1 fundamentaltheoremofsolutionstodiﬀerential equationsee16b/sp20/homework/sol2.pdf ·...

EECS 16B Spring 2020 Homework 2

1 Fundamental Theorem of Solutions to DifferentialEquations

In this question, you will discover the power of the fundamental theorem ofsolutions to differential equations. For convenience, we shall restate the theoremhere.

Theorem. Consider a differential equation of the form,

d=H

dC=(C) + =−1

d=−1H

dC=−1(C) + . . . + 1

dH

dC(C) + 0H(C) = 0

Given = initial conditions of the form,

H(C0) = 00 ,dH

dC(C0) = 01 , . . . ,

d=−1H

dC=−1(C0) = 0=−1 ,

there exists a unique solution (say, 5 ).

a) Consider the following 2 functions.

)1(G) = 4G , )2(G) =∞∑==0

G=

=!

Prove that )1(G) = )2(G) by showing that both functions satisfy thefollowing differential equation:

d 5

dG(G) = 5 (G)with 5 (0) = 1

Side note: Assume 00 = 1.

SolutionLet’s first look at )1(G).

)1(G = 0) = 40 = 1

3)1

3G(G) = 4G

So )1 satisfies the conditions. Now let’s look at )2(G)

)2(G = 0) =∞∑==0

0=

=!

Removing the first term from the series, we get

)2(G = 0) = 00

1+∞∑==1

0=

=!

1

Homework 2 @ 2020-02-13 20:29:31-08:00

Zero to any positive finite power is zero, so we can say

∞∑==1

0=

=!= 0

)2(G = 0) = 00 = 1

To find the derivative of )2(G), we can say

3

3G

∞∑==0

5 (G) =∞∑==0

35

3G(G)

35

3G(G) = 3

3G(G

=

=!) = =G=−1

=!

3)2

3G(G) =

∞∑==0

=G=−1

=!

If we remove the first term of the series, we get

3)2

3G(G) = 0G−1

1+∞∑==1

=G=−1

=!

3)2

3G(G) =

∞∑==1

=G=−1

=!

Using the fact that

=

=!=

1

(= − 1)!

We can say

3)2

3G(G) =

∞∑==1

G=−1

(= − 1)!

If we use substitution and let : = (= − 1), we get

3)2

3G(G) =

∞∑:=0

G:

:!= )2(G)

Both )1 and )2 meet the conditions, therefore they must be the samefunction.

2


b) Consider the following 2 functions.

)1(G) = cos(G), )2(G) = cos(−G)

Prove that )1(G) = )2(G) by showing that both functions satisfy thefollowing differential equation:

d2 5

dG2(G) = − 5 (G)with 5 (0) = 1,

d 5

dG(0) = 0

SolutionInitial conditions:

cos(0) = cos(−0) = 1

So both )1 and )2 satisfy 5 (0) = 1. Taking the first derivative of each:

3)1

3G= − sin(G)

3)2

3G= sin(−G)

Plugging in 0 for x:

− sin(0) = 0

sin(−0) = 0

Both functions satisfy 35

3G (0) = 0. Taking the second derivative of each:

32)1

3G2(G) = − cos(G) = −)1(G)

32)2

3G2(G) = − cos(−G) = −)2(G)

Both functions satisfy all three conditions, so )1(G) = )2(G).

2 IC Power Supply

Digital integrated circuits (ICs) often have very non-uniform current require-ments which can cause voltage noise on the supply lines. If one IC is adding alot of noise to the supply line, it can affect the performance of other ICs that use

3

Homework 2 @ 2020-02-13 20:29:31-08:00

the same power supply, which can hinder performance of the entire device. Forthis reason, it is important to take measures to mitigate, or “smooth out”, thepower supply noise that each IC creates. A common way of doing this is to adda “supply capacitor” between each IC and the power supply. (If you look at acircuit board, and the supply capacitor is the small capacitor next to each IC.)

Here’s a simple model for a power supply and digital circuit:

The current source is modeling the “spiky,” non-uniform nature of digitalcircuit current consumption. The resistor represents the sum of the sourceresistance of the supply and any wiring resistance between the supply and theload.

