Question

Problem 3 (10 pts). Let f(x)-δ(z + a) + δ(z-a); Ict r(z) and h(z) be functions de- scribed in Fig. 2 below. As discussed in class, one can show thatof(u)r u)du Assume that f(x) and h(x) are known but r(x) has been lost. Recover r(x) f(x) r(x) h(x) co 1f -bl b -a FIG. 2: This refers to Problem 3.

Answer 1

r(x) can be obtained by the scaled convolution of f(x) and h(x).

$r(x)=\frac{1}{2} \int_{-\infty}^{+\infty} h(u)f(x-u)du$

graph of f(x-u)

for x<-b

The overlapping graph of f(x-u) with h(u) is

from the graph, it is evident that the value of the integral

$r(x)=\frac{1}{2} \int_{-\infty}^{&plus;\infty} h(u)f(x-u)du= \frac{1}{2}[h(-a-x)&plus;h(a-x)]= \frac{1}{2}[0&plus;0]= 0$

for this case (x<-b) is 0.

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case ii : -b<x<b

The value of integral is

$r(x)=\frac{1}{2} \int_{-\infty}^{&plus;\infty} h(u)f(x-u)du= \frac{1}{2}[h(-a-x)&plus;h(a-x)]= \frac{1}{2}[1&plus;1]= 1$

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for the case when x>b

$r(x)=\frac{1}{2} \int_{-\infty}^{&plus;\infty} h(u)f(x-u)du= \frac{1}{2}[h(-a-x)&plus;h(a-x)]= \frac{1}{2}[0&plus;0]= 0$

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Hence ,

from the three cses discussed above

$\\ r(x)= 0 \ \ for \ x<-b \\ r(x)= 1\ \ for \ -b<x<b \\ r(x)= 0 \ \ for \ b<x$

Hence we recover r(x)