ee 435 homework 6 solutions spring 2021 problem 1
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EE 435 Homework 6 Solutions Spring 2021
Problem 1
Problem 2
Part A By inspection for the first stage:
π΄01 =ππ2/2
π02ππ4ππ4
+ππ6ππ6
ππ8 + π πΆπΆ
By inspection for the second stage:
π΄02 =ππ9
π010ππ9ππ9
+ππ11ππ11
ππ12 + π πΆπΏ
For the overall amplifier:
π΄0 =ππ2ππ9/2
(π02ππ4ππ4
+ππ6ππ6
ππ8 + π πΆπΆ) (π010ππ9ππ9
+ππ11ππ11
ππ12 + π πΆπΏ)
Part B By inspection:
|π1| =π02
ππ4ππ4
+ππ6ππ6
ππ8
πΆπΆ
|π2| =π010
ππ9ππ9
+ππ11ππ11
ππ12
πΆπΏ
Part C To achieve a maximally flat step response with no peaking, we want π = 2π½π΄0. Using the poles calculated in Part B, we can solve for the πΆπΆ needed to obtain this pole spread:
π =π2
π1=
π010ππ9ππ9
+ππ11ππ11
ππ12πΆπΆ
πΆπΏ (π02ππ4ππ4
+ππ6ππ6
ππ8)=
π½ππ2ππ9
(π02ππ4ππ4
+ππ6ππ6
ππ8) (π010ππ9ππ9
+ππ11ππ11
ππ12)
πΆπΆ =π½πΆπΏππ2ππ9
(π010ππ9ππ9
+ππ11ππ11
ππ12)2
Part D To achieve a response with fast rise time and no overshoot, we want π = 4π½π΄0. Using the poles calculated in Part B, we can solve for the πΆπΆ needed to obtain this pole spread:
π =π2
π1=
π010ππ9ππ9
+ππ11ππ11
ππ12πΆπΆ
πΆπΏ (π02ππ4ππ4
+ππ6ππ6
ππ8)=
2π½ππ2ππ9
(π02ππ4ππ4
+ππ6ππ6
ππ8) (π010ππ9ππ9
+ππ11ππ11
ππ12)
πΆπΆ =2π½πΆπΏππ2ππ9
(π010ππ9ππ9
+ππ11ππ11
ππ12)2
Problem 3
Part A Start by finding an expression for the gain of each stage separately:
π΄1(π ) =30ππ΅
(π
10 + 1)=
31.62π
10 + 1
π΄2(π ) =40ππ΅π
30000 + 1=
100π
30000 + 1
When cascading two stages together in the fashion that we are doing so for this problem, we can simply multiple the two stage expressions together to get the overall gain expression:
π΄(π ) =3162
(π
10 + 1) (π
30000 + 1)
Part B
A 2nd-order all-pole low-pass system is maximally flat when π = 1/β2 or when the pole separation, π, is equal to 2π½π΄0. Because we know the pole locations, we can find π:
π =π2
π1=
30000
10= 3000
We also know π΄0, so we can solve for π½: π = 3000 = [2][π½][3162]
π½ =3000
6324β 0.47
The maximum value of π½ which will allow our amplifier to be maximally flat is π½ = 0.47.
Part C Start by finding an expression for the feedback amplifierβs gain:
π΄πΉπ΅ =π΄(π )
1 + π½π΄(π )=
3162
(π
10 + 1) (π
30000 + 1)
1 + π½3162
(π
10 + 1) (π
30000 + 1)
=948600000
π 2 + 30010π + 95160000
Recall that the denominator of the transfer function of a 2nd-order all-pole system can be represented like this:
π·(π ) = π 2 +π0
ππ + π0
2
Where π0 is the first pole location. If we can get our feedback gain expression to have a denominator of this form, we can find π easily. Letβs do that:
β951600000
π= 30010 β π = 0.325
Part D Start by finding an expression for the feedback amplifierβs gain:
π΄πΉπ΅ =π΄(π )
1 + π½π΄(π )=
3162
(π
10 + 1) (π
30000 + 1)
1 + π½3162
(π
10 + 1) (π
30000 + 1)
=948600000
π 2 + 30010π + 474600000
Recall that the denominator of the transfer function of a 2nd-order all-pole system can be represented like this:
π·(π ) = π 2 +π0
ππ + π0
2
Where π0 is the first pole location. If we can get our feedback gain expression to have a denominator of this form, we can find π easily. Letβs do that:
β474600000
π= 30010 β π = 0.725
Problem 4
Part A Begin by finding a transfer function for the open loop amplifier.