The capacitor is added to try to minimize the noise on +�� . Assuming that+B = 3V, ' = 1Ω, 80 = 1A, ) = 10ns, and C? = 1ns,

a) Sketch the voltage +�� vs. time for one or two periods T assuming that� = 0.

Solution

If � = 0, then this circuit will respond instantaneously to changes in thecurrent; thus we may break this down into two segments, wherein thecurrent source ��(C) equals 0 (and thus +33 = +B), and where the currentsource ��(C) equals 80 (and thus +33 = +B − 80' = 2V). These will followthe current source’s flips precisely. With that in mind, your sketch shouldlook something like this:

4


0 5 10 15 20 25 30−0.5

00.51

1.5

C (ns)

� ��(C)(A)

Sketch of transient current drawn by IC

0 5 10 15 20 25 301.52

2.53

3.5

C (ns)

+��(C)(V)

Sketch of transient IC supply voltage

In the above sketch, we have the first current spike at C = 10 ns. Yoursdoesn’t have to align with that: for example, if you had the first currentspike at C = 0, that’s okay. However, what doesmatter is that you have thetiming between the current spikes drawn correctly.

b) Give expressions for and sketch the voltage +�� vs. time for one or twoperiods T for each of three different capacitor values for �: 1pF , 1nF, 1�F.(1pF = 10−12F, 1nF = 10−9F, 1�F = 10−6F)

Solution

Since the current through the source is a series of pulses, it will be easiestif we solve for +33(C) assuming a piecewise constant �� . Starting withKVL:

+( = +' ++33 (1)

+( = (�� + �3

3C+33)' ++33 (2)

3

3C+33 =

1

'�(+( − '�� −+33) (3)

Where �� is the piecewise constant value of the current flowing throughthe digital circuit. At this point we can use substitution with +̃ =

(+( − '�� −+33).

5

Homework 2 @ 2020-02-13 20:29:31-08:00

3

3C+̃ = − 3

3C+33 (4)

3

3C+33 =

+̃

'�(5)

3

3C+̃ = − +̃

'�(6)

+̃(C) = �4− C'� (7)

Substituting back to solve for +33(C) we get the following general expres-sion for the voltage during any one piecewise constant time slice:

+33(C) = +( − '�� − �4−C'� (8)

From here, based on the value of the piecewise current �� and the initialconditions imposed by previous time segments we can simplify the +33expression and solve for �. At the start, we will assume a convenientinitial condition which corresponds with the behavior of +33 if �� = 0 fora long time before C = 0. In this case, +33 = +( and there is zero currentflowing through the circuit. Things get exciting once the first currentpulse starts such that �� = 80. The initial condition for this piecewisesection is the final voltage +33 from the previous section, +(.

+( = +( − '80 − �40 (9)� = −'80 (10)

+33(C) = +( − '80(1 − 4−C'� ) (11)

You can compute the voltage at the end of the pulse by plugging in C = C?and the ' and � values for your scenario. This voltage will serve as theinitial condition for the next piecewise constant section. The process ofsimplifying the general piecewise differential equation and solving for� can be performed repeatedly to determine the shape of the plot forfurther pulses.In general: eachof the three curveswill tend towards a final value+33 = +B ,growing exponentially slower towards this goal as time progresses. How-ever, on each time interval C? , the current source will start drawing chargefrom both +B–whose current decreases as time proceeds–and �–whosecharge, and therefore whose potential to contribute voltage, tends toincrease with time. With each C? , +33 decreases nonlinearly, as there areboth exponential and linear factors contributing to the rise and fall of thevoltage.

6


With a lower capacitance, we will see the capacitor charge and dischargefaster with time–this means that +33 will fluctuate more/change moredrastically on C? ; as you increase capacitance, this fluctuation is lessevident, as � has more charge to pull from, and will thus be less affectedby the change of charge incurred by the current source.

