Calculating the time taken for the step response of a system

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MattH150197
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Homework Statement


I have worked out the transfer function of a block diagram in 1st order form To = (K/1+ts) and the 2nd part of the question asks to calculate the time for the output temperature to change by 80% when k1=1, k2=2, k3=3 and k4=4

Homework Equations


where K = (k1k2)/(k1k2k4+k2k3) and t= 1/(k1k2k4+k2k3)

The Attempt at a Solution


So K = 0.2 and t = 0.1 and i thought that the equation to do this would be y=K(1-e^(-T/t)) giving 0.8=0.2(1-e^(-T/0.1)) but this doesn't work can anyone see where I am going wrong? Is the y=K(1-e^(-T/t)) right? Thanks!
 
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Presuming that your transfer function represents the impulse response, then your solution form is correct for the step response.

Check your values for ##K## and ##\tau##. If your k values are as you've specified and the formulas for ##K## and ##\tau## are correct, I don't get the same values that you do.

The final (steady state) value for y will be equal to ##K##. So you're looking for the time when y is 80% of that final value.
 
sorry K4 = 2 and if the solution is correct how come i get T = 0.109 seconds, on the booklet it says it should be 0.16 seconds however it doesn't show the methodology of it, but i noticed taking K completely out of the equation would give that answer, is that just a coincidence
 
MattH150197 said:
sorry K4 = 2 and if the solution is correct how come i get T = 0.109 seconds, on the booklet it says it should be 0.16 seconds however it doesn't show the methodology of it, but i noticed taking K completely out of the equation would give that answer, is that just a coincidence
I didn't say that your solution was correct, I said that its form was correct. I should have been more clear about that, for which I apologize.

The steady state value for y is not 1, so 0.8 does not represent 80% of the final value...
 
ah okay i got it now, thanks!