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128 CHAPTER 3 ANALYSIS TECHNIQUES

3.72 In the circuit Fig. P3.72, what value of Rs would resultin maximum power transfer to the 10-� load resistor?

2 A RLRs 10 Ω

Figure P3.72: Circuit for Problem 3.72.

Section 3-7: Bipolar Junction Transistor

∗3.73 The two-transistor circuit in Fig. P3.73 is known as acurrent mirror. It is useful because the current I0 controls thecurrent IREF regardless of external connections to the circuit.In other words, this circuit behaves like a current-controlledcurrent source. Assume both transistors are the same size suchthat IB1 = IB2 . Find the relationship between I0 and IREF.(Hint: You do not need to know what is connected above orbelow the transistors. Nodal analysis will suffice.)

I0

C2

E2

C1

E1

BTransistor 2Transistor 1

IREF

Figure P3.73: A simple current mirror (Problem 3.73).

3.74 The circuit in Fig. P3.74 is a BJT common collectoramplifier. Find both the voltage gain (AV = Vout/Vin) and thecurrent gain (AI = Iout/Iin). Assume Vin � VBE.

3.75 The circuit in Fig. P3.75 is identical with the circuit inFig. P3.74, except that the voltage source Vin is more realisticin that it has an associated resistance Rin. Find both the voltagegain (AV = Vout/Vin) and the current gain (AI = Iout/Iin).Assume Vin � VBE.

C

E(Power supply)

BIin

RL

V0

Vin

Vout

+

_

Iout

+_

+_


C

E(Power supply)

BIin

RL

Rin

V0

Vin

Vout

+

_

Iout

+_

+_


3.76 The circuit in Fig. P3.76 is a BJT common emitteramplifier. Find Vout as a function of Vin.

(Power supply)

RL

RsV0

Vin

Vout

+

_

+_

+_


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150 CHAPTER 4 OPERATIONAL AMPLIFIERS

Table 4-3: Summary of op-amp circuits.

Op-Amp Circuit Block Diagram

Rs Rf

vo

vs

+

−

Inverting Amp

vs vo = GvsG = − Rs

Rf

Inverting Summer

v2 vo = G1v1 + G2v2 + G3v3

v1

v3

G2 = − Rf /R2

G1 = − Rf /R1

G3 = − Rf /R3

+

R1 Rf

vo

v1

+

−R2

v2

R3v3

Subtracting Amp

v1

vo = G1v1 + G2v2

v2

+

( )R4R3 + R4( )R1

R1 + R2G2 =

R2R1

G1 = −R1 R2

vo

v1

+

−R3

R4

v2

Voltage Follower(vo independent of Rs)

vs vo = vsG = 1Rsvs

vo+

−

Noninverting Amp(vo independent of Rs)

vs vo = GvsG =R2

R1 + R2

R1

Rs

R2

vsvo

+

−


174 CHAPTER 4 OPERATIONAL AMPLIFIERS

4.6 The inverting-amplifier circuit shown in Fig. P4.6 uses aresistor Rf to provide feedback from the output terminal to theinverting-input terminal.

(a) Use the equivalent-circuit model of Fig. 4-4 to obtain anexpression for the closed-loop gain G = vo/vs in terms ofRs, Ri, Ro, RL, Rf , and A.

(b) Determine the value of G for Rs = 10 �, Ri = 10 M�,Rf = 1 k�, Ro = 50 �, RL = 1 k�, and A = 106.

(c) Simplify the expression for G obtained in (a) by lettingA → ∞, Ri → ∞, and Ro → 0 (ideal op-amp model).

(d) Evaluate the approximate expression obtained in (c) andcompare the result with the value obtained in (b).

RL

RfRs

vp

vnvo

vs

+_

+_


4.7 For the circuit in Fig. P4.7:

(a) Use the op-amp equivalent-circuit model to develop anexpression for G = vo/vs.

(b) Simplify the expression by applying the ideal op-ampmodel parameters, namely A → ∞, Ri → ∞, andRo → 0.

RL

vo

vs

+_

+_


4.8 The op-amp circuit shown in Fig. P4.8 has a constant dcvoltage of 6 V at the noninverting input. The inverting input isthe sum of two voltage sources consisting of a 6-V dc sourceand a small time-varying signal vs.

(a) Use the op-amp equivalent-circuit model given in Fig. 4-4to develop an expression for vo.