π΄(π ) =100
π π1
+ 1
100π
10π + 1=
10000
(π
π1+ 1) (
π 10 Γ 106 + 1)
If the open loop amplifierβs gain is infinitely large, then the gain of the closed-loop amplifier will be approximately equal to 1/π½. The gain of the amplifier that weβve given is not infinite, but it is large, so we expect this will still be true. Letβs see:
π΄πΉπ΅(π = 0) =10000
1 + 10000π½= 5 β π½ = 0.1999 β 0.2
Now, for the amplifier to be as fast as possible, the pole π needs to be equal to 1/2. Find an expression for the closed-loop amplifier gain.
π΄πΉπ΅ =1 Γ 1012 β π1
π 2 + π (1 Γ 107 + π1) + 2.001 Γ 1010 β π1
Find the π1 needed to obtain the desired π:
β2.001 Γ 1010 β π1
1/2= 1 Γ 107 + π1
π1 = 1249.69 or π1 = 8.002 Γ 1010 If the first pole location is to be smaller than the second pole, then we want π1 =1249.69 πππ/π ππ.
Part B To find the πΊπ΅π, first find the DC gain and then multiple it by the first pole:
π΄πΉπ΅(π = 0) =10000
(0π1
+ 1) (0
10 Γ 106 + 1)= 10000
πΊπ΅π = π΄πΉπ΅(π = 0)π1 = 10000 β 1249.69 = 12.49π πππ/π ππ
Problem 5
Part A Recall from lecture that:
πΆπΆ =πΆπΏπ½
π2
ππππππ
(πππ β π½πππ)2
Where πππ = ππ1 and πππ = ππ5. Calculate each small-signal parameter.
ππ1 =2πΌπ1
ππΈπ΅1=
2 β1ππ
5β
12 β
12
0.1= 0.001
ππ5 =2πΌπ5
ππΈπ΅5=
2 β1ππ
5β
12
0.1= 0.002
Therefore:
πΆπΆ =10ππΉ β 1
22
[0.002][0.001]
(0.002 β 0.001)2= 5ππΉ
Part B By inspect, the amplifierβs DC gain is given by:
π΄0 =ππ1
ππ2 + ππ4
ππ5
ππ5 + ππ6
From Part A, we know ππ1 and ππ5. Letβs calculate ππ5 and ππ2. Assume ππ = ππ = 0.01πβ1.
ππ5 = ππ6 = ππΌπ·π = [0.01][πΌπ5] = [0.01] [1ππ
5β
1
2] = 1ππ
ππ2 = ππ4 = ππΌπ·π = [0.01][πΌπ2] = [0.01] [1ππ
5β
1
2β
1
2] = 500ππ
So:
π΄0 =0.001
1π
0.002
2π= 100000010 = 120ππ΅
Part C To find the gain-bandwidth, recall:
πΊπ΅ βπππ
πΆπΆ=
0.001
5ππΉ= 200π πππ/π ππ
Problem 6
Part A Recall from lecture that:
πΆπΆ =πΆπΏπ½
π2
ππππππ
(πππ β π½πππ)2
Where πππ = ππ1 and πππ = ππ5. Calculate each small-signal parameter.
ππ1 =2πΌπ1
ππΈπ΅1=
2 β1ππ
5β
12 β 0.1
0.1= 0.0002
ππ5 =2πΌπ5
ππΈπ΅5=
2 β1ππ
5β 0.9
0.1= 0.0036
Therefore:
πΆπΆ =10ππΉ β 1
22
[0.0036][0.0002]
(0.0036 β 0.0002)2= 155.7ππΉ
To find the gain-bandwidth:
πΊπ΅ βπππ
πΆπΆ=
0.0002
155.7ππΉ= 1.28πΊ πππ/π ππ
Part B Recall from lecture that:
πΆπΆ =πΆπΏπ½
π2
ππππππ
(πππ β π½πππ)2
Where πππ = ππ1 and πππ = ππ5. Calculate each small-signal parameter.
