Digital Signal Processing

Properties of Z Transform

Question 1
Marks : +2 | -2
Pass Ratio : 100%
If x(n) is causal, then \\(\\lim_{z\\rightarrow\\infty}\\) X(z)=?
x(-1)
x(1)
x(0)
Cannot be determined
Explanation:
According to the initial value theorem, X(z)=x(0)+x(1)z-1+x(2)z-2+….
Question 2
Marks : +2 | -2
Pass Ratio : 100%
If X(z) is the z-transform of the signal x(n) then what is the z-transform of anx(n)?
X(az)
X(az-1)
X(a-1z)
None of the mentioned
Explanation:
We know that from the definition of z-transform
Question 3
Marks : +2 | -2
Pass Ratio : 100%
What is the z-transform of the signal x(n)=sin(jω0n)u(n)?
\\(\\frac{z^{-1} sin\\omega_0}{1+2z^{-1} cos\\omega_0+z^{-2}}\\)
\\(\\frac{z^{-1} sin\\omega_0}{1-2z^{-1} cos\\omega_0-z^{-2}}\\)
\\(\\frac{z^{-1} cos\\omega_0}{1-2z^{-1} cos\\omega_0+z^{-2}}\\)
\\(\\frac{z^{-1} sin\\omega_0}{1-2z^{-1} cos\\omega_0+z^{-2}}\\)
Explanation:
By Euler’s identity, the given signal x(n) can be written as
Question 4
Marks : +2 | -2
Pass Ratio : 100%
What is the signal whose z-transform is given as X(z)=\\(\\frac{1}{2Ï€j} \\oint X_1 (v) X_2 (\\frac{z}{v})v^{-1} dv\\)?
x1(n)*x2(n)
x1(n)*x2(-n)
x1(n).x2(n)
x1(n)*x2*(n)
Explanation:
From the convolution property in z-domain we have,
Question 5
Marks : +2 | -2
Pass Ratio : 100%
What is the z-transform of the signal x(n)=an(sinω0n)u(n)?
\\(\\frac{az^{-1} sin\\omega_0}{1+2 az^{-1} cos\\omega_0+a^2 z^{-2}}\\)
\\(\\frac{az^{-1} sin\\omega_0}{1-2 az^{-1} cos\\omega_0- a^2 z^{-2}}\\)
\\(\\frac{(az)^{-1} cos\\omega_0}{1-2 az^{-1} cos\\omega_0+a^2 z^{-2}}\\)
\\(\\frac{az^{-1} sin\\omega_0}{1-2 az^{-1} cos\\omega_0+a^2 z^{-2}}\\)
Explanation:
Question 6
Marks : +2 | -2
Pass Ratio : 100%
What is the convolution x(n) of the signals x1(n)={1,-2,1} and x2(n)={1,1,1,1,1,1}?
{1,1,0,0,0,0,1,1}
{-1,-1,0,0,0,0,-1,-1}
{-1,1,0,0,0,0,1,-1}
{1,-1,0,0,0,0,-1,1}
Explanation:
Question 7
Marks : +2 | -2
Pass Ratio : 100%
What is the signal x(n) whose z-transform X(z)=log(1+az-1);|z|>|a|?
\\((-1)^n.\\frac{a^n}{n}.u(n-1)\\)
\\((-1)^n.\\frac{a^n}{n}.u(n+1)\\)
\\((-1)^{n-1}.\\frac{a^n}{n}.u(n-1)\\)
\\((-1)^{n-1}.\\frac{a^n}{n}.u(n+1)\\)
Explanation:
Given X(z)=log(1+az-1)
Question 8
Marks : +2 | -2
Pass Ratio : 100%
What is the z-transform of the signal defined as x(n)=u(n)-u(n-N)?
\\(\\frac{1+z^N}{1+z^{-1}}\\)
\\(\\frac{1-z^N}{1+z^{-1}}\\)
\\(\\frac{1+z^{-N}}{1+z^{-1}}\\)
\\(\\frac{1-z^{-N}}{1-z^{-1}}\\)
Explanation:
Question 9
Marks : +2 | -2
Pass Ratio : 100%
X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal nx(n)?
\\(-z\\frac{dX(z)}{dz}\\)
\\(z\\frac{dX(z)}{dz}\\)
\\(-z^{-1}\\frac{dX(z)}{dz}\\)
\\(z^{-1}\\frac{dX(z)}{dz}\\)
Explanation:
Question 10
Marks : +2 | -2
Pass Ratio : 100%
If X(z) is the z-transform of the signal x(n), then what is the z-transform of x*(n)?
X(z*)
X*(z)
X*(-z)
X*(z*)
Explanation:
According to the conjugation property of z-transform, we have