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I understand that the model for a first order system is $$G(s)=\frac{k_m*e^{-\tau s}}{T_ms+1}$$

In this case $\tau=3$ and the system gain $k_m=10$, correct?

I need to calculate the transfer function and after that I can calculate the poles by solving for $s$ in the transfer function's denominator. Am I on the right track?

How can I calculate the transfer function given this graph? I don't have access to a book.

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  • $\begingroup$ Do you know what the (transient) response would look like for such a system? $\endgroup$ – fibonatic May 31 '15 at 17:03
  • $\begingroup$ Also because you define the delay as $e^{-\tau s}$ then $\tau$ should be positive. $\endgroup$ – fibonatic May 31 '15 at 17:05
  • $\begingroup$ No I dont. $\tau$ must be negative since it's a delay. I corrected. My mistake. $\endgroup$ – Rachel May 31 '15 at 21:02
  • $\begingroup$ For a first order, $T_m$ is well know. It is given by various methods. For example, it is the time at 63.2% of final value, or is given by the tangent at $t=\tau$ with your notations. You should be able to find numerous online lessons. $\endgroup$ – TZDZ Jun 1 '15 at 13:31
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The response of a linear time invariant system can be split up into a steady state and transient response. For a stable system (all poles of the system have negative real parts) the transient response will go to zero when the time goes to infinity and thus the response would only contain the steady state response. If you would apply some periodic input signal to the system the steady state response would also contain the frequencies present in the input, however for a step response, of step size 1, the steady state would just be a constant defined by the gain of the transfer function of the system at $s=0$.

The steady state response in the given graph seems to tends to 10, so $\left|G(0)\right|=10$ indeed yields $k_m=10$. Your value for the delay $\tau$ is correct as well, assuming the step function jumps to one at $t=0$, be remember to add the units of seconds to this parameter (same units as the time axis). Similar if the y-axis of the figure would have been given an unit, then $k_m$ would require to have this unit as well, but in this case it is dimensionless.

The transient response can be found with the help of the poles of the system, namely for distinct poles the transient response, $y_t(t)$, can be expressed as

$$ y_t(t)=\sum C_i e^{p_it}, $$

where each $p_i$ is a pole of the system and all $C_i$ are constants, which can be solved for with boundary conditions.

I will leave finding the pole of the system up to you, but if you have any questions feel free to ask.

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  • $\begingroup$ I don't know how to find the poles. Can you help me? I can only do it if I have the TF. $\endgroup$ – Rachel Jun 1 '15 at 9:26
  • $\begingroup$ @Rachel How many poles does this system have and what would the transient response look like symbolically? $\endgroup$ – fibonatic Jun 1 '15 at 11:31
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As already figured out, $ \tau = 3$ and $ k_m = 10 $.

The time constant $ T_m $ is the time (ignoring the delay) the first-order system takes to reach $63.2 \%$ of its final value. From the plot it seems to be around $2$.

Thus the transfer function appears to be $ \frac{10 e^{-3 s}}{2 s+1} $

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