첫 번째 질문

[eqn: 문제13.1)# # 임펄스`응답이``h(t)=e ^{-t} `u(t)``이고``입력신호가``v _{i} (t)= LEFT [ u(t)-u(t-1) RIGHT ] ``일``때,# # 컨벌루션``적분을``이용해서``출력신호``v _{o} (t)=h(t) AST v _{i} (t)을``구하는``경우# # v _{o} (t)= int _{- INF } ^{INF } {h(t- tau )`v _{i} ( tau )d tau }# # ```````````````= int _{- INF } ^{0} {h(t- tau )`v _{i} ( tau )d tau + int _{0} ^{t} {h(t- tau )`v _{i} ( tau )d tau + int _{t} ^{INF } {h(t- tau )`v _{i} ( tau )d tau }}}# # ``````````````````= int _{0} ^{t} {h(t- tau )v _{i} ( tau )d tau ```}# # < BECAUSE tau <0;``입력이`일어나기`전이므로````v _{i} ( tau )=u( tau )-u( tau -1)=0```그리고``인과성에``의해`# # ````````````````시스템의``임펄스``응답``h(t- tau )=0```` RARROW ` int _{- INF } ^{0} {h(t- tau )`v _{i} ( tau )d tau =0,}# # `````````````````t< tau `;``인과성에``의해``현재의`출력``v _{o} (t)이```미래의``입력v _{i} ( tau )에``의해``# # ```````````````````일어날``수``없기``때문에``미래의``입력의``영향``v _{i} ( tau )=0`` RARROW int _{t} ^{INF } {h(t- tau )v _{i} ( tau )d tau } =0># # THEREFORE v _{o} (t)= int _{0} ^{t} {h(t- tau )x( tau )d tau ``:``0 LEQ tau LEQ t;``입력이``출력보다``먼저``일어나므로```}# ````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````인과성에``위배되지``않는다.]

입력이 일어나기 전의 상황과 인과성에 의해 [eqn: THEREFORE v _{o} (t)= int _{0} ^{t} {h(t- tau )x( tau )d tau]과 같이

적분 범위를 줄여 보았는데 논리전개 과정에서 틀린 부분이 있으면 말씀해 주십시오!

두 번째 질문

[eqn: 만약``위의``전개과정이``맞다면``v _{o} (t)= int _{0} ^{t} {h(t- tau )`v _{i} ( tau )d tau } ``을``이용해서``# # 문제13.1)``(c)의``v _{o} (t)를``구해도``되지``않나요?# # v _{o} (t)= int _{0} ^{t} {h(t- tau )`v _{i} ( tau )d tau = int _{0} ^{t} {e ^{tau -t} `u(t- tau )` LEFT [ u( tau )-u( tau -1) RIGHT ] d tau }}# # ``````````````````= int _{0} ^{t} {e ^{tau -t}} ` LEFT [ u( tau )-u( tau -1) RIGHT ] d tau ``< BECAUSE u(t- tau )=1``(t> tau )># # ````````````````````=e ^{-t} int _{0} ^{t} {e ^{tau } ` LEFT [ u( tau )-u( tau -1) RIGHT ] d tau }# # i)``0 LEQ t LEQ 1;``v _{o} (t)=e ^{-t} int _{0} ^{t} {e ^{tau } `d tau } ``< BECAUSE 0 LEQ tau LEQ t LEQ 1`` RARROW ``u( tau )-u( tau -1)=1># # `````````v _{o} (t)=e ^{-t} int _{0} ^{t} {e ^{tau } d tau =e ^{-t}} ` LEFT [ e ^{tau } RIGHT ] _{0}^{t} =e ^{-t} `(e ^{t} -1)=1-e ^{-t}# # `````````` THEREFORE v _{o} (t)=(1-e ^{-t} )` LEFT [ u(t)-u(t-1) RIGHT ] [V]# # ii)`t GEQ 1;`v _{o} (t)=e ^{-t} LEFT [ int _{0} ^{1} {e ^{tau }} `d tau +0 RIGHT ] ``< BECAUSE t GEQ 1;``u(t)-u(t-1)=0`` RARROW int _{1} ^{t} {e ^{tau }} LEFT [ u(t)-u(t-1) RIGHT ] d tau =0># # ````````````````````````````````````````````````=e ^{-t} int _{0} ^{1} {e ^{tau } `d tau =e ^{-t}} ` LEFT [ e ^{tau } RIGHT ] _{0}^{1} =(e-1)e ^{-t}# # ```````````````````````````````` THEREFORE v _{o} (t)=(e-1)e ^{-t} `u(t-1)`[V]# # `i),``ii)에``의해``v _{o} (t)=(1-e ^{-t} )` LEFT [ u(t)-u(t-1) RIGHT ] +(e-1)e ^{-t} `u(t-1)# # `````````````````````````````````````````````````````````````````=(1-e ^{-t} )`u(t)-(1-e ^{-(t-1)} )`u(t-1)````````]

[eqn: THEREFORE v _{o} (t)=(1-e ^{-t} )`u(t)-(1-e ^{-(t-1)} )`u(t-1)`[V]]

위의 풀이 과정에서 잘못된 부분이 있으면 수정해 주시면 고맙겠습니다.