The final sketches should look something like this:

0 5 10 15 20 25 30−0.5

00.51

1.5

C (ns)

� ��(C)(A)

Sketch of transient current drawn by IC

0 5 10 15 20 25 301.52

2.53

3.5

C (ns)

+��(C)(V)

+��(C)with 1 pF capacitor

0 5 10 15 20 25 301.52

2.53

3.5

C (ns)

+��(C)(V)

+��(C)with 1 nF capacitor

0 5 10 15 20 25 301.52

2.53

3.5

C (ns)

+��(C)(V)

+��(C)with 1 �F capacitor

Note that if your solutions contain just the correct plots without muchverbose, textual explanation of the plots, then you still deserve full credit.

The idea here is to see the effect that the capacitors have on +��(C) whenviewed at the time scale of the current spikes.

7

Homework 2 @ 2020-02-13 20:29:31-08:00

• the 1 pF capacitor causes the RC circuit to have a time constant of� = '� = 1ps, that is 1picosecond = 10−12 seconds, and the effectthat this has on +��(C) is invisible at the nanosecond time scale. Forthis reason, we can conclude that the 1 pF capacitor would not beadequate to mitigate the noise that the IC will put on the powersupply.

• the 1 nF capacitor causes the RC circuit to have a time constant of� = '� = 1ns. This is a long enough time scale that the effect on+��(C)will be visible. At the end of the 1 ns current spike, +��(C)will have dropped from 3 V to 2+ exp(−1) ≈ 2.37 V. This means thatthe 1 nF capacitor is actually reducing the power supply noise a littlebit, but not much.

• the 1 �F capacitor causes the RC circuit to have a time constant of� = '� = 1�s. This time constant is 1000 times longer than theduration of the current spike. At the end of the current spike,+��(C)will have dropped by only onemillivolt, so at the scale at which thesesketches are drawn, there is no visible change. The 1 �F capacitorhas almost totally removed the power supply noise.

c) Launch the attached Jupyter notebook to interact with a simulated versionof this IC power supply. Try to simulate the scenarios outlined in theprevious parts. For one of these scenarios, keep the RC time constant fixed,but vary the relative value of R vs. C (e.g. compare ' = 1, � = 24 − 9 tothe case where ' = 2, � = 14 − 9). Is it better to have a lower R or lowerC value for a fixed RC time constant when attempting to minimizesupply noise? Give an intuitive explanation for why this might be thecase.

Solution

A lower resistance and higher capacitance leads to smaller variation in thesupply voltage with each current spike. One intuitive way to see this is tothink about where the charge comes from whenenver the current sourceturns on. The charge comes from the capacitor and from the voltagesource through the resistor. By & = �+ for a constant amount of chargedrawn, a larger capacitor results in lower voltage change. By + = �' for aconstant amount of current drawn through the resistor, a larger resistorleads to a larger voltage drop.

In the case where ' = 1 and � = 24 − 9, we get the following plot:

8


In the case where ' = 2 and � = 14 − 9, we get the following plot:

Notice that the shape of the +33 curves is the same becuase the '�constant is the same. However they drop to different voltages by the end

9

Homework 2 @ 2020-02-13 20:29:31-08:00

of each pulse.

3 Simple scalar differential equations driven by an input

In class, you learned that the solution for C ≥ 0 to the simple scalar first-orderdifferential equation

3

3CG(C) = �G(C) (12)

with initial conditionG(C = 0) = G0 (13)

is given for C ≥ 0 byG(C) = G04�C . (14)

In an earlier homework, you proved that these solutions are unique– that is, thatG(C) of the form in (14) are the only possible solutions to the equation (12) withthe specified initial condition (13).

In this question, we will extend our understanding to differential equationswith inputs.

In particular, the scalar differential equation

3

3CG(C) = �G(C) + D(C) (15)

where D(C) is a known function of time from C = 0 onwards.

a) Suppose that you are given an G6(C) that satisfies both (13) and (15) forC ≥ 0.

Show that if H(C) also satisfies (13) and (15) for C ≥ 0, then it must bethat H(C) = G6(C) for all C ≥ 0.

(HINT: You already used ratios in an earlier HW to prove that two things werenecessarily equal. This time, you might want to use differences. Be sure toleverage what you already proved earlier instead of having to redo all that work.)

