| written 9.2 years ago by | • modified 9.2 years ago |
$(i) \space x[n]=n(\dfrac {-1^n}4)u(n) * (\dfrac {-1^{-n}}6)u(-n) \ (ii) \space x[n] =u[n-5]-u[n-10]$
Mumbai University > EXTC > Sem 4 > Signals and Systems
Marks : 10
Year : DEC 2014
| written 9.2 years ago by | • modified 9.2 years ago |
$(i) \space x[n]=n(\dfrac {-1^n}4)u(n) * (\dfrac {-1^{-n}}6)u(-n) \ (ii) \space x[n] =u[n-5]-u[n-10]$
Mumbai University > EXTC > Sem 4 > Signals and Systems
Marks : 10
Year : DEC 2014
| written 9.2 years ago by |
$$(i)\space x[n] = n(\dfrac {-1^n}4)u(n)*(\dfrac {-1^{-n}}6)u(-n) $$
Convolution in z-transform results in multiplication of the two,
By using Time Scaling property of z-transform,
$a^n u(-n)↔^z \dfrac 1{1-\dfrac za} \\ ∴x_1 [n]=(\dfrac {-1^{-n}}6)u(-n)=(-6)^n u(-n)= \dfrac 1{1-\dfrac z{-6}} \\ ∴x_1 (z)=\dfrac 6{z+6} ……….. (1) $
Similarly, $na^n u(n)↔^z \dfrac { az}{(z-a)^2} \\ …