# Why can't a first-order block have an oscillatory step response?

From what I understand, multiple-order blocks can have step responses that oscillate.

It's apparent that a first-order block cannot have an oscillatory step response.

How come?

Is it due to the mathematics revolving around it?

• Is it due to the mathematics revolving around it? Hmm, Mathematics are used to describe certain behavior, so how can mathematics be the cause of anything ? We can use mathematics to illustrate or describe why a certain behavior occurs but the mathematics never cause a certain behavior. – Bimpelrekkie Sep 5 '17 at 13:42
• As yourself: What is needed to get an oscillatory response ? What kind of responses have that behavior ? What kind of systems have such a response ? Hint: some systems are stable, others are not. What makes then stable (or not) ? – Bimpelrekkie Sep 5 '17 at 13:46
• In short: yes, it's due to the math, and in particular, damping. If you haven't already, take a read through en.wikipedia.org/wiki/Damping_ratio - it links directly into second-order differential equations. – Reinderien Sep 5 '17 at 14:01
• Systems need positive feedback(ie 180 degrees phase difference) to start oscillating. First order systems produce at most 90 degrees phase difference between input and output, not enough to provide the required feedback. – Bart Sep 5 '17 at 14:07
• first order system can not maintain a "velocity" to form the overshoot, ("du/dt" in the case of voltage signal). To have a overshoot the system needs a non-zero velocity at the equilibrium-corssing point. If your system is first order, e.g. RC or RL only, or a spring with a damper but without mass, the velocity is "stateless" by itself. The system must have velocity as one of its states to be able to overshoot, and first order system doesn't have it. – user3528438 Sep 5 '17 at 14:14

It's apparent that a first-order block cannot have an oscillatory step response.

How come?

A first order system will reach a state of equilibrium in the fullness of time and the energy transfers between components will also reach equilibrium and remain stable in the fullness of time.

An underdamped second order system will not have the component energy transfers expired when its output apparently reaches equilibrium and, because of this, the output will overshoot.

As an example, a low pass LC circuit will charge the capacitor up when a step voltage is applied to the input and, when the capacitor voltage equals the magnitude of the step input, there is still energy left in the inductor that will cause the capacitor to carry on being charged to a higher voltage. Eventually this will die out (in a series of ever-decreasing oscillations) due to circuit losses producing heat and thus getting rid of the excess energy.

• "in the fullness of time" is a curious phrase. Do you mean as the time elapsed approaches infinity? – Reinderien Sep 6 '17 at 21:51
• @R Infinity is a long time so to be realistic, the fullness of time makes sense to me! – Andy aka Sep 7 '17 at 9:07

A simple metaphor:

A first-order, continuous time system is like an ice cube that you pull off the freezer and drop on the table. It will warm up, melt, then reach equilibrium, and then nothing happens.

Since it is continuous-time and first-order, the equations which govern it are something like that, expressed in enginneer mathematics (that's mathless math):

• heat flux is proportional to the difference in temperature between the ice cube and ambient
• temperature is the integral of heat flux (ie, heat flux is the derivative of temperature).

The differential equation has only the first derivative, so it is first order. The idea here is that things only go one way. There is no way for the sign of the derivative to reverse. Once the ice cube is a puddle on the table, it won't heat up further and "oscillate".

Now, second order systems can oscillate, but only when they are powered by an energy source.

For example, if you pluck a guitar string, it will go through damped oscillations, and ultimately go quiet. It is a stable system, only underdamped. It is second-order (position gives spring force which gives acceleration which is second derivative of position). Its damping is dependent on speed vs air drag. It is very likely that a guitar wouldn't work underwater, for this reason.

If you put a balloon on top of your head, this is also second-order, ultimately it will fall off (system is unstable) but the ground will stop it, without external power it won't keep bouncing forever.

In order to have sustained oscillations, you need a source of energy, a time delay (phase lag) and some gain. Let's consider things from the point of view of a Google Car, whose cpu now has a time lag of three seconds, due to the OS deciding to install Windows Update(tm) because you just drove near a McDonalds' WiFi and it picked it up.

• OMG we're going too crash in the right safety rail! Turn the steering wheel to the left three turns.
• (Windows sound) Installing updates... hourglass...
• OMG we're going too crash in the left safety rail! Turn the steering wheel to the right three turns.