ππ1 =2πΌπ1
ππΈπ΅1=
2 β1ππ
5β
12 β 0.9
0.1= 0.0018
ππ5 =2πΌπ5
ππΈπ΅5=
2 β1ππ
5β 0.1
0.1= 0.0004
Therefore:
πΆπΆ =10ππΉ β 1
22
[0.0004][0.0018]
(0.0004 β 0.0018)2= 918.3ππΉ
To find the gain-bandwidth:
πΊπ΅ βπππ
πΆπΆ=
0.0018
918.3ππΉ= 1.96πΊ πππ/π ππ
Part C When the power is split evenly, the GBW is dramatically lower than when most of the power is split into either the first or second branch. The compensation capacitor size is also dramatically decreased.
Problem 7
Circuit A The pole approximation method referenced in this question is discussed in Lecture 13, Slide 31. For the first circuit, let the set of small capacitors (high frequency poles) be {100ππΉ} and the set of large capacitors (low frequency poles) be {5ππΉ, 1ππΉ}. Now, start by finding the low frequency poles. Replace all the voltage sources with short circuits and the current sources with open circuits. Replace the small capacitors with open circuits. Now, look at each of the large capacitors independently. For the 1ππΉ capacitor, if all other large capacitors are short circuits, the capacitor sees a load resistance of 1πΞ©. So this capacitor is responsible for a pole located at
β1
[1πΞ©][1ππΉ]= β1000 πππ/π ππ. For the 5ππΉ capacitor, if all other large capacitors are short
circuits, the capacitor sees a load resistance of 500Ξ©. It is responsible for a pole located at β400 πππ/π ππ. Now find the high frequency poles. Replace the voltage sources with short circuits, the current sources with open circuits, and the low-frequency capacitors with short circuits. The 100ππΉ capacitor now sees a load of 50πΞ©. This creates a high frequency pole located at β200π πππ/π ππ.
We claim that there are three poles located at β400 πππ/π ππ, β1000 πππ/π ππ, and β200π πππ/π ππ. We can show this mathematically by performing a hand analysis or computer simulation. Using MATLAB, create a new Simulink model and build the circuit in it.
Then, right click the line from the step block and select Linear Analysis Point β Input Perturbation. Right click on the line coming from the PS-Simulink Converter and select Linear Analysis Point β Output Measurement. Then, in the top menu, select Analysis, Control Design, Linear Analysis. Run a Step analysis.
Drag linsys1 into the MATLAB Workspace to add it as a variable. In MATLAB, type sys=tf(linsys1) to view the transfer function of the linearized system.
Type pole(sys) in MATLAB to receive a listing of the poles. For this circuit, three poles will be returned: β200020 πππ/π ππ, β1238.42 πππ/π ππ, and β161.48 πππ/π ππ. These poles match closely with the poles that were approximated by hand. Some error exists, but it is small enough to allow this approximation to be useful for first-pass designs, gaining intuition, etc.
Circuit B Repeat the same process as used in the previous part. From this, we approximate the pole locations to be at β400 πππ/π ππ, β1000 πππ/π ππ, and β200π πππ/π ππ. A computer analysis yields poles at β200020 πππ/π ππ, β1238.42 πππ/π ππ, and β161.48 πππ/π ππ, which is similar.
Problem 8
Part A Recall that for a 2nd-order all-pole low-pass filter, we can find phase margin using π as follows:
ππ = cosβ1 (β1 +1
4π2β
1
2π2 )
So, we only need to find the feedback amplifierβs π to find the phase margin. Start by finding the transfer function for the feedback amplifier.
π΄πΉπ΅ =π΄(π )
1 + π½π΄(π )=
108
π 2 + 5002π + 2.001 Γ 107
To find the π:
β2.001 Γ 107
π= 5002
π = 0.894 Using the equation:
ππ = 58.63Β°
Part B
To achieve a feedback amplifier with a magnitude response without peaking, we want π =1
β2.
We can do so by finding the feedback transfer function assuming π1 is unknown. Then, we can look at the denominator of the transfer function.
π·(π ) = π 2 + π (5000 + π1) + 10005000π1
To find the π1 needed to obtain a π of 1/β2:
β10005000π1
1/β2= 5000 + π1
β2β10005000π1 = 5000 + π1
π1 = 1.25 πππ/π ππ or π1 = 20 Γ 106 πππ/π ππ For π1 to be the dominant pole, it must be 1.25 πππ/π ππ and not 20 Γ 106 πππ/π ππ.
