# u(-t) = or not = to - u(t) ?

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#### wcz

##### Member level 2
according to the subject, could anyone show me a proof on that?
thanks.

#### Jayson

##### Full Member level 4
This isn't really a mathematic proof, but does explain.

u(-t) is always 0 for all positive values of t, and 1 for all negative values of t.
-u(t) is always -1 for all positive values of t, and 0 for all negative values of t.

- Jayson

#### freewilly30

##### Member level 3
this is not an even function to say that u(t)=u(-t)
revise the plots of those functions.....
you will find the answer yourself..

#### amraldo

in general:
f(-t) = -f(t) in case of odd functions only & u(t) is not an odd function.

#### suvendu

##### Full Member level 3
1>in case of u(-t),the independent variable is reversed but in case of -u(t),the function itself reversed.
2> unit step function is not an odd function.so we can not write u(-t)=-u(t)
3>if u plot these functions independtenly then u will see that u(-t) lies in second quadrant and -u(t) lies in fourth quadrant.

#### tantoun2004

##### Member level 1
Re: u(-t) = or not = to - u(t) ? the proof u need

Hi,
This relation is not correct.
The proof is very easy:
write u(t)=1 as t>0 ,0 as t<0
put -t instead of t you get:
u(-t)=1 as -t>0 i.e. t<0 and 0 as -t<0 i.e. t>0
So we get that:
u(-t)= 1 as t<0 and 0 as t>0
while -u(t)= -1 as t>0,and 0 as t<0
so it's evident that u(-t) is not equal to -u(t)
Regards,

#### sxg

##### Junior Member level 3
according the opinion of Singal and System,the y=u(-t) is roated the singal
with the axis of y,but the y=-u(t) is roated with axis of t.

#### the_jackal

##### Member level 4
Consider the four quadrants that we study with reference to trignometry.
For y(t) = u(t) ---> sketch is in the 1st quadrant
y(t) = -u(t) ----> sketch is in the 4th quadrant i.e. reflection of u(t) about the x-axis
y(t) = u(-t) ----> sketch is in the 2nd quadrant
y(t) = -u(-t) ----> sketch is in the 3rd quadrant.

#### Highlander-SP

##### Member level 3
it's a case of odd function, like u(t)=t:
for t=1
u(-t)= u(-(1)) = -1
-u(t)= -(1) = -1

for t= -1
u(-t)= u(-(-1)) = u(1) = 1
-u(t)= -u(-1) = -(-1) = 1

for t= 1/2
u(-t)= u(-(1/2)) = -1/2
-u(t)= -(1/2) = -1/2

and so on....

u(-t)=1-u(t)

#### A.Anand Srinivasan

they are not equal because u(-t) is on the left side and has a value 1 throughout but -u(t) is on the right side and a value of -1 throughout
if you mean to say in non signal terms then it is equal only for odd functions

#### afti_khan

##### Member level 2
in case of even function, it's symmetrical abt the y-axis and odd function is symmetric abt the origin

#### vivek_raj_verma

##### Junior Member level 2
Hi,
If u want a proof. I think the proof given by Tantoun2004 is correct. Otherwise there is no doubt that they r diff. as u can see them from their plot easily.

vicky

##### Junior Member level 2
u(-t) = - u(t) if u(t) is odd. But no one sided function can be odd!

#### hamidr_karami

##### Banned
hi

u(t) != u(-t)

beacuse u(t) is not an evev function => u(t)!=u(t)

u(t) = 1-u(-t)

bye[/img]

##### Junior Member level 3
it depends on the nature of the functon.in case of odd funstion u(-t)=-u and vice versa.first you should check for even or odd nature.

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