# How can Haar basis be written as the impulse response of an LTI system?

Haar wavelets are defined as:

$$\phi_{0,0}(t) = \begin{cases} 1, & \text{ for } 0

Where the mother wavelet is $$\phi_{k,n}(t) = 2^{k/2} \phi_{0,0}(2^k t -n)$$

I know that this wavelet set is orthonormal and as such the coefficients $$c_{k,n}$$ in:

$$x(t) = \sum_{k=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} c_{k,n} \phi_{k,n} (t)$$

can be found as $$\langle x(t),\phi_{k,n(t)}\rangle$$.

However, I have a question which asks:

Let $$x_k(t) = \sum_{n=-\infty}^{\infty} c_{k,n} \phi_{k,n}(t)$$ for a given $$k$$.

Show that the $$c_{k,n}'s$$ can be found as: Find $$h_k(t)$$.

Now clearly $$h_k(t)$$ is some combination of the basis function.

My attempt:

\begin{align*} y(t) &= \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau \quad \text{Desired Form}\\ c_{k,n} &= \int_{-\infty}^{\infty} x(t) \phi_{k,n}(t) dt\\ &= 2^{k/2} \int_{-\infty}^{\infty} x(t)\phi_{0,0}(2^k t -n) dt \end{align*}

But now if I sample $$t=\frac{n}{2^k}$$ I will be left with:

$$2^{k/2} \int_{-\infty}^{\infty} x(\frac{n}{2^k})\phi_{0,0}(n-n) dt$$

and I don't know how to change the integral to a summation in this case. Plus $$\phi_{0,0}(n-n)$$ makes no sense.

How can I proceed? Any help is appreciated!

A very important idea to grasp here is that convolution is nothing but the inner product of $$x(\tau)$$ and $$h(t-\tau)$$ where the variable of integration is $$\tau$$ and $$t$$ is fixed. Therefore, for each $$t$$ (for each output function instant), there will be another inner product.
What you need to identify is the corresponding $$h(\tau)$$ and the values of $$t$$ in the convolution which is an inner product. So, write your inner product of $$x(\tau)$$ with the wavelet basis $$\psi(\frac{\tau -b}{a}) = \psi(\frac{\tau}{a} - \frac{b}{a})$$ and then see that if you choose $$h(\tau)=\psi(-\frac{\tau}{a})$$ then you are done for each convolution output at $$t=\frac ba$$.