a signal x(n) then is this concept of shirting and folding correct?

x(n) is given

need x(-n+3)

so to solve it:

• first advance the x(n) signal by 3 units(time)
• then fold 开发者_StackOverflow社区it, or make a reflection of it

are the above steps correct or is the following correct

• first fold the x(n) signal
• then advance the signal by 3 units

?

Yes, this is a common source of confusion when learning about signals. Here's what I usually do.

Let y[n] = x[-n+3]. Because of -n, y[n] is obviously a time-reversed version of x[n]. But the question about the shift remains.

Notice that y[3] = x[0]. Therefore, y[n] is achieved by first reflecting x[n] about n=0 and then delaying the reflected signal by 3.

For example, let x[n] be the unit step function u[n]. Draw x[n], then draw y[n].

Actually here is what I do:

Let

x(n) = {1,-1,2,4,-3,0,6,-3,-1,2,7,9,-7,5}
                      ^

Suppose origin or n=0 is 6. Note that the ^ symbol indicates the origin. First, we find the folder sequence of x(-n) from x(n). So first we fold or we can say reverse the form of x(n), we get,

The folder sequence of x(-n) from x(n) is

x(-n) = {5,-7,9,7,2,-1,-3,6,0,-3,4,2,-1,1}
                          ^

then shift the sequence of x(-n) towards right hand side by 3 units, we will get

x(-n+3) = {5,-7,9,7,2-1,-3,6,0,-3,4,2,-1,1}
                                  ^

Now, the sample 4 is at the origin.

Above steps are correct. The following steps can be corrected too if these are followed like: first fold the x(n) signal then delay the signal by 3 units this will yield x(-n+3).