Esercizi sulla Serie di Fourier

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Esercizi sulla Serie di Fourier

Concetti Chiave

La serie di Fourier è un metodo per rappresentare una funzione periodica come somma di funzioni sinusoidali (seno e coseno). La forma generale della serie di Fourier per una funzione periodica f(t)f(t)f(t) con periodo TTT è data da:

f(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{2\pi n}{T} t\right) + b_n \sin\left(\frac{2\pi n}{T} t\right) \right) f(t)=a0+∑n=1∞(ancos⁡(2πnTt)+bnsin⁡(2πnTt)) f(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{2\pi n}{T} t\right) + b_n \sin\left(\frac{2\pi n}{T} t\right) \right)

dove:

  • a_0a0a_0 è il termine costante, calcolato come:
    a_0 = \frac{1}{T} \int_0^T f(t) \, dt a0=1T∫0Tf(t) dt a_0 = \frac{1}{T} \int_0^T f(t) \, dt
  • a_nana_n e b_nbnb_n sono i coefficienti di Fourier, calcolati come:
    a_n = \frac{2}{T} \int_0^T f(t) \cos\left(\frac{2\pi n}{T} t\right) \, dt an=2T∫0Tf(t)cos⁡(2πnTt) dt a_n = \frac{2}{T} \int_0^T f(t) \cos\left(\frac{2\pi n}{T} t\right) \, dt
    b_n = \frac{2}{T} \int_0^T f(t) \sin\left(\frac{2\pi n}{T} t\right) \, dt bn=2T∫0Tf(t)sin⁡(2πnTt) dt b_n = \frac{2}{T} \int_0^T f(t) \sin\left(\frac{2\pi n}{T} t\right) \, dt

Esercizi

Esercizio 1: Serie di Fourier di una Funzione Rettangolare

Considera una funzione rettangolare definita come:

f(t) = \begin{cases} 1 & \text{se } 0 < t < T/2 \\ 0 & \text{se } T/2 < t < T \end{cases} f(t)={1se 0<t<T/20se T/2<t<T f(t) = \begin{cases} 1 & \text{se } 0 < t < T/2 \\ 0 & \text{se } T/2 < t < T \end{cases}

Calcola i coefficienti di Fourier a_0a0a_0, a_nana_n e b_nbnb_n.

Soluzione:

  1. Calcolo di a_0a0a_0:

a_0 = \frac{1}{T} \int_0^T f(t) \, dt = \frac{1}{T} \left( \int_0^{T/2} 1 \, dt + \int_{T/2}^T 0 \, dt \right) = \frac{1}{T} \cdot \frac{T}{2} = \frac{1}{2} a0=1T∫0Tf(t) dt=1T(∫0T/21 dt+∫T/2T0 dt)=1T⋅T2=12 a_0 = \frac{1}{T} \int_0^T f(t) \, dt = \frac{1}{T} \left( \int_0^{T/2} 1 \, dt + \int_{T/2}^T 0 \, dt \right) = \frac{1}{T} \cdot \frac{T}{2} = \frac{1}{2}

  1. Calcolo di a_nana_n:

a_n = \frac{2}{T} \int_0^T f(t) \cos\left(\frac{2\pi n}{T} t\right) \, dt = \frac{2}{T} \int_0^{T/2} \cos\left(\frac{2\pi n}{T} t\right) \, dt an=2T∫0Tf(t)cos⁡(2πnTt) dt=2T∫0T/2cos⁡(2πnTt) dt a_n = \frac{2}{T} \int_0^T f(t) \cos\left(\frac{2\pi n}{T} t\right) \, dt = \frac{2}{T} \int_0^{T/2} \cos\left(\frac{2\pi n}{T} t\right) \, dt

Calcoliamo l’integrale:

\int \cos\left(\frac{2\pi n}{T} t\right) \, dt = \frac{T}{2\pi n} \sin\left(\frac{2\pi n}{T} t\right) ∫cos⁡(2πnTt) dt=T2πnsin⁡(2πnTt) \int \cos\left(\frac{2\pi n}{T} t\right) \, dt = \frac{T}{2\pi n} \sin\left(\frac{2\pi n}{T} t\right)

