You are at Basic Concepts - Time Response - Second Order Systems -  Effect of Pole Location on Second Order System Response
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        It is important to understand the response of second order systems because the response of higher order systems is composed of responses that look like first order and second order responses.  Second order responses are more complex than first order responses, and you need to spend a little more time examining them.  We begin by examining the response of a particular second order system.  Here is the differential equation of the system we will use.

This system is the simplest second order system because the transfer function of this system has no zeroes.  (The RHS of the differential equation is a constant multiplying the input, u(t).)

        We are going to focus on the step response of this system.  If the input is a unit step, the expression for the step response is given below.  This expression is valid when the damping ratio, z, is less than one.

That response would look something like the one shown below.  The overshoot might be more or less, the oscillation frequency might not be what is shown, and the DC gain could change, but this shows the general form of this step response.

• There is a constant, G_dc, in the response, and when the rest of the response dies out, that is what remains.  If the input is a unit step, then it should be obvious that G_dc is the DC gain.
• The rest of the response dies out.  The rest of the response is a sinusoid multiplied ("modulated") by a decaying exponential.
• The frequency of the sinusoid is close to the undamped natural frequency, w_n.
• The exact frequency of oscillation is:
• 
• The decaying exponential has a "time constant" of 1/zw_n.
• There will be some overshoot in the response.  The overshoot is a function of the damping ratio - and nothing else.  Here is a graph of the overshoot as a function of the damping ratio.  (This link takes you to the explanation for this function.)
• 

• The poles of the system are given by:
• 
• The poles of the system give you information about how the system responds because the pole encode all of the information about the natural frequency and the damping ratio.
• Here is a plot of the poles in the s-plane.
• 
• Note the following.
• The vertical location of the pole is the frequency of the oscillations in the response.
• The horizontal location of the pole is the reciprocal of the time constant of the exponential decay.
• So, the further the pole is to the left in the s-plane, the faster the response dies out.
• Get familiar with measures of response time.
• The distance of the pole from the origin in the s-plane is the undamped natural frequency.  (Or, you can interpret that as the length of the vector from the origin to the pole.)
• The damping ratio is given by:
• z = cos (Angle of the pole off the horizontal)  (That angle is f in the diagram above.)

Let us examine a particular case.

If you use the quadratic formula on this polynomial, you should find that the roots are:

• s = -4 + 4j
• s = -4 - 4j

• Since the poles are at a 45^o angle off the horizontal, the damping ratio is 0.7
• Since the damping ratio is 0.7, we should expect about a 5% overshoot for a step input.
• Since the natural frequency is around 5. 7 rad/sec or just a little under 1 Hz at 0.9 Hz, we expect oscillations with a period of around 1.1 second period, but we don't expect to see them real well since they die out so quickly.  (Remember that 5% overshoot?)

That response could look something like the one shown below.  It looks something like a single time constant system, but if you examine it closely it starts out with zero derivatie.  That's something for you to explore.