Find Signal x(n) - Hint for Solving x(n)+x(-n)

[robert25pl]

I have to find the signal if x(n) = nv(n-1) for -infinite < n < + infinite

x(n)+x(-n)

Can I get hint with this problem so I can do rest of them. Thanks

 

[dimensionless]

x(n) is a vector function where n is all the intergers from -infinity to +infinity. nv(n-1) is n*v(n-1) which is the multiplication of two vectors. Again n is all integers from -infinity to +infinity and v(n-1) is the vector function v(n) after all the elements have been shifted over one position to the right.

At least that's what it looks like form my position.

 

[ravenprp]

Sure, no problem! To find the signal x(n), we can use the hint given, which is x(n)+x(-n). Using this hint, we can rewrite the equation for x(n) as x(n) = x(n)+x(-n)-x(-n). Now, we can see that the first two terms on the right side cancel out, leaving us with x(n) = -x(-n). From here, we can use this relationship to solve for x(n). Since x(n) is equal to -x(-n), we can substitute -x(-n) in place of x(n) in the original equation. This gives us -x(-n) = nv(n-1). Solving for x(-n), we get x(-n) = -nv(n-1). Therefore, the signal x(n) can be written as x(n) = -nv(n-1). I hope this helps and you are able to solve the rest of the problems!