Quindi:

a_n = \frac{2}{T} \cdot \left[ \frac{T}{2\pi n} \sin\left(\frac{2\pi n}{T} t\right) \right]_0^{T/2} = \frac{2}{T} \cdot \frac{T}{2\pi n} \left( \sin\left(\pi n\right) - \sin(0) \right) = 0 an=2T⋅[T2πnsin⁡(2πnTt)]0T/2=2T⋅T2πn(sin⁡(πn)−sin⁡(0))=0 a_n = \frac{2}{T} \cdot \left[ \frac{T}{2\pi n} \sin\left(\frac{2\pi n}{T} t\right) \right]_0^{T/2} = \frac{2}{T} \cdot \frac{T}{2\pi n} \left( \sin\left(\pi n\right) - \sin(0) \right) = 0

  1. Calcolo di b_nbnb_n:

b_n = \frac{2}{T} \int_0^T f(t) \sin\left(\frac{2\pi n}{T} t\right) \, dt = \frac{2}{T} \int_0^{T/2} \sin\left(\frac{2\pi n}{T} t\right) \, dt bn=2T∫0Tf(t)sin⁡(2πnTt) dt=2T∫0T/2sin⁡(2πnTt) dt b_n = \frac{2}{T} \int_0^T f(t) \sin\left(\frac{2\pi n}{T} t\right) \, dt = \frac{2}{T} \int_0^{T/2} \sin\left(\frac{2\pi n}{T} t\right) \, dt

Calcoliamo l’integrale:

\int \sin\left(\frac{2\pi n}{T} t\right) \, dt = -\frac{T}{2\pi n} \cos\left(\frac{2\pi n}{T} t\right) ∫sin⁡(2πnTt) dt=−T2πncos⁡(2πnTt) \int \sin\left(\frac{2\pi n}{T} t\right) \, dt = -\frac{T}{2\pi n} \cos\left(\frac{2\pi n}{T} t\right)

Quindi:

b_n = \frac{2}{T} \cdot \left[-\frac{T}{2 \pi n} \cos\left(\frac{2\pi n}{T} t\right)\right]_0^{T/2} bn=2T⋅[−T2πncos⁡(2πnTt)]0T/2 b_n = \frac{2}{T} \cdot \left[-\frac{T}{2 \pi n} \cos\left(\frac{2\pi n}{T} t\right)\right]_0^{T/2}

Sostituendo i limiti:

b_n = \frac{2}{T} \cdot \left[-\frac{T}{2\pi n} \left( \cos\left(\pi n\right) - \cos(0) \right)\right] bn=2T⋅[−T2πn(cos⁡(πn)−cos⁡(0))] b_n = \frac{2}{T} \cdot \left[-\frac{T}{2\pi n} \left( \cos\left(\pi n\right) - \cos(0) \right)\right]

= \frac{2}{T} \cdot \left[-\frac{T}{2\pi n} \left((-1)^n - 1\right)\right] =2T⋅[−T2πn((−1)n−1)] = \frac{2}{T} \cdot \left[-\frac{T}{2\pi n} \left((-1)^n - 1\right)\right]

= -\frac{1}{\pi n} \left((-1)^n - 1\right) =−1πn((−1)n−1) = -\frac{1}{\pi n} \left((-1)^n - 1\right)

Quindi, i coefficienti di Fourier per la funzione rettangolare sono:

  • a_0 = \frac{1}{2}a0=12a_0 = \frac{1}{2}
  • a_n = 0an=0a_n = 0
  • b_n = -\frac{1}{\pi n} \left((-1)^n - 1\right)bn=−1πn((−1)n−1)b_n = -\frac{1}{\pi n} \left((-1)^n - 1\right)

Esercizio 2: Serie di Fourier di una Funzione Triangolare

Considera una funzione triangolare definita su [-L, L][−L,L][-L, L] con un’altezza di HHH e una base di 2L2L2L.

Calcola i coefficienti di Fourier a_0a0a_0, a_nana_n e b_nbnb_n.

Soluzione:

  1. Calcolo di a_0a0a_0:

a_0 = \frac{1}{2L} \int_{-L}^{L} f(t) \, dt a0=12L∫−LLf(t) dt a_0 = \frac{1}{2L} \int_{-L}^{L} f(t) \, dt

La funzione triangolare è simmetrica, quindi possiamo calcolare solo l’integrale da 000 a LLL e moltiplicarlo per 2:

= \frac{1}{2L} \cdot 2 \int_0^{L} f(t) \, dt = \frac{1}{L} \int_0^{L} f(t) \, dt =12L⋅2∫0Lf(t) dt=1L∫0Lf(t) dt = \frac{1}{2L} \cdot 2 \int_0^{L} f(t) \, dt = \frac{1}{L} \int_0^{L} f(t) \, dt

