1. Signal conditioning
Noisy

2. Amplification Filter Attenuation Isolation Linearization
Key Functions of Signal Conditioning:
Amplification
Filter
Attenuation
Isolation
Linearization

3. Many sensors develop extremely low-level output signals
AMPLIFICATION
Many sensors develop extremely low-level output signals
Two common examples of low-level sensors are thermocouples and strain-gauge (less than 50 mV)
Requires amplification
Using Operational Amplifier (Op-Amp)
Inverting Amplifier
Non-Inverting Amplifier
Differential Amplifier
Instrumentation Amplifier

4. INVERTING AND NON-INVERTING AMPLIFIERS
NOTE: The maximum input signal that the amplifier can handle without damage is usually about 2 V less than the supply voltage. For example, when the supply is ±15 VDC, the input signal should not exceed ±13 VDC.

5. Gain = -10
Vo = - 5 V
Gain = 11
Vo = 5.5 V

6. DIFFERENTIAL AMPLIFIERS
Differential-input amplifiers offer some advantages over inverting and non-inverting amplifiers.
The output signal of the differential input amplifier responds only to the differential voltage that exists between the two input terminals.
Example to use differential amplifier is when thermocouple is used.
Both Rf must be equal and both Ri must also be equal
g is referred to as gain

7. 𝑉 𝑎 = 𝑅 𝑓 𝑅 𝑖 + 𝑅 𝑓 𝑠𝑜, 𝑉 𝑏 = 𝑅 𝑓 𝑅 𝑖 + 𝑅 𝑓 𝑠𝑜, 𝑉 𝑏 = 𝑅 𝑓 𝑥 Va = Vb
No current going into the op amp b
𝑉 𝑎 = 𝑅 𝑓 𝑅 𝑖 + 𝑅 𝑓
𝑉 1 a
𝑠𝑜, 𝑉 𝑏 = 𝑅 𝑓 𝑅 𝑖 + 𝑅 𝑓
𝑉 1
Let (𝑅 𝑖 + 𝑅 𝑓 ) = x
Nodal analysis at node b
𝑠𝑜, 𝑉 𝑏 = 𝑅 𝑓 𝑥
𝑉 1
𝑉 2 − 𝑉 𝑏 𝑅 𝑖 = 𝑉 𝑏 − 𝑉 𝑜 𝑅 𝑓
𝑉 2 − 𝑅 𝑓 𝑥 𝑉 1 𝑅 𝑖 = 𝑅 𝑓 𝑥 𝑉 1 − 𝑉 𝑜 𝑅 𝑓

8. 𝑉 2 − 𝑅 𝑓 𝑥 𝑉 1 𝑅 𝑖 = 𝑅 𝑓 𝑥 𝑉 1 − 𝑉 𝑜 𝑅 𝑓
𝑥𝑉 2 − 𝑅 𝑓 𝑉 1 𝑥 𝑅 𝑖 = 𝑅 𝑓 𝑉 1 − 𝑥𝑉 𝑜 𝑥𝑅 𝑓
𝑥𝑉 2 − 𝑅 𝑓 𝑉 1 𝑅 𝑖 = 𝑅 𝑓 𝑉 1 − 𝑥𝑉 𝑜 𝑅 𝑓
Using superposition theorem
Set V2 = 0 V
Vo = Vo1 + Vo2
Set V1 = 0 V
𝑥𝑉 2 − 𝑅 𝑓 𝑉 1 𝑅 𝑖 = 𝑅 𝑓 𝑉 1 − 𝑥𝑉 𝑜2 𝑅 𝑓
𝑥𝑉 2 − 𝑅 𝑓 𝑉 1 𝑅 𝑖 = 𝑅 𝑓 𝑉 1 − 𝑥𝑉 𝑜 𝑅 𝑓
𝑉 𝑜 =− 𝑉 2 𝑅 𝑓 𝑅 𝑖 + 𝑉 1 𝑅 𝑓 𝑅 𝑖
0− 𝑅 𝑓 𝑉 1 𝑅 𝑖 = 𝑅 𝑓 𝑉 1 − 𝑥𝑉 𝑜2 𝑅 𝑓
𝑥𝑉 2 −0 𝑅 𝑖 = 0−𝑥 𝑉 𝑜1 𝑅 𝑓
𝑉 𝑜 = 𝑉 1 −𝑉 2 𝑅 𝑓 𝑅 𝑖
− 𝑅 𝑓 𝑉 1 𝑅 𝑖 𝑅 𝑓 = 𝑅 𝑓 𝑉 1 − 𝑥𝑉 𝑜2
𝑥𝑉 2 𝑅 𝑖 = −𝑥 𝑉 𝑜1 𝑅 𝑓
𝑥𝑉 𝑜2 = 𝑅 𝑓 𝑉 1 𝑅 𝑖 𝑅 𝑓 +𝑅 𝑓 𝑉 1
𝑉 𝑜1 =− 𝑉 2 𝑅 𝑓 𝑅 𝑖
𝑥𝑉 𝑜2 = 𝑅 𝑓 𝑅 𝑖 +1 𝑅 𝑓 𝑉 1
𝑉 𝑜2 = 𝑅 𝑓 +𝑅 𝑖 𝑅 𝑖
𝑅 𝑓 𝑉 1 𝑥
𝑉 𝑜2 = 𝑉 1 𝑅 𝑓 𝑅 𝑖

