For the following signal : \(x(n)=\left(\dfrac{1}{2}\right)^{n}u(n)+\left(\dfrac{1}{3}\right)^{n}u(n)\) Z transforms and ROC has been given in below statements : (a) \(Z[x(n)]=\dfrac{Z}{Z-\frac{1}{2}}+\dfrac{Z}{Z-\frac{1}{3}}\) (b) ROC : \(|Z| \gt \dfrac{1}{2}\) and \(|Z| \gt \dfrac{1}{3}\) (c) \(Z[x(n)]=\dfrac{Z}{Z+\frac{1}{2}}+\dfrac{Z}{Z+\frac{1}{3}}\) (d) ROC : \(|Z| \lt \dfrac{1}{3}\) and \(|Z| \lt \dfrac{1}{3}\) Out of the above given statements which are correct ?
(a) and (b) are correct
The standard pair to work from.
\(a^{n}u(n)\ \longleftrightarrow\ \dfrac{Z}{Z-a}, \qquad \text{ROC}:\ |Z| \gt |a|\)
The transform is linear, so each term is handled separately and the results added.
Step 1 — transform each term. With \(a=\tfrac12\) and \(a=\tfrac13\),
\(X(Z)=\dfrac{Z}{Z-\frac{1}{2}}+\dfrac{Z}{Z-\frac{1}{3}}\)
which is statement (a). Statement (c) has both signs reversed — it would be the transform of \((-\tfrac12)^{n}u(n)+(-\tfrac13)^{n}u(n)\), an alternating sequence, not the given one.
The sign rule worth fixing. The pole sits at the base of the exponential, not at its negative. Since both bases here are positive, both poles are positive: \(Z=\tfrac12\) and \(Z=\tfrac13\), which requires \(Z-a\) in the denominator.
Step 2 — the region of convergence. Both terms are right-sided (each carries u(n)), so each ROC is the exterior of a circle through its pole. The ROC of the sum is the intersection:
\(\left(|Z| \gt \tfrac12\right)\cap\left(|Z| \gt \tfrac13\right)=|Z| \gt \tfrac12\)
Statement (b) lists both conditions, which is correct as written — and their intersection is dominated by the larger pole. So (a) and (b) survive, giving option 2.
Why statement (d) is impossible on two counts. It gives \(|Z| \lt \tfrac13\), an interior region, which belongs to a left-sided sequence — but u(n) makes this signal strictly right-sided. And it quotes 1/3 twice, dropping the 1/2 pole altogether. Options 3 and 4 both carry it.
The general rule this illustrates.
| Sequence type | ROC |
|---|---|
| Right-sided, \(u(n)\) | Outside the outermost pole |
| Left-sided, \(u(-n-1)\) | Inside the innermost pole |
| Two-sided | An annulus between two poles |
Why the ROC must always be quoted. The algebraic expression alone is ambiguous: \(Z/(Z-\tfrac12)\) is the transform of the causal \((\tfrac12)^{n}u(n)\) for \(|Z| \gt \tfrac12\) and of the anti-causal \(-(\tfrac12)^{n}u(-n-1)\) for \(|Z| \lt \tfrac12\). Only the ROC distinguishes them — and, since stability requires the unit circle to lie inside it, only the ROC tells you whether the system is stable.
Hence, the correct statements are (a) and (b).
Similar Questions
The poles of an analog system are related to the corresponding poles of the digital system by the selection z = eST.
A. Analog system poles in the left half of S-plane map on to digital system poles inside the circle |z| = 1.
B. Analog system zeros in the left half of S-plane map on to digital system zeros inside the circle |z| = 1.
C. Analog system poles on the imaginary axis of s-plane map onto digital system zeros that on the unit circle |z| = 1.
D. Analog system zeros on the imaginery axis of s-plane map on to digital system zeros on the unit circle |z| = 1.
E. Analog system zeros on the real axis of s-plane map on to digital system zero on the circle |z| = 2.
Choose the correct answer from the options given below :
Match the following :
List - I List - II (a) U(t) (i) \(\dfrac{TZ}{(Z-1)^{2}}\) (b) t (ii) \(\dfrac{Z}{Z-e^{-aT}}\) (c) t2 (iii) \(\dfrac{Z}{Z-1}\) (d) e-at (iv) \(\dfrac{T^{2}Z(Z+1)}{(Z-1)^{3}}\) Codes :
Important Questions from Z Transform
The z transform of the following real exponential sequence
x(n) = {a n ;n >= 0} , {= 0 ; n < 0} and a > 0 is given by
The causal signal with z-transform z 2(z - a) -2 is
(u[n] is the unit step signal)
The z transform of e −t sampled at 10 Hz will be:
Consider the z-transform X (z) = 5z2 +4z-1 + 3; 0 < |z| < ∞. The inverse z-transform x[n] is
The ROC of a system is the