(b) Simplify the expression by applying the ideal op-ampmodel, which lets A → ∞, Ri → ∞, and Ro → 0.

RL

vo

vs

6 V6 V

+_

+_+

_+_


Sections 4-3 and 4-4: Ideal Op-Amp and Inverting Amp

Assume all op amps to be ideal from here on forward.∗4.9 The supply voltage of the op amp in the circuit of Fig. P4.9is 16 V. If RL = 3 k�, assign a resistance value to Rf so thatthe circuit would deliver 75 mW of power to RL.

RLRf

Vcc = 16 V

3 V

50 Ω

4 kΩ

+_

+_


4.10 In the circuit of Fig. P4.10, a bridge circuit is connectedat the input side of an inverting op-amp circuit.

(a) Obtain the Thevenin equivalent at terminals (a, b) for thebridge circuit.

(b) Use the result in (a) to obtain an expression for G = vo/vs.

(c) Evaluate G for R1 = R4 = 100 �, R2 = R3 = 101 �,and Rf = 100 k�.

vo

vs

R2

b

a

Rf

R1

R4R3

+_

+_



198 CHAPTER 5 RC AND RL FIRST-ORDER CIRCUITS

C2 ‖ C3 = C2 + C3

= 1 μF + 5 μF

= 6 μF.

Ceq = C1 × 6 × 10−6

C1 + 6 × 10−6

=(

12 × 6

12 + 6

)× 10−6

= 4 μF.

The equivalent circuit is shown Fig. 5-18(b).

Example 5-6: Voltage Division

Figure 5-19(a) contains two resistors R1 and R2 connected inseries to a voltage source vs. In Chapter 2, we demonstratedthat the voltage vs is divided among the two resistors and, forexample, v1 is given by

v1 =(

R1

R1 + R2

)vs. (5.43)

Derive the equivalent voltage-division equation for the seriescapacitorsC1 andC2 in Fig. 5-19(b). Assume that the capacitorshad no charge on them before they were connected to vs.

Solution: From the standpoint of the source vs, it “sees” anequivalent, single capacitor C given by the series combinationof C1 and C2, namely

C = C1C2

C1 + C2. (5.44)

The voltage across C is vs. The law of conservation of energyrequires that the energy that would be stored in the equivalent

C1C2

+

+

_

_

vs v2

v1

q1q2

−q2

−q1

+_

R1

R2+

+

_

_

vs v2

v1

+_+_

+_

Voltage Division

(a) v1 =(

R1

R1 + R2

)vs

v2 =(

R2

R1 + R2

)vs

(b) v1 =(

C2

C1 + C2

)vs

v2 =(

C1

C1 + C2

)vs

Figure 5-19: Voltage-division rules for (a) in-series resistors and (b)in-series capacitors.

capacitor C be equal to the sum of the energies stored in C1and C2. Hence, application of Eq. (5.29) gives

1

2Cv2

s = 1

2C1v

21 + 1

2C2v

22 . (5.45)

Upon replacing C with the expression given by Eq. (5.44) andreplacing the source voltage with vs = v1 + v2, we have

1

2

(C1C2

C1 + C2

)(v1 + v2)

2 = 1

2C1v

21 + 1

2C2v

22, (5.46)

which reduces to

C1v1 = C2v2. (5.47)

Using v2 = vs − v1 in Eq. (5.47) leads to

C1v1 = C2(vs − v1)

or

v1 =(

C2

C1 + C2

)vs. (5.48)

We note that in the voltage-division equation for resistors, v1 isdirectly proportional to R1, whereas in the capacitor case, v1 isdirectly proportional to C2 (instead of to C1). Additionally, inview of the relationship given by Eq. (5.47), application of thebasic definition for capacitance, namely C = q/v, leads to

q1 = q2. (5.49)

This result is exactly what one would expect when viewing thecircuit from the perspective of the voltage source vs.

Review Question 5-11: Compare the voltage-divisionequation for two capacitors in series with that for tworesistors in series. Are they identical or different in form?

Review Question 5-12: Two capacitors are connected inseries between terminals (a, b) in a certain circuit withcapacitor 1 next to terminal a and capacitor 2 next toterminal b. How does the magnitude and polarity ofcharge q1 on the plate (of capacitor 1) near terminal a

compare with charge q2 on the plate (of capacitor 2) nearterminal b?


