Module 4 : Time Response of discrete time systems

Lecture 1 : Time Response of discrete time systems

Solution: The open loop transfer function is:

Taking Z-transform

Steady state error for step input $ =\dfrac{1}{1+K_p}$ where $ K_p=\lim_{z\rightarrow 1}G(z) = \infty$. $ \Rightarrow e^{step}_{ss}=0$.

Steady state error for ramp input $ =\dfrac{1}{K_v}$ where $ K_v=\frac{1}{T}\lim_{z\rightarrow 1}\left[(z-1)G(z)\right]=2$. $ \Rightarrow e^{ramp}_{ss}=0.5$.

Steady state error for parabolic input $ =\dfrac{1}{K_a}$ where $ K_a=\frac{1}{T}\lim_{z\rightarrow 1}\left[(z-1)^2G(z)\right]=0$. $ \Rightarrow e^{para}_{ss}=\infty$.