# Time-shifting operation post the time-reversal operation when performing convolution

I'm confused with the time-shifting operation post the time-reversal operation when performing convolution. Let's say we were to convolve $$x(t)$$ and $$h(t)$$, so I would have the term $$x(k)$$ and $$h(t-k)$$ to solve for. In $$h(t-k)$$, we first reverse $$h(k)$$ to $$h(-k)$$ and then shift it to $$t$$ units. Ideally, we should shift t units to the left but I find examples where they have moved $$t$$ units to the right instead.

I'm confused because, for any $$x(t+n)$$, we shift $$n$$ units to the left of $$t$$, so for $$x(-t+n)$$, why we are moving n units to the right of $$-t$$?

Just think of the value of $$t$$ where $$x(0)$$ appears. For $$x(t+t_0)$$ it is at $$t=-t_0$$, which corresponds to a left shift if $$t_0>0$$. However, for $$x(-t+t_0)$$ the value $$x(0)$$ occurs at $$t=t_0$$, which is to the right of its original position if $$t_0>0$$.
You can think of deriving $$x(-t+t_0)$$ from $$x(t)$$ in two different ways:
1. invert the time axis: $$x(-t)$$
2. replace $$t$$ by $$t-t_0$$, i.e., shift to the right if $$t_0>0$$: $$x(-(t-t_0))=x(-t+t_0)$$
1. shift $$x(t)$$ to the left (for $$t_0>0$$): $$x(t+t_0)$$
2. invert the time axis: $$x(-t+t_0)$$: this flips the shifted function from the left to the right of the original function (again, if $$t_0>0$$).