L’area del triangolo è:

\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{altezza} = \frac{1}{2} \cdot 2L \cdot H = LH Area=12⋅base⋅altezza=12⋅2L⋅H=LH \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{altezza} = \frac{1}{2} \cdot 2L \cdot H = LH

Quindi:

a_0 = \frac{1}{L} \cdot LH = H a0=1L⋅LH=H a_0 = \frac{1}{L} \cdot LH = H

  1. Calcolo di a_nana_n:

a_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos\left(\frac{n\pi t}{L}\right) \, dt an=1L∫−LLf(t)cos⁡(nπtL) dt a_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos\left(\frac{n\pi t}{L}\right) \, dt

Poiché la funzione triangolare è dispari, il termine coseno (che è pari) darà un contributo nullo:

a_n = 0 an=0 a_n = 0

  1. Calcolo di b_nbnb_n:

b_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) \, dt bn=1L∫−LLf(t)sin⁡(nπtL) dt b_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) \, dt

Poiché la funzione triangolare è simmetrica rispetto all’asse yyy, possiamo calcolare solo l’integrale da 000 a LLL e moltiplicarlo per 2:

b_n = \frac{2}{L} \int_0^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) \, dt bn=2L∫0Lf(t)sin⁡(nπtL) dt b_n = \frac{2}{L} \int_0^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) \, dt

Calcoliamo l’integrale. La funzione triangolare può essere espressa come:

f(t) = \frac{H}{L} t \quad \text{per } 0 < t < L f(t)=HLtper 0<t<L f(t) = \frac{H}{L} t \quad \text{per } 0 < t < L

Quindi:

b_n = \frac{2}{L} \int_0^{L} \frac{H}{L} t \sin\left(\frac{n\pi t}{L}\right) \, dt bn=2L∫0LHLtsin⁡(nπtL) dt b_n = \frac{2}{L} \int_0^{L} \frac{H}{L} t \sin\left(\frac{n\pi t}{L}\right) \, dt

b_n = \frac{2H}{L^2} \left[ -\frac{L}{n\pi} t \cos\left(\frac{n\pi t}{L}\right) \right]_0^{L} + \frac{2H}{L^2} \cdot \frac{L^2}{(n\pi)^2} \left[ \sin\left(\frac{n\pi t}{L}\right) \right]_0^{L} bn=2HL2[−Lnπtcos⁡(nπtL)]0L+2HL2⋅L2(nπ)2[sin⁡(nπtL)]0L b_n = \frac{2H}{L^2} \left[ -\frac{L}{n\pi} t \cos\left(\frac{n\pi t}{L}\right) \right]_0^{L} + \frac{2H}{L^2} \cdot \frac{L^2}{(n\pi)^2} \left[ \sin\left(\frac{n\pi t}{L}\right) \right]_0^{L}

Continuando con il calcolo di b_nbnb_n:

b_n = \frac{2H}{L^2} \left[ -\frac{L}{n\pi} L \cos\left(n\pi\right) + 0 \right] + \frac{2H}{L^2} \cdot \frac{L^2}{(n\pi)^2} \left[ \sin\left(n\pi\right) - 0 \right] bn=2HL2[−LnπLcos⁡(nπ)+0]+2HL2⋅L2(nπ)2[sin⁡(nπ)−0] b_n = \frac{2H}{L^2} \left[ -\frac{L}{n\pi} L \cos\left(n\pi\right) + 0 \right] + \frac{2H}{L^2} \cdot \frac{L^2}{(n\pi)^2} \left[ \sin\left(n\pi\right) - 0 \right]

Poiché \cos(n\pi) = (-1)^ncos⁡(nπ)=(−1)n\cos(n\pi) = (-1)^n e \sin(n\pi) = 0sin⁡(nπ)=0\sin(n\pi) = 0, otteniamo:

b_n = \frac{2H}{L^2} \left[ -\frac{L^2}{n\pi} (-1)^n \right] + 0 bn=2HL2[−L2nπ(−1)n]+0 b_n = \frac{2H}{L^2} \left[ -\frac{L^2}{n\pi} (-1)^n \right] + 0

Semplificando:

b_n = \frac{2H L}{n\pi} (-1)^{n+1} bn=2HLnπ(−1)n+1 b_n = \frac{2H L}{n\pi} (-1)^{n+1}

Risultati Finali

Per la funzione triangolare definita su [-L, L][−L,L][-L, L] con altezza HHH, i coefficienti di Fourier sono:

  • a_0 = Ha0=Ha_0 = H
  • a_n = 0an=0a_n = 0
  • b_n = \frac{2H L}{n\pi} (-1)^{n+1}bn=2HLnπ(−1)n+1b_n = \frac{2H L}{n\pi} (-1)^{n+1}

English version

Fourier Series Exercises

Key Concepts

Fourier series is a method of representing a periodic function as a sum of sinusoidal functions (sine and cosine). The general form of the Fourier series for a periodic function f(t)f(t)f(t) with period TTT is given by:

f(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{2\pi n}{T} t\right) + b_n \sin\left(\frac{2\pi n}{T} t\right) \right) f(t)=a0+∑n=1∞(ancos⁡(2πnTt)+bnsin⁡(2πnTt)) f(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{2\pi n}{T} t\right) + b_n \sin\left(\frac{2\pi n}{T} t\right) \right)

where:

  • a_0a0a_0 is the constant term, computed as:
    a_0 = \frac{1}{T} \int_0^T f(t) \, dt a0=1T∫0Tf(t) dt a_0 = \frac{1}{T} \int_0^T f(t) \, dt
  • a_nana_n and b_nbnb_n are the Fourier coefficients, computed as:
    a_n = \frac{2}{T} \int_0^T f(t) \cos\left(\frac{2\pi n}{T} t\right) \, dt an=2T∫0Tf(t)cos⁡(2πnTt) dt a_n = \frac{2}{T} \int_0^T f(t) \cos\left(\frac{2\pi n}{T} t\right) \, dt
    b_n = \frac{2}{T} \int_0^T f(t) \sin\left(\frac{2\pi n}{T} t\right) \, dt bn=2T∫0Tf(t)sin⁡(2πnTt) dt b_n = \frac{2}{T} \int_0^T f(t) \sin\left(\frac{2\pi n}{T} t\right) \, dt

Exercises

Exercise 1: Fourier Series of a Rectangular Function

Consider a rectangular function defined as:

f(t) = \begin{cases} 1 & \text{if } 0 < t < T/2 \\ 0 & \text{if } T/2 < t < T \end{cases} f(t)={1if 0<t<T/20if T/2<t<T f(t) = \begin{cases} 1 & \text{if } 0 < t < T/2 \\ 0 & \text{if } T/2 < t < T \end{cases}

Calculate the Fourier coefficients a_0a0a_0, a_nana_n and b_nbnb_n.

Solution:

  1. Calculation of a_0a0a_0:

a_0 = \frac{1}{T} \int_0^T f(t) \, dt = \frac{1}{T} \left( \int_0^{T/2} 1 \, dt + \int_{T/2}^T 0 \, dt \right) = \frac{1}{T} \cdot \frac{T}{2} = \frac{1}{2} a0=1T∫0Tf(t) dt=1T(∫0T/21 dt+∫T/2T0 dt)=1T⋅T2=12 a_0 = \frac{1}{T} \int_0^T f(t) \, dt = \frac{1}{T} \left( \int_0^{T/2} 1 \, dt + \int_{T/2}^T 0 \, dt \right) = \frac{1}{T} \cdot \frac{T}{2} = \frac{1}{2}

  1. Calculation of a_nana_n:

a_n = \frac{2}{T} \int_0^T f(t) \cos\left(\frac{2\pi n}{T} t\right) \, dt = \frac{2}{T} \int_0^{T/2} \cos\left(\frac{2\pi n}{T} t\right) \, dt an=2T∫0Tf(t)cos⁡(2πnTt) dt=2T∫0T/2cos⁡(2πnTt) dt a_n = \frac{2}{T} \int_0^T f(t) \cos\left(\frac{2\pi n}{T} t\right) \, dt = \frac{2}{T} \int_0^{T/2} \cos\left(\frac{2\pi n}{T} t\right) \, dt

Let’s calculate the integral:

\int \cos\left(\frac{2\pi n}{T} t\right) \, dt = \frac{T}{2\pi n} \sin\left(\frac{2\pi n}{T} t\right) ∫cos⁡(2πnTt) dt=T2πnsin⁡(2πnTt) \int \cos\left(\frac{2\pi n}{T} t\right) \, dt = \frac{T}{2\pi n} \sin\left(\frac{2\pi n}{T} t\right)