Solution

We had already used ratios in the earlier homework, and so the naturalthing to look at is the difference. Consider I(C) = H(C) − G6(C). We knowthat

I(0) = H(0) − G6(0) (16)= G0 − G0 (17)= 0. (18)

10


Furthermore,

3

3CI(C) = 3

3CH(C) − 3

3CG6(C) (19)

= �H(C) + D(C) − (�G6(C) + D(C)) (20)= �(H(C) − G6(C)) + (D(C) − D(C)) (21)= �I(C) (22)

By (22) and (18), we know that I(C) satisfies the exact conditions for whichwe proved uniqueness in an earlier homework. (This is why we caredso much about getting the zero initial condition case correct withouthandwaving.) So we know that I(C) = 04�C = 0 for all C ≥ 0.This means that H(C) − G6(C) = 0 and hence H(C) = G6(C). This successfullyproves the uniqueness of solutions to simple scalar differential equationswith inputs.

b) Suppose that the given D(C) starts at C = 0 (it is zero before that) and isa nicely integrable function (feel free to assume bounded and continu-ously differentiably with bounded derivative — whatever conditions youassumed in your calculus course when considering integration and thefundamental theorem of calculus). Let

G2(C) = G04�C +∫ C

0

D(�)4�(C−�)3� (23)

for C ≥ 0.Show that the G2(C) defined in (23) indeed satisfies (15) and (13).Note: the � here in (23) is just a dummy variable of integration. We couldhave used any letter for that local variable. We just used � because itvisually reminds us of C while also looking different. If you think theylook too similar in your handwriting, feel free to change the dummyvariable of integration to another symbol of your choice.(HINT: Remember the fundamental theorem of calculus that you proved in yourcalculus class and manipulate the expression in (23) to get it into a form whereyou can apply it along with other basic calculus rules.)

SolutionChecking (13) just involves plugging in C = 0 into (23).

G2(0) = G040 +∫ 0

0

D(�)4�(0−�)3� = G0 + 0 = G0. (24)

The real action is in checking that this satisfies the differential equationwith the given input. To do this, there are many approaches. One would

11

Homework 2 @ 2020-02-13 20:29:31-08:00

be to use the big hammer of the full Fundamental Theorem of Calculusin Leibniz form. However, in this case, we can do this in a simpler way byfirst simplifying the integral by pulling out factors that do not vary withthe variable of integration. Notice that:

G2(C) = G04�C +∫ C

0

D(�)4�(C−�)3� (25)

= G04�C + 4�C

∫ C

0

D(�)4−��3�. (26)

Taking a derivative we have:

3

3CG2(C) = �G04

�C + �4�C∫ C

0

D(�)4−��3� + 4�C 33C

∫ C

0

D(�)4−��3� (27)

= �G2(C) + 4�C3

3C

∫ C

0

D(�)4−��3� (28)

= �G2(C) + 4�CD(C)4−�C (29)= �G2(C) + D(C) (30)

wherewe used the product rule in (27) and the basic fundamental theoremof calculus in (29).

Because we have checked that the solution satisfies the conditions of thedifferential equation, and we have shown uniqueness in the previouspart, we know that this must be the only solution.

The discussion and the supported course notes show how an argumentbuilding from approximating by piecewise constant functions and thenReimann integration lets you naturally guess this form of the solution.

c) Use the previous part to get an explicit expression for G2(C) for C ≥ 0when D(C) = 4 BC for some constant B, when B ≠ � and C ≥ 0.

Solution

In this part we are given input D(C) = 4 BC . We have to solve a differentialequation of the form:

3

3CG(C) = �G(C) + 4 BC (31)

In the previous part we were given equation (23) to solve for G2(C) forC ≥ 0, for a nonhomogeneous ode of the form (15) with any input D(C).

