Pole-Zero Plot and ROC for LTI System

[hxluo]

Consider a discrete-time LTI system with transfer function
H(z) =(1-3z^-1) / ( 2-z^-1)
(a) Sketch the pole-zero plot of H(z).
(b) Suppose the system is stable. Determine all possible regions of convergence (ROC) for H(z) under this condition, or state that none exists.
(c) Repeat part (b) assuming the system is causal instead of stable.
(d) Can this system have a causal and stable inverse? If so, determine
H^(-1)(z) including its ROC. If not, explain why not.

 

[LeBrad]

Consider a discrete-time LTI system with transfer function
H(z) =(1-3z^-1) / ( 2-z^-1)
(a) Sketch the pole-zero plot of H(z).
(b) Suppose the system is stable. Determine all possible regions of convergence (ROC) for H(z) under this condition, or state that none exists.
(c) Repeat part (b) assuming the system is causal instead of stable.
(d) Can this system have a causal and stable inverse? If so, determine
H^(-1)(z) including its ROC. If not, explain why not.

1.) Find the poles and zeros of H. They will be easy to see if you multiply the top and bottom by z.

2.) Find how stability and causality relate to ROC of your plot. Stability is related to the unit circle and causality is related to inside or outside of the extreme poles.

3.) Draw a pair of axes. Draw a circle centered at the origin. Use knowledge gained in steps 1 and 2 to draw the pole-zero diagram.

 

[piercas]

(a) The pole-zero plot for the given transfer function is shown below:

[image of pole-zero plot]

(b) Since the system is stable, all poles must lie within the unit circle in the z-plane. Therefore, the possible regions of convergence for H(z) are: |z| > 1 and |z| < 0.

(c) If the system is causal, then all poles must lie outside the unit circle. Therefore, the possible regions of convergence for H(z) are: 0 < |z| < 1 and |z| > 2.

(d) Yes, this system can have a causal and stable inverse. The inverse transfer function, H^(-1)(z), can be found by interchanging the poles and zeros of H(z) and taking the reciprocal. Therefore, H^(-1)(z) = (2-z^-1) / (1-3z^-1) and its ROC is |z| < 1 and |z| > 2. This is because the poles and zeros of H(z) have been interchanged, so the ROCs for H(z) and H^(-1)(z) are also interchanged.