$\newcommand{\ud}{\mathrm{d}}$ $\newcommand{\fou}{\mathscr{F}}$ I have to compute the Fourier transform of the function: $$ x(t) = \int_{-\infty}^{+\infty} g(\xi) \int_{-\infty}^{t - 3\xi} h(s - 3) \ud s \ud \xi $$ Fourier transforms are typeset in capital letters.
Focussing on the second integral and introducing the step function $u$, it can be rewritten as: $$ w(t, \xi) = \int_{-\infty}^{t - 3\xi} h(s - 3) \ud s = \int_{-\infty}^{+\infty} h(s - 3) u (t - 3\xi - s) \ud s $$ Substituting the variable $\alpha = t - 3 \xi$, the integral becomes: $$ w(\alpha) = \int_{-\infty}^{+\infty} h(s - 3) u(\alpha - s) \ud s $$ Let $h_1(s) = h(s - 3)$, it's possible to write: $$ w(\alpha) = \int_{-\infty}^{+\infty} h_1(s) u(\alpha - s) \ud s = \big( h_1 (\cdot) * u (\cdot) \big) (\alpha) $$ Applying the product rule of the Fourier transform and the sampling property of the Dirac delta, it is: $$ W(f) = \fou\{w(\alpha)\}(f) = H_1(f) U(f) = H(f) e^{-j 6 \pi f} U(f) = \frac{H(f)e^{-j 6 \pi f}}{j 2 \pi f} + \frac{H(0) \delta(f)}{2} $$ The initial function can now be rewritten as: $$ x(t) = \int_{-\infty}^{+\infty} g(\xi) w(\alpha) \ud \xi = \int_{-\infty}^{+\infty} g(\xi) w(t - 3\xi) \ud \xi $$ Introducing the substitution $\xi_1 = 3\xi$, the last relationship becomes: $$ x(t) = \int_{-\infty}^{+\infty} g\left( \frac{\xi_1}{3} \right) w(t - \xi_1) \frac{\ud \xi_1}{3} = \frac{1}{3} \left( g\left( \frac{1}{3} \cdot \right) * w(\cdot) \right) (t) $$ Applying the product rule of the Fourier transform and the sampling property of the Dirac delta, it is: $$ X(f) = \fou\{x(t)\}(f) = \frac{1}{3} 3 G(3f) W(f) = \frac{G(3f) H(f) e^{-j 6 \pi f}}{j 2 \pi f} + \frac{G(0) H(0) \delta(f)}{2} $$
I have two questions about my solution:
- is it formally correct to write the second integral as a function of the two variables $t, \xi$, namely $w(t, \xi)$?
- is the substitution $\alpha = t - 3\xi$ formally legit?