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
图解法计算卷积举例例f (t) ,h(t) 如图所示,求yzs(t)= h(t) * f (t) 。[解] 采用图形卷积 。 f ( t -τ)f (τ)反折f (-τ)平移t① t0时 , f ( t -τ)向左移f ( t -τ) h(τ) = 0,故yzs(t) = 0② 0≤t ≤1 时, f ( t -τ)向右移③ 1≤t ≤2时⑤ 3≤t 时f ( t -τ) h(τ) = 0,故yzs(t) = 0④ 2≤t ≤3 时0
图解法计算卷积举例例f (t) ,h(t) 如图所示,求yzs(t)= h(t) * f (t) 。[解] 采用图形卷积。 f ( t -τ)f (τ)反折f (-τ)平移t① t0时 , f ( t -τ)向左移f ( t -τ) h(τ) = 0,故yzs(t) = 0② 0≤t ≤1 时, f ( t -τ)向右移③ 1≤t ≤2时⑤ 3≤t 时f ( t -τ) h(τ) = 0,故yzs(t) = 0④ 2≤t ≤3 时0
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
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)
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)
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