Time Scaling/Shifting of Signals - Order of Operations

Discussion in 'Physics' started by Chiver, Aug 31, 2012.

1. Chiver Thread Starter New Member

Mar 1, 2012
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I am having troubles with time shifting and scaling of signals, concerning the order of operations.

In this example the scale is done first, then inversion and finally shifting. But I have seen it done differently.

For example if I had this:

and I wanted to do something like x(-2t+5)

What should I do first? or Could it be done different ways?

EDIT: This is what I did:

Last edited: Aug 31, 2012

Dec 28, 2014
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3. WBahn Moderator

Mar 31, 2012
23,384
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In this case shankar123578's suggestion works, but it isn't a universal answer.

You need to do the manipulations in the reverse order as indicated by the expression.

The proof is in the pudding. Let's take an example x(t) and see what we get:

x(t) = 3t + 12

z(t) = x(-2t+5) = 3(-2t+5) + 12 = -6t + 27

That's our gold standard.

There are eight possible ways to order the three operations of scale (S), shift (s), and reversal (R). So let's try some of them out.

First, let's do what shankar123578 suggested, which is (sSR):

(s) - shift by +5: w(t) = x(t+5) = 3(t+5) + 12 = 3t + 27
(S) - scale by 2: y(t) = w(2t) = 3(2t) + 27 = 6t + 27
(R) - time reversal: z(t) = y(-t) = 6(-t) + 27 = -6t + 27

Which works. But is that the only way to do it? And, if so, why?

Let's try doing them in the order indicated, which is (SRs):

S - scale by 2: w(t) = x(2t) = 3(2t) + 12 = 6t + 12
R - time reversal: y(t) = w(-t) = 6(-t) + 12 = -6t + 12
s - shift by +5: z(t) = y(t+5) = -6(t+5) + 12 = -6t - 18

Well, that didn't work. But why not?

Let's look at our gold standard a bit more closely.

z(t) = x(-2t+5) = 3(-2t+5) + 12 = -6t + 27

Lets replace the minus sign with R (which is either +1 or -1), the 2 with S, and the 5 with s:

z(t) = x(RSt+s) = 3(RSt+s) + 12 = 3RSt + 3s + 12

What we see is that the shift is not scaled or reversed, indicating that it should be done first. So let's try doing them in the opposite order indicated, which is (sRS):

s - shift by +5: w(t) = x(t+5) = 3(t+5) + 12 = 3t + 27
R - time reversal: y(t) = w(-t) = 3(-t) + 27 = -3t + 27
S - scale by 2: z(t) = y(2t) = -3(2t) + 27 = -6t + 27

So that worked, too.

So can we always do either of the two that worked, namely always do (sSR) or do them in the opposite order indicated by the expression? Let's try an equivalent expression.

-2t+5 = -(2t-5) = -((2t)-5) Here the order is scaling, shifting, and then reversal, so the opposite order would be (RsS). Let's see what happens:

R - time reversal: y(t) = w(-t) = 3(-t) + 12 = -3t + 12
s - shift by -5: y(t) = w(t-5) = -3(t-5) + 12 = -3t + 27S - scale by 2: z(t) = y(2t) = -3(2t) + 27 = -6t + 27

So that worked. How about if we use the (sSR) order?

(s) - shift by -5: w(t) = x(t-5) = 3(t-5) + 12 = 3t - 3
(S) - scale by 2: y(t) = w(2t) = 3(2t) - 3 = 6t - 3
(R) - time reversal: z(t) = y(-t) = 6(-t) - 3 = -6t - 3

It didn't work here. So why did it work before?

Notice that in the expression we had before,

z(t) = x(RSt+s) = 3(RSt+s) + 12 = 3RSt + 3s + 12

that RS appear in one place multiplied together. So it doesn't matter what order they are done in.

But if we do the same with our reordered expression:

z(t) = x(-((2t)-5)) = x(R((St)+s)) where R=-1, S=2, and s=-5, we get:

z(t) = 3(R((St)+s)) + 12 = 3RSt + 3Rs + 12

And now the R appears someplace other than just multiplied by S. So order does matter. But do we do R or s first? Well, we know that (RsS) worked, but how about (sRS)? It seems like it should work. Let's find out:

(s) - shift by -5: w(t) = x(t-5) = 3(t-5) + 12 = 3t - 3
(R) - time reversal: z(t) = w(-t) = 3(-t) - 3 = -3t - 3
(S) - scale by 2: y(t) = w(2t) = -3(2t) - 3 = -6t - 3

Why not? If we think about it, the shift is affected by the scaling and/or the reversal only if they are done before the shifting. Hence we need to do the shifting after those operations that affect the shift and before the operations that don't. The key to why we do things in the opposite order is that later operations effect the result of earlier ones and do not affect the result of later ones.

So let's try one more and see if this still holds.

-2t+5 = 2((-t)+2.5) would be reversal then shifting then scaling, so the opposite order would be (SsR).

(S) - scale by 2: w(t) = x(2t) = 3(2t) + 12 = 6t + 12
(s) - shift by -5: y(t) = w(t+2.5) = 6(t+2.5) + 12 = 6t + 27
(R) - time reversal: z(t) = y(-t) = 6(-t) + 27 = -6t + 27

Yep, it worked.

So do the manipulations in the opposite order that the expression indicates.

BTW: The time reversal you did in Post #1 is wrong. The reversal mirrors about the y-axis. You mirrored about y=0.5.

GeoDer likes this.
4. Harsha234 New Member

Aug 1, 2017
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How have you done scaling

5. Khalid1234 New Member

Nov 29, 2017
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There are two ways to do this.
Standard and easy approach is to shift first and scale later.
If you want to scale first. Then take the scaling factor common and then perform the resulting shift operation.
like for example you want to perform x(-2t + 5).
1. The easy and standard approach is to shift x(t) to left by 5 units (Advanced signal). Then scale the resulting signal, x(t+5), by -2 i.e. compress and reverse.
2. If you want to scale first, then slight mathematical manipulation would be required.
x(-2t +5) = x( -2 ( t - 5/2) )
It means that scale x(t) to x(-2t) then make the shift of 5/2 to the right (delayed signal).
By doing so the result would be the same.
NOTE: Second method is employed wile calculating convolution of two signals.

Nov 29, 2017
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Nov 29, 2017
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8. MrAl AAC Fanatic!

Jun 17, 2014
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Hi,

This MIGHT work so this is a side note really...

Take the two expressions:
z(t)
z(2-0.5*t)

and the two endpoints:
(-1,1)
(1,-1)

set the two expressions equal:
z(t)=z(2-0.5*t)

but with different time values:
z(t1)=z(2-0.5*t2)

replace t1 with -1:
z(-1)=z(2-0.5*t2)

and since they both must equal 1 we have:
-1=2-0.5*t2

solve for t2 we get:
t2a=2

then replace t1 with 1:
z(1)=z(2-0.5*t2)

and since they both must equal -1 we have:
-1=2-0.5*t2

and we get:
t2b=6

so the new wave spans from 2 to 6 as in the example.

Did not check this for any other inputs or functions so you should check this better if you are interested.

There is an additional fact that helps here and that is that the basic wave shape does not change even though it may reverse, stretch, and shift.
There may have to be additional refinements such as computing the amplitudes.
The numerical variational method might be replaced with a symbolic form.