic family.

4 What should now feel automatic

Before moving to FS02, a learner should be able to do the following without guessing:

test a candidate period by substitution,
find the fundamental period of a single sinusoid,
convert among T, f, and ω,
convert same-frequency sine and cosine terms to amplitude-phase form,
identify harmonic numbers from frequency multiples,
determine common fundamentals for commensurate frequencies,
prove the periodicity of a finite harmonic sum,
explain why a common period need not be fundamental,
prove that a simple irrational frequency ratio prevents exact periodic repetition,
interpret a partial sum as a finite harmonic waveform.

Figure 8. FS01E1 completes the arithmetic and algebra needed before function-space ideas begin. FS02 will reinterpret harmonic extraction as projection using an inner product of functions.

5 Further practice

Find the fundamental period of sin(14t).
Determine the fundamental period of cos(6t) + sin(10t).
Find the common fundamental angular frequency of 18, 30, and 42 rad∕s.
Determine whether sin(2t) + cos(πt) is periodic.
Convert −8 cos(ωt) + 6 sin(ωt) to one cosine using atan2 for the phase quadrant.
Construct a finite harmonic sum containing a DC term, the second harmonic, and the fifth harmonic only.
Give an example where T₀ is a period of every term but the sum has fundamental period T₀∕3.
Evaluate S₅(π∕4) for the square-wave preview formula.

6 Summary

Periodic-function work begins with the definition

f(t + T ) = f(t).

(200)

For one sinusoid,

T = 2-π, f = 1-, ω = 2πf . 0 ω 0 T0 0 0

(201)

Integer harmonics have frequencies

ωn = n ω0.

(202)

Commensurate component frequencies can share a finite common period. Incommensurate frequencies generally do not.

A finite harmonic sum is periodic because every harmonic gains an integer multiple of 2π in phase after one common period.

The main unresolved question is no longer which frequencies fit together. It is how to extract the amount of one harmonic from a general function.

That requires a notion of projection for functions, which begins in FS02.

References

[1] PhysicsLibrary, FS01: Periodic Functions, Harmonics, and Why Fourier Series Exist.

[2] PhysicsLibrary, FS00: Fourier Series Roadmap, Motivation, and Learning Path.

"Periodic-Function and Harmonic-Building Exercises" is owned by bloftin.
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