## Digital Signal Processing Questions and Answers Part-9

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)

Explanation: 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

Explanation: 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) $$\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}$$

Explanation: 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 Time shifting property of 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)

Explanation: 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) $$\frac{1-z^{-N}}{1-z^{-1}}$$

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

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

Explanation: 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

Explanation: 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. 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

Explanation: From the definition of z-transform, we have
Z{x(-n)}=$$\sum_{n=-\infty}^{\infty} x(-n) z^{-n}=\sum_{n=-\infty}^{\infty} x(-n) (z^{-1})^{-(-n)}=X(z^{-1})$$

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

Explanation:
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)}.

10. 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)=$$-z\frac{dX(z)}{dz} = \frac{az^{-1}}{(1-az^{-1})^2}$$