Input/output signal

  • Thread starter EvLer
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  • #1
EvLer
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Not even sure how to title this thread. I am saying honestly: we did not cover this in class but he gave us this in homework!
Two identical linear systems are connected in cascade. When the input signal to the first stage is x(t) the output signal from the first stage is y(t), the output of the second stage is z(t). If an input signal, tu(t) is applied to a single stage, the output signal of that stage is : [tex] (e^{-2t}-1)u(t)[/tex] find z(t) if x(t) = u(t).

u(t) is a step function, u(t) = 1 if t >=0, and 0 otherwise;

what i was thinking of doing is converting the e...-expression to frequency domain and also the tu(t) and then seeing how input is modified to get the output and apply the same thing to u(t) which is the input into cascaded system, but TA said that there's multiplication involved in s-domain. I am sort of clueless...
If someone could explain or direct to a source, that would be very much appreciated, 'coz my book does not these examples and i don't even know what to seach for on Google! :cry:
maybe i am not making a connection to something...the latest we covered was circuit analysis in frequency domain doing nodal analysis.
 
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Answers and Replies

  • #3
EvLer
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sorry, should have given more detail: i understand laplace transform.
What i don't understand is how to make use of the given info that input is one function and output signal is a different function and what to do in s-domain (laplace "world") to apply that to u(t) which is the input signal to the 2-stage linear system.
 
  • #4
EvLer
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EDIT: deleted my previous garbage...
EDIT: hold on...have another idea...would this work?:
[tex]H(s) = \frac{L(output)}{L(input)}[/tex]

then when i find this factor i will have to multiply my given L(input) by it in s-domain and take L-1 to get the time-domain signal...
would that work?
 
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