ENGG 1015 Tutorial
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Op Amps and Signals
19 Nov
Learning Objectives
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News
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Analyze circuits with operational amplifiers
Convert between representations of systems
HW2 deadline (19 Nov 23:55)
Revision tutorial (TBD)
Ack.: MIT OCW 6.01, 6.003
1
Simple Op Amps (1)
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Assume the op-amps are “ideal”
Determine the current Ix
when V1 =1V and V2 =2V.
Determine the voltage VA
when V1 =1V and 2 =2V.
Determine a general expression
for VA in terms of V1 and V2.
2
Simple Op Amps (2)
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When V1 =1V and V2 =2V, Ix =1A
When V1 =1V and V2 =2V, VA = 4V
2
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A general expression for VA:
𝑉𝐴 = 𝑉2 + 𝐼𝑥 × 2
𝑉2 − 𝑉1
= 𝑉2 +
×2
1
= 𝑉2 + 2 𝑉2 − 𝑉1
= −2𝑉1 + 3𝑉2
4
2
1
-1
1
3
Non-inverting Amplifier
Vn
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Use a single op-amp and
resistors to make a circuit
that is equivalent to the
following circuit.
𝑉𝑛
𝑅2
= 1+
𝑉𝑖
𝑅1
𝑉𝑜 𝑉𝑜 𝑉𝑛
𝑅3
=
= 1+
𝑉𝑖 𝑉𝑛 𝑉𝑖
𝑅4
=1
𝑅2
1+
𝑅1
𝑅1 𝑅4 +𝑅2 𝑅3 +𝑅2 𝑅4
+
𝑅1 𝑅3
4
Voltage-controlled Current Source
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Use the ideal op-amp model (V+ = V-) to determine an
expression for the output current Io in terms of the input
voltage Vi and resistors R1 and R2.
𝑣𝑥 = 𝑣𝑖 + 𝑣𝑥
𝑅2
⇒ 𝑣𝑥 = 𝑣𝑖
𝑅1
𝑅2
𝑅1 + 𝑅2
vi +vx
vi +vx
vx
𝑣𝑥
1 𝑅2 𝑣𝑖
𝐼𝑜 =
=
𝑣𝑖
=
𝑅2 𝑅2 𝑅1 𝑅1
5
Op Amp Configurations (1)
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Fill in the values of R1 and R2 required to satisfy the
equations in the left column of the following table. The
values must be non-negative (i.e., in the range [0,∞])
R1
R2
Vo = 2V2 - 2V1
Vo = V2-V1
Vo = 4V2 - 2V1
6
Op Amp Configurations (2)
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𝑉+ =
𝑅2
𝑉2
10𝑘+𝑅2
𝑉𝑜 =
10𝑘+𝑅1
10𝑘+𝑅2
×
= 𝑉− =
𝑅2
10𝑘
× 𝑉2 −
Negative R 
i.e. Impossible
 3rd:
𝑅1
𝑉1
10𝑘+𝑅1
𝑅1
10𝑘
+
10𝑘
𝑉𝑜
10𝑘+𝑅1
× 𝑉1
R1
R2
Vo=2V2-2V1
20kΩ
20kΩ
Vo=V2-V1
20kΩ
20kΩ
Vo=4V2-2V1
Impossible
Impossible
7
Unusual Op Amp Configurations
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What is Vo?
Vo = 0
Vo = V1 – V2
8
Motor Control (1)
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The following circuit is a proportional controller that
regulates the current through a motor by setting the
motor voltage VC to VC = K(Id − Io) where K is the gain
(notice that its dimensions
are ohms), Id is the desired
motor current, and Io is the
actual current through the
motor.
9
Motor Control (2)
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Consider the circuit inside the dotted rectangle.
Determine V1 as a function of Io.
V+ = 1/2 x Io = VV- = 100/(100+9900) x V1
V1 = 1/2 x Io x 100
Determine the gain K and
desired motor current Id.
KCL at -ve input to right op-amp:
𝑉𝑐 − 2.5 2.5 − 0.5 × 𝐼𝑜 × 100
=
⇒ 𝑉𝑐 = 50 0.1 − 𝐼𝑜
10000
10000
10
Multiple Representations of Systems
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State machines:
computationally efficient.
Difference equations:
mathematically compact.
Signal flow graphs:
illustrate signal flow paths.
