How to properly scale, time-reverse and shift a signal?

Question

I'm not sure if this one is on-topic here, but I'll give it a shot anyway.

I know that if I have some sort of signal, like for example \$y(t)=x(t)\$, and I want a signal with twice the frequency, I can write that as \$z(t)=x(2t)\$.

I also know that if I have signal \$y(t)=x(t)\$ and I want to shift it to right, I can write the shifted signal as \$z(t)=x(t-1)\$.

I also know that if I have signal \$y(t)=x(t)\$ and I want to time-reverse it, I can write the time-reversed signal as \$z(t)=x(-t)\$.

What confuses me is if I have a signal which has already been shifted, time-reversed or scaled and I need to shift it or scale it again.

For example I have a signal which is \$y(t)=x(2t-1)\$ and I need a signal which has been time-reversed, should that be \$z(t)=x(-2t-1)\$ or \$z(t)=x(-2t+1)\$? Same thing for scaling: If I have \$y(t)=x(2t-1)\$ and want to scale it by a factor of 2, should the result be \$z(t)=x(4t-1)\$ or \$z(t)=x(4t-2)\$?

Note that this is a homework question, so I'm looking for an answer which will explain the principle behind the transformations.

Answer 1

I think what might help is if we start to use different symbols to avoid confusion. When you're doing this shifts, inverts and so on what's happening is that the variable t is being replaced by some transform of t (t = -1*t for mirroring, t=(t-2) for shifting, etc). You can see how ungainly and confusing the equations above are: there's t's on both sides.

So when you do these transforms, figure out how t is being affected. If we're doing a right shift of x(t), then we're replacing t with (t-n). Define a new variable (a temporary one, to keep things straight) tau and the relationship with t: tau = t-2. So now the transformed equation is x(tau) and tau = t-2, so to simplify, the transformed equation is x(t-2).

One step is plain and simple, but applying another shift, invert or scale makes the reasoning behind the approach more clear. To shift x(tau) all we do is treat tau like t was treated before. So this time, y(tau) = x(tau-1) (right shift by one). Then substitute t for tau from the above equation and you get this: y(t) = x( (t-2) -1 ) == x(t-3)

If instead we mirrored it, y(tau) = x(-tau) then we simplify again and get this: y(t) = x(-1 * (t-2) ) == x( -t + 2 )

Taking it in small steps and using substitution will help keep things straight and prevent what basically boils down to algebra mistakes. The trick is that you're always replacing t and only t with something else. Good luck!