TECHNOLOGY BRIEF 11: SUPERCAPACITORS 199

Technology Brief 11: Supercapacitors

According to Section 5-2.1, the energy (in joules) stored in a capacitor is given by w = 12 CV2, where C is the capacitance

and V is the voltage across it. Why then do we not charge capacitors by applying a voltage across them and then usethem instead of batteries in support of everyday gadgets and systems? To help answer this question, we refer thereader to Fig.TF11-1, whose axes represent two critical attributes of storage devices. It is the combination (intersection)

Figure TF11-1: Energy and power densities of modern energy-storage technologies. Even though supercapacitors store lesscharge than batteries, they can discharge their energy more quickly, making them more suitable for hybrid cars. (Science,Vol. 313, p. 902.)


200 TECHNOLOGY BRIEF 11: SUPERCAPACITORS

of these attributes that determines the type of applications best suited for each of the various energy devices displayedin the figure. Energy density W ′ is a measure of how much energy a device or material can store per unit weight.[Alternatively, energy density can be defined in terms of volume (instead of weight) for applications where minimizingthe volume of the energy source is more important than minimizing its weight.] Even though the formal SI unit forenergy density is (J/kg), a more common unit is the watt-hour/kg (Wh/kg) with 1 Wh = 3600 J. The second dimensionin Fig.TF11-1 is the power density P ′ (W/kg), which is a measure of how fast energy can be added to or removed froman energy-storage device (also per unit weight). Power is defined as energy per unit time as P ′ = dW ′/dt. According toFig.TF11-1, fuel cells can store large amounts of energy, but they can deliver that energy only relatively slowly (severalhours). In contrast, conventional capacitors can store only small amounts of energy—several orders of magnitude lessthan fuel cells—but it is possible to charge or discharge a capacitor in just a few seconds—or even a fraction of asecond. Batteries occupy the region in-between fuel cells and conventional capacitors; they can store more energy perunit weight than the ordinary capacitor by about three orders of magnitude, and they can release their energy fasterthan fuel cells by about a factor of 10. Thus, capacitors are partly superior to other energy devices because they canaccomodate very fast rates of energy transfer, but the amount of energy that can be “packed into” a capacitor is limitedby its size and weight. To appreciate what that means, let us examine the relation w = 1

2 CV2. To increase w, we needto increase either C or V. For a parallel-plate capacitor, C = εA/d, where ε is the permittivity of the material betweenthe plates, A is the area of each of the two plates, and d is the separation between them. The material between theplates should be a good insulator, and for most such insulators, the value of ε is in the range between ε0 (permittivityof vacuum) and 6ε0 (for mica), so the choice of material can at best increase C by a factor of 6. Making A largerincreases both the volume and weight of the capacitor. In fact, since the mass m of the plates is proportional directlyto A, the energy density W ′ = w/m is independent of A. That leaves d as the only remaining variable. Reducing d willindeed increase C, but such a course will run into two serious obstacles: (a) to avoid voltage breakdown (arcing), Vhas to be reduced along with d such that V/d remains greater than the breakdown value of the insulator; (b) Eventuallyd approaches subatomic dimensions, making it infeasible to construct such a capacitor. Another serious limitationof the capacitor as an energy storage device is that its voltage does not remain constant as energy is transferred toand from it.

Supercapacitor Technology

A new generation of capacitor technologies, termed supercapacitors or ultracapacitors, is narrowing the gap betweencapacitors and batteries. These capacitors can have sufficiently high energy densities to approach within 10 percentof battery storage densities, and additional improvements may increase this even more. Importantly, supercapacitorscan absorb or release energy much faster than a chemical battery of identical volume. This helps immensely duringrecharging. Moreover, most batteries can be recharged only a few hundred times before they are degraded completely;supercapacitors can be charged and discharged millions of times before they wear out. Supercapacitors also havea much smaller environmental footprint than conventional chemical batteries, making them particularly attractive forgreen energy solutions.

