# Digital Signal Processing Questions and Answers – One Sided Z Transform

This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “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

Explanation: The z-transform of the x(n) whose definition exists in the range n=-∞ to +∞ is known as 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 Unilateral or one sided z-transform.

2. For what kind of signals one sided z-transform is unique?
a) All signals
b) Anti-causal signal
c) Causal signal
d) None of the mentioned

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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.
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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

Explanation: 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) $$\frac{z^{-2}}{1+az^{-1}} + a^{-1}z^{-1} + a^{-2}$$

Explanation:
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

Explanation: 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

Explanation: 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) $$\frac{a}{1-a}$$

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

11. What is the step response of the system y(n)=ay(n-1)+x(n) -1<a<1, when the initial condition is y(-1)=1?
a) $$\frac{1}{1-a}$$(1+an+2)u(n)
b) $$\frac{1}{1+a}$$(1+an+2)u(n)
c) $$\frac{1}{1-a}$$(1-an+2)u(n)
d) $$\frac{1}{1+a}$$(1-an+2)u(n)

Explanation: By taking the one sided z-transform of the given equation, we obtain
Y+(Z)=a[z-1Y+(z)+y(-1)]+X+(z)
Upon substitution for y(-1) and X+(z) and solving for Y+(z), we obtain the result
Y+(z)=$$\frac{a}{1-az^{-1}} + \frac{1}{(1-az^{-1})(1-z^{-1})}$$
By performing the partial fraction expansion and inverse transforming the result, we have
y(n)=$$\frac{1}{(1-a)}(1-a^{n+2})u(n)$$.

Sanfoundry Global Education & Learning Series – Digital Signal Processing.

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