Discrepancies in Step Response

I'm going through a solution of a problem from a problem set. However, I don't know how to get the step response of this transfer function
properly: G(s) = 10/(s+1)/(s+10). Using the Laplace transform tables, I got a transfer function of g(t) = 10/9*(exp(-t) - exp(-10*t))*u(t), with the graph shown here (
http://imagebin.ca/img/EBd-Li.jpg ). I have entered this system's transfer function and plotted its step response in Matlab (
http://imagebin.ca/img/SRLhIXU.jpg ). However, even the Matlab plot disagrees with the textbook solutions (http://imagebin.ca / img/Y6BqRb.png). I have no idea how the textbook solutions plot is computed. It would be great to hear your opinions.
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ssylee wrote:

I see room for confusion. With three division signs in a row, what divides what? Matlab and the textbook seem to agree except for an offset of one in the time axix. Since Matlab forces the origins of its arrays to 1, that doesn't surprise me.
Matlab's and the textbook's plots show the asymptotic response (t->00) to be unity. The other shows it to be zero. Understanding the order of operations in your expression, you should be able to check that.
Jerry
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On Sat, 30 Jan 2010 12:06:19 -0800, ssylee wrote:

"Using the Laplace transform tables, I got a transfer function of g(t) ="
Transfer functions (in the context of linear systems theory) do not exist in the time domain. Think about that. Then think about what someone means when they say "step response". What exactly is a "step response" of a system -- don't give it in terms of some regurgitated Laplace-domain stuff, tell me what is the system input in the time domain, and what do you expect out of it?
Perhaps then you'll see your error.
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On Sat, 30 Jan 2010 12:06:19 -0800, ssylee wrote:

And the graph in the book is shifted over by one -- it's probably a typo. Matlab's plot is correct.
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Did you forget the step forcing function 1/s? Assuming G(s) is supposed to be 10/((s+1)(s+10)) the response should look like a first order response except there will be a slight inflection due to the second pole. Your plot looks like you forgot the step forcing function 1/s. Now your transfer function is (1/s)*(10/((s+1)(s+10)). The simple thing to do is to use wxMaxima or some other CAS to do the inverse Laplace transform. I get f(t)=-10*%e^-t/9+%e^-(10*t)/9+1
Peter Nachtwey
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