π1 = 1.25 πππ/π ππ
Part C Because we know the systemβs π, we can use the expression from Part A to find the phase margin.
ππ = 35.84Β°
Part D The π was found in Part B.
Problem 9
Part A Begin by finding an expression for the gain of the feedback amplifier formed using π΄(π ):
π΄πΉπ΅ =π΄(π )
1 + π½π΄(π )
=10π + 10000
(π
π1+ 1) (
π 5000
+ 1) (π
10000 + 1) (π΅(10π + 10000)
(π
π1+ 1) (
π 5000
+ 1) (π
10000 + 1)+ 1)
=(1 Γ 109)(π + 1000)
π 3 + 15002π 2 + 250030000π + 200100 Γ 106
To find phase margin, first recall that phase margin can be thought of as how far away the phase is from 180Β° when the gain of the transfer function is 1. Now that the transfer function has been obtained, find its magnitude and solve for the frequency which results in a gain of 1:
|π΄πΉπ΅| = |(1 Γ 109)(π + 1000)
π 3 + 15002π 2 + 250030000π + 200100 Γ 106|
=β(1 Γ 1012)2 + (π Γ 109)2
β(200100000000 β 15002π2)2 + (βπ3 + 250030000π)2= 1
π = Β±33450.8 Now, find the transfer function phase at π. Using MATLAB, it is found that the phase is β152Β°. Phase margin is the βdistanceβ away from β180Β° that the phase is at the unity-gain frequency, so subtract to find phase margin:
ππ = 180Β° β 152Β° β 28Β°
Part B
To achieve a feedback amplifier with a magnitude response without peaking, we want π =1
β2.
From Part A, we know the feedback amplifierβs transfer function assuming π½ and π1 are unknown. Substitute in π½:
π΄πΉπ΅ =5π Γ 108(π + 1000)
π 3 + π 2(π + 15000) + π (π1.00015 Γ 108 + 50 Γ 106) + (π1.0005 Γ 1011)
To find the system poles, one could solve for the denominatorβs roots. Doing this by hand would not be practical, so MATLAB or some other form of computer anlaysis would be necessary. This would yield three poles in terms of π1 (not shown here due to length). One of the poles is a real pole and the other two form a complex conjugate pair. The pair of poles forming a complex conjugate will determine our π Specifically, by definition:
π =ππ
2ππ
MATLAB is unable to find an explicit solution to this equation, so an approximation can be used.
It is found that a pole location of 0.6 πππ/π ππ yields a π of approximately 1/β2.
Part C Repeating the process performed in Part A will yield a phase margin of ππ = 77.56Β°.
Part D
By design, the pole π is 1/β2.
Problem 10
Part A
Amplifier 1
Amplifier 2
Amplifier 3
Part B
Amplifier 1 The open-loop transfer function of amplifier 1 is:
π΄(π ) =104
(π
5000+ 1) (
π 10 + 1)
If π = 0: π΄(0) = 104
Noninverting.
Amplifier 2 The open-loop transfer function of amplifier 1 is:
π΄π΄(π ) =104
(π
5000+ 1) (
π 10
β 1)
If π = 0:
π΄(0) = β104 Inverting.
Amplifier 3 The open-loop transfer function of amplifier 1 is:
π΄(π ) =104
(π
5000+ 1) (
π 40 β 1)
If π = 0: π΄(0) = β104
Inverting.
Part C
Amplifier 1 The open-loop transfer function of amplifier 1 is:
π΄(π ) =500 Γ 106
π 2 + 5010π + 1 Γ 108
If π = 0: π΄(0) = 5
Noninverting.
Amplifier 2 The open-loop transfer function of amplifier 1 is:
π΄π΄(π ) =50 Γ 106
π 2 + 4990π + 9.95 Γ 106
If π = 0:
π΄(0) = 5.025
Noninverting.
Amplifier 3 The open-loop transfer function of amplifier 1 is:
π΄(π ) =50 Γ 106
π 2 + 4960π + 9.8 Γ 106
If π = 0: π΄(0) = 5.1
Noninverting.
Part D When open-loop, amplifier one has two LHP poles, indicating it is stable across all frequencies. Amplifiers two and three, in comparison, each have a RHP pole. This indicates a positive real frequency which will make each amplifier unstable. When closed-loop, all three amplifiers have two LHP poles, indicating they are stable across all frequencies.
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