Sketching the Periodic Extension of x(t) for y(t) with Ramp and Step Functions

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The discussion centers on the incorrect sketch of the periodic extension of the signal y(t) derived from x(t) = r(t) u(2-t). The user initially presented a flawed graph, which lacked the expected peaks, and sought clarification on their mistake. Key feedback emphasized the importance of correctly multiplying the ramp function r(t) with the step function u(2-t) to accurately represent x(t). Further guidance suggested that understanding the behavior of the step function is crucial, as it effectively turns the ramp signal on and off based on its argument. The user ultimately grasped the concept after receiving clarification and encouragement from others in the thread.
ankyk91
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I sketched the signal for y(t) and my answer is shown pic1.jpg
but apparently it's wrong and the right answer from the tutorial answers is shown in pic2.jpg. So could someone tell me what I have done wrong. Thanks

Homework Statement



Consider the signal
x(t) = r(t) u(2-t)
where r(t) is the ramp function and u(t) is the step function.
The signal y(t) is formed as the periodic extension of x(t) with period 5

a. Clearly sketch the signal y(t).

Homework Equations

The Attempt at a Solution

 

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Welcome to PF!

The flat "tooths" are obviously wrong. Try sketch the two signals r(t) and u(2-t) in the same diagram and see if that helps you spot where you went wrong. Notice that when you multiply a signal with the step function you get the original signal on the "one side" of the step and zero on the "zero side".
 
hey Filip!

thanks for the reply, i drew both the functions in the same graph (shown in pic3.jpg) but I still am not getting the peak like the answer.
 

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If you still get the same flat tooth I guess you wrongly must be taking the minimum of the signals, or something similar.

You need to multiply the two signals, that is, for each t multiply the value for r(t) with the value for u(2-t). For instance, at t = 1.5 you will have r(1.5) = 1.5 and u(2-1.5) = u(0.5) = 1, so x(1.5) = r(1.5)*u(2-1.5) = 1*1.5 = 1.5 (and not 1 as you had in your first figure).

And when you get your multiplication under control you should notice, like I mentioned before, that multiplication of a step function with another signal effectively means that the unit step "turns on" a signal in some region (namely where the argument to the step function is positive) and turns it off everywhere else (namely where the argument is negative).
 
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ankyk91 said:
hey Filip!

thanks for the reply, i drew both the functions in the same graph (shown in pic3.jpg) but I still am not getting the peak like the answer.

There's no problems with that graph Ankyk. Now you just need to learn how to multiply by one.

Lets start with the ones times table

0 x 1 = 0
1 x 1 = 1
2 x 1 = 2
3 x 1 = 3
4 x 1 = 4
...

Keep practicing it, eventually you'll crack it :-p
 
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O I get it now.

Sweet! Thank you guys. Really appreciate it =)
 
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