Expressions for the Trigonometric Fourier Series Coefficients

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,

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

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.,

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

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

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,

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

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

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.,

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

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

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,

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

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.,

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

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

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,

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

Manish Kumar Saini

Updated on: 08-Dec-2021

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