# One Sided Z Transform MCQ Quiz – Objective Question with Answer for One Sided Z Transform

1. The z-transform of a signal x(n) whose definition is given by $$X(z)=\sum_{n=0}^{\infty} x(n)z^{-n}$$ is known as ________

A. Unilateral z-transform
B. Bilateral z-transform
C. Rational z-transform
D. None of the mentioned

The z-transform of the x(n) whose definition exists in the range n=-∞ to +∞ is known as a bilateral or two-sided z-transform. But in the given question the value of n=0 to +∞. So, such a z-transform is known as a Unilateral or one-sided z-transform.

2. For what kind of signals is one-sided z-transform unique?
A. All signals
B. Anti-causal signal
C. Causal signal
D. None of the mentioned

One-sided z-transform is unique only for causal signals because only these signals are zero for n<0.

3. What is the one sided z-transform X+(z) of the signal x(n)={1,2,5,7,0,1}?

A. z2+2z+5+7z-1+z-3
B. 5+7z+z3
C. z-2+2z-1+5+7z+z3
D. 5+7z-1+z-3

Since the one sided z-transform is valid only for n>=0, the z-transform of the given signal will be X+(z)= 5+7z-1+z-3.

4. What is the one-sided z-transform of x(n)=δ(n-k)?

A. z-k
B. zk
C. 0
D. 1

Since the signal x(n)= δ(n-k) is a causal signal i.e., it is defined for n>0 and x(n)=1 at z=k
So, from the definition of one-sided z-transform X+(z)=z-k.

5. What is the one-sided z-transform of x(n)=δ(n+k)?

A. z-k
B. 0
C. zk
D. 1

Since the signal x(n)=δ(n+k) is an anti causal signal i.e., it is defined for n<0 and x(n)=1 at z=-k. Since the one-sided z-transform is defined only for causal signal, in this case X+(z)=0.

6. If X+(z) is the one sided z-transform of x(n), then what is the one sided z-transform of x(n-k)?

A. z-k X+(z)

B. zk X+(z-1)

C. z-k $$[X^+(z)+\sum_{n=1}^k x(-n)z^n]$$; k>0

D. z-k $$[X^+(z)+\sum_{n=0}^k x(-n)z^n]$$; k>0

From the definition of one sided z-transform we have,

Z+{x(n-k)}=$$z^{-k}[\sum_{l=-k}^{-1} x(l) z^{-l}+\sum_{l=0}^{\infty} x(l)z^{-l}]$$

=$$z^{-k}[\sum_{l=-1}^{-k} x(l) z^{-l}+X^+ (z)]$$

By changing the index from l to n= -l, we obtain

Z+{x(n-k)}=$$z^{-k}[X^+(z)+\sum_{n=1}^k x(-n)z^n]$$ ;k>0

7. If x(n)=an, then what is one sided z-transform of x(n-2)?

A. $$\frac{z^{-2}}{1-az^{-1}} + a^{-1}z^{-1} + a^{-2}$$

B. $$\frac{z^{-2}}{1-az^{-1}} – a^{-1}z^{-1} + a^{-2}$$

C. $$\frac{z^{-2}}{1-az^{-1}} + a^{-1}z^{-1} – a^{-2}$$

D. $$\frac{z^{-2}}{1+az^{-1}} + a^{-1}z^{-1} + a^{-2}$$

Given x(n)=an=>X+(z)=$$\frac{1}{1-az^{-1}}$$

We will apply the shifting property for k=2. Indeed we have

Z+{x(n-2)}=z-2[X+(z)+x(-1)z+x(-2)z2]

=z-2 X+(z)+x(-1)z-1+x(-2)

Since x(-1)=a-1 and x(-2)=a-2, we obtain

X1+(z)=$$\frac{z^{-2}}{1-az^{-1}}$$ + a-1z-1 + a-2

8. If x(n)=an, then what is one sided z-transform of x(n+2)?

A. $$\frac{z^{-2}}{1-az^{-1}}$$ + a-1z-1 + a-2

B. $$\frac{z^{-2}}{1-az^{-1}}$$ – a-1z-1 + a-2

C. $$\frac{z^2}{1-az^{-1}}$$ + a z + z2

D. $$\frac{z^2}{1+az^{-1}}$$ – z2 – az

We will apply the time advance theorem with the value of k=2.We obtain,
Z+{x(n+2)}=z2 X+(z)-x(0)z2-x(1)z

=>X1+(z)=$$\frac{z^2}{1+az^{-1}}$$ – z2 – az.

9. If X+(z) is the one sided z-transform of the signal x(n), then
$$\lim_{n \rightarrow \infty} x(n)=\lim_{z\rightarrow 1}(z-1) X^+(z)$$ is called Final value theorem.

A. True
B. False

In the above theorem, we are calculating the value of x(n) at infinity, so it is called as final value theorem.

10. The impulse response of a relaxed LTI system is h(n)=anu(n), |a|<1. What is the value of the step response of the system as n→∞?

A. $$\frac{1}{1+a}$$

B. $$\frac{1}{1-a}$$

C. $$\frac{a}{1+a}$$

D. $$\frac{a}{1-a}$$

The step response of the system is y(n)=x(n)*h(n) where x(n)=u(n)
On applying z-transform on both sides, we get

Y(z)=$$\frac{1}{1-az^{-1}} \frac{1}{1-z^{-1}}=\frac{z^2}{(z-1)(z-A.}$$ ROC |z|>|a|

Now

(z-1)Y(z)=$$\frac{z^2}{(z-A.}$$ ROC |z|>|a|

Since |a|<1 the ROC of (z-1)Y(z) includes the unit circle. Consequently by applying the final value theorem

$$\lim_{n\rightarrow\infty}y(n)=\lim_{z\rightarrow 1}⁡ \frac{z^2}{z-a}=\frac{1}{1-a}$$

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