History and Design

Supercapacitors are a special class of capacitor known as an electrochemical capacitor. This should not be confusedwith the term electrolytic capacitor, which simply refers to a specific way of fabricating a conventional capacitor.Electrochemical capacitors work by making use of a special property of water solutions (and some polymers andgels). When a metal electrode is immersed in water and a potential is applied, the water molecules (and any dissolvedions) immediately align themselves to the charges present at the surface of the metal electrode, as illustrated inFig. TF11-2(a). This re-arrangement generates a thin layer of organized water molecules (and ions), called a double


TECHNOLOGY BRIEF 11: SUPERCAPACITORS 201

layer, that extends over the entire surface of the metal. The very high charge density, separated by a tiny distanceon the order of a few nanometers, effectively looks like a capacitor (and a very large one: capacitive densities on theorder of ∼ 10 μF/cm2 are common for water solutions). This phenomenon has been known to physicists and chemistssince the work of von Helmholtz in 1853, and later Guoy, Chapman, and Stern in the early 20th century. In order tomake capacitors useful for commercial applications, several technological innovations were required. Principal amongthese were various methods for increasing the total surface area that forms the double layer. The first working capacitorbased on the electrochemical double layer (patented by General Electric in 1957) used very porous conductive carbon.Modern electrochemical capacitors employ carbon aerogels, and more recently carbon nanotubes have been shownto effectively increase the total double layer area (Fig. TF11-2(b)).

Supercapacitors are beginning to see commercial use in applications ranging from transportation to low-powerconsumer electronics. Several bus lines around the world now run with buses powered with supercapacitors; trainsystems are also in development. Supercapacitors intended for small portable electronics (like your MP3 player) arein the pipeline as well!

(a)

(b)

Activated carbon

5-10 nm

Solvated ion

and hydration

(water) sheet

Outer

Helmholtz

Plane (OHP)

Separator

Electrodes

Figure TF11-2: (a) Conceptual illustration of the waterdouble layer at a charged metal surface; (b) conceptualillustration of an electrochemical capacitor.


5-4 RESPONSE OF THE RC CIRCUIT 215

(a) Actual circuit

(b) Circuit during 0 ≤ t ≤ 10 s

(c)

Circuit after t = 10 s

v2C1+_

R2

vC1

2

+_

+_

t = 10 s

t = 0

V1

R2R1

+_

v1C

2

+_

+_V1

R2R1

+_

Figure 5-32: After having been in position 1 for a long time, theswitch is moved to position 2 at t = 0 and then returned to position 1at t = 10 s (Example 5-10).

to

v1(t) = v1(∞) + [v1(0) − v1(∞)]e−t/τ1

= 20(1 − e−0.04t ) V (for 0 ≤ t ≤ 10 s).

Time Segment 2: t ≥ 10 s

Voltage v2(t), corresponding to the second time segment(Fig. 5-32(c)), is given by Eq. (5.98) with a new time constant τ2as

v2(t) = v2(∞) + [v2(10) − v2(∞)]e−(t−10)/τ2 .

The new time constant is associated with the capacitor circuitremaining after returning the switch to position 1,

τ2 = R2C

= 20 × 103 × 0.25 × 10−3 = 5 s.

The initial voltage v2(10) is equal to the capacitor voltage v1 atthe end of time segment 1, namely

v2(10) = v1(10) = 20(1 − e−0.04×10)

= 6.59 V.

With no voltage source present in the R2C circuit, the chargedcapacitor will dissipate its energy into R2, exhibiting a naturalresponse with a final voltage of v2(∞) = 0. Consequently,

v2(t) = v2(10) e−(t−10)/τ2

= 6.59e−0.2(t−10) V (for t ≥ 10 s).

Example 5-11: RC-Circuit Response to Rectangular

Pulse

Determine the voltage response of a previously unchargedRC circuit to a rectangular pulse vi(t) of amplitude Vs andduration T0, as depicted in Fig. 5-33(a). Evaluate and plot theresponse for R = 25 k�, C = 0.2 mF, Vs = 10V, and T0 = 4 s.

Solution: According to Example 5-2, a rectangular pulse isequivalent to the sum of two step functions. Thus

vi(t) = Vs[u(t − T1) − u(t − T2)],where u(t − T1) accounts for the rise in level from 0 to 1 att = T1 and the second term (with negative amplitude) serves tocounteract (cancel) the first term after t = T2. For the presentproblem, T1 = 0, and T2 = 4 s. Hence, the input pulse can bewritten as

vi(t) = Vs u(t) − Vs u(t − 4).

Since the circuit is linear, we can apply the superpositiontheorem to determine the capacitor response v(t). Thus,

v(t) = v1(t) + v2(t),

where v1(t) is the response to Vs u(t) acting alone and,similarly, v2(t) is the response to −Vs u(t−4) also acting alone.

