This set of Tough Digital Signal Processing Questions focuses on “Properties of Z Transform-2”.
1. What is the signal x(n) whose z-transform X(z)=log(1+az-1);|z|>|a|?
a) \((-1)^n.\frac{a^n}{n}.u(n-1)\)
b) \((-1)^n.\frac{a^n}{n}.u(n+1)\)
c) \((-1)^{n-1}.\frac{a^n}{n}.u(n-1)\)
d) \((-1)^{n-1}.\frac{a^n}{n}.u(n+1)\)
View Answer
Explanation: Given X(z)=log(1+az-1)
By taking first order derivative of X(z) on both sides
=>\(\frac{dX(z)}{dz}=\frac{-az^{-2}}{1+az^{-1}}\)
Thus \(-z.\frac{dX(z)}{dz}=az{-1}[\frac{1}{1-(-a) z^{-1}}]\);|z|>|a|
now, we obtain nx(n)=a.(-a)n-1u(n-1)
=>x(n)=\((-1)^{n-1}.\frac{a^n}{n}.u(n-1)\)
2. If Z{x1(n)}=X1(z) and Z{x2(n)}=X2(z) then Z{x1(n)*x2(n)}=?
a) X1(z).X2(z)
b) X1(z)+X2(z)
c) X1(z)*X2(z)
d) None of the mentioned
View Answer
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.
3. What is the convolution x(n) of the signals x1(n)={1,-2,1} and x2(n)={1,1,1,1,1,1}?
a) {1,1,0,0,0,0,1,1}
b) {-1,-1,0,0,0,0,-1,-1}
c) {-1,1,0,0,0,0,1,-1}
d) {1,-1,0,0,0,0,-1,1}
View Answer
Explanation:
Given x1(n)={1,-2,1}=>X1(z)=1-2z-1+z-2
x2(n)={1,1,1,1,1,1}=>X2(z)=1+z-1+z-2+z-3+z-4+z-5
Now X1(z).X2(z)=1-z-1-z-6+z-7
According to the convolution property of z-transforms,
Z{x1(n)*x2(n)}= X1(z).X2(z)=1-z-1-z-6+z-7
=> x1(n)*x2(n)={1,-1,0,0,0,0,-1,1}.
4. If Z{x1(n)}=X1(z) and Z{x2(n)}=X2(z) then what is the z-transform of correlation between the two signals?
a) X1(z).X2(z-1)
b) X1(z).X2(z-1)
c) X1(z).X2(z)
d) X1(z).X2(-z)
View Answer
Explanation: We know that rx1x2(l)=x1(l)*x2(-l)
Now Rx1x2(z)=Z{x1(l)}.Z{x2(-l)}=X1(z).X2(z-1).
5. If x(n) is causal, then \(\lim_{z\rightarrow\infty}\) X(z)=?
a) x(-1)
b) x(1)
c) x(0)
d) Cannot be determined
View Answer
Explanation: According to the initial value theorem, X(z)=x(0)+x(1)z-1+x(2)z-2+….
When \(z\rightarrow\infty\), z-n tends to 0 because n>0.
So \(\lim_{z\rightarrow\infty}\)=x(0).
6. 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)=?
a) \(\lim_{z \rightarrow 1} [(z-1) X(z)] \)
b) \(\lim_{z \rightarrow 0} [(z-1) X(z)] \)
c) \(\lim_{z \rightarrow -1} [(z-1) X(z)] \)
d) \(\lim_{z \rightarrow 1} [(z+1) X(z)] \)
View Answer
Explanation: According to the Final Value theorem of z-transform we have,
\(\lim_{n \rightarrow \infty} x(n) = \lim_{z \rightarrow 1} [(z-1) X(z)]\)
7. What is the z-transform of the signal x(n)=δ(n-n0)?
a) zn0
b) z-n0
c) zn-n0
d) zn+n0
View Answer
Explanation: From the definition of z-transform,
X(z) = \(\sum_{n=-\infty}^{\infty} \delta (n-n_0)z^{-n}\) = z-n|n=n0 = z-n0.
8. If X(z) is the z-transform of the signal x(n), then what is the z-transform of x*(n)?
a) X(z*)
b) X*(z)
c) X*(-z)
d) X*(z*)
View Answer
Explanation: According to the conjugation property of z-transform, we have
Z{x*(n)}=X*(z*).
9. If x(n) is an imaginary sequence, then the z-transform of the real part of the sequence is?
a) \(\frac{1}{2}\)[X(z)+X*(z*)]
b) \(\frac{1}{2}\)[X(z)-X*(z*)]
c) \(\frac{1}{2}\)[X(-z)-X*(z*)]
d) \(\frac{1}{2}\)[X(-z)+X*(z*)]
View Answer
Explanation: If x(N) is an imaginary sequence, then the real part of x(n) is given as
Real{x(n)}=\(\frac{1}{2}[x(n)+x*(n)]\).
