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Question about rect () function

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markwilest

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Hi,

For what range of t is x(t) = rect [(t/T)-(1/2)] equal to one ? Like for exapmple rect (t) = 1 for |t|<1/2. Please help me to understand this. Thanks

Mark
 

eecs4ever

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markwilest said:
Hi,

For what range of t is x(t) = rect [(t/T)-(1/2)] equal to one ? Like for exapmple rect (t) = 1 for |t|<1/2. Please help me to understand this. Thanks

Mark
rect(t) = 1 if |t| < 1/2
so rect ( any number) = 1 if |any number| < 1/2

So if | (t/T)-(1/2) | < 1/2
then x(t) = 1 .

so | (t/T)-(1/2) | < 1/2 implies

-1/2 < t/T - 1/2 < 1/2 which implies

0 < t / T < 1 which implies

0 < t < T

so this function is simply a shifted rectangle thats only 1 if t is between 0 and T.
x(t) = 1 when 0<t <T . and x(t) = 0 everywhere else.
 

    markwilest

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purnapragna

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Hi note that \[rect(\frac{t}{T})\] is \[1\] for \[[-\frac{T}{2},\frac{T}{2}]\]. \[rect(\frac{t}{T}-1)\] is just shifted version of original signal \[rect(\frac{t}{T})\] by \[\frac{1}{2}\] to the left. So the signal will be \[1\] for \[[-frac{T}{2}+\frac{1}{2},rac{T}{2}+\frac{1}{2}]\]. Hope you got it.

thnx

purna!

Added after 47 seconds:

Hi note that \[rect(\frac{t}{T})\] is \[1\] for \[[-\frac{T}{2},\frac{T}{2}]\]. \[rect(\frac{t}{T}-1)\] is just shifted version of original signal \[rect(\frac{t}{T})\] by \[\frac{1}{2}\] to the left. So the signal will be \[1\] for \[[-\frac{T}{2}+\frac{1}{2},\frac{T}{2}+ \frac{1}{2}]\]. Hope you got it.

thnx

purna!

Added after 1 minutes:

Hi note that \[rect(\frac{t}{T})\] is \[1\] for \[[-\frac{T}{2},\frac{T}{2}]\]. \[rect(\frac{t}{T}-1)\] is just shifted version of original signal \[rect(\frac{t}{T})\] by \[1\] to the left. So the signal will be \[1\] for \[[-\frac{T}{2}+1,\frac{T}{2}+ 1]\]. Hope you got it.

thnx

purna!
 
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v_c

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One thing that I always keep in mind about the \[rect()\] function is that it can be constructed using two step functions. In other words,
\[rect(t) = u(t+1/2) - u(t-1/2)\]
So if you know how a step function works -- you know how the \[rect()\] function works. It is always good to decompose complex functions down to simpler functions to do analyze, find their transforms, etc.

Best regards,
v_c
 
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andreas1982

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It is one for:
½-(T/2) e ½+(T/2)
 

sssr

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Hi Ive got a rec function as well - how do I express this in unit step functions

x(t)= Π[(t-3)/6] + Π[(t-4)/2]

thanks
 

v_c

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sssr said:
Hi Ive got a rec function as well - how do I express this in unit step functions

x(t)= ?[(t-3)/6] + ?[(t-4)/2]

thanks
For the first rect function you would get u[(t-3)/6 + 1/2] - u[(t-3)/6 - 1/2].
For the second rect function you get u[(t-4)/2 + 1/2] - u[(t-4)/2 - 1/2].
Just add these four step function and you should be set. Notice that all I did was
to use the decomposition of the rect function into the step functions as I showed
in my previous post. So rect(anything) = u(anything + 1/2) - u(anything - 1/2)
in your case "anything" was the function inside the parantheses of your rect function.
Sometimes the rect function is called "gate" function and denoted (as you have) by the Greek uppercase letter pi since it looks like a gate.

Best regards,
v_c
 

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