Operator representations /
z-transforms: analyze
systems as polynomials.
11
From Blocks to Equations (1)
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Determine the difference equation that
relates x[·] and y[·].
12
From Blocks to Equations (1)
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Assign names to all signals. Replace Delay with R.
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Express relations among signals algebraically.
E = X +W ; Y = RE ; W = RY
Solve: Y = RE = R(X +W) = R(X + RY )
 RX = Y − R2Y
Difference equation: y[n] = x[n − 1] + y[n − 2]
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13
Operator Algebra (1)
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Start with X, subtract 2 times a right-shifted version of X,
and add a double-right-shifted version of X
14
Operator Algebra (2)
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15
Quick Checking
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How many of the following systems are equivalent?
3
16
Geometric Growth
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The value of p0 determines the rate of growth.
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p0 > 1: magnitude diverges monotonically
0 < p0 < 1: magnitude converges monotonically
−1 < p0 < 0: magnitude converges, alternating sign
p0 < −1: magnitude diverges, alternating sign
17
Computing Responses (1)
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A given linear, time-invariant system turns the
unit pulse into the triangle:
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The system is given the following input signal:
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Sketch the output signal.
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Computing Responses (2)
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Consider the system described by
y[n] = x[n] + y[n − 1] + 2y[n − 2] .
Assume that the system starts at
rest and that the input x[n] is the
unit-step signal u[n].
We can compute the
response through the
iterating the difference
equation.
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Analysis of a Simple System (1)
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Consider a system H which transforms
an input signal X into an output signal Y
Suppose the input signal is a unit sample:
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and for that input, the samples of
the output response signal are:
Block diagram:
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20
Analysis of a Simple System (2)
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Consider the input
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Let y[n] be the output response to x[n], as given in part 1,
above, and let y′[n] be the output response to input x′[n].
y′[n] = y[n] − 16y[n − 4]
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y′[0] = 1
y′[1] = -2
y′[2] = 4
y′[3] = −8
y′[4] = 16 − 16 = 0
21
Difference Equation Construction (1)
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Newton’s law of cooling states that:
The change in an object’s temperature from one time
step to the next is proportional to the difference (on the
earlier step) between the temperature of the object and
the temperature of the environment, as well as to the
length of the time step.
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Let o[n] be temperature of object, s[n] be temperature of
environment, T be the duration of a time step, K be the
constant of proportionality
22
Difference Equation Construction (2)
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The difference equation for Newton’s law of cooling
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i) the difference (on the earlier step) between the temperature of
the object and the temperature of the environment
ii) as well as to the length of the time step
Be sure the signs are such that the temperature of the object will
eventually equilibrate with that of the environment.
The system function corresponding to this equation
23
Signals and Feedback Control (1)
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X : Desired temperature Tset
Y : Actual hot water temperature THW
The sensor reads THW with gain ks, but has a unit delay.
The valve also takes a unit delay to respond, and has
some persistence about its motion, characterized by kv.
The proportional controller,
with setting k, converts the
temperature difference
between Tset and the sensed
temperature to a signal input
to the valve.
24
Signals and Feedback Control (2)
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The difference equation:
y[n] = kax[n − 1] − kvy[n − 1] − kskay[n − 2]
Assume the system starts at rest, and the input is a unit
sample signal x[n] (i.e. x[n] = [1,0,0,0,…]), suppose kv =
0.9 and ka = 0.5
When ks = 0
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y[0] = 0
y[1] = 0.5 (1) – 0.9 (0) – 0 = 0.5
y[2] = 0.5 (0) – 0.9 (0.5) – 0 = – 0.45
y[3] = 0.5 (0) – 0.9 (– 0.45) – 0 = 0.405
y[4] = 0.5 (0) – 0.9 (0.405) – 0 = – 0.3645
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Signal and Feedback Control (3)
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When ks = – 0.35
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y[n] = 0.5 x[n − 1] − 0.9 y[n − 1]
− 0.5 ks y[n − 2]
y[0] = 0
y[1] = 0.5 (1) – 0.9 (0) – (– 0.35)(0.5) (0) = 0.5
y[2] = 0.5 (0) – 0.9 (0.5) – (– 0.35)(0.5) (0) = – 0.45
y[3] = 0.5 (0) – 0.9 (– 0.45) – (– 0.35)(0.5) (0.5) = 0.4925
y[4] = 0.5 (0) – 0.9 (0.4925) – (– 0.35)(0.5) (– 0.45) = – 0.522
When ks = 1.5
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y[0] = 0
y[1] = 0.5 (1) – 0.9 (0) – (1.5)(0.5) (0) = 0.5
y[2] = 0.5 (0) – 0.9 (0.5) – (1.5)(0.5) (0) = – 0.45
y[3] = 0.5 (0) – 0.9 (– 0.45) – (1.5)(0.5) (0.5) = 0.03
y[4] = 0.5 (0) – 0.9 (0.03) – (1.5)(0.5) (– 0.45) = 0.3105
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Grow, baby, grow (1)
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In each time period, every cell in the bioreactor divides to
yield itself and one new daughter cell. However, due to
aging, half of the cells die after reproducing.
