Fourier Properties

Question

If I have a function $h(x)$ defined as $f(x)\ast g(x)-n(x)$, can I just say that $H(\omega)=F(\omega)G(\omega)-N(\omega)$ Where the capital letters are the fourier transforms. I am familiar with the convolution property, but I am not sure how this works when I have the additional $n(x)$ . Thanks

Answer 1

Assume that $f$ is a signal that passes through some signal processing device, $g$ is the impulse response of the device and $n$ is for example transistor noise term. Because Fourier transform is somewhat limited, we cannot transform for example constant function $1(t)$, but constant distribution $1(\varphi)$. The transform of constant \begin{equation} \mathbb{1}: \mathcal{S} \rightarrow \mathbb{C}, \ \mathbb{1}(\varphi) = \int_{-\infty}^\infty 1(t) \varphi(t) dt \end{equation} is then \begin{eqnarray} \mathcal{F}\mathbb{1}(\varphi) & = & \mathbb{1}(\mathcal{F}\varphi) = \int_{-\infty}^\infty 1(\omega) \mathcal{F}\varphi(\omega) d\omega = \int_{-\infty}^\infty \mathcal{F}\varphi(\omega) e^{i\omega 0} d\omega \\ & = & \mathcal{F}^{-1} \mathcal{F} \varphi(0) = \varphi(0) = \delta_0(\varphi) \\ \\ \mathcal{F}\mathbb{1} & = & \delta_0 \ . \end{eqnarray} Here we have applied Fourier inversion theorem with Fourier transform \begin{eqnarray} \mathcal{F}\varphi(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty \varphi(t) e^{-i\omega t} dt \end{eqnarray} and the definition of Fourier transform of a distribution \begin{equation} \mathcal{F}\Lambda(\varphi) = \Lambda (\mathcal{F}\varphi) \ . \end{equation} In a similar way with the definition of $1(\varphi)$ we can define for any suitable function $f(t)$ a distribution \begin{eqnarray} F: \mathcal{S} \rightarrow \mathbb{C}, \ F(\varphi) = \int_{-\infty}^\infty f(t) \varphi(t) dt \ . \end{eqnarray} Then the definition \begin{eqnarray} (g*\Lambda)(\varphi) = \Lambda((g\circ(-I))*\varphi) \ , \end{eqnarray} where $I(t) = t$, satisfies \begin{eqnarray} (g*F)(\varphi) = \int_{-\infty}^\infty (g*f)(t) \varphi(t) dt \ . \end{eqnarray} This can be shown by interchanging the order of integration. We have applied the convolution \begin{equation} (f*g)(x) = \frac{1}{2\pi} \int_{-\infty}^\infty f(t) g(x-t) dt \ . \end{equation} Note that $g$ has to be chosen s.t. $(g\circ (-I))*\varphi \in \mathcal{S}$. This is true for example if $g(t) = \begin{cases} A e^{-\alpha t} \cos(\omega_0 t), & t \geq 0 \\ 0, & t < 0 \end{cases}$. Then the convolution can be differentiated by differentiating the function $\varphi$. Hence $D^k((g \circ(-I))*\varphi)$ is smooth. Its transform is $\mathcal{F}g(-\omega)\mathcal{F}D^k\varphi(\omega)$. Both $\mathcal{F}D^k\varphi$ and $\mathcal{F}g$ are smooth because $D^k\varphi$ and $g$ are rapidly decreasing. Hence the transform is also smooth. It follows that $D^k((g \circ (-I))*\varphi)$ is rapidly decreasing and is hence in $\mathcal{S}$. We now have \begin{eqnarray} \mathcal{F}(g*F)(\varphi) & = & (g*F)(\mathcal{F}\varphi) = \int_{-\infty}^\infty (g*f)(t)\mathcal{F}\varphi(t) dt = \int_{-\infty}^\infty \mathcal{F}(g*f)(\omega)\varphi(\omega) d\omega \\ & = & \int_{-\infty}^\infty \mathcal{F}g(\omega)\mathcal{F}f(\omega) \varphi(\omega) d\omega = \int_{-\infty}^\infty f(t) \mathcal{F}(\mathcal{F}g \varphi)(t) dt = F(\mathcal{F}(\mathcal{F}g \varphi)) \\ \\ \\ & = & \mathcal{F}F(\mathcal{F}g \varphi) = (\mathcal{F}g \mathcal{F}F)(\varphi) \\ \\ \\ \mathcal{F}(g*F) & = & \mathcal{F}g \mathcal{F}F \ , \end{eqnarray} where we have applied multiplication of a distribution by a pointwise defined function $\psi$ that is smooth with derivatives s.t. for each of them there is a polynomial