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

Question

Haar wavelets are defined as:

$$\phi_{0,0}(t) = \begin{cases} 1, & \text{ for } 0<t< 1/2\\ -1, & \text{ for } 1/2<t<1 \\ 0, & \text{ otherwise } \end{cases} $$

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!

Answer 1

This seems like a HW question so I will abstain from my previous habit of giving full solutions to HW questions and rather just explain, in a general way, how to approach this problem.

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 $t=\frac ba$.