The response v1(t) is given by Eq. (5.96) with v1(0) = 0,v1(∞) = Vs, and τ = RC. Hence,

v1(t) = v1(∞) + [v1(0) − v1(∞)]e−t/τ

= Vs(1 − e−t/τ ) (for t ≥ 0).

For Vs = 10 V and τ = RC = 25 × 103 × 0.2 × 10−3 = 5 s,

v1(t) = 10(1 − e−0.2t ) V (for t ≥ 0).

The second step function has an amplitude of−Vs and is delayedin time by 4 s. Upon reversing the polarity of Vs and replacingt with (t − 4), we have

v2(t) = −10[1 − e−0.2(t−4)] V (for t ≥ 4 s).



(a)Pulse excitation

(d)

Vs+_v1

v2

i

C

R

+_

During 0 ≤ t ≤ 4 s

(b)

+_C

Ri

After t = 4 s

(c)

+_

vi = v

i

C

R

t = 0 t = 4 s

Vs = 10 V +_

+_+_

v (V)

t (s)0 4 10 20

0

2

4

6Forcedresponse

Natural response

Figure 5-33: RC-circuit response to a 4-s-long rectangular pulse.

The total response for t ≥ 0 therefore is given by

v(t) = v1(t) + v2(t)

= 10[1 − e−0.2t ] − 10[1 − e−0.2(t−4)] u(t − 4) V,

(5.100)

where we introduced the time-shifted step function u(t − 4)

to assert that the second term is zero for t ≤ 4 s. The plot ofv(t) displayed in Fig. 5-33(d) shows that v(t) builds up to amaximum of 5.5 V by the end of the pulse (at t = 4 s) and thendecays exponentially back to zero thereafter. The build-up partis due to the external excitation and often is called the forcedresponse. In contrast, during the time period after t = 4 s, v(t)

exhibits a natural decay response as the capacitor dischargesits energy into the resistor. During this latter time segment, i(t)flows in a counterclockwise direction.

Review Question 5-20: What are the three quantitiesneeded to establish v(t) across a capacitor in an RCcircuit?

Review Question 5-21: If Vs2 < Vs1 in the circuit ofFig. 5-30, what would you expect the direction of thecurrent to be after the switch is moved from position 1to 2? Analyze the process in terms of charge accumulationon the capacitor.

Exercise 5-15: Determine v1(t) and v2(t) for t ≥ 0, giventhat in the circuit of Fig. E5.15 C1 = 6 μF, C2 = 3 μF,R = 100 k�, and neither capacitor had any charge priorto t = 0.

v1C1

R

+_+_12 V

v2C2

t = 0

Figure E5.15

Answer: v1(t) = 4(1 − e−5t ) V, for t ≥ 0,v2(t) = 8(1 − e−5t ) V, for t ≥ 0. (See )

5-5 Response of the RL Circuit

With RC circuits, we developed a first-order differentialequation for v(t), the voltage across the capacitor, and thenwe solved it (subject to initial and final conditions) to obtaina complete expression for v(t). By applying i = C dv/dt ,p = iv, and w = 1

2Cv2, we were able to determine thecorresponding current passing through the capacitor, the powergetting transferred to it, and the net energy stored in it. We nowwill follow an analogous procedure for the RL circuit, but ouranalysis will focus on the current i(t) through the inductor.

5-5.1 Natural Response of the RL Circuit

After having been in the closed position for a long time, theswitch in the RL circuit of Fig. 5-34(a) was opened at t = 0,thereby disconnecting the RL circuit from the current source Is.What happens to the current i flowing through the inductor afterthe sudden change caused by opening the switch? That is, whatis the waveform of i(t) for t ≥ 0? To answer this question, wefirst note that at t = 0− (just before opening the switch), theRL circuit can be represented by the circuit in Fig. 5-34(b), inwhich the inductor has been replaced with a short circuit. This



C1

R1 +

_C1

R1

C1

R1

C1

R1

C1

R1

+_

vout1

C2

R2 +

_C2

R2

C2

R2

C2

R2

C2

R2

vout2vs(t)

Figure P5.70: Circuit for Problem 5.70 with R1 = R2 = 10 �, C1 = 7 pF, and C2 = 5 pF.

