Trigonometric Fourier Series Coefficients

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The infinite series of sine and cosine terms of frequencies $0,\:\omega_{0},\:2\omega_{0},\:3\omega_{0},\:.... \:k\omega_{0}$ is known as trigonometric Fourier series and can written as,

$$\mathrm{x(t)\:=\:a_{0}\:+\:\sum_{n=1}^{\infty}a_{n}\:cos\:n\omega_{0}t\:+\:b_{n}\:sin\:n\:\omega_{0}\:t\:\:\dotso\: (1)}$$

Here, the constant $a_{0},\:a_{n}$ and $b_{n}$ are called trigonometric Fourier series coefficients.

Evaluation of a₀

To evaluate the coefficient $a_{0}$, we shall integrate the equation (1) on both sides over one period, i.e.,

$$\mathrm{\int_{t_{0}}^{(t_{0}+T)}x(t)\:dt\:=\:a_{0}\int_{t_{0}}^{(t_{0}+T)}dt \:+\: \int_{t_{0}}^{(t_{0}+T)} \left(\sum_{n=1}^{\infty}a_{n}\:cos\:n\:\omega_{0} t \:+\:b_{n}\:sin\:n\:\omega_{0} t\right)dt}$$

$$\mathrm{\Rightarrow\:\int_{t_{0}}^{(t_{0}+T)}x(t)\:dt \:=\: a_{0}T \:+\: \sum_{n=1}^{\infty}a_{n} \int_{t_{0}}^{(t_{0}+T)}\:cos\:n\omega_{0} t\:dt+\sum_{n=1}^{\infty}b_{n}\int_{t_{0}}^{(t_{0}+T)}sin\:n\omega_{0} t\:dt \:\:\dotso\: (2)}$$

As we know that the net areas of sinusoids over complete periods are zero for any non-zero integer n and any time $t_{0}$. Therefore,

$$\mathrm{\int_{t_{0}}^{(t_{0}+T)}cos\:n\omega_{0} t\:dt \:=\: 0\:\:and\:\:\int_{t_{0}}^{(t_{0}+T)}sin\:n\omega_{0} t\:dt\:=\:0}$$

Hence, from equation (2), we get,

$$\mathrm{\int_{t_{0}}^{(t_{0}+T)}x(t)\:dt \:=\: a_{0}T}$$

$$\mathrm{\therefore\:a_{0} \:=\: \frac{1}{T}\int_{t_{0}}^{(t_{0}+T)}x(t)\:dt\:\:\dotso\: (3)}$$

Using equation (3), we can obtain the value of the Fourier coefficient $a_{0}$.

Evaluation of a_n

To evaluate the Fourier coefficient $a_{n}$, multiply both sides of the equation (1) by $cos\:m\omega_{0}t\:dt$ and then integrate over one period, i.e.,

$$\mathrm{\int_{t_{0}}^{(t_{0}+T)}x(t)\:cos\:m\:\omega_{0}\:t\:dt}$$

$$\mathrm{=\:a_{0}\int_{t_{0}}^{(t_{0}+T)}cos\:m\omega_{0}t\:dt \:+\: \sum_{n=1}^{\infty}a_{n}\int_{t_{0}}^{(t_{0}+T)} cos(n\omega_{0} t)\:cos(m\omega_{0} t)dt \:+\:\sum_{n=1}^{\infty}b_{n}\int_{t_{0}}^{(t_{0}+T)}sin(n\omega_{0} t)\:cos(m\omega_{0} t)dt\:\:\dotso\: (4)}$$

When m = n, then the first and third integrals in the equation (4) are equal to zero and the second integral is equal to $\left(\frac{T}{2}\right)$. Therefore,

$$\mathrm{\int_{t_{0}}^{(t_{0}+T)}x(t)\:cos\:m\omega_{0} t\:dt \:=\:a_{m}\left(\frac{T}{2}\right)}$$

Since m = n,

$$\mathrm{\therefore\:a_{n}\:=\:\frac{2}{T}\int_{t_{0}}^{(t_{0}+T)}x(t)\:cos\:n\omega_{0} t\:dt\:\:\dotso\: (5)}$$

Evaluation of b_n

To evaluate the Fourier coefficient $b_{n}$, multiply both sides of the equation (1) by $sin\:m\omega_{0} t$and then integrate over one period, i.e.,

$$\mathrm{\int_{t_{0}}^{(t_{0}+T)}x(t)\:sin\:m\omega_{0}t\:dt}$$

$$\mathrm{=\:a_{0}\int_{t_{0}}^{(t_{0}+T)}sin\:m\omega_{0}t\:dt \:+\: \sum_{n=1}^{\infty}a_{n}\int_{t_{0}}^{(t_{0}+T)} cos(n\omega_{0} t)\:sin(m\omega_{0} t)dt \:+\: \sum_{n=1}^{\infty}b_{n}\int_{t_{0}}^{(t_{0}+T)}sin(n\omega_{0} t)\:sin(m\omega_{0} t)dt\:\:\dotso\: (6)}$$

When m = n, then the first and second integrals in the equation (6) are equal to zero and the third integral is equal to $\left(\frac{T}{2} \right)$. Therefore,

$$\mathrm{\int_{t_{0}}^{(t_{0}+T)}x(t)\:sin\:m\omega_{0} t\:dt\:=\:b_{m}\left(\frac{T}{2}\right)}$$

Since m = n,

$$\mathrm{\therefore\:b_{n}\:=\:\frac{2}{T}\int_{t_{0}}^{(t_{0}+T)}x(t)\:sin\:n\omega_{0} t\:dt\:\:\dotso\: (7)}$$