Match the following : Codes :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) \(\dfrac{T^{2}Z(Z+1)}{(Z-1)^{3}}\)
a-(iii), b-(i), c-(iv), d-(ii)
Sort the four transforms by the order of their pole at Z = 1 — that single observation matches three of them at once.
| Signal | Sampled sequence | Pole structure | Transform |
|---|---|---|---|
| U(t) | 1, 1, 1, … | single pole at Z = 1 | (iii) |
| t | 0, T, 2T, … | double pole at Z = 1 | (i) |
| t2 | 0, T2, 4T2, … | triple pole at Z = 1 | (iv) |
| e−at | 1, e−aT, e−2aT, … | pole at \(Z=e^{-aT}\) | (ii) |
So the answer is a-(iii), b-(i), c-(iv), d-(ii) — option 3.
Why the pole order rises with the power of t. Each extra factor of t is generated by differentiating with respect to Z, and every differentiation raises the order of the pole by one. Starting from the step,
\(Z\{u(nT)\}=\sum_{n=0}^{\infty}Z^{-n}=\dfrac{1}{1-Z^{-1}}=\dfrac{Z}{Z-1}\)
a geometric series, convergent for |Z| > 1. Applying the multiplication-by-n property \(nx(n)\leftrightarrow-Z\dfrac{dX}{dZ}\) gives the ramp with its squared denominator, and again gives the parabola with a cubed one. Counting the exponent in the denominator therefore identifies the signal immediately.
The exponential is the odd one out, and the easiest to place. It is the only transform whose pole is not at Z = 1. Summing the geometric series with ratio \(e^{-aT}Z^{-1}\),
\(Z\{e^{-anT}\}=\dfrac{1}{1-e^{-aT}Z^{-1}}=\dfrac{Z}{Z-e^{-aT}}\)
Spotting the lone \(e^{-aT}\) in item (ii) pairs it with (d) at a glance and removes options 1, 2 and 4 in one step.
The dimensional check. Note how the sampling period T appears exactly as often as the power of t: none for the step, one T for the ramp, T2 for the parabola. Since the transform of t must carry the units of time, this is another way to confirm the pairing.
The mapping to remember. A pole at Z = 1 is the discrete counterpart of a pole at s = 0, and in general \(z=e^{sT}\) — which is why \(s=-a\) becomes \(z=e^{-aT}\), and why the stable left half plane maps into the unit circle.
Hence, the correct matching is a-(iii), b-(i), c-(iv), d-(ii).
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 :
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 ?
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