s.t. the derivative doesn't grow faster than the polynomial. The product is defined by \begin{eqnarray} (\psi \Lambda)(\varphi) = \Lambda(\psi \varphi) \ . \end{eqnarray} The calculation above assumes that $f \in L^1$. Note that \begin{eqnarray} (\psi F)(\varphi) & = & \int_{-\infty}^\infty (\psi f)(t) \varphi(t) dt \ , \\ (\mathcal{F}F)(\varphi) & = & \int_{-\infty}^\infty \mathcal{F}f(\omega) \varphi(\omega) d\omega \ . \end{eqnarray} The latter can be shown by change of order of integration. The separation of convolution can be done without transforming $f$ under the integral. Definition of convolution of a distribution with a function implies \begin{eqnarray} \mathcal{F}(g*F)(\varphi) & = & (g*F)(\mathcal{F}\varphi) = F((g \circ(-I))*\mathcal{F}\varphi) = F(\mathcal{F}\mathcal{F}^{-1}((g \circ(-I))*\mathcal{F}\varphi)) \\ & = & F(\mathcal{F}(\frac{1}{2\pi}\mathcal{F}^{-1}(g \circ(-I)) \mathcal{F}^{-1}\mathcal{F}\varphi)) = F(\mathcal{F}(\mathcal{F}g \varphi)) = \mathcal{F}F(\mathcal{F}g \varphi) \\ & = & (\mathcal{F}g\mathcal{F}F)(\varphi) \\ \\ \mathcal{F}(g*F) & = & \mathcal{F}g \mathcal{F}F \ . \end{eqnarray} This is the same result, that was obtained using transform of $f$. Define \begin{eqnarray} N(\varphi) = \int_{-\infty}^\infty n(t) \varphi(t) dt \ . \end{eqnarray} The distribution $N$ is well-defined if there is a polynomial s.t. $n$ doesn't increase more rapidly than the polynomial. Note that for example gaussian random signal may get any value in an infinite time. In the case of transistor noise it can be assumed that $n$ is bounded. This is also a reasonable assumption because an unbounded voltage will break the device. Definitions \begin{eqnarray} h(t) & = & (g*f)(t) - n(t) \ , \\ H(\varphi) & = & \int_{-\infty}^\infty h(t)\varphi(t) dt \end{eqnarray} imply \begin{eqnarray} H(\varphi) & = & \int_{-\infty}^\infty ((g*f)(t) - n(t))\varphi(t) dt = \int_{-\infty}^\infty (g*f)(t)\varphi(t) dt - \int_{-\infty}^\infty n(t)\varphi(t) dt \\ & = & (g*F)(\varphi) - N(\varphi) = (g*F-N)(\varphi) \\ \\ H & = & g*F-N \\ \\ \\ \mathcal{F}H(\varphi) & = & \mathcal{F}(g*F-N)(\varphi) = (\mathcal{F}(g*F) - \mathcal{F}N)(\varphi) = (\mathcal{F}g\mathcal{F}F - \mathcal{F}N)(\varphi) \\ \\ \mathcal{F}H & = & \mathcal{F}g\mathcal{F}F - \mathcal{F}N \ , \end{eqnarray} where operator addition \begin{equation} (\Lambda_1+\Lambda_2)(\varphi) = \Lambda_1(\varphi) + \Lambda_2(\varphi) \end{equation} and linearity of Fourier transform of distributions \begin{eqnarray} \mathcal{F}(\Lambda_1+\Lambda_2)(\varphi) & = & (\Lambda_1+\Lambda_2)(\mathcal{F}\varphi) = \Lambda_1(\mathcal{F}\varphi) + \Lambda_2(\mathcal{F}\varphi) = \mathcal{F}\Lambda_1(\varphi) + \mathcal{F}\Lambda_2(\varphi) \\ & = & (\mathcal{F}\Lambda_1 + \mathcal{F}\Lambda_2)(\varphi) \\ \mathcal{F}(\Lambda_1+\Lambda_2) & = & \mathcal{F}\Lambda_1 + \mathcal{F}\Lambda_2 \end{eqnarray} are applied. Hence the expression of $h$ has a distribution transform. However it has to be remembered, that two distributions don't have a product. The transfer function has to be a pointwise defined function.

Answer 2

You are right if all functions involved are deterministic. However, looking at your model, I assume that $n(x)$ is a random noise term. In this case the Fourier transform of $n(x)$ does not exist. Wide-sense stationary random processes are characterized by their power spectral density, which is defined as the Fourier transform of their autocorrelation function ($E[\cdot]$ denotes expectation):

$$R_n(\tau)=E[n(t+\tau)n^*(t)]\\ S_n(\omega)=\mathcal{F}\{R_n(\tau)\}=\int_{-\infty}^{\infty}R_n(\tau)e^{-j\omega\tau}d\tau$$