This question is related to Weiner filter. However my doubt is not about the Wiener Filter itself but the calculations from one of the exercise.
I was given: $$H(f) = \frac{S_v(f)}{S_v(f)+S_x(f)}$$ where, $$S_v(f)=(1-|f|)I_{[-1,1]}(f)$$ $$S_x(f)=1-I_{[-1,1]}(f)$$ Hence, I have: $$H(f) = \frac{(1-|f|)I_{[-1,1]}(f)}{1-|f|I_{[-1,1]}(f)}$$ But the answer further simplifies it to: $$H(f) = I_{[-1,1]}(f)$$
How do I achieve the last step? It doesn't make sense to me.
It looks like you want to show that
$$ \frac{(1-|f|)I_{[-1,1]}(f)}{1-|f|I_{[-1,1]}(f)} = I_{[-1,1]}(f). \tag{$\star$}$$
You can show this by cases. (Note that technically the LHS here is undefined if $f=\pm 1$ since you get $0/0$.)
Case 1: $-1< f < 1$ (i.e. $|f|<1$)
In this case, the RHS of $(\star)$ is $1$ and the LHS is also $\frac{(1-|f|)\times 1}{1 -|f|\times 1} = \frac{1-|f|}{1-|f|}=1$.
Case 2: $|f| > 1$
In this case, the RHS is $0$ and the LHS is also $\frac{(1-|f|)\times 0}{1 -|f|\times 0} = \frac{0}{1}=0$.
