For a causal discrete-time LTI system with transfer function
$H(z) = \frac{2z^2+3}{(z+\frac{1}{3})(z-\frac{1}{3})}$,
which of the following statements is/are true?
Given the transfer function for a causal discrete-time LTI system: $H(z) = \frac{2z^2+3}{(z+\frac{1}{3})(z-\frac{1}{3})} = \frac{2z^2+3}{z^2 - \frac{1}{9}}$
System Stability Analysis
A causal discrete-time LTI system is BIBO stable if all its poles lie strictly inside the unit circle ($|z| < 1$).
- The poles are the roots of the denominator $z^2 - \frac{1}{9} = 0$.
- The poles are $z_1 = \frac{1}{3}$ and $z_2 = -\frac{1}{3}$.
- Checking magnitudes: $|z_1| = \frac{1}{3} < 1$ and $|z_2| = \frac{1}{3} < 1$.
Since both poles are inside the unit circle, the system is stable.
Minimum Phase System Check
A causal system is minimum phase if all its poles and zeros lie strictly inside the unit circle.
- Poles: $z = \pm \frac{1}{3}$ (inside the unit circle).
- Zeros are roots of $2z^2 + 3 = 0 \implies z^2 = -\frac{3}{2} \implies z = \pm j\sqrt{\frac{3}{2}}$.
- Magnitude of zeros: $| \pm j\sqrt{\frac{3}{2}} | = \sqrt{\frac{3}{2}} \approx 1.22 > 1$.
Since there are zeros outside the unit circle, the system is not minimum phase.
Initial Value of Impulse Response
For a causal system, the initial value of the impulse response, $h[0]$, is found by evaluating $H(z)$ as $z \to \infty$.
- Calculate the limit: $\lim_{z \to \infty} H(z) = \lim_{z \to \infty} \frac{2z^2+3}{z^2 - \frac{1}{9}}$
- Divide by the highest power of $z$ ($z^2$): $\lim_{z \to \infty} \frac{2 + \frac{3}{z^2}}{1 - \frac{1}{9z^2}} = \frac{2+0}{1-0} = 2$.
The initial value of the impulse response, $h[0]$, is 2.
Final Value of Impulse Response
The final value, $h[\infty]$, is determined using the final value theorem. This theorem applies if all poles of $z H(z)$ lie strictly inside the unit circle.
- Consider $z H(z) = z \frac{2z^2+3}{z^2 - \frac{1}{9}} = \frac{2z^3+3z}{z^2 - \frac{1}{9}}$.
- The poles of $z H(z)$ are the same as the poles of $H(z)$, which are $z = \pm \frac{1}{3}$.
- Since $| \pm \frac{1}{3} | < 1$, all poles of $z H(z)$ are strictly inside the unit circle.
Therefore, the final value of the impulse response, $h[\infty]$, is 0.
Conclusion
Based on the analysis:
- Statement 1 (The system is stable) is true.
- Statement 2 (The system is a minimum phase system) is false.
- Statement 3 (The initial value of the impulse response is 2) is true.
- Statement 4 (The final value of the impulse response is 0) is true.
The correct statements are 1, 3, and 4.
Important Questions from Z Transform
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: For a given sequence $X[n]=(\frac{1}{2})^n u(n)$ and its Z-transform $X(z)=\frac{1}{1-\frac{1}{2}z^{-1}}, \quad |z|>\frac{1}{2}$ is a stable system.
Reason R: For a stable system ROC (Region of Convergence) must include the unit circle.
In the light of the above statements, choose the most appropriate 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 :
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 e −t sampled at 10 Hz will be:
Consider a discrete-time sequencebr
$x[n] = \begin{cases} (0.2)^n, & 0 \le n \le 7 \\ 0, & \text{otherwise} \end{cases}$
The region of convergence of $X(z)$, the z-transform of $x[n]$, consists of
