Line 39: | Line 39: | ||
---- | ---- | ||
===Answer 1=== | ===Answer 1=== | ||
− | + | <math> 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)</math> | ||
+ | --[[User:Cmcmican|Cmcmican]] 16:05, 15 January 2011 (UTC) | ||
===Answer 2=== | ===Answer 2=== | ||
write it here. | write it here. |
Revision as of 12:05, 15 January 2011
Contents
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)
Answer 2
write it here.
Answer 3
write it here.