Linear convolutions in DSP

Consider relaxed LTI system. A relaxed system means if input x (n) is zero then output y (n) = 0 is zero. Let us say unit impulse 𝛿 (n) is applied to this system then its output is denoted by h (n). h (n) we called as the impulse response of the system.

Step 1 :

         T
𝛿 (n) → y (n) = h (n)

Step 2 :

             T
𝛿 (n-k)  → y (n) = h (n-k)

Step 3 :

                       T

x (k) 𝛿 (n-k)    →  y (n) = x (k) h(n-k) 

Step 4 :

 ∞                                       T   ∞  

∑        = x (k) 𝛿 (n-k) =   y (n) →  ∑      x (k) h(n-k)

k= -∞                                             k= -∞

y (n) = 

 ∞                                        

∑        = x (k) h (n-k)  

k= -∞           

                   
                          ∞

x (n) * h (n)  =   ∑      x (k) h(n-k)

                          k= -∞                                            

                          

Consider relaxed LTI system. A relaxed system means if input x (n) is zero then output y (n) = 0 is zero. Let us say unit impulse 𝛿 (n) is applied to this system then its output is denoted by h (n). h (n) we called as the impulse response of the system.

Step 1 :

         T
𝛿 (n) → y (n) = h (n)

Step 2 :

             T
𝛿 (n-k)  → y (n) = h (n-k)

Step 3 :

                       T

x (k) 𝛿 (n-k)    →  y (n) = x (k) h(n-k) 

Step 4 :

 ∞                                       T   ∞  

∑        = x (k) 𝛿 (n-k) =   y (n) →  ∑      x (k) h(n-k)

k= -∞                                             k= -∞

y (n) = 

 ∞                                        

∑        = x (k) h (n-k)  

k= -∞           

                   
                          ∞

x (n) * h (n)  =   ∑      x (k) h(n-k)

                          k= -∞