5.70 A step voltage source vs(t) sends a signal down twotransmission lines simultaneously (Fig. P5.70). In Multisim,the step voltage may be modeled as a 1-V square wave with aperiod of 10 ns. Model the circuit in Multisim and answer thefollowing questions:

(a) If a detector registers a signal when the output voltagereaches 0.75 V, which signal arrives first?

(b) By how much?

Hint: When using cursors in the Grapher View, select a trace,then right-click on a cursor and select Set Y Value, and enter750 m. This will give you the exact time point at which thattrace equals 0.75 V.

5.71 Consider the delta topology in Fig. P5.71. Use Multisimto generate response curves for va , vb, and vc. Apply TransientAnalysis with TSTOP = 3 × 10−10 s.

5.72 Use Multisim to generate a plot for current i(t) in thecircuit in Fig. P5.72 from 0 to 15 ms.

5.73 Construct the integrator circuit shown in Fig. P5.73,using a 3-terminal virtual op amp. Print the outputcorresponding to each of the following input signals:

(a) vin(t) is a 0-to-1-V square wave with a period of 1 ms anda 50 percent duty cycle. Plot the output from 0 to 10 ms.

(b) vin(t) = −0.2t V. Plot the output from 0 to 50 ms.

1 MΩ 3.5 kΩ

10 kΩ

1 V

5 fF

2 fF

1 fF

+_ va

vb

vct = 0

+

_

+_


voutvin

R1

C1

+

_100 Ω

100 μF


220 Ω90 Ω 500 mH

R1

i

L1

R2+_vs(t) = [−5u(−t) + 5u(0.003 − t)] V 0.1u(t − 0.003) A



6-1 INITIAL AND FINAL CONDITIONS 249

Table 6-1: Modes of behavior of a capacitor C and an inductor L. For each circuit, A represents the configuration before the sudden change(at t = 0−) and B the configuration after the change.

Capacitor

Circuit

Inductor

C = open circuit L = short circuit

C = open circuit L = short circuit

C

a

b

Circuit1

vC

iC

L

c

d

Circuit2

vL

iL

C

a

b

Circuit1-A

iC = 0

iL(0−)vL = 0 L

c

d

Circuit2-A

C = voltage source vC(0−) L = current source iL(0−)

C = short circuit if vC(0−) = 0 L = open circuit if iL(0−) = 0

vC(0−)

vC(0−)

a

b

Circuit1-B

iC(0)

+_ iL(0−)

c

d

Circuit2-B

vL(0)

C

a

b

Circuit1-B

iC(0)

vL(0−) L

c

d

iL(0) = 0

Circuit2-B

vC( )

a

b

Circuit1-B

iC = 0

8

c

d

vL = 0Circuit2-B

iL( )8

At t = 0−:

At t = 0:

At t = :8

��

(b) In circuits containing dc sources, the steady-statecondition of the circuit (after all transients have died out)is such that no currents flow through capacitors and novoltages exist across inductors, allowing us to representcapacitors as open circuits and inductors as short circuits.

Table 6-1 outlines the modes of behavior of a capacitor C

contained in a circuit labeled Circuit 1 and an inductor L

contained in Circuit 2 at three distinct instants in time. It isassumed that sudden changes in the circuits occur at t = 0,and that by t = 0−, the circuits already had reached a steadystate. To distinguish between the capacitor circuit configuration


TECHNOLOGY BRIEF 12: MICROMECHANICAL SENSORS AND ACTUATORS 269

Technology Brief 12: Micromechanical Sensors and Actuators

Energy is stored in many different forms in the world around us. The conversion of energy from one form to another iscalled transduction. Each of our five senses, for example, transduces a specific form of energy into electrochemicalsignals: tactile transducers on the skin convert mechanical and thermal energy; the eye converts electromagneticenergy; smell and taste receptors convert chemical energy; and our ears convert the mechanical energy of pressurewaves. Any device, whether natural or man-made, that converts energy signals from one form to another is atransducer.

Most modern man-made systems are designed to manipulate signals (i.e., information) using electrical energy.Computation, communication, and storage of information are examples of functions performed mostly with electricalcircuits. Most systems also perform a fourth class of signal manipulation: the transduction of energy from theenvironment surrounding them into electrical signals that their circuits can use in support of their intended application.If the transducer converts external signals into electrical signals, it is called a sensor. The charge-coupled device(CCD) chip on your camera is a sensor that converts electromagnetic energy (light) into electrical signals that can beprocessed, stored, and communicated by your camera circuits. Some transducers perform the reverse function, namelyto convert a circuit’s electrical signal into an environmental excitation. Such a transducer then is called an actuator.The components that deploy the airbag in your car are actuators: given the right signal from the car’s microcontroller,the actuators convert electrical energy into mechanical energy and the airbag is released and inflated.

