= Solution
The filter is a real even <triangular function>. Its <Fourier transform> is
$$
\widetilde f(\omega)=2\int_0^1(1-t)\cos(\omega t)\,dt=\boxed{\frac{2(1-\cos\omega)}{\omega^2}=\left[\frac{\sin(\omega/2)}{\omega/2}\right]^2},
$$
with $\widetilde f(0)=1$. Since the transmitted pulse is the <convolution> $b=a*f$, the <convolution theorem> gives
$$
\boxed{\widetilde b(\omega)=\widetilde a(\omega)\widetilde f(\omega)=\frac{4\sin\omega(1-\cos\omega)}{\omega^3},\qquad\widetilde b(0)=2.}
$$
The normalized filter preserves the time <integral> of the pulse, while its extra inverse-frequency factors suppress high-frequency content. It is a smoothing filter rather than an exact compact-bandwidth cutoff.
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