# Convolution property of the Fourier Transform

Whatsapp
Recall from Chapter 23 that if $x(t)$ is the input excitation to a circuit with an impulse function of $h(t)$, then the output response $y(t)$ is given by the convolution integral
$$y(t)=h(t) * x(t)=\int_{-\infty}^{\infty} h(\lambda) x(t-\lambda) d \lambda \tag{1}$$
If $X(\omega), H(\omega)$, and $Y(\omega)$ are the Fourier transforms of $x(t), h(t)$, and $y(t)$, respectively, then
$$Y(\omega)=\mathcal{F}[h(t) * x(t)]=H(\omega) X(\omega)$$
which indicates that convolution in the time domain corresponds with multiplication in the frequency domain.
To derive the convolution property, we take the Fourier transform of both sides of Eq. (1) to get
$$Y(\omega)=\int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty} h(\lambda) x(t-\lambda) d \lambda\right] e^{-j \omega t} d t$$
Exchanging the order of integration and factoring $h(\lambda)$, which does not depend on $t$, we have
$$Y(\omega)=\int_{-\infty}^{\infty} h(\lambda)\left[\int_{-\infty}^{\infty} x(t-\lambda) e^{-j \omega t} d t\right] d \lambda$$
For the integral within the brackets, let $\tau=t-\lambda$ so that $t=\tau+\lambda$ and $d t=d \tau$. Then,
\begin{aligned}Y(\omega) &=\int_{-\infty}^{\infty} h(\lambda)\left[\int_{-\infty}^{\infty} x(\tau) e^{-j \omega(\tau+\lambda)} d \tau\right] d \lambda \\&=\int_{-\infty}^{\infty} h(\lambda) e^{-j \omega \lambda} d \lambda \int_{-\infty}^{\infty} x(\tau) e^{-j \omega \tau} d \tau=H(\omega) X(\omega)\end{aligned}
as expected. This result expands the phasor method beyond what was done with the Fourier series in the previous chapter.
To illustrate the convolution property, suppose both $h(t)$ and $x(t)$ are identical rectangular pulses, as shown in Fig. 1(a) and 1(b). According to the convolution property, the product of the sinc functions should give us the convolution of the rectangular pulses in the time domain.
Fig. 1: Graphical illustration of the convolution property.
Thus, the convolution of the pulses in Fig. 1(e) and the product of the sinc functions in Fig. 1(f) form a Fourier pair.
In view of the duality property, we expect that if convolution in the time domain corresponds with multiplication in the frequency domain, then multiplication in the time domain should have a correspondence in the frequency domain. This happens to be the case. If $f(t)=f_{1}(t) f_{2}(t)$, then
$$F(\omega)=\mathcal{F}\left[f_{1}(t) f_{2}(t)\right]=\frac{1}{2 \pi} F_{1}(\omega) * F_{2}(\omega)$$
or
$$F(\omega)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F_{1}(\lambda) F_{2}(\omega-\lambda) d \lambda$$
which is convolution in the frequency domain.
If we replace $x(t)$ with the unit step function $u(t)$ and $h(t)$ with $f(t)$ in Eq. (1), then
$$\int_{-\infty}^{\infty} f(\lambda) u(t-\lambda) d \lambda=f(t) * u(t) \tag{2}$$
But by the definition of the unit step function,
$$u(t-\lambda)=\left\{\begin{array}{ll}1, & t-\lambda > 0 \\0, & t-\lambda > 0\end{array}\right.$$
We can write this as
$$u(t-\lambda)=\left\{\begin{array}{ll}1, & \lambda < t \\0, & \lambda > t\end{array}\right.$$
Substituting this into Eq. (2) makes the interval of integration change from $[-\infty, \infty]$ to $[-\infty, t]$, and thus Eq. (2) becomes
$$\int_{-\infty}^{t} f(\lambda) d \lambda=u(t) * f(t)$$
Taking the Fourier transform of both sides yields
$$\mathcal{F}\left[\int_{-\infty}^{t} f(\lambda) d \lambda\right]=U(\omega) F(\omega) \tag{3}$$
The Fourier transform of the unit step function is
$$U(\omega)=\frac{1}{j \omega}+\pi \delta(\omega)$$
Substituting this into Eq. (3) gives
\begin{aligned}\mathcal{F}\left[\int_{-\infty}^{t} f(\lambda) d \lambda\right] &=\left(\frac{1}{j \omega}+\pi \delta(\omega)\right) F(\omega) \\&=\frac{F(\omega)}{j \omega}+\pi F(0) \delta(\omega)\end{aligned} \tag{4}
Note that in Eq. (4), $F(\omega) \delta(\omega)=F(0) \delta(\omega)$, since $\delta(\omega)$ is only nonzero at $\omega=0$.

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