- #1
Toyona10
- 31
- 0
Greetings,
We just had 1.5 classes about this topic (due to some unavoidable circumstances in our country...) so our teacher just scribbled and roughly explained them on the board...So after i got home i couldn't get some steps he did for convolution in z-transform:
find inverse z-transform of z^2/(z-2)(z-5)
so here, inverse of z/(z-2) is 2^n which is Un and inverse of z/(z-5) is 5^n which is Vn
so Vn*Un = Ʃ(from m=0 to n) Un-mVm
=Ʃ(from m=0 to n)2n-m5m
=Ʃ(from m=0 to n)2n(5m/2m
=2^nƩ(from m=0 to n)[5/2]m
=2^n[1+(5/2)+(5/2)^2 + (5/2)^3...(5/2)^n]
By convolution-
=2^n{[1(5/2)n+1 - 1]/(5/2 -1)}
=2^n[(5n+1-2n+1)/2n+1×3/2]
=1/3[5n+1-2n+1]
I don't get the last 3 steps, I would REALLY appreciate it if somebody would explain that ^^
We just had 1.5 classes about this topic (due to some unavoidable circumstances in our country...) so our teacher just scribbled and roughly explained them on the board...So after i got home i couldn't get some steps he did for convolution in z-transform:
find inverse z-transform of z^2/(z-2)(z-5)
so here, inverse of z/(z-2) is 2^n which is Un and inverse of z/(z-5) is 5^n which is Vn
so Vn*Un = Ʃ(from m=0 to n) Un-mVm
=Ʃ(from m=0 to n)2n-m5m
=Ʃ(from m=0 to n)2n(5m/2m
=2^nƩ(from m=0 to n)[5/2]m
=2^n[1+(5/2)+(5/2)^2 + (5/2)^3...(5/2)^n]
By convolution-
=2^n{[1(5/2)n+1 - 1]/(5/2 -1)}
=2^n[(5n+1-2n+1)/2n+1×3/2]
=1/3[5n+1-2n+1]
I don't get the last 3 steps, I would REALLY appreciate it if somebody would explain that ^^