Revision as of 13:08, 20 January 2011

Cascade a time delay and a time scaling

Consider the following two systems:

$ x(t) \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{system 1} & \\ & & \end{array}\right] \rightarrow y(t)=x(t+2) $

$ x(t) \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{system 2} & \\ & & \end{array}\right] \rightarrow y(t)=x(5t) $

Obtain a simple expression for the output of the following cascade:

$ x(t) \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{system 1} & \\ & & \end{array}\right] \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{system 2} & \\ & & \end{array}\right] \rightarrow y(t) $

(Sorry, I don't know how to make a real "box" to represent a system. If somebody knows, please help. -pm)

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Answer 1

$ x(t) \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{system 1} & \\ & & \end{array}\right] \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{system 2} & \\ & & \end{array}\right] \rightarrow y(t) = x(5(t + 2)) = x(5t + 10) $ --Cmcmican 16:05, 15 January 2011 (UTC)

Instructors comments: Unfortunately, the answer is not correct. (I actually expected that mistake to happen. Almost everybody does it the first time.) Try to carefully write the output after each step of the cascade, and change the variable explicitely. -pm

Answer 2

$ x(t) \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{system 1} & \\ & & \end{array}\right] \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{system 2} & \\ & & \end{array}\right] \rightarrow y(t) = x(5t + {2 \over {5}}) $

--Rgieseck 11:47, 18 January 2011 (UTC)

Draw an example signal and apply the systems graphically to logically deduce this answer.

Instructors comments: Unfortunately, this answer is not correct either. I know the graphical approach is tempting, but there is a point where problems become too complicated to be handled that way. I think ECE301 is such a point. Let me suggest another approach below. -pm

$ x(t) \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{system 1} & \\ & & \end{array}\right] \rightarrow z(t)=x(t+2) \rightarrow \left[ \begin{array}{ccc} & & \\ & \text{system 2} & \\ & & \end{array}\right] \rightarrow w(t)= z(5t) = x(5t + 2 ) $

Does everybody understand why z(5t)=x(5t+2)? -pm

No, I am confused as to why this is true. Could you please explain in more detail? -RG

:Certainly. Actually, let me try to clarify every step slowly for the benefit of everybody. First, the fact that the output after the first system is x(t+2) should be clear to everyone (by definition of System 1). I did not call this output y(t) to avoid confusion (there are just too many different "outputs" in this explanation, and so none of them will get to be called with the canonical y(t)): so I gave it the name z(t). Now the overall system's output is easy to write in terms of z(t), because System 2 transform its input z(t) into z(5t). So the only tricky part is to find out what is z(5t). To do it correctly, I recommend rewriting z(t) using a new independent variable, say u. (This will avoid confusion when we replace the t inside later on.) So we have z(t)=x(t+2), or z(u)=x(u+2) if we use u as the independent variable. Now z(5t) is z(u) with the u replaced by the expression "5t". So z(5t)=x(5t+2). Does that help a bit? I cannot over-emphasize the importance of understanding how to do these cascades in ECE301. -pm

Answer 3

video tutorial on how to cascade transformations of the independent variable

Back to ECE301 Spring 2011 Prof. Boutin