Math 414-501 — Spring 2020

Assignment 7 - Due Monday, 4/6/2020.

• Read section 4.2, 4.3.1 and 4.3.2
• Problems.
  1. Chapter 4: 2, 6, 7

  2. Let $d>1$. Consider the running average of a function: $L[f](t) =\frac{1}{d}\int_{t-d}^t f(\tau)d\tau$.
     1. Using the definition of a linear time invariant filter, show that $L$ is LTI; that is, show directly from the formula for $L[f]$ that $L[f_a]=(L[f])_a$.
     2. Use $L[\delta](t)=h(t)$ to find the impuse response function.

  3. Let f(t) be a signal that is 0 when t < 0 or t > 1. Show that, for the Butterworth filter, one has

     \[ L[f] = Ae^{-\alpha t} \int_0^{\min(1,t)} e^{\alpha \tau} f(\tau)d\tau, \ \text{if }t\ge 0, \ \text{and } L[f] = 0 \ \text{if } t<0. \]

  4. Consider a function $f=\begin{cases}\cos(\pi x/4) &|x| \le 2\\ 0&\text{otherwise}\end{cases}$. Find the projection of $f$ onto $V_0$, $W_0$ and $V_1$. Do the coefficients satisfy the formulas invoving a's and b's in Theorem 4.12?

     Solution The projection of $f$ onto $V_j$ is \[ \text{Proj}_{V_j}(f) =\sum_{k=-\infty}^\infty \langle f, 2^{j/2}\phi(2^jt-k)\rangle \,2^{j/2}\phi(2^jx-k)=\sum_{k=-\infty}^\infty a^j_k \phi(2^jx-k), \] where $a_k^j=2^j\int_{-\infty}^\infty f(x) \phi(2^jx-k)dx$. Since $ \phi(2^jx-k)dx=1$ for all $x$ such that $2^{-j}k \le x\e 2^{-j}(k+1)$, and $0$ otherwise, we have that \[ a^j_k =2^j\int_{2^{-j} k}^{2^{-j}(k+1)} f(x)dx \] Pretty much the same calculation shows that \[b^j_k =2^j\int_{-\infty}^\infty \psi(2^j x -k) f(x)dx = 2^j\int_{2^{-j} k}^{2^{-j}(k+1/2)} f(x)dx - 2^j\int_{2^{-j} (k+1/2)}^{2^{-j}(k+1)} f(x)dx\] For $j=0$, $a^0_k =\int_{k}^{k+1} f(x)dx$. Since $f(x)=0$ outside of the interval $[-2,2]$, the only $k$'s that contribute are $-2,-1, 0, 1$. This is also true for $b_k^0$. For $a_k^1$, the values of $k$ that contribute are $-4$ through $3$. To complete the problem, one should do the integrals. This I leave to the reader.