Answer
See the detailed answer below.
Work Step by Step
To find the range of frequencies, or bandwidth, needed to create a 1.0-ns radar pulse, we need to use the concept of the uncertainty principle for signals.
$$\Delta f \approx \frac{1}{\Delta t}\tag 1$$
The pulse duration is given by
$$ \Delta t =1.0\;\rm ns $$
Plug into (1),
$$\Delta f \approx \frac{1}{1.0\times 10^{-9}}=\bf 1.0\times 10^{0}\;\rm Hz\tag 2$$
And we are given that the period of this pulse is $T=0.1\;\rm ns$.
So the oscillation frequency is
$$f=\dfrac{1}{T}=\dfrac{1}{0.1\times 10^{-9}}=\bf 1.0\times 10^{10}\;\rm Hz\tag 3$$
Hence the range of frequencies that is needed to be superimposed to create a 1.0-ns-long radar pulse is given by
$$ \left[f-\frac{1}{2}\Delta f \right] \leq f\leq\left[ f+\frac{1}{2}\Delta f \right]$$
Plug from (2) and (3);
$$ \left[(10^{10})-\frac{1}{2}(10^{9}) \right] \leq f\leq \left[(10^{10})+\frac{1}{2}(10^{9}) \right]$$
$$ \left[ 9.5\times 10^9\right] \leq f\leq \left[ 10.5\times 10^9 \right]$$
$$ {\bf 9.5\;\rm GHz} \leq f\leq{\bf10.5\;\rm GHz}$$
