Suppose you would like to measure a signal, which can be modelled as a Gaussian random variable with $1.0V$ mean and $0.2V$ standard deviation. If you measure this signal with your instrument, it will be subject to the instrument noise and the measured signal will be sum of the noise and the signal. Noise of your instrument can also be modelled as a Gaussian random variable with $0.1V$ mean and $0.1V$ standard deviation.
Write out exact expression to represent the pdf of the measured signal $Z$
My attempt:
Let $X$ represent the random variable for signal & $Y$ represent the random variable for noise then the measured signal $Z$ is given by:
$Z = X+Y$
To find PDF of $Z$, we first find $CDF$ and then take the derivative of it to get the PDF
$CDF$ is given by:
$F_Z(z)=F_Z(Z\leq z)=F_Z(X+Y\leq z)$
I don't know how to go further. Any help and guidance will be appreciated
Best Answer
If you assume that signal $X$ and noise $Y$ are independent, then the characteristic function of $Z=X+Y$ is $$\phi_Z(\omega)=\mathbb{E}[e^{i\omega Z}]$$ from which we can compute $$\phi_Z(\omega)=\mathbb{E}[e^{i\omega (X+Y)}]\underbrace{=}_{\textrm{indep.}}\mathbb{E}[e^{i\omega X}]\mathbb{E}[e^{i\omega Y}]=\phi_X(\omega)\phi_Y(\omega)$$ it is known that the characteristic function of a Gaussian density is $$\phi_j(\omega)=e^{i\mu_j-0.5\omega^2\sigma^2_j}$$ So if $X \sim N(\mu_X,\sigma_X^2)$ and $Y \sim N(\mu_Y,\sigma_Y^2)$ $$\phi_Z(\omega)=e^{i(\mu_X+\mu_Y)-0.5\omega^2(\sigma^2_X+\sigma^2_Y)}$$ which implies $Z \sim N(\mu_X+\mu_Y,\sigma^2_X+\sigma^2_Y)$. So the pdf of $Z=X+Y$ is $$f_Z(z)=\frac{1}{\sqrt{2\pi(\sigma^2_X+\sigma^2_Y)}}e^{-\frac{(z-(\mu_X+\mu_Y))^2}{2(\sigma^2_X+\sigma^2_Y)}}$$