- #1
Brian Moughtin
- 5
- 0
Hi
I need to "plot the output response of the system shown to the input stimulus given"
I suppose it would help if i knew where to start, but I'll try and explain as best I can.
There are two drawings.
Drawing 1 - shows a rectangle with 1 / 1 +0.5s inside and a line drawn horizontally from the centre of the left vertical side labelled Input, another line is drawns from the centre of the right vertical side saying Output.
Drawing 2 - Shows a graph and a curve starting at 10 on the Y axis with the figure 10e^-0.2t the curve flattens at 3.27 seconds on the X axis.
Now being the worlds worst student of maths, this is what I've got so far.
f(t) = 10e^-0.2t : 0 < t 3.27
f(t) = 5.19 : t>3.27
( Found 5.19 by 10e^(-0.2 x 3.27) = 5.19
Therefore f(t) = 10e^-0.2t - 10e^0.2t H(t-3.27) + 5.19H(t-3.27)
e^-3.27 = 0.038
5.19 x 0.038 = 0.19722
Thus
f(t) = 10e^0.2t - 0.197H(t-3.27) e^-(ta-3.27) +0.197H(t-3.27)
To give
F(s) = 10/(S + 1) - 0.197 x (e^3.27s) /(S+1) = (0.197 x e^3.27s)/S
So far this is probably easily understandable to some but makes me want to put a round in the chamber !
I've no confidence that what I've done is right so far, so would appreciate any comments and if anyone is an expert in this stuff ...the answer
As I've no access to any maths software and have never used it a plain english step by step explanation would be really appreciated!
Thanks
Brian
I need to "plot the output response of the system shown to the input stimulus given"
I suppose it would help if i knew where to start, but I'll try and explain as best I can.
There are two drawings.
Drawing 1 - shows a rectangle with 1 / 1 +0.5s inside and a line drawn horizontally from the centre of the left vertical side labelled Input, another line is drawns from the centre of the right vertical side saying Output.
Drawing 2 - Shows a graph and a curve starting at 10 on the Y axis with the figure 10e^-0.2t the curve flattens at 3.27 seconds on the X axis.
Now being the worlds worst student of maths, this is what I've got so far.
f(t) = 10e^-0.2t : 0 < t 3.27
f(t) = 5.19 : t>3.27
( Found 5.19 by 10e^(-0.2 x 3.27) = 5.19
Therefore f(t) = 10e^-0.2t - 10e^0.2t H(t-3.27) + 5.19H(t-3.27)
e^-3.27 = 0.038
5.19 x 0.038 = 0.19722
Thus
f(t) = 10e^0.2t - 0.197H(t-3.27) e^-(ta-3.27) +0.197H(t-3.27)
To give
F(s) = 10/(S + 1) - 0.197 x (e^3.27s) /(S+1) = (0.197 x e^3.27s)/S
So far this is probably easily understandable to some but makes me want to put a round in the chamber !
I've no confidence that what I've done is right so far, so would appreciate any comments and if anyone is an expert in this stuff ...the answer
As I've no access to any maths software and have never used it a plain english step by step explanation would be really appreciated!
Thanks
Brian