Finding upper and lower bound superposition frequencies of ultrasound pulses

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SUMMARY

The discussion focuses on calculating the upper and lower bound superposition frequencies of ultrasound pulses transmitted at 1.000 MHz into water with a sound speed of 1500 m/s and a pulse length of 12 mm. It is established that each pulse contains 8 complete cycles, derived from the time duration of the pulse calculated as Δt = 8 * 10^-6 seconds. The frequency range necessary for superposition is determined to be 125,000 Hz, centered around 1.000 MHz, leading to a lower bound of 875,000 Hz and an upper bound of 1,125,000 Hz.

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  • Knowledge of frequency and period calculations
  • Familiarity with the speed of sound in different media
  • Basic mathematical skills for solving equations
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  • Learn about the Fourier Transform and its application in signal processing
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Mugen112
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Homework Statement


Ultrasound pulses of with a frequency of 1.000 MHz are transmitted into water, where the speed of sound is 1500m/s . The spatial length of each pulse is 12 mm.

a) How many complete cycles are in each pulse?

b) What is the lower bound of the range of frequencies must be superimposed to create each pulse?

c)What is the upper bound of the range of frequencies must be superimposed to create each pulse?

Homework Equations


ΔX=ΔtV
f=1/T


The Attempt at a Solution



a) Δt= 12mm/1500 = 8 * 10^-6
T = 1/f
T = 1/10^6 = 10^-6

So... (8 * 10^-6)/(10^-6) = 8 full cycles in each pulse

b) and c) I have no idea... All I have is..

Δf = 1/Δt = 125,000 Hz.. How do I find the upper/lower bound?
 
Physics news on Phys.org
It looks like the range of frequencies will cover a 125,000 Hz range, centered on 1.000 Mhz. From that you can figure out the lowest and highest frequencies in the range.
 

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