12


Plugging into equation (23) we get:

G2(C) = G04�C +∫ C

0

4 B�4�(C−�)3� (32)

= G04�C + 4�C

∫ C

0

4�(B−�)3� (33)

= G04�C + 4�C

[ 1

B − � 4(B−�)�]�=C

�=0(34)

= G04�C + 4�C

[ 1

B − � 4(B−�)C − 1

B − � 4(B−�)0] (35)

= G04�C + 4�C

[ 1

B − � 4(B−�)C − 1

B − �]

(36)

= G04�C + 4�C 1

B − � 4(B−�)C − 4�C 1

B − � (37)

= G04�C + 1

B − � 4BC − 1

B − � 4�C (38)

= (G0 −1

B − � )4�C + 4 BC

B − � (39)

Thus we get G2(C) = (G0 − 1B−� )4�C + 4 BC

B−� .

d) Similarly, what is G2(C) for C ≥ 0 when D(C) = 4�C for C ≥ 0.(HINT: Don’t worry if this seems too easy.)

SolutionSimilar to the previous part here we are given another exponential 4 BC asan input. However here B = �. Just like the previous part we will proceedby plugging our new input into (15):

G2(C) = G04�C +∫ C

0

4��4�(C−�)3� (40)

= G04�C + 4�C

∫ C

0

4�(�−�)3� (41)

= G04�C + 4�C

∫ C

0

13� (42)

= G04�C + 4�C

[�]�=C�=0

(43)

= G04�C + C4�C (44)

Thus we get G2(C) = G04�C + C4�C .This basic pattern can be continued. We can plug in C4�C as an input andget further polynomials in C multiplying the same exponential.

13

Homework 2 @ 2020-02-13 20:29:31-08:00

4 Op-Amp Stability

In this question we will revisit the basic op-amp model that was introduced inEE16A and we will add a capacitance �>DC to make the model more realistic(refer to figure 1). Now that we have the tools to do so, we will study thebehavior of the op-amp in positive and negative feedback (refer to figure 2).Furthermore, we will begin looking at the integrator circuit (refer to figure 3) tosee how a capacitor in the negative feedback can behave. In the next homework,you will see why it ends up being close to an integrator.

−+

Rout

Cout

G∆V

V_

V+

Figure 1: Op-amp model: Δ+ = ++ −+−

−

+

VoutVin

(a) Buffer in negative feedback

−

+

Vout

Vin

(b) “Buffer” in positive feedback that doesn’tactually work as a buffer.

Figure 2: Op-amp in buffer configuration

14


−

+

Vout

Vin

R

C+

+

-

-

Figure 3: Integrator circuit

−+

Rout

Cout+

−

VinVout

R

Ci1

i2

i1 + i2

G∆V

∆V= V+-V-

+ −

+

+

+ −

−

−

Figure 4: Integrator circuit with Op-amp model

a) Using the op-amp model in figure 1 and the buffer in negative-feedbackconfiguration in figure 2a, draw a combined circuit. Remember thatΔ+ = ++ −+−, the voltage difference between the positive and negativelabeled input terminals of the op-amp.

(HINT: Look at figure 4 to see how this was done for the integrator. That mighthelp.)

Note: here, we have used the Thevenin-equivalent model for the op-ampgain to be compatible with what you have seen in 16A. In more advancedanalog circuits courses, it is traditional to use a controlled current sourcewith a resistor in parallel instead.

15

Homework 2 @ 2020-02-13 20:29:31-08:00

SolutionPlease refer to Figure 5 for the completed circuit.

Cout

G∆V

∆V

+

−

−+

Rout

Vin Vout

Figure 5: Negative-feedback buffer configuration using given op-amp model

b) Let’s look at the op-amp in negtive feedback. From our discussions inEE16A, we know that the buffer in figure 2a should work with +>DC ≈ +8=by the golden rules. Write a differential-equation for +>DC by replacingthe op-amp with the given model and show what the solution will beas a function of time for a static+8= . What does it converge to as C →∞?Note: We assume the gain � > 1 for all parts of the question.