So:

a_n = \frac{2}{T} \cdot \left[ \frac{T}{2\pi n} \sin\left(\frac{2\pi n}{T} t\right) \right]_0^{T/2} = \frac{2}{T} \cdot \frac{T}{2\pi n} \left( \sin\left(\pi n\right) - \sin(0) \right) = 0 an=2T⋅[T2πnsin⁡(2πnTt)]0T/2=2T⋅T2πn(sin⁡(πn)−sin⁡(0))=0 a_n = \frac{2}{T} \cdot \left[ \frac{T}{2\pi n} \sin\left(\frac{2\pi n}{T} t\right) \right]_0^{T/2} = \frac{2}{T} \cdot \frac{T}{2\pi n} \left( \sin\left(\pi n\right) - \sin(0) \right) = 0

  1. Calculation of b_nbnb_n:

b_n = \frac{2}{T} \int_0^T f(t) \sin\left(\frac{2\pi n}{T} t\right) \, dt = \frac{2}{T} \int_0^{T/2} \sin\left(\frac{2\pi n}{T} t\right) \, dt bn=2T∫0Tf(t)sin⁡(2πnTt) dt=2T∫0T/2sin⁡(2πnTt) dt b_n = \frac{2}{T} \int_0^T f(t) \sin\left(\frac{2\pi n}{T} t\right) \, dt = \frac{2}{T} \int_0^{T/2} \sin\left(\frac{2\pi n}{T} t\right) \, dt

Let’s calculate the integral:

\int \sin\left(\frac{2\pi n}{T} t\right) \, dt = -\frac{T}{2\pi n} \cos\left(\frac{2\pi n}{T} t\right) ∫sin⁡(2πnTt) dt=−T2πncos⁡(2πnTt) \int \sin\left(\frac{2\pi n}{T} t\right) \, dt = -\frac{T}{2\pi n} \cos\left(\frac{2\pi n}{T} t\right)

So:

b_n = \frac{2}{T} \cdot \left[-\frac{T}{2 \pi n} \cos\left(\frac{2\pi n}{T} t\right)\right]_0^{T/2} bn=2T⋅[−T2πncos⁡(2πnTt)]0T/2 b_n = \frac{2}{T} \cdot \left[-\frac{T}{2 \pi n} \cos\left(\frac{2\pi n}{T} t\right)\right]_0^{T/2}

Substituting the limits:

b_n = \frac{2}{T} \cdot \left[-\frac{T}{2\pi n} \left( \cos\left(\pi n\right) - \cos(0) \right)\right] bn=2T⋅[−T2πn(cos⁡(πn)−cos⁡(0))] b_n = \frac{2}{T} \cdot \left[-\frac{T}{2\pi n} \left( \cos\left(\pi n\right) - \cos(0) \right)\right]

= \frac{2}{T} \cdot \left[-\frac{T}{2\pi n} \left((-1)^n - 1\right)\right] =2T⋅[−T2πn((−1)n−1)] = \frac{2}{T} \cdot \left[-\frac{T}{2\pi n} \left((-1)^n - 1\right)\right]

= -\frac{1}{\pi n} \left((-1)^n - 1\right) =−1πn((−1)n−1) = -\frac{1}{\pi n} \left((-1)^n - 1\right)

So, the Fourier coefficients for the rectangular function are:

  • a_0 = \frac{1}{2}a0=12a_0 = \frac{1}{2}
  • a_n = 0an=0a_n = 0
  • b_n = -\frac{1}{\pi n} \left((-1)^n - 1\right)bn=−1πn((−1)n−1)b_n = -\frac{1}{\pi n} \left((-1)^n - 1\right)

Exercise 2: Fourier Series of a Triangular Function

Consider a triangular function defined on [-L, L][−L,L][-L, L] with a height of HHH and a base of 2L2L2L.

Compute the Fourier coefficients a_0a0a_0, a_nana_n and b_nbnb_n.

Solution:

  1. Calculating a_0a0a_0:

a_0 = \frac{1}{2L} \int_{-L}^{L} f(t) \, dt a0=12L∫−LLf(t) dt a_0 = \frac{1}{2L} \int_{-L}^{L} f(t) \, dt

The triangular function is symmetric, so we can only calculate the integral from 000 to LLL and multiply it by 2:

= \frac{1}{2L} \cdot 2 \int_0^{L} f(t) \, dt = \frac{1}{L} \int_0^{L} f(t) \, dt =12L⋅2∫0Lf(t) dt=1L∫0Lf(t) dt = \frac{1}{2L} \cdot 2 \int_0^{L} f(t) \, dt = \frac{1}{L} \int_0^{L} f(t) \, dt

The area of ​​the triangle is:

