0
1.2kviews
Calculate z-transform of the following signal:

$(i) \space x[n]=n(\dfrac {-1^n}4)u(n) * (\dfrac {-1^{-n}}6)u(-n) \ (ii) \space x[n] =u[n-5]-u[n-10]$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : DEC 2014

0
4views

$$(i)\space x[n] = n(\dfrac {-1^n}4)u(n)*(\dfrac {-1^{-n}}6)u(-n)$$

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, $na^n u(n)↔^z \dfrac { az}{(z-a)^2} \\ ∴x_2 [n]=n(\dfrac {-1^n}4)u(n) = \dfrac {(-1/4)z}{(z-(\dfrac {-1}4))^2 } \\ ∴x_2 (z)= \dfrac {(-1/4)}z{(z+(1/4))^2} ………. (2)$

From (1) and (2) we get,

$X (z) = x_1 (z) x_2 (z) \\ X(z)= \dfrac {-3/2}{(z+6)} \dfrac z{(z+(1/4))^2}$

$$(ii) x[n] = u [n-5]-u[n-10]$$

By using Time Shifting property of z-transform that states

If $x[n] ↔^z x(z) \\ \text {Then } \space x[n-k]↔^z z^{-k} x(z) \\ X (z) = z^{-5} \dfrac z{z-1} - z^{-10} \dfrac z{z-1}\\ =\dfrac z{z-1} [\dfrac 1{z^5} -\dfrac 1{z^{10}}] \\ =\dfrac z{z-1} \dfrac 1{z^5} [ 1-\dfrac 1{z^5} ] \\ =\dfrac 1{z-1} \dfrac 1{z^4} [\dfrac {z^5-1}{z^5} ] \\ X(z) = \dfrac z{z-1} [\dfrac {z^5-1}{z^9} ]$