# LTI System Analysis MCQ Quiz – Objective Question with Answer for LTI System Analysis

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

X(z)=$$\sum_{n=-\infty}^{\infty} x(n) z^{-n}$$
$$\frac{dX(z)}{dz}=\sum_{n=-\infty}^{\infty} (-n) x(n) z^{-n-1}=-z^{-1} \sum_{n=-\infty}^{\infty}nx(n) z^{-n}=-z^{-1}Z\{nx(n)\}$$
Therefore, we get $$-z\frac{dX(z)}{dz}$$ = Z{nx(n)}.