\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2L \cdot H = LH Area=12⋅base⋅height=12⋅2L⋅H=LH \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2L \cdot H = LH

So:

a_0 = \frac{1}{L} \cdot LH = H a0=1L⋅LH=H a_0 = \frac{1}{L} \cdot LH = H

  1. Calculation of a_nana_n:

a_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos\left(\frac{n\pi t}{L}\right) \, dt an=1L∫−LLf(t)cos⁡(nπtL) dt a_n = \frac{1}{L} \int_{-L}^{L} f(t) \cos\left(\frac{n\pi t}{L}\right) \, dt

Since the triangular function is odd, the cosine term (which is even) will give a null contribution:

a_n = 0 an=0 a_n = 0

  1. Calculation of b_nbnb_n:

b_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) \, dt bn=1L∫−LLf(t)sin⁡(nπtL) dt b_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) \, dt

Since the triangular function is symmetric about the axis yyy, we can just calculate the integral from 000 to LLL and multiply it by 2:

b_n = \frac{2}{L} \int_0^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) \, dt bn=2L∫0Lf(t)sin⁡(nπtL) dt b_n = \frac{2}{L} \int_0^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) \, dt

Let’s calculate the integral. The triangular function can be expressed as:

f(t) = \frac{H}{L} t \quad \text{for } 0 < t < L f(t)=HLtfor 0<t<L f(t) = \frac{H}{L} t \quad \text{for } 0 < t < L

So:

b_n = \frac{2}{L} \int_0^{L} \frac{H}{L} t \sin\left(\frac{n\pi t}{L}\right) \, dt bn=2L∫0LHLtsin⁡(nπtL) dt b_n = \frac{2}{L} \int_0^{L} \frac{H}{L} t \sin\left(\frac{n\pi t}{L}\right) \, dt

b_n = \frac{2H}{L^2} \left[ -\frac{L}{n\pi} t \cos\left(\frac{n\pi t}{L}\right) \right]_0^{L} + \frac{2H}{L^2} \cdot \frac{L^2}{(n\pi)^2} \left[ \sin\left(\frac{n\pi t}{L}\right) \right]_0^{L} bn=2HL2[−Lnπtcos⁡(nπtL)]0L+2HL2⋅L2(nπ)2[sin⁡(nπtL)]0L b_n = \frac{2H}{L^2} \left[ -\frac{L}{n\pi} t \cos\left(\frac{n\pi t}{L}\right) \right]_0^{L} + \frac{2H}{L^2} \cdot \frac{L^2}{(n\pi)^2} \left[ \sin\left(\frac{n\pi t}{L}\right) \right]_0^{L}

Continuing with the calculation of b_nbnb_n:

b_n = \frac{2H}{L^2} \left[ -\frac{L}{n\pi} L \cos\left(n\pi\right) + 0 \right] + \frac{2H}{L^2} \cdot \frac{L^2}{(n\pi)^2} \left[ \sin\left(n\pi\right) - 0 \right] bn=2HL2[−LnπLcos⁡(nπ)+0]+2HL2⋅L2(nπ)2[sin⁡(nπ)−0] b_n = \frac{2H}{L^2} \left[ -\frac{L}{n\pi} L \cos\left(n\pi\right) + 0 \right] + \frac{2H}{L^2} \cdot \frac{L^2}{(n\pi)^2} \left[ \sin\left(n\pi\right) - 0 \right]

Since \cos(n\pi) = (-1)^ncos⁡(nπ)=(−1)n\cos(n\pi) = (-1)^n and \sin(n\pi) = 0sin⁡(nπ)=0\sin(n\pi) = 0, we get:

b_n = \frac{2H}{L^2} \left[ -\frac{L^2}{n\pi} (-1)^n \right] + 0 bn=2HL2[−L2nπ(−1)n]+0 b_n = \frac{2H}{L^2} \left[ -\frac{L^2}{n\pi} (-1)^n \right] + 0

Simplifying:

b_n = \frac{2H L}{n\pi} (-1)^{n+1} bn=2HLnπ(−1)n+1 b_n = \frac{2H L}{n\pi} (-1)^{n+1}

Final Results

For the triangular function defined on [-L, L][−L,L][-L, L] with height HHH, the Fourier coefficients are:

  • a_0 = Ha0=Ha_0 = H
  • a_n = 0an=0a_n = 0
  • b_n = \frac{2H L}{n\pi} (-1)^{n+1}bn=2HLnπ(−1)n+1b_n = \frac{2H L}{n\pi} (-1)^{n+1}

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