test 3- beam former ppt Flashcards by Miss Nancy

Ultrasound systems in use today use ______________
method

pulse-echo

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These can be thought of as subdivision names

t= transducers

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Beam Former
◼ Responsible for:
 Generating voltages that energize the transducer
elements
 Determining the PRF, coding, frequency, and
intensity
 Scanning, focusing, and apodizing the transmitted
beam
 Amplifying the returning echo voltages
 Compensating for attenuation
 Digitizing the reflected echo signals
 Directing, focusing, and apodizing the received echo
beam

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beam former goes down what happens to image

no image produced

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beam former

what are the “houses” inside the “subdivision”

t/r= transmit/receive switches

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t/r switch “house”

is like the regulator
monitor the amount coming in and going out
makes sure sound is going the right way

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◼ Where electrical impulses begin
◼ Produces electrical voltage pulses that drive
the transducer
 Short impulses for shock excitation
 A cycle or two of voltage for burst excitation
◼ Causes piezo-electric element to expand and
contract, producing sound waves

in the pulser “house” which is in the beam former “subdivision”

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Number of pulses produced per second (kHz)

PRF pulse repetition frequency

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what is the range of PRF

Ranges from 1 - 10 kH

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Ultrasound PRF is equal to the _____________PRF

voltage

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◼ If PRF is too high, range ambiguity results

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PRP equation

1/PRF

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Range Ambiguity

All echoes from one pulse must be received before the
next pulse is emitted

If imaging deeper structures, must use lower PRF (lower
frame-rate)

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deeper image uses (higher/lower) PRF

lower

which also means lower frame rate

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If a 6 MHz transducer images to a depth of 10 cm, what is
the maximum PRF the pulser can produce while
avoiding range ambiguity?

13 us X 10 cm = 130 us(microseconds)

PRF = 1/PRP
X = 1/130
X = .0077 MHz

or 7.7 kHz

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PRF equation

1/PRP

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If a 2 MHz transducer images to a depth of 15 cm, what is
the maximum PRF the pulser can produce while avoiding
range ambiguity?

13 us x 15cm= 195us (microseconds)

PRF= 1/PRP
PRF= 1/ 195
=.0051 MHz

or 5.1 kHz

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◼ When penetration is multiplied by:
 Number of focuses
 Number of scan lines per frame
 Frame rate
…..the number must not exceed 77,000
Also—please note that the 3 factors above, when
multiplied together = THE PRF !!

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= # focuses x lines per frame x frame rate

PRF

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focuses x lines per frame x frame rate < ______________

77,000

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Why must PRF be ≤77,000 Hz

To avoid range ambiguity

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Also note that these are the same formulas you
would use to answer a question as to how many
focal zones you could have or what the limit is!!

PRF= # focuses x lines per frame x frame rate

focuses x lines per frame x frame rate < 77,000

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what is beam intensity controlled by

the pulser “house” in the beam former “subdivision”

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meaning of -3db

minus 50% output

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-10db

10% output

lower acoustic output will result in slight (higher/ lower) penetration

lower (decrease 50% output = decrease 5% penetration) – can compensate by upping gain

Pulses supplied by pulser are controlled by the ________________-

Pulse Delays (beam former)

◼ Pulse delays control timing delay operations necessary for sequencing and phasing when using arrays

 Scanning  Steering  Focusing  Dynamic aperture  Apodization (creating firing with non-uniform voltages to minimize grating lobes)

Specific pulse-code-patterns for maximum penetration and resolution may be encoded together creating a long, wide-bandwidth pulse (i.e. more power/penetration while maintaining maximum resolution) The secret knock!!

Coded Excitation (beam former)

◼ Serves to improve signal-to-noise ratio and/or penetration depth

Coded Excitation

◼ Uses complicated driving voltage-pulse forms to encode a unique signature on the sound beam by repeating a specific pattern of pulses

Coded Excitation

Coded Excitation ◼ Accomplishes:

 Penetration  Multiple transmit focuses  Separation of harmonic echoes from transmitted pulses  Increased penetration  Reduction of acoustic speckle  Improved contrast resolution  Gray-scale imaging of blood flow (b-flow)

◼ A device used to direct current from one conductor to another

T/R Switches (beam former)

◼ Directs the driving voltages between the pulse delays and the transducer

T/R Switches

◼ Each independent element and pulse delay = ____________-

channel

◼ (higher/lower) number of channels allows more precise control of beam characteristics

higher

echo reception

how to improve low channels

sequencing

Convert small voltages received from the transducer to larger ones suitable for processing and storage

amplifier Amplification = gain

◼ The ratio of output voltages to input voltages  usually expressed in decibels (dB)  equal to the voltage ratio squared

gain ◼ EXAMPLE:  Input voltage is 2 mV & output voltage is 200 mV  voltage ratio is 200/2 or 100  Gain is (100)2 or 10,000  Gain of 10,000 is equal to 40 dB

