91Ó°ÊÓ

.a) The number of complete cycles in one pulse can be obtained by dividing the spatial length of the pulse by the wavelength of the ultrasound pulse, where the wavelength can be calculated as follows

$\mathrm{λ}=\frac{v}{f}=\frac{1500\mathrm{m}/\mathrm{s}}{1×{10}^6\mathrm{Hz}}=1.5×{10}^{-3}\mathrm{m}$

Number of cycle complete in one pulse

$n=\frac{\mathrm{Δ}x}{\mathrm{λ}}=\frac{12×{10}^{-3}\mathrm{m}}{1.5×{10}^{-3}\mathrm{m}}=8$

first, we need to find the pulse duration $\left(\mathrm{Δ}t\right)$, and that can be done by finding the period of the wave and multiply it by the number of complete cycles in one pulse. The period is

$T=\frac{1}{f}=\frac{1}{1×{10}^6\mathrm{Hz}}=1×{10}^{-6}\mathrm{s}$

thus, the duration of the pulse (the wave packet ) is

$\mathrm{Δ}t=n×T=8×1×{10}^{-6}\mathrm{s}=8×{10}^{-6}\mathrm{s}$

now $\mathrm{Δ}f$can be calculated as

$\mathrm{Δ}f=\frac{1}{\mathrm{Δ}t}=\frac{1}{8×{10}^{-6}\mathrm{s}}=1.25×{10}^5\mathrm{Hz}$

Finally, the range of frequencies that must be superimposed to create the given pulse is

$\left(1\mathrm{MHz}-\frac{\mathrm{Δ}f}{2}\right)≤f≤\left(1\mathrm{MHz}+\frac{\mathrm{Δ}f}{2}\right)$

$\left(1×{10}^6\mathrm{Hz}-\frac{1.25×{10}^5\mathrm{Hz}}{2}\right)≤f≤\left(1×{10}^6\mathrm{Hz}+\frac{1.25×{10}^5\mathrm{Hz}}{2}\right)$

$9.38×{10}^5\mathrm{Hz}≤f≤10.63×{10}^5\mathrm{Hz}$

$0.938\mathrm{MHz}≤f≤1.063\mathrm{MHz}$