# Z Transform Properties MCQ Quiz – Objective Question with Answer for Z Transform Properties

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)
B. x(n)+y(n) ↔ X(z)+Y(z)
C. x(n)y(n) ↔ X(z)+Y(z)
D. x(n)y(n) ↔ X(z)Y(z)

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

Let us divide the given x(n) into x1(n)=3(2n)u(n) and x2(n)= 4(3n)u(n)
and x(n)=x1(n)-x2(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. $$\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}$$

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)
B. z-kX(z)
C. X(z-k)
D. X(z+k)

According to the definition of Z-transform

X(z)=$$\sum_{n=-\infty}^{\infty} x(n) z^{-n}$$
=>Z{x(n-k)}=$$X^1(z)=\sum_{n=-\infty}^{\infty} x(n-k) z^{-n}$$

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. $$\frac{1-z^{-N}}{1-z^{-1}}$$

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)
B. X(az-1)
C. X(a-1z)
D. None of the mentioned

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
B. |a|r1>|z|>|a|r2
C. |a|r1<|z|>|a|r2
D. |a|r1>|z|<|a|r2

Given ROC of X(z) is r1<|z|<r2
Then ROC of X(a-1z) will be given by r1<|a-1z |<r2=|a|r1<|z|<|a|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.$$\frac{az^{-1} sin\omega_0}{1-2 az^{-1} cos\omega_0+a^2 z^{-2}}$$

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)
B. X(z-1)
C. X-1(z)
D. None of the mentioned

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. $$z^{-1}\frac{dX(z)}{dz}$$

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)}.

11. What is the z-transform of the signal x(n)=nanu(n)?

A. $$\frac{(az)^{-1}}{(1-(az)^{-1})^2}$$

B. $$\frac{az^{-1}}{(1-(az)^{-1})^2}$$

C. $$\frac{az^{-1}}{(1-az^{-1})^2}$$

D. $$\frac{az^{-1}}{(1+az^{-1})^2}$$

We know that Z{anu(n)}=$$\frac{1}{1-az^{-1}}$$=X(z) (say)
Now the z-transform of nanu(n)=$$-z\frac{dX(z)}{dz} = \frac{az^{-1}}{(1-az^{-1})^2}$$

12. Sampling rate conversion by the rational factor I/D is accomplished by what connection of interpolator and decimator?

A. Parallel
C. Convolution
D. None of the mentioned

A sampling rate conversion by the rational factor I/D is accomplished by cascading an interpolator with a decimator.

13. Which of the following has to be performed in sampling rate conversion by rational factor?

A. Interpolation
B. Decimation
C. Either interpolation or decimation
D. None of the mentioned

We emphasize that the importance of performing the interpolation first and decimation second is to preserve the desired spectral characteristics of x(n).

14. Which of the following operation is performed by the blocks given in the figure below?

A. Sampling rate conversion by a factor I
B. Sampling rate conversion by a factor D
C. Sampling rate conversion by a factor D/I
D. Sampling rate conversion by a factor I/D

In the diagram given, an interpolator is in a cascade with a decimator which together performs the action of sampling rate conversion by a factor I/D.

15. The Nth root of unity WN is given as _____________

A. ej2πN
B. e-j2πN
C. e-j2π/N
D. ej2π/N

X(k)=$$\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N}=\sum_{n=0}^{N-1} x(n) W_N^{kn}$$