SolutionWe have Δ+ = +8= − +>DC . Next, we can write the following branchequations:

8 = �>DC3

3C+>DC

�Δ+ = +' ++>DC

�(+8= −+>DC) = '�3

3C+>DC ++>DC

Simplifying the last line, we get:

�+8= = '�3

3C+>DC + (1 + �)+>DC

3

3C+>DC +

1 + �'�

+>DC −�

'�+8= = 0

16


Solving the above differential equation, with the substitution +̃>DC =+>DC − �

�+1+8= , we get

+̃>DC(C) = :4−�+1'� C .

Substituting for the initial condition +>DC(0) = 0, we get:

+>DC =�+8=

� + 1(1 − 4− �+1'� C

).

Since � > 1, the exponent is negative, hence as C →∞, the solution willconverge to +>DC → �

�+1+8= .

c) Next, let’s look at the op-amp in positive feedback. We know that theconfiguration given in figure 2b is unstable and +>DC will just rail. Again,using the op-amp model in figure 1, show that +>DC does not convergeand hence the output will rail. For positive DC input +8= > 0, will +>DCrail to the positive or negative side? Explain.

SolutionThis time, we have Δ+ = +>DC − +8= . Next, we can write the followingbranch equations:

8 = �>DC3

3C+>DC

�Δ+ = +' ++>DC

�(+>DC −+8=) = '�3

3C+>DC ++>DC

Simplifying the last line, we get:

− �+8= = '�3

3C+>DC + (1 − �)+>DC

3

3C+>DC +

1 − �'�

+>DC +�

'�+8= = 0

Solving the above differential equation, with the substitution +̃>DC =+>DC − �

�−1+8= , we get:

+̃>DC = :4− 1−�

'� C

= :4�−1'� C

Substituting for the hypothetical initial condition +>DC(0) = 0, we get:

+>DC = −�+8=

� − 1(4�−1'� C − 1

)17

Homework 2 @ 2020-02-13 20:29:31-08:00

Since � > 1, the exponent is positive, hence as C →∞, the solution will beunbounded, and +>DC → −∞. Of course, it can’t grow to negative infinity,and so we can conclude that +>DC will rail to the negative side if +>DC hadstarted at zero. The same exact story would hold if +>DC started anywherebelow +8= .The case of +>DC starting out significantly greater than +8= deserves somemention, although not necessary for full credit on this question. In thiscase, the initial condition for +̃>DC would be positive, and would proceedto unstably attempt to run away to positive infinity. It would be stoppedat the positive rail, where it would stay.

Essentially, all that matters is the initial condition of +̃>DC — start outpositive, then we rail to a positive rail for+>DC . Start out negative, then werail to the negative rail for +>DC . An op-amp in positive feedback retainsits comparator-like character.

d) For an ideal op-amp, we can assume that it has an infinite gain, i.e.,� → ∞. Under these assumptions, show that the op-amp in negativefeedback behaves as an ideal buffer, i.e., +>DC = +8= .

SolutionTaking the limit of our solution in part (a),

+>DC = lim�→∞

�+8=

� + 1(1 − 4− �+1'� C

)= +8=

The coeffecient of the exponent goes to 1 whereas the exponent itself goesto −∞, and hence 1 − 4−∞ → 1.

5 (OPTIONAL) Write Your Own Question And Provide aThorough Solution.

Writing your own problems is a very effective way to really learn material.Having some practice at trying to create problems helps you study for examsmuch better than simply solving existing practice problems. This is becausethinking about how to create an interesting problem forces you to reallyconsolidate your understanding of the course material.

6 Homework Process and Study Group

Citing sources and collaborators are an important part of life, including beinga student! We also want to understand what resources you find helpful andhow much time homework is taking, so we can change things in the future ifpossible.

18


a) What sources (if any) did you use as you worked through the home-work?

b) If you worked with someone on this homework, who did you workwith? List names and student ID’s. (In case of homework party, you canalso just describe the group.)

c) How did you work on this homework? (For example, I first worked bymyself for 2 hours, but got stuck on problem 3, so I went to office hours. Then Iwent to homework party for a few hours, where I finished the homework.)

d) Roughly how many total hours did you work on this homework?

19

1 fundamentaltheoremofsolutionstodiﬀerential equationsee16b/sp20/homework/sol2.pdf ·...

Documents