Digital Signal Processing

Properties of Z Transform

Question 1
Marks : +2 | -2
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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
Question 2
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If Z{x1(n)}=X1(z) and Z{x2(n)}=X2(z) then Z{x1(n)*x2(n)}=?
X1(z).X2(z)
X1(z)+X2(z)
X1(z)*X2(z)
None of the mentioned
Explanation:
According to the convolution property of z-transform, the z-transform of convolution of two sequences is the product of their respective z-transforms.
Question 3
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If X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal x(-n)?
X(-z)
X(z-1)
X-1(z)
None of the mentioned
Explanation:
From the definition of z-transform, we have
Question 4
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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 5
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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 6
Marks : +2 | -2
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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 7
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Pass Ratio : 100%
If x(n) is an imaginary sequence, then the z-transform of the real part of the sequence is?
\\(\\frac{1}{2}\\)[X(z)+X*(z*)]
\\(\\frac{1}{2}\\)[X(z)-X*(z*)]
\\(\\frac{1}{2}\\)[X(-z)-X*(z*)]
\\(\\frac{1}{2}\\)[X(-z)+X*(z*)]
Explanation:
If x(N) is an imaginary sequence, then the real part of x(n) is given as
Question 8
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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 9
Marks : +2 | -2
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If Z{x(n)}=X(z) and the poles of X(z) are all inside the unit circle, then the final value of x(n) as \\(n\\rightarrow\\infty\\) is given by i.e., \\(\\lim_{n\\rightarrow\\infty}\\)x(n)=?
\\(\\lim_{z \\rightarrow 1} [(z-1) ⁡ X(z)] \\)
\\(\\lim_{z \\rightarrow 0} [(z-1) ⁡ X(z)] \\)
\\(\\lim_{z \\rightarrow -1} [(z-1) X(z)] \\)
\\(\\lim_{z \\rightarrow 1} [(z+1) ⁡ X(z)] \\)
Explanation:
According to the Final Value theorem of z-transform we have,
Question 10
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Pass Ratio : 100%
What is the z-transform of the signal x(n)=[3(2n)-4(3n)]u(n)?
\\(\\frac{3}{1-2z^{-1}}-\\frac{4}{1-3z^{-1}}\\)
\\(\\frac{3}{1-2z^{-1}}-\\frac{4}{1+3z^{-1}}\\)
\\(\\frac{3}{1-2z}-\\frac{4}{1-3z}\\)
None of the mentioned
Explanation:
Let us divide the given x(n) into x1(n)=3(2n)u(n) and x2(n)= 4(3n)u(n)