Total Response of a DEQ

The graph below shows the response of a system where the input is a step function. The equation shows the transfer function multiplied by the input step function - recall that ℒ μ(t) = 1/s. The poles (x) and zeros (o) of the transfer function are plotted on the complex plane. In addition the system is multiplied by an adjustable gain (in green).

Use the controls to change the order of the numerator (number of zeros) and denominator (number of poles). Drag the poles and zeros on the complex plane to change their location. As you change the transfer function the gain is automatically adjusted so the steady-state response to the step input is 1 (remind me). After changing the transfer function you can readjust the gain.

Questions:

1. Adjust the system gain. How does the system gain affect steady state response? (answer)
   The system response at t→0 is linearly proportional to the gain.
   
   You should verify that the steady state output corresponds to the system gain.
2. How does the addition of zeros affect the system response? (answer)
   Zeros tend to amplify (or attenuate) the effect of a pole
3. How does a zero in the right half plane affect the system? (answer)
   A zero in the right half plane causes the system to initially move in a opposite direction.
4. How does a zero affect the contribution of a particular pole to the systems transient response? (answer)
   When a zero is near a pole it cancels the effect of that pole. This should make sense. When a zero matches a pole exactly you can factor them both out the transfer function (zero/pole=1).
   
   Create a transfer function with no zeros (1 in the numerator). Then add zeros and move them near and away from particular poles. Notice how the system response changes.