Analog Signal Processing for EMG Flashcards
Where do you place electrodes?
2 positions:
- middle (meaty part) of the muscle
- at the end of the muscle
Briefly describe the steps for EMG signal processing
- obtain raw EMG signal from electrodes
- Instrument amplifier
- amplify signal from micro to mili - band-pass filter
- remove interfering inputs (noise and DC level)
- the low-pass filter part of the band-pass filter gets rid of the DC level - Precision Rectifier
- calculate abs value of EMG signals - Integrator
- quantify amount of EMG activity (area under EMG signal) - Display
Describe EMG signal characteristics
Frequency:
ranges from 25 Hz to several kHz
- we want to remove motion artifact (low frequencies)
Amplitude
ranges from 100uV to 90mV
Compare and contrast skin-surface vs intramuscular electrodes
Skin-surface
- low amplitude signals
- low impedance (200 to 5000 ohms)
Intramuscular electrodes
- larger amplitude signals
- high impedence (due to smaller surface area)
What is the significance of impedence for electrodes and EMG signals
Impedence: amt of opposition to electrical current
We want electrodes to have less impedence.
Why? –> so that more of the signal (voltage) gets pass through to the rest of the processing system
(smaller cross-sec area = larger impedence)
how does a bandpass filter help process raw EMG signals? (4) `
Low-pass component:
- removes baseline drift
- removes DC offset
High-pass component:
- Removes high-freq noise
- prevent aliasing in sampled signal
!!! noise can cause aliasing !!!
What properties must an instrument amplifer have?
- Large CMRR
(Common-mode rejection ratio)
- large CMRR = more effective rectifier - Adjutable differential gain
- different muscles / have diff raw EMG signal amplitudes
(therefore this amplifer must be adaptable to all possible use cases) - Large input resistance
(input resistance is the resistance of the amplifier)
- we don’t need a lot of signal for the rest of the processing - small output resistance
(output resistance is the resistance of the electrode)
- allows amplifier to receive more of the signal
What are some sources of common-mode interferences in biopotentials? (3)
- electric power sources
- electromagnetic devices
- EMG from distant muscles
How does an amplifer amplify biopotential signals?
Biopotential signals measure the difference between the middle electrode and the end electrode
V1 = Vcm - (Vd / 2)
V2 = Vcm + (Vd / 2)
Vo = AdVd + AcmVcm
(where A = gain)
Common-mode rejection ratio:
CMRR = 20 log (|Ad| / |Acm|)
we want Acm to decrease for the rectifier to be effective (amplifies biopotential signals)
Compare and contrast a single-opamp difference amplifier and an instrumentation amplifier
Single-opamp difference amplifier:
Gain:
Vo = (R4 / R3) (V2 - V1)
Ad = (R4 / R3)
- many drawbacks:
1. requires R4 and R3 to be exactly matching to have a large CMRR
2. Can’t have large differential gain and large input resistance
(b/c large differential gain requires small R3 but that also means small input resistance)
3. Unable to vary the differential gain
instrumentation amplifier:
Gain:
Vo = (R4 / R3) (1 + (R2 / R1))Vin
Ad = (R4 / R3) (1 + (R2 / R1))
- positives:
1. doesn’t need resistors of op amp A1 and A2 to match
2. large input resistance (infinite when using ideal op amps)
3. Easy to vary gain
How do diodes behave like rectiviers?
Diodes pass only positive signals after a specified voltage
- made from P (pos) and N (neg) type doped semiconductors (P have holes, N has electrons)
- P-N junction shrinks w/ forward biased voltage –> allows conduction of charge at high enough voltages)
- expands w/ reverse biased voltage (large reverse voltage –> diode burns and becomes unusable)
Cons: forward voltage inhibits small signals
Precision Half-wave rectifier (w/ 1 diode)
- describe firing and off stages
How to make it suitable for small signals?
- add op-amp to diode
- Case 1 (diode firing)
a) assume v1 = V0 = 0
b) increase in V1 –> (Vp - Vn) > 0
c) Va = A(Vp - Vn) > diode forward voltage [diode acts like a wire]
d) use virtual short assumption:
Vp = Vn –> V0 = V1 - Case 2 (diode off)
a) assume v1 = V0 = 0
b) decrease in V1 –> (Vp - Vn) < 0
c) Va = A(Vp - Vn) < diode forward voltage [diode acts like a break]
d) CANNOT use virtual short assumption:
why? current = 0
therefore: V0 = RI = 0
What is a virtual short assumption in Opamps?
V1 = V2 only if there is a negative feedback from output to input
Precision Half-wave rectifier (w/ 1 diode)
- limitations
- output saturates with slightly negative input
- if input becomes positive again, op-amp needs to get out of the saturated stated before it can amplify the positive signal (no sudden switch between positive and negative)
- dictated by op-amp slew rate - because change takes time, circuit is undesirable for higher freq signals
Precision Half-wave rectifier (w/ 2 diode)
- describe firing and off stages
2 diodes make circuit suited for higher frequencies
- Case 1 (D1 off, D2 firing)
a) assume all voltages = 0
b) increase in V1 –> (Vp - Vn) < 0
c) Va = A(Vp - Vn) < D1 forward voltage
d) use virtual short assumption:
Vp = Vn = 0 –> VA = 0 - Case 2 (D2 off, D1 firing)
a) assume all voltages = 0
b) decrease in V1 –> (Vp - Vn) > 0
c) Va = A(Vp - Vn)»_space; diode forward voltage
d) use virtual short assumption:
V0 = - R2 / R1 (v1)
Precision full-wave rectifier
- describe firing and off stages
- has 2 op-amps and 2 diodes
- Case 1 (top op-amp fires)
a) assume all voltages = 0
b) increase in V1 –> (Dtop fires and Dbottom off)
c) use virtual short assumption:
V0 = v1 - Case 2 (bottom op-amp works )
a) assume all voltages = 0
b) decrease in V1 –> (Dtop off and Dbottom fires)
c) use virtual short assumption:
V0 = - R2 / R1 (v1)
Slew rate
quantifies how fast output of op-amp can change
