Derivation of Linearity for CT signals by Xiaodian Xie
Suppose z(t) = {ax(t)+by(t)}, then the fourier transform of z is z(w)=\int\limits_{-\infty}^{\infty}(ax(t)+by(t))e^{(-\jmath wt)}dt z(w)=\int\limits_{-\infty}^{\infty}ax(t)e^{(-\jmath wt)}dt+\int\limits_{-\infty}^{\infty}by(t)e^{(-\jmath wt)}dt z(w)=a\int\limits_{-\infty}^{\infty}x(t)e^{(-\jmath wt)}dt+b\int\limits_{-\infty}^{\infty}y(t)e^{(-\jmath wt)}dt Because x(w)=\int\limits_{-\infty}^{\infty}x(t)e^{(-\jmath wt)}dt (Same for Y(w));So we can say that z(w)=a*x(w)+b*y(w)
If z(t) = {a*x(t)+b*y(t)}, then the \mathcal{F} is Z(w)=\int\limits_{-\infty}^{\infty}(a*x(t)+b*y(t))e^{(-\jmath wt)}dt
* Z(w)=\int\limits_{-\infty}^{\infty}a*x(t)e^{(-\jmath wt)}dt+\int\limits_{-\infty}^{\infty}b*y(t)e^{(-\jmath wt)}dt
o Z(w)=a\int\limits_{-\infty}^{\infty}x(t)e^{(-\jmath wt)}dt+b\int\limits_{-\infty}^{\infty}y(t)e^{(-\jmath wt)}dt
+ Since X(w)=\int\limits_{-\infty}^{\infty}x(t)e^{(-\jmath wt)}dt (Same for Y(w))
# Therefore, Z(w) = a * X(w) + b * Y(w)
