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Question Number 158182 by Eric002 last updated on 31/Oct/21

Commented by Eric002 last updated on 31/Oct/21

$$findtheevenandoddcomponentsof$$$$X\left(t\right).$$

Answered by aleks041103 last updated on 05/Nov/21

$$givenafunctionf\left(x\right)wecandecompose$$$$itinanevenandoddfunctionse\left(x\right)ando\left(x\right):$$$$f\left(x\right)=e\left(x\right)+o\left(x\right)$$$$\Rightarrow f(-x)=e(-x)+o(-x)=e\left(x\right)-o\left(x\right)$$$$\Rightarrow e\left(x\right)=\frac{1}{2}(f\left(x\right)+f(-x))$$$$o\left(x\right)=\frac{1}{2}(f\left(x\right)-f(-x))$$$$Now:$$$$x\left(t\right)=\{\begin{array}{l}0,t\in (-\mathrm{\infty },-1]\cup [2,\mathrm{\infty })\\ 1+t,t\in (-1,0]\\ 1,t\in (0,2)\end{array}=\{\begin{array}{l}0,t\in (-\mathrm{\infty },-2]\cup [2,\mathrm{\infty })\\ 0,t\in (-2,-1]\\ 1+t,t\in (-1,0)\\ 1,t\in [0,1)\\ 1,t\in [1,2)\end{array}$$$$and$$$$x(-t)=\{\begin{array}{l}0,t\in (-\mathrm{\infty },-2]\cup [1,\mathrm{\infty })\\ 1-t,t\in [0,1)\\ 1,t\in (-2,0)\end{array}=\{\begin{array}{l}0,t\in (-\mathrm{\infty },-2]\cup [2,\mathrm{\infty })\\ 1,t\in (-2,-1]\\ 1,t\in (-1,0)\\ 1-t,t\in [0,1)\\ 0,t\in [1,2)\end{array}$$$$Then:$$$$e\left(t\right)=\frac{1}{2}(x\left(t\right)+x(-t))$$$$e\left(t\right)=\frac{1}{2}(\{\begin{array}{l}0,t\in (-\mathrm{\infty },-2]\cup [2,\mathrm{\infty })\\ 0,t\in (-2,-1]\\ 1+t,t\in (-1,0)\\ 1,t\in [0,1)\\ 1,t\in [1,2)\end{array}+\{\begin{array}{l}0,t\in (-\mathrm{\infty },-2]\cup [2,\mathrm{\infty })\\ 1,t\in (-2,-1]\\ 1,t\in (-1,0)\\ 1-t,t\in [0,1)\\ 0,t\in [1,2)\end{array})=$$$$=\frac{1}{2}\{\begin{array}{l}0+0,t\in (-\mathrm{\infty },-2]\cup [2,\mathrm{\infty })\\ 0+1,t\in (-2,-1]\\ 1+t+1,t\in (-1,0)\\ 1+1-t,t\in [0,1)\\ 1+0,t\in [1,2)\end{array}=$$$$=\{\begin{array}{l}0,t\in (-\mathrm{\infty },-2]\cup [2,\mathrm{\infty })\\ 1/2,t\in (-2,-1]\\ 1+t/2,t\in (-1,0)\\ 1-t/2,t\in [0,1)\\ 1/2,t\in [1,2)\end{array}=\{\begin{array}{l}1-\frac{t}{2},t\in [0,1)\\ 1/2,t\in [1,2)\\ 0,t\in [2,\mathrm{\infty })\\ e(-t),t<0\end{array}$$$$\Rightarrow e\left(t\right)=\{\begin{array}{l}1-\frac{t}{2},t\in [0,1)\\ 1/2,t\in [1,2)\\ 0,t\in [2,\mathrm{\infty })\\ e(-t),t<0\end{array}$$$$o\left(t\right)=\frac{1}{2}(x\left(t\right)-x(-t))=$$$$=\frac{1}{2}(\{\begin{array}{l}0,t\in (-\mathrm{\infty },-2]\cup [2,\mathrm{\infty })\\ 0,t\in (-2,-1]\\ 1+t,t\in (-1,0)\\ 1,t\in [0,1)\\ 1,t\in [1,2)\end{array}+\{\begin{array}{l}0,t\in (-\mathrm{\infty },-2]\cup [2,\mathrm{\infty })\\ 1,t\in (-2,-1]\\ 1,t\in (-1,0)\\ 1-t,t\in [0,1)\\ 0,t\in [1,2)\end{array})$$$$=\frac{1}{2}\{\begin{array}{l}0-0,t\in (-\mathrm{\infty },-2]\cup [2,\mathrm{\infty })\\ 0-1,t\in (-2,-1]\\ 1+t-1,t\in (-1,0)\\ 1-1+t,t\in [0,1)\\ 1-0,t\in [1,2)\end{array}=$$$$=\{\begin{array}{l}0,t\in (-\mathrm{\infty },-2]\cup [2,\mathrm{\infty })\\ -1/2,t\in (-2,-1]\\ t/2,t\in (-1,0)\\ t/2,t\in [0,1)\\ 1/2,t\in [1,2)\end{array}=\{\begin{array}{l}\frac{t}{2},t\in [0,1)\\ 1/2,t\in [1,2)\\ 0,t\in [2,\mathrm{\infty })\\ -o(-t),t<0\end{array}$$$$\Rightarrow o\left(t\right)=\{\begin{array}{l}\frac{t}{2},t\in [0,1)\\ 1/2,t\in [1,2)\\ 0,t\in [2,\mathrm{\infty })\\ -o(-t),t<0\end{array}$$

Commented by Eric002 last updated on 05/Nov/21

$$welldonesir$$

Commented by aleks041103 last updated on 05/Nov/21

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