9. For a gain of 10, where:
Rf = 100 kΩ and Ri = 10 kΩ
Prove by derivation that the output equation, Vo = 10 (V1 – V2)
What is the output voltage if the voltage difference is 50 mV

10. 𝑉 𝑎 = 𝑉 𝑏 =
𝑉 1
𝑉 𝑏 =
𝑉 1
𝑉 2 − 𝑉 = 𝑉 1 − 𝑉 𝑜 100
11𝑉 2 − 10𝑉 = 10𝑉 1 − 11𝑉 𝑜 1100
110𝑉 2 − 100𝑉 = 10𝑉 1 − 11𝑉 𝑜 1100
110𝑉 2 − 100𝑉 1
=10𝑉 1 − 11𝑉 𝑜
110𝑉 2 − 110𝑉 1
=− 11𝑉 𝑜
10𝑉 2 − 10𝑉 1
=− 𝑉 𝑜
𝑉 𝑜 =10𝑉 1 − 10𝑉 2
𝑉 𝑜 =10(𝑉 1 − 𝑉 2 )

11. INSTRUMENTATION AMPLIFIERSVA VB I
= 𝑹 𝟑 𝑹 𝟐
( 𝑽 𝑩 − 𝑽 𝑨 ) I I
𝑽 𝒐𝒖𝒕 = 𝑹 𝟑 𝑹 𝟐
( 𝑽 𝟐 − 𝑽 𝟏 )
𝟐( 𝑹 𝟏 𝑹 𝒈𝒂𝒊𝒏 )+𝟏
DIFFERENTIAL AMPLIFIER
ideal for measuring low-level signals in noisy environments without error, and amplifying small signals
the gain increases even more than just using differential amplifier
Using a potentiometer can vary the gain

12. EXAMPLE
An instrumentation amplifier has a gain of 20. If R3 = 10 k and R2 = 1 k. The current through R1 is 4 mA and V’ is 1V. VA = -2V.
Draw Schematic
Find Rgain & R1.
𝑉 ′ − 𝑉 𝐴 𝑅 1 =4𝑚𝐴
1−(−2) 𝑅 1 =4𝑚𝐴 VA
𝑅 1 = 3 4𝑚𝐴 =0.750𝑘 V’

13. 𝑮𝒂𝒊𝒏= 𝑹 𝟑 𝑹 𝟐
𝟐( 𝑹 𝟏 𝑹 𝒈𝒂𝒊𝒏 )+𝟏
𝟐𝟎= 𝟏𝟎 𝟏
𝟐( 𝟎.𝟕𝟓𝟎 𝑹 𝒈𝒂𝒊𝒏 )+𝟏 𝟐=
𝟐( 𝟎.𝟕𝟓𝟎 𝑹 𝒈𝒂𝒊𝒏 )+𝟏
𝟐= 𝟏.𝟓 𝑹 𝒈𝒂𝒊𝒏 +𝟏
𝟏= 𝟏.𝟓 𝑹 𝒈𝒂𝒊𝒏
𝑹 𝒈𝒂𝒊𝒏 =𝟏.𝟓𝒌

