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)
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)
用定义计算卷积举例例:f (t) = e t,(-∞t∞),h(t) = (6e-2t – 1)ε(t),求yzs(t)。解: yzs(t) = f (t) * h(t)当t τ,即τ t时,ε(t -τ) = 0
用定义计算卷积举例例:f (t) = e t,(-∞t∞),h(t) = (6e-2t – 1)ε(t),求yzs(t)。解: yzs(t) = f (t) * h(t)当t τ,即τ t时,ε(t -τ) = 0
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)
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
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
不进位乘法求卷积和例例 f1(k) ={0, 2 , 1 , 5,0} ↑k=1 f2(k) ={0, 3 , 4,0,6,0} ↑k=03 ,4,0,62 , 1 , 5解×15 ,20, 0, 303 ,4,0,66 ,8,0, 12+ 6 ,11,19,32,6,30求f(k) = f1(k)* f2(k)f(k) = {0,6 ,11,19,32,6,30}↑k=1
×不进位乘法求卷积和例例 f1(k) ={0, 2 , 1 , 5,0} ↑k=1 f2(k) ={0, 3 , 4,0,6,0} ↑k=03 ,4,0,62 , 1 , 5解15 ,20, 0, 303 ,4,0,66 ,8,0, 12+ 6 ,11,19,32,6,30求f(k) = f1(k)* f2(k)f(k) = {0,6 ,11,19,32,6,30}↑k=1
违法有害信息,请在下方选择原因提交举报