Difference between revisions of "HW 3.3 - Xiaodian Xie ECE301 Summer2009" - Rhea

Revision as of 18:01, 8 July 2009

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)=d^(-1)/dt((ax(t)+by(t))*exp(-jwt))

so z(w)=d^(-1)/dt(ax(t)*exp(-jwt)))+d^(-1)/dt(ay(t)*exp(-jwt)))

So z(w)=ax(w)+by(w)

@@ Line 1: / Line 1: @@
 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
+Suppose z(t) = {ax(t)+by(t)}, then the fourier transform of z is z(w)=d^(-1)/dt((ax(t)+by(t))*exp(-jwt))
-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)
+so z(w)=d^(-1)/dt(ax(t)*exp(-jwt)))+d^(-1)/dt(ay(t)*exp(-jwt)))
+So z(w)=ax(w)+by(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)