Po : The number of cells at each time step.
 Po[n] = 2Po[n−1] − 0.5Po[n−2]
Suppose that Po[0]= 10 and Po[n]= 0 if n<0.
Po[0] = 10
Po[1] = 2Po[0] − 0.5Po[-1] = 20 + 0 = 20
Po[2] = 2Po[1] − 0.5Po[0] = 40 − 5 = 35
Po[3] = 2Po[2] − 0.5Po[1] = 70 − 10 = 60
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Grow, baby, grow (2)
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Your goal is to create a constant population of cells, that
is, to keep Po constant at some desired level Pd. You are
to design a proportional controller that can add or
remove cells as a function of the difference between the
actual and desired number of cells.
Assume that any additions/deletions at time n are based
on the measured number of cells at time n−1. Denote the
number of cells added or removed at each step Pinp.
The difference equations
Po[n] = 2Po[n − 1] − 0.5Po[n − 2] + Pinp[n]
Pinp[n] = k(Pd[n] − Po[n − 1])
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Grow, baby, grow (3)
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Draw a block diagram that represents the system
Po[n] = 2Po[n − 1] − 0.5Po[n − 2] + Pinp[n]
Pinp[n] = k(Pd[n] − Po[n − 1])
Delays +
Adders +
Gains
29
Stepping Up and Down (1)
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Use a small number of delays, gains, and 2-input adders
(and no other types of elements) to implement a system
whose response (starting at rest) to a unit-step signal
[1,1,1,…]
is
[1,2,3,1,2,3,1,…]
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Draw a block diagram of your system.
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Stepping Up and Down (3)
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First find a system whose unit-sample response is the
desired sequence. The periodicity of 3 suggests that y[n]
depends on y[n − 3].
The resulting difference equation is
y[n] = y[n − 3] + w[n] + 2w[n − 1] + 3w[n − 2].
[1,0,0,…]
[1,2,3,1,2,3,1,…]
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Stepping Up and Down (3)
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Next compute w[n] which is the first difference of x[n]:
w[n] = x[n] − x[n − 1].
The result is the cascade of the first difference and the
previous result.
[1,0,0,…]
[1,2,3,1,2,3,1,…]
[1,1,1,…]
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Summary
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Response
Difference equations
Signal flow graphs
Operator representations
Signal flow graphs  Difference equations
Operator representations  Signal flow
graphs
33
Appendix: Partial Fractions
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1
𝑓 𝑥 = 2
𝑥 + 2𝑥 − 3
The denominator splits into two distinct linear
factors: 𝑞 𝑥 = 𝑥 2 + 2𝑥 − 3 = 𝑥 + 3 𝑥 − 1
1
𝑥 2 +2𝑥−3
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⇒𝑓 𝑥 =
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Multiplying through by x2 + 2x - 3, we have the
polynomial identity 1 = 𝐴 𝑥 − 1 + 𝐵 𝑥 + 3
Substituting x = -3 into this equation gives A = -1/4, and
substituting x = 1 gives B = 1/4, so that
1
−1/4
1/4
𝑓 𝑥 = 2
=
+
𝑥 + 2𝑥 − 3 𝑥 + 3 𝑥 − 1
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=
𝐴
𝐵
+
𝑥+3
𝑥−1
34
Appendix: Complex Numbers (1)
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Complex number z = x (Real) + yi (Imaginary)
Complex conjugate of z = x + yi is defined to be x − yi
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Appendix: Complex Numbers (2)
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Polar expressions
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Argument φ and Modulus/magnitude/absolute value r
The argument or phase of z is the angle of the radius OP
with the positive real axis
36