Now I have one function $f(x)$, and its support interval is $x\in[T_0,T_1]$. The support interval of the mother wavelet $\psi(x)$ is $x\in[-\Delta,+\Delta]$. And we also know that the continuous wavelet transform is $$ W(a,b) = \frac{1}{\sqrt{a}}\int_{-\infty}^{+\infty} f(t)\Psi(\frac{t-b}{a})\text{d}t $$
According to the formula above, It seems like you need to do the integration from $-\infty$ to $+\infty$. However, we don't have to do it since that $f(x)$ and $\psi(x)$ are compact support. So my question is How do we determine the appropriate integral limits?
- Here is my attempt:
According to "the support interval of the mother wavelet $\psi(x)$ is $x\in[-\Delta,+\Delta]$", we know that $$ -\Delta\leq\frac{t-b}{a}\leq\Delta\, \,\Rightarrow \, \, b-a\Delta\leq t \leq b+a\Delta $$
And because that the support interval of $f(t)$ is $[T_0,T_1]$ , we have $$ T_0\leq t \leq T_1 $$
But I don't know how what I should do next. Whether could we determine the integral limits by these two inequalities above?
It looks like your integration limits should just be the intersection of the intervals you specified. That is, the product $f(t)\Psi\left(\frac{t-b}{a}\right)$ is zero whenever $$t\notin A \triangleq\left([b−aΔ, b+aΔ]\bigcap[T_0, T_1]\right)$$ So $$ \frac{1}{\sqrt{a}}\int_{-\infty}^{+\infty}f(t)\Psi\left(\frac{t-b}{a}\right)dt = \frac{1}{\sqrt{a}}\int_{\min A}^{\max A}f(t)\Psi\left(\frac{t-b}{a}\right)dt $$ Example: For Δ=10, a=1, b=2, $T_0=-3$, $T_1=35$, $\min A=-3$ and $\max A=12$