Convolution Property of Continuous-Time Fourier Series

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, 𝐶_𝑛 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 𝐶_𝑛 and 𝐷_𝑛. 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

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

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

Manish Kumar Saini

Updated on: 06-Dec-2021

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