# Is a LTI system with dead-time equivalent with a LTI system with no dead-time but delayed input signal?

I wonder if the system parameters would remain the same after all. According to the definitions, the gain is the change of the output induced by the change of the input if we assume that the gain is a constant number then it should remain the same.

But what about the time constant or rise time?

According the differential equations governing the first and second order dynamic systems, the paramater time constant is independent of the input signal, so I would say this also remains the same, but according to the other definition of the time constant which states that time-consnat is a measure of how fast a system goes from one steady state to another, then I would say no, and here is why:

If we feed a system with delayed input or a system has an intrinsic delay or dead-time (Dead time is the delay from when a controller output (CO) signal is issued until when the measured process variable (PV) first begins to respond.), then the time in which system changes its state between the regiems would prolong with the dead-time.

To clarify the concepts, I consider a RC circuit with a step input function, in this case a 1.5 volt battery, here you can find the circuit: The $$R_i$$ resistor represents the losses in the capacitor.The differential equation governs this circuit:

$$RC \frac{dv_c}{dt} + v_c=v$$

in terms of controle engineering:

$$\tau \frac{dy}{dt} + y = Ku$$.

With $$\tau$$, I mean time constant and with K the so called gain. $$u$$ is the input signal and $$y$$ is the output.

I have all the the out put voltages and inputs all together in an excel file, they collected all the data via NiDAQ, and I want to measure the parameters $$K$$ and $$\tau$$.

Here is the graphs, the orange curve is the input, or in this case the battery voltage and the blue curve is the output voltage over the capacitor: This circuit or system has no intrinsic dead time, I don't know where this delay is coming from.

My question is: Does this system show equivalent behaviour as the exact same system, but with the input chopped to zero up to the starting time or in mathematical terms, does this two inputs $$u(t)$$ and $$u(t)= 0$$ if $$t < 3.5$$ $$u(t) = 1.5$$ if $$t >3.5$$results in the same gain and time constant?

• You use some unusual terms without properly defining them. Can you please post references to your definition of "process gain" and "dead time" ? – Hilmar Apr 10 at 13:21
• @Hilmar Unfortunately I don't have a reference besides my college slides, but I replaced the process gain by "Gain" and tried to clarify the term "Dead-time". – Sam Farjamirad Apr 10 at 13:35
• creating some graphs of your system and the input output responses in each case may help clarify this. – Dan Boschen Apr 10 at 13:55
• "Dead time" is actually pretty common in process control. – TimWescott Apr 10 at 17:38
• Well since your plot shows the output given a step; time can start at 0 or can start at t = 1000000 and we would see the exact same result. Time = 0 is just a reference point. – Dan Boschen Apr 10 at 19:52