A level: Ultrasound Questions

Q1.

(a) A piezoelectric ultrasound transducer is made from a thin slice of an artificial ceramic such as lead zirconate titanate (PZT).

(i) Describe what happens when an alternating voltage is applied to a PZT transducer so that ultrasound is produced.

The electric field across a transducer causes a change in physical dimensions. If and alternating voltage is applied the transducer vibrates at the same frequency

(ii) Under what conditions will maximum energy conversion into ultrasound occur?

Maximum energy transfer occurs when the frequency of the applied voltage is equal to the natural mechanical vibration frequency of the transducer causing resonance to occur.

4 marks

(b) The diagram below shows the oscilloscope display of pulse amplitude against time for an ultrasound A-scan through a person's abdomen. Assume that the weaker echoes come from internal organs.

(i) Describe the procedures which are used to obtain this type of scan.

The ultrasound transmitter or receiver or transducer is placed in contact with skin. Transmission is improved by using a contact gel layer between transducer and skin, this allows the pulses of ultrasound to betransmitted into the body more effectively.

(ii) Explain how the spacing of the pulses is interpreted.

Echoes reflected back to transducer appear as voltage peaks or pulses on c.r.o. The spacing of the peaks gives the time delay between transmitted pulse and echoes.

Distance s between transmitter and a reflecting surface is given by

s = 1/2 c t

where c = speed of ultrasound in tissue,

t = time delay between pulses

(iii) Give two reasons why the amplitude of the reflected pulses varies.

The amplitude of pulses is attenuated as it penetrates deeper (caused by absorption and dispersion). The amplitude of the echo pulses depends on the proportions of ultrasound which are reflected and transmitted at each surface or boundary (or depends on the acoustic impedences of the two materials at the interface)

(iv) If the speed of ultrasound through water and soft tissue is about 1500 ms–1, estimate the distance between the front of the patient's abdomen and the spinal column.

see diagram: t = 0.32 – 0.08

= 0.24 ms

s = c t

= 0.5 × 1500 × 0.24 × 10–3

= 0.18 m

8 marks

Total 12 marks