If you have to choose between increasing power or increasing gain….

always select gain

Output power adjusts the ______________ of the sound sent into the body

strength

Receiver Gain changes the _____________ of the signal by amplifying the signal after sound enters the receiver (no possible bioeffects

brightness

 Time-gain compensation (TGC)  Depth gain compensation (DGC)

interchangeable terms

go write attenuation coef equation and understand it

deals with compensation

◼ Digitizer ◼ Converts voltage from complex analog signal to digital form (numerical values representing echo strength)

Analog-to-Digital Converters (beam former)

◼ After amplification and digitization, echo voltages pass through digital delays to accomplish dynamic focusing and steering functions

Echo Delays (beam former)

◼ All individual channel signals are added together to produce the resulting scan-line ◼ Reception apodization and dynamic aperture are also accomplished here

Summer (beam former)

◼ Signal processor(subdivision) performs the following functions:

 Bandpass filtering (rejection)  Amplitude detection (RF to video)  Compression (dynamic range reduction) Signal Processor

◼ Eliminates the smaller-amplitude voltages produced by electronic “noise”

filtering (house inside signal processor subdivision)

◼ “_______________” filter passes a range of frequencies (its bandwidth) and rejects those above and below the acceptance bandwidth

Bandpass part of filtering

◼ Tissue harmonic imaging operates by transmitting a fundamental (original) beam that has a lower frequency. ◼ This fundamental pulse, as it propagates through tissue INSIDE THE BODY, generates the higher frequency harmonic sound. ◼ The key to understanding tissue harmonic imaging is that when you have harmonics ON, THE IMAGE is formed ONLY FROM THE HIGHER FREQUENCY HARMONIC SOUND ◼ Echoes from the fundamental (ORIGINAL) frequency are rejected and thus, are not used in making the image.

Filtering & Harmonics

Obviously, sophisticated transmit beam formation and ________________ is required to produce good quality harmonic images.

signal detection

 Only a small part of the returning sound involves these higher frequency harmonics because they are much (higher/lower) in amplitude than the reflected fundamental beam.

lower

________________to remove the fundamental frequency is the technique currently used most commonly to produce tissue harmonic images

Filtration

The major benefit of tissue harmonic imaging is ________________---.

artifact reduction

All of the following are requirements of contrast agents:

◼ positive risk-benefit relationship ◼ long persistence ◼ strong reflector of sound

Detection aka

demodulation

◼ Process of converting the voltages delivered to the receiver form one form to another  i.e. radio frequency to video  Echo voltages are produced in radio-frequency which is difficult to store and display  Only the amplitude of each echo is needed for gray-scale display  RF is converted to video form (retains the amplitude information)

detection "house" in signal processing "subdivision"

◼ Demodulation removes the transmit signal from the echo information and only leaves information caused by sound’s interaction with tissue. ◼ Remaining information looks like spikes/like A- mode ◼ Accomplished through rectification and smoothing

Detection

Turning all of the negative voltages into positive ones

Rectification

◼ putting an envelope around the “bumps” to even out the rough edges

Smoothing

◼ The ratio of the largest to the smallest amplitude (power) that a system can handle ◼ Expressed in dB

Dynamic Range

◼ Function of COMPRESSION

dynamic range

◼ Dynamic range for display is typically 20 dB ◼ Dynamic range after compensation is typically 40-70 dB ◼ Must compress the dynamic range to an acceptable ratio for display ◼ The human eye can only perceive ~ 64 shades of gray= 36dB at best

Process of decreasing the differences between the smallest and the largest voltage amplitudes, while maintaining the relationships of the reflected echoes

Compression

compression affects _______________

dynamic range

Presented as a “dynamic range” or “log compression” control

compression

This may be called Receiver Operations

Sound Reception

Sound Reception aka receiver operations 5 operations

 Amplification’  Compensation  Compression  Demodulation  Reject Remember that five operations must be performed in the appropriate order for the system to function properly

◼ Output power (pulser) ◼ Frequency if multihertz capable (pulser) ◼ Depth (pulser) ◼ Focus and steering (pulse delays) ◼ Overall gain (amplifier) ◼ Time-Gain compensation (amplifier) ◼ Dynamic Range (signal processor)

Which house and subdivision ◼ Output power

Pulser

Which house and subdivision ◼ Frequency if multihertz capable

pulser, beam former

Which house and subdivision ◼ Depth

pulser, beam former

Which house and subdivision ◼ Focus and steering

pulse delays, beam former

Which house and subdivision ◼ Overall gain

amplifier, beam former

Which house and subdivision ◼ Time-Gain compensation

amplifier, beam former

Which house and subdivision ◼ Dynamic Range

signal processor, beam former