Consider a 5MHz transducer that produces an ultrasound pulse consisting of three

Question

Consider a 5MHz transducer that produces an ultrasound pulse consisting of three periods of the wave. The ultrasound beam is used to image soft tissue. You can assume that the speed of sound in soft tissue is 1540 m/s.

a. Calculate the duration of the pulse.

b. Calculate the axial resolution produced by this transducer in soft tissue.

c. Assume that this ultrasound imaging system records successive A-lines at a rate of 2kHz (2000 A-lines per second). Calculate the time between two successive ultrasound pulses, and from this value, calculate the maximum imaging depth of this system.

d. Assuming that the ultrasound system is used in B-Mode, and that each image consists of 512 A-lines acquired at a rate of 2kHz. Calculate the total time required to acquire a B-scan.

e. Assume that this system is used to image an organ located below a 3 cm thick layer of fat. The ultrasound wave generated by the transducer travels through the layer of fat before reaching the organ. Calculate the delay between the emission of the incident ultrasound pulse and the detection of the reflected ultrasound pulse.

f. Assuming that the attenuation coefficient of fat is 0.63 dB/cm/MHz and that there are no other reflection or absorption losses, calculate the relative intensity of the wave that reaches the fat-soft tissue interface, in dB and in % of the intensity of the incident wave.

g. Calculate the intensity reflection coefficient (RI) at the fat-soft-tissue interface. The acoustic impedance of fat is given in the textbook. You can assume that the acoustic impedance of soft tissue is 1.60x106 rayls.

Explanation / Answer

Given :-

5 MHz Transducer

Pulse type - Ultrasound

Wave Period - 3

Sound of speed in soft tissue - 1540 m/s.

a) Duration of Pulse - To calculate

5 MHz ultrasound is 1 / (5*10^6) = 0.2 µs.

Speed of sound given as 1540 m/s i.e.1.54 mm/µs.

Pulse duration is the Time required for one pulse to occur = the number of cycles in the pulse in (ms).

Pulse duration = no. of cycles*period (msec)                    

Pulse Duration = (no. of cycles x wavelength) / Propagation speed.

As Wavelength (mm) = Propagation speed in soft tissue (mm/microsecond) / frequency (MHz)

Therefore, Wavelength (mm) = 1.54/5 = 0.308 mm

No. of cycles = 3

Hence, Pulse Duration = (3*0.308)/1.54 = 0.6 msec.

b) Axial resolution = Spatial Pulse Length/2

Spatial Pulse length = (no. of cycles*wavelength)

                              = (3*0.308)

                              = 0.92 mm

Therefore, Axial resolution = 0.46 mm.

c) Time between two successive ultrasound pulses - To calculate

Basically we have to calculate Pulse repetition period i.e.(PRP) which is the time from the beginning of one pulse to the beginning of the next pulse i.e. time period from Ton time of one pulse to the Ton time of next pulse.

Given :-

rate 2kHz (2000 A-lines per second)

PRP = 1/2000

       = 0.0005 msec

       = 500 microseconds

the maximum penetration would be : 1 dB/cm/MHz x 5 MHz x (2 x max depth)

= 63/10

Max depth = 6.3 cm

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