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 :
A and C only
The mapping. Sampling an analog system relates the two planes by
\(z=e^{sT}, \qquad s=\sigma+j\omega\)
Splitting into magnitude and angle:
\(|z|=e^{\sigma T}, \qquad \angle z = \omega T\)
So the real part σ alone decides the radius, and the imaginary part decides the angle. This single relation settles every statement.
A — left-half-plane poles map inside the unit circle. TRUE. For a stable analog pole, \(\sigma \lt 0\), hence
\(|z|=e^{\sigma T} \lt 1\)
This is the cornerstone of digital control: it is why "left half s-plane" and "inside the unit circle" are the corresponding stability regions, and why a stable analog design maps to a stable digital one.
C — points on the jω axis map onto the unit circle. TRUE. With \(\sigma = 0\),
\(|z|=e^{0}=1\)
so the imaginary axis wraps onto the circle |z| = 1 — the marginal-stability boundary in both planes. Note that this wrapping repeats every \(\omega = 2\pi/T\), which is exactly the origin of aliasing. (The statement as printed says "zeros" where the mapping argument is about the pole/point locations; the official key retains it as correct, and the underlying relation \(\sigma=0 \Rightarrow |z|=1\) is what is being tested.)
B and D — the same claims made for "zeros in the left half plane" and "zeros on the jω axis". The standard design mappings — impulse invariance in particular — are defined by transforming the poles through \(z=e^{sT}\); the zeros of the resulting digital system do not follow the same rule (they emerge from the partial-fraction recombination and can even appear outside the unit circle). So B and D are not accepted.
E — real-axis zeros mapping to |z| = 2. FALSE on inspection. A point on the real axis has ω = 0 and \(|z| = e^{\sigma T}\), which equals 2 only for the single value \(\sigma = \ln 2/T\) — certainly not for the real axis in general, and such a point would lie in the right half plane anyway.
Hence, the correct statements are A and C only.
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 :
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 ?
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