Convolution Property of Continuous-Time Fourier Series

Signals and SystemsElectronics & ElectricalDigital Electronics

Fourier Series

If 𝑥(𝑡) is a periodic function with period T, then the continuous-time Fourier series of the function is defined as,

$$\mathrm{x(t)=\sum_{n=-\infty}^{\infty}C_ne^{jn\omega_{0}t}\:\:\:\:\:.....(1)}$$

Where, 𝐶𝑛 is the exponential Fourier series coefficient, that is given by

$$\mathrm{C_n=\frac{1}{T}\int_{t_0}^{t_0+T}x(t)e^{-jn\omega_0t}dt\:\:\:\:\:.....(2)}$$

Convolution Property of Fourier Series

According to the convolution property, the Fourier series of the convolution of two functions 𝑥1(𝑡) and 𝑥2(𝑡) in time domain is equal to the multiplication of their Fourier series coefficients in frequency domain.

If 𝑥1(𝑡) and 𝑥2(𝑡) are two periodic functions with time period T and with Fourier series coefficients 𝐶𝑛 and 𝐷𝑛. Then, if

$$\mathrm{x_1(t)\overset{FS}{\leftrightarrow}C_n}$$

$$\mathrm{x_2(t)\overset{FS}{\leftrightarrow}D_n}$$

Then, the convolution property of continuous time Fourier series states that

$$\mathrm{x_1(t)*x_2(t)\overset{FS}{\leftrightarrow}TC_nD_n}$$

Proof

By the definition of the Fourier series, we get,

$$\mathrm{FS[x_1(t)*x_2(t)]=\frac{1}{T}\int_{t_0}^{t_0+T}[x_1(t)*x_2(t)]e^{-jn\omega_0 t}dt}$$

Since [0 to T] or [𝑡0to (𝑡0 + 𝑇)] will have the same period, thus,

$$\mathrm{\Rightarrow FS[x_1(t)*x_2(t)]=\frac{1}{T}\int_{0}^{T}[x_1(t)*x_2(t)]e^{-jn\omega_0 t}dt\:\:\:\:.....(3)}$$

But from the definition of the convolution integral for a periodic signal, we obtain,

$$\mathrm{x_1(t)*x_2(t)=\int_{0}^{T}x_1(\tau)x_2(t-\tau)d\tau\:\:\:\:.....(4)}$$

$$\mathrm{x_1(t)*x_2(t)=\int_{0}^{T}x_1(t-\tau)x_2(\tau)d\tau\:\:\:\:.....(5)}$$

By substituting value of [𝑥1(𝑡) ∗ 𝑥2(𝑡)] from equation (4) into eqn. (3), we have,

$$\mathrm{FS[x_1(t)*x_2(t)]=\frac{1}{T}\int_{0}^{T}(\int_{0}^{T}x_1(\tau)x_2(t-\tau)d\tau)e^{-jn\omega_{0} t}dt\:\:\:\:......(6)}$$

Rearranging the order of integration in eqn. (6), we get

$$\mathrm{FS[x_1(t)*x_2(t)]=\frac{1}{T}\int_{0}^{T}x_1(\tau)(\int_{0}^{T}x_2(t-\tau)e^{-jn\omega_{0} t}dt)d\tau\:\:\:\:......(7)}$$

Substituting (𝑡 − 𝜏) = 𝑡0 in RHS of equation (7), then 𝑑𝑡 = 𝑑𝑡0, we get,

$$\mathrm{FS[x_1(t)*x_2(t)]=\frac{1}{T}\int_{0}^{T}x_1(\tau)(\int_{-\tau}^{T-\tau}x_2(t_0)e^{-jn\omega_{0} (t_0+\tau)}dt_{0})d\tau}$$

$$\mathrm{\Rightarrow FS[x_1(t)*x_2(t)]=T(\frac{1}{T}\int_{0}^{T}x_1(\tau)e^{-jn\omega_{0}t}d\tau)(\frac{1}{T}\int_{-\tau}^{T-\tau}x_2(t_0)e^{-jn\omega_{0} t_0}dt_{0})\:\:\:\:.....(8)}$$

On comparing equation (8) with equation (2), we get

$$\mathrm{ FS[x_1(t)*x_2(t)]=T[C_n][D_n]}$$

$$\mathrm{\therefore x_1(t)*x_2(t)\overset{FS}{\leftrightarrow}TC_n D_n\:\:\:(Hence,\:Proved)}$$

raja
Updated on 06-Dec-2021 13:15:15

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