性质求卷积和例例1 复合系统中h1(k) = ε(k), h2(k) = ε(k – 5),求复合系统的单位序列响应h (k) 。 解 根据h(k)的定义,有h(k)= [δ(k)* h1(k) –δ(k)* h2(k) ]* h1(k)= [h1(k) – h2(k) ]* h1(k) = h1(k) * h1(k) –h2(k) * h1(k) = ε(k)* ε(k) – ε(k – 5)
性质求卷积和例例1 复合系统中h1(k) = ε(k), h2(k) = ε(k – 5),求复合系统的单位序列响应h (k) 。 解 根据h(k)的定义,有h(k)= [δ(k)* h1(k) –δ(k)* h2(k) ]* h1(k)= [h1(k) – h2(k) ]* h1(k) = h1(k) * h1(k) –h2(k) * h1(k) = ε(k)* ε(k) – ε(k – 5)
卷积性质例1例1:f1(t) 如图, f2(t) = e–tε(t),求f1(t)* f2(t) 解: f1(t)* f2(t) = f1’(t)* f2(–1)(t)f1’(t) =δ (t) –δ (t –2) f1(t)* f2(t)=(1- e–t)ε(t) – [1- e–(t-2)]ε(t-2) 注意:当 f1(t)=1 , f2(t) = e–tε(t),套用 f1(t)* f2(t
卷积性质例2图(a)系统由三个子系统构成,已知各子系统的冲激响应 如图(b)所示。求复合系统的冲激响应 ,并画出它的波形。 (a)(b)解:如图(c)所示 (c)
卷积性质例3例:f1(t), f2(t)如图,求f1(t)* f2(t) 解: f1(t) = 2ε (t) –2ε (t –1)f2(t) = ε (t+1) –ε (t –1) f1(t)* f2(t) = 2 ε (t)* ε (t+1) –2 ε (t)* ε (t –1) –2ε (t –1)* ε (t+1) +2ε (t –1)* ε (t –1) 由于ε (t)* ε (t) =
卷积性质例3例:f1(t), f2(t)如图,求f1(t)* f2(t) 解: f1(t) = 2ε (t) –2ε (t –1)f2(t) = ε (t+1) –ε (t –1) f1(t)* f2(t) = 2 ε (t)* ε (t+1) –2 ε (t)* ε (t –1) –2ε (t –1)* ε (t+1) +2ε (t –1)* ε (t –1) 由于ε (t)* ε (t) =
卷积性质例1例1:f1(t) 如图, f2(t) = e–tε(t),求f1(t)* f2(t) 解: f1(t)* f2(t) = f1’(t)* f2(–1)(t)f1’(t) =δ (t) –δ (t –2) f1(t)* f2(t)=(1- e–t)ε(t) – [1- e–(t-2)]ε(t-2) 注意:当 f1(t)=1 , f2(t) = e–tε(t),套用 f1(t)* f2(t
卷积性质例2图(a)系统由三个子系统构成,已知各子系统的冲激响应 如图(b)所示。求复合系统的冲激响应 ,并画出它的波形。 (a)(b)解:如图(c)所示
用定义求卷积和例例:f (k) = a kε(k),h(k) = b kε(k) ,求yzs(k)。解: yzs(k) = f (k) * h(k)当i0,ε(i) = 0;当ik时,ε(k - i) = 0ε(k)*ε(k) = (k+1)ε(k)
图解法求卷积和例例:f1(k)、 f2(k)如图所示,已知f(k) = f1(k)* f2(k),求f(2) =?解:(1)换元(2) f2(i)反转得f2(– i)(3) f2(–i)右移2得f2(2–i)(4) f1(i)乘f2(2–i)(5)求和,得f(2) = 45
违法有害信息,请在下方选择原因提交举报