According to linearity property of z-transform, we get
Z{Real{x(n)}}=\(\frac{1}{2}[X(z)+X*(z*)]\).
10. What is the signal whose z-transform is given as X(z)=\(\frac{1}{2πj} \oint X_1 (v) X_2 (\frac{z}{v})v^{-1} dv\)?
a) x1(n)*x2(n)
b) x1(n)*x2(-n)
c) x1(n).x2(n)
d) x1(n)*x2*(n)
View Answer
Explanation: From the convolution property in z-domain we have,
Z{x1(n).x2(n)}=\(\frac{1}{2 \pi j} \oint X_1 (v) X_2 (\frac{z}{v})v^{-1} dv\)
11. What is the z-transform of the signal x(n)= x1(n).x2*(n)?
a) \(\frac{1}{2πj} \oint X_1(v) X_2 (\frac{z}{v})v^{-1} dv\)
b) \(\frac{1}{2πj} \oint X_1(v) X_2^* (\frac{z^*}{v^*})v^{-1} dv\)
c) \(\frac{1}{2πj} \oint X_1(v) X_2^* (\frac{z}{v})v^{-1} dv\)
d) None of the mentioned
View Answer
Explanation:
We know that Z{x*(n)}=X*(z*)
Now from the multiplication property in time domain we get,
Z{x1(n).x2*(n)}=\(\frac{1}{2 \pi j} \oint X_1(v) X_2^* (\frac{z^*}{v^*})v^{-1} dv\)
12. If x1(n)={1,2,3} and x2(n)={1,1,1}, then what is the convolution sequence of the given two signals?
a) {1,2,3,1,1}
b) {1,2,3,4,5}
c) {1,3,5,6,2}
d) {1,2,6,5,3}
View Answer
Explanation: Given x1(n)={ 1,2,3}=>X1(z)=1+2z-1+3z-2
x2(n){1,1,1}=>X2(z)=1+z-1+z-2
Now from the convolution in time domain property of z-transform, we have
Z{x1(n)* x2(n)}=X1(z). X2(z)
=> X(z)=1+2z-1+6z-2+5z-3+3z-4 => x(n)={1,2,6,5,3}.
13. What is the z-transform of the signal x(n)=cos(jω0n)u(n)?
a) \(\frac{z^{-1} sin\omega_0}{1+2z^{-1} cosω_0+z^{-2}}\)
b) \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cosω_0-z^{-2}}\)
c) \(\frac{1-z^{-1} cos\omega_0}{1-2z^{-1} cosω_0+z^{-2}}\)
d) \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cosω_0+z^{-2}}\)
View Answer
Explanation: By Euler’s identity, the given signal x(n) can be written as x(n)=\(cos(j\omega_0n)u(n)=\frac{1}{2}j[e^j\omega_0n u(n)+e^{-j\omega_0n} u(n)]\)
Thus X(z)=\(\frac{1}{2j}[\frac{1}{1-e^{j\omega_0} z^{-1}}+\frac{1}{1-e^{-j\omega_0} z^{-1}}]\)
On simplification, we obtain
=> X(z)=\(\frac{1-z^{-1} cos\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\).
14. What is the z-transform of the signal defined as x(n)=u(n)-u(n+N)?
a) \(\frac{1+z^N}{1+z^{-1}}\)
b) \(\frac{1-z^N}{1-z^{-1}}\)
c) \(\frac{1+z^{-N}}{1+z^{-1}}\)
d) \(\frac{1-z^{-N}}{1-z^{-1}}\)
View Answer
Explanation:
We know that Z{u(n)}=\(\frac{1}{1-z^{-1}}\)
And by the time shifting property, we have Z{x(n-k)}=z-k.Z{x(n)}
=>Z{u(n+N)}=\(z^N.\frac{1}{1-z^{-1}}\)
=>Z{u(n)-u(n-N)}=\(\frac{1-z^N}{1-z^{-1}}\).
15. What is the z-transform of the signal x(n)=[5(3n)-9(7n)]u(n)?
a) \(\frac{5}{1-3z^{-1}}-\frac{9}{1-7z^{-1}}\)
b) \(\frac{5}{1+3z^{-1}}-\frac{9}{1+7z^{-1}}\)
c) \(\frac{5}{1-3z}-\frac{9}{1-7z}\)
d) None of the mentioned
View Answer
Explanation: Let us divide the given x(n) into x1(n)= 5(3n)u(n) and x2(n)=9(7n)u(n)
and x(n)=x1(n)-x2(n)
From the definition of z-transform X1(z)=\(\frac{5}{1-3z^{-1}}\) and X2(z)=\(\frac{9}{1-7z^{-1}}\) So, from the linearity property of z-transform
X(z)=X1(z)-X2(z)
=> X(z)=\(\frac{5}{1-3z^{-1}}-\frac{9}{1-7z^{-1}}\)
Sanfoundry Global Education & Learning Series – Digital Signal Processing.
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