1. Which of the following justifies the linearity property of z-transform?[x(n)↔X(z)].
A. x(n)+y(n) ↔ X(z)Y(z)
Show Explanation Answer: B
According to the linearity property of z-transform, if X(z) and Y(z) are the z-transforms of x(n) and y(n) respectively then, the z-transform of x(n)+y(n) is X(z)+Y(z).
2. What is the z-transform of the signal x(n)=[3(2n)-4(3n)]u(n)?
A. \(\frac{3}{1-2z^{-1}}-\frac{4}{1-3z^{-1}}\)
B. \(\frac{3}{1-2z^{-1}}-\frac{4}{1+3z^{-1}}\)
C. \(\frac{3}{1-2z}-\frac{4}{1-3z}\)
D. None of the mentioned
Show Explanation Answer: A
Let us divide the given x(n) into x1(n)=3(2n)u(n) and x2(n)= 4(3n)u(n)
From the definition of z-transform
X1(z)=\(\frac{3}{1-2z^{-1}}\) and X2(z)=\(\frac{4}{1-3z^{-1}}\)
So, from the linearity property of z-transform
X(z)=X1(z)-X2(z)
=> X(z)=\(\frac{3}{1-2z^{-1}}-\frac{4}{1-3z^{-1}}\).
3. What is the z-transform of the signal x(n)=sin(jω0n)u(n)?
A. \(\frac{z^{-1} sin\omega_0}{1+2z^{-1} cos\omega_0+z^{-2}}\)
B. \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0-z^{-2}}\)
C. \(\frac{z^{-1} cos\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\)
D. \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\)
Show Explanation Answer: D
By Euler’s identity, the given signal x(n) can be written as
x(n) = sin(jω0n)u(n)=\(\frac{1}{2j}[e^{jω_0n} u(n)-e^{-jω_0n} u(n)]\)
Thus X(z)=\(\frac{1}{2j}[\frac{1}{1-e^{jω_0} z^{-1}}-\frac{1}{1-e^{-jω_0} z^{-1}}]\)
On simplification, we obtain
=> \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\).
4. According to the Time-shifting property of the z-transform, if X(z) is the z-transform of x(n) then what is the z-transform of x(n-k)?
A. zkX(z)
Show Explanation Answer: B
According to the definition of Z-transform
X(z)=\(\sum_{n=-\infty}^{\infty} x(n) z^{-n}\)
Let n-k=l
=> X1(z)=\(\sum_{l=-\infty}^{\infty} x(l) z^{-l-k}=z^{-k}.\sum_{l=-\infty}^{\infty} x(l) z^{-l}= z^{-k}X(z)\)
5. 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. \(\frac{1-z^{-N}}{1-z^{-1}}\)
Show Explanation Answer: D
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)}=\(\frac{1-z^{-N}}{1-z^{-1}}\).
6. If X(z) is the z-transform of the signal x(n) then what is the z-transform of anx(n)?
A. X(az)
Show Explanation Answer: C
We know that from the definition of z-transform
Z{anx(n)}=\(\sum_{n=-\infty}^{\infty}a^n x(n) z^{-n}=\sum_{n=-\infty}^{\infty}x(n)(a^{-1}z)^{-n}=X(a^{-1}z)\).
7. If the ROC of X(z) is r1<|z|<r2, then what is the ROC of X(a-1z)?
A. |a|r1<|z|<|a|r2
Show Explanation Answer: A
Given ROC of X(z) is r1<|z|<r2
8. What is the z-transform of the signal x(n)=an(sinω0n)u(n)?
A.\(\frac{az^{-1} sin\omega_0}{1+2 az^{-1} cos\omega_0+a^2 z^{-2}}\)
B.\(\frac{az^{-1} sin\omega_0}{1-2 az^{-1} cos\omega_0- a^2 z^{-2}}\)
C.\(\frac{(az)^{-1} cos\omega_0}{1-2 az^{-1} cos\omega_0+a^2 z^{-2}}\)
D.\(\frac{az^{-1} sin\omega_0}{1-2 az^{-1} cos\omega_0+a^2 z^{-2}}\)
Show Explanation Answer: D
we know that by the linearity property of z-transform of sin(ω0n)u(n) is
X(z)=\(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\)
Now by the scaling in the z-domain property, we have z-transform of an (sin(ω0n))u(n) as
X(az)=\(\frac{az^{-1} sin\omega_0}{1-2az^{-1} cosω_0+a^2 z^{-2}}\)
9. If X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal x(-n)?
A. X(-z)
Show Explanation Answer: B
From the definition of z-transform, we have
Z{x(-n)}=\(\sum_{n=-\infty}^{\infty} x(-n) z^{-n}=\sum_{n=-\infty}^{\infty} x(-n) (z^{-1})^{-(-n)}=X(z^{-1})\)
10. X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal nx(n)?
A. \(-z\frac{dX(z)}{dz}\)
B. \(z\frac{dX(z)}{dz}\)
C. \(-z^{-1}\frac{dX(z)}{dz}\)
D. \(z^{-1}\frac{dX(z)}{dz}\)
Show Explanation Answer: A
From the definition of z-transform, we have
X(z)=\(\sum_{n=-\infty}^{\infty} x(n) z^{-n}\)
On differentiating both sides, we have
\(\frac{dX(z)}{dz}=\sum_{n=-\infty}^{\infty} (-n) x(n) z^{-n-1}=-z^{-1} \sum_{n=-\infty}^{\infty}nx(n) z^{-n}=-z^{-1}Z\{nx(n)\}\)
Therefore, we get \(-z\frac{dX(z)}{dz}\) = Z{nx(n)}.
