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6.5: Continuous Time Circular Convolution and the CTFS

  • Page ID
    22876
  • Introduction

    This module relates circular convolution of periodic signals in the time domain to multiplication in the frequency domain.

    Signal Circular Convolution

    Given a signal \(f(t)\) with Fourier coefficients \(c_n\) and a signal \(g(t)\) with Fourier coefficients \(d_n\), we can define a new signal, \(v(t)\), where \(v(t)=f(t) \circledast g(t)\). We find that the Fourier Series representation of \(v(t)\), \(a_n\), is such that \(a_n=c_nd_n\). \(f(t) \circledast g(t)\) is the circular convolution (Section 7.5) of two periodic signals and is equivalent to the convolution over one interval, i.e. \(f(t) \circledast g(t)=\int_{0}^{T} \int_{0}^{T} f(\tau) g(t-\tau) d \tau d t\).

    Note

    Circular convolution in the time domain is equivalent to multiplication of the Fourier coefficients.

    This is proved as follows

    \[\begin{align}
    a_{n} &=\frac{1}{T} \int_{0}^{T} v(t) e^{-\left(j \omega_{0} n t\right)} \mathrm{d} t \nonumber \\
    &=\frac{1}{T^{2}} \int_{0}^{T} \int_{0}^{T} f(\tau) g(t-\tau) \mathrm{d} \tau e^{-\left(\omega j_{0} n t\right)} \mathrm{d} t \nonumber \\
    &=\frac{1}{T} \int_{0}^{T} f(\tau)\left(\frac{1}{T} \int_{0}^{T} g(t-\tau) e^{-\left(j \omega_{0} n t\right)} \mathrm{d} t\right) \mathrm{d} \tau \nonumber \\
    &=\forall \nu, \nu=t-\tau:\left(\frac{1}{T} \int_{0}^{T} f(\tau)\left(\frac{1}{T} \int_{-\tau}^{T-\tau} g(\nu) e^{-\left(j \omega_{0}(\nu+\tau)\right)} \mathrm{d} \nu\right) \mathrm{d} \tau\right) \nonumber \\
    &=\frac{1}{T} \int_{0}^{T} f(\tau)\left(\frac{1}{T} \int_{-\tau}^{T-\tau} g(\nu) e^{-\left(j \omega_{0} n \nu\right)} \mathrm{d} \nu\right) e^{-\left(j \omega_{0} n \tau\right)} \mathrm{d} \tau \nonumber \\
    &=\frac{1}{T} \int_{0}^{T} f(\tau) d_{n} e^{-\left(j \omega_{0} n \tau\right)} \mathrm{d} \tau \nonumber \\
    &=d_{n}\left(\frac{1}{T} \int_{0}^{T} f(\tau) e^{-\left(j \omega_{0} n \tau\right)} \mathrm{d} \tau\right) \nonumber \\
    &=c_{n} d_{n}
    \end{align}\]

    Exercise

    Take a look at a square pulse with a period of \(T\).

    Figure \(\PageIndex{1}\)

    For this signal

    \[c_{n}=\left\{\begin{array}{l}
    \frac{1}{T} \text { if } n=0 \\
    \frac{1}{2} \frac{\sin \left(\frac{\pi}{2} n\right)}{\frac{\pi}{2} n} \text { otherwise }
    \end{array}\right.\]

    Take a look at a triangle pulse train with a period of \(T\).

    Figure \(\PageIndex{2}\)

    This signal is created by circularly convolving the square pulse with itself. The Fourier coefficients for this signal are \(a_{n}=c_{n}^{2}=\frac{1}{4} \frac{\sin ^{2}}{\left(\frac{\pi}{2} n\right)}\).

    Exercise \(\PageIndex{1}\)

    Find the Fourier coefficients of the signal that is created when the square pulse and the triangle pulse are convolved.

    Answer

    \(a_{n}=\left\{\begin{array}{ll}
    \text { undefined } & n=0 \\
    \frac{1}{8} \frac{\sin ^{3}\left(\frac{\pi}{2} n\right)}{\left(\frac{\pi}{2} n\right)^{3}} & \text { otherwise }
    \end{array}\right.\)

    Conclusion

    Circular convolution in the time domain is equivalent to multiplication of the Fourier coefficients in the frequency domain.

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