14. The three most common types of filters are called Butterworth
FILTERING
The three most common types of filters are called
Butterworth
fairly flat response in the pass-band
a steep attenuation rate,
a non-linear phase response
Chebyshev
steeper rate of attenuation, develop some ripple in the pass band.
The phase response is much more non-linear than the Butterworth.
Bessel
have the best step response and phase linearity. But requires a number of stages or orders

18. BESSEL
Sharper frequency cutoff
Better step response in time domain, less ripple
BUTTERWORTH
CHEBYSHEV
Step response in time domain shows the characteristic of the system for the output to go from 0 – 1 at the point of the cutoff frequency

19. Step response in Time Domain

20. LOW-PASS FILTER A low-pass filter allows for easy passage of low-frequency signals from source to load, and difficult passage of high-frequency signals. The cutoff frequency for a low-pass filter is that frequency at which the output (load) voltage equals 70.7% of the input (source) voltage. Above the cutoff frequency, the output voltage is lower than 70.7% of the input, and vice versa.
𝑋 𝐶 = 1 2𝑓𝐶
First order low pass filter

22. HIGH-PASS FILTER
A high-pass filter allows for easy passage of high-frequency signals from source to load, and difficult passage of low-frequency signals.
The cutoff frequency for a high-pass filter is that frequency at which the output (load) voltage equals 70.7% of the input (source) voltage. Above the cutoff frequency, the output voltage is greater than 70.7% of the input, and vice versa.
First order high pass filter

24. known commonly as a Band Pass Filter
By connecting or “cascading” together a single High Pass Filter circuit with a Low Pass Filter circuit, we can produce another type of passive RC filter that passes a selected range or “band” of frequencies that can be either narrow or wide while attenuating all those outside of this range.
known commonly as a Band Pass Filter
To set the first band pass frequency (HP)
To set the second band pass frequency (LP)

26. EXAMPLE 1
A second-order band pass filter is to be constructed using RC components that will only allow a range of frequencies to pass above 1kHz (1,000Hz) and below 30kHz (30,000Hz). Assuming that both the resistors have values of 10kΩ´s, calculate the values of the two capacitors required
Answers:
C1 = 530 pF
C2 = 15.8 nF

27. BAND-STOP FILTER
combine the low and high pass filter to produce another kind of RC filter network
that can block or at least severely attenuate a band of frequencies within these two cut-off frequency points.

28. Band-pass filters are constructed by combining a low pass filter in series with a high pass filter
Band stop filters are created by combining together the low pass and high pass filter sections in a “parallel” type configuration as shown.
Cutoff frequency is 200Hz
Cutoff frequency is 800Hz

30. Passive filters use R, L and C while Active Filters use R, L and C with combination of an active component such as Op-Amp.
Main disadvantage of passive filters is that the amplitude of the output signal is less than that of the input signal
NON - INVERTING CONFIGURATION
Gain, AV = 𝑅 2 𝑅 1

31. At low frequency (Pass Band)
𝑉 𝑜𝑢𝑡 𝑉 𝑖
=𝐺𝑎𝑖𝑛, 𝐴𝑣
At cutoff frequency
𝑉 𝑜𝑢𝑡 𝑉 𝑖
= 𝐴𝑣

32. Design a non-inverting active low pass filter circuit that has a gain of ten at low frequencies, given that the resistor of the filter is 10 k , the feedback resistor is 9 k and a high frequency cut-off or corner frequency of 159 Hz
10 k
9 k
R1 = 1 k
C1 = 100 nF

33. NON - INVERTING CONFIGURATION
Gain, AV = 𝑅 2 𝑅 1

34. A first order active high pass filter has a pass band gain of two and a cut-off corner frequency of 1 kHz. If the input capacitor has a value of 10 nF, calculate the value of the resistor of the filter and the resistors of the amplifier.
R3 = 16 k
R1 = R2

35. BAND PASS - ACTIVE FILTER
200 
100 nF
10 sin t
16 k
1 k
100 nF
1 k
NON - INVERTING CONFIGURATION

36. Output after the high pass filter
Output after the low pass filter

37. 8 k
100 nF
10 sin t
100 nF
2 k
The band stop filter can be configured using 2 op-amps as non-inverting amplifiers (in this case as voltage followers) and their outputs are then connected using the summing amplifier configuration

38. Output after the high pass filter
Output after the low pass filter
Summing amplifier output