Micro- and nanofabrication technology have begun to revolutionize many aspects of sensor and actuator design.Humans increasingly are able to embed transducers at very fine scales all over their environment. This is leadingto big changes, as our computational elements are becoming increasingly aware of their environment. Shippingcontainers that track their own acceleration profiles (allowing the customer to detect falling and damage), laptops thatscan fingerprints for routine login, cars that detect collisions, and even office suites that modulate energy consumptionbased on human activity are all examples of this transduction revolution. In this technology brief, we will focus on aspecific type of microscale transducers that lend themselves to direct integration with silicon ICs. Collectively, devicesof this type are called Microelectromechanical Systems (MEMS) or Microsystems Technologies (MST); the twonames are used interchangeably.

A Capacitive Sensor: The MEMS Accelerometer

According to Eq. (5.21), the capacitance C of a parallel plate capacitor varies directly with A, the effective area ofoverlap between its two conducting plates, and inversely with d, the spacing between the plates. By capitalizing onthese two attributes, capacitors can be made into motion sensors that can measure velocity and acceleration along x,y, and z.

Figure TF12-1 illustrates two mechanisms for translating motion into a change of capacitance. The first generally iscalled the gap-closing mode, while the second one is called the overlap mode. In the gap-closing mode, A remainsconstant, but if a vertical force is applied onto the upper plate, causing it to be displaced from its nominal position atheight d above the lower plate to a new position (d − z), then the value of capacitance Cz will change in accordancewith the expression given in Fig. TF12-1(a). The sensitivity of Cz to the vertical displacement is given by dCz/dz.

The overlap mode (Fig. TF12-1(b)) is used to measure horizontal motion. If a horizontal force causes one of theplates to shift by a distance y from its nominal position (where nominal position corresponds to a 100 percent overlap),the decrease in effective overlap area will lead to a corresponding change in the magnitude of capacitance Cy. In thiscase, d remains constant, but the width of the overlapped areas changes from w to (w − y). The expression for Cygiven in Fig.TF12-1(b) is reasonably accurate (even though it ignores the effects of the fringing electric field betweenthe edges of the two plates) so long as y � w. To measure and amplify changes in capacitance, the capacitor canbe integrated into an appropriate op-amp circuit whose output voltage is proportional to C. As we shall see shortly, acombination of three capacitors, one to sense vertical motion and two to measure horizontal motion along orthogonalaxes, can provide complete information on both the velocity and acceleration vectors associated with the applied force.The capacitor configurations shown in Fig TF12-1 illustrate the basic concept of how a capacitor is used to measuremotion, although more complex capacitor geometries also are possible, particularly for sensing angular motion.


270 TECHNOLOGY BRIEF 12: MICROMECHANICAL SENSORS AND ACTUATORS

Gap-Closing Mode

Cz =

dCz

dz=

Capacitance:

Sensitivity:

z F

d

w

Nominal position

(no force)

Metal

plates

Overlap Mode

dCy =

d

dCy

dy

Capacitance:

Sensitivity:

Metal

plates

F

d

y

w

Figure TF12-1: Basic capacitive measurement modes. For (b), the expressions hold only for small displacements such thaty � w.

To convert the capacitor-accelerometer concept into a practical sensor—such as the automobile accelerometer thatcontrols the release of the airbag—let us consider the arrangement shown in Fig. TF12-2(b). The lower plate is fixedto the body of the vehicle, and the upper plate sits on a plane at a height d above it. The upper plate is attached tothe body of the vehicle through a spring with a spring constant k. When no horizontal force is acting on the upperplate, its position is such that it provides a 100 percent overlap with the lower plate, in which case the capacitance willbe a maximum at Cy = εW�/d. If the vehicle accelerates in the y-direction with acceleration ay, the acceleration forceFacc will generate an opposing spring force Fsp of equal magnitude. Equating the two forces leads to an expressionrelating the displacement y to the acceleration ay, as shown in the figure. Furthermore, the capacitance Cy is directlyproportional to the overlap area �(w−y) and therefore is proportional to the acceleration ay. Thus, by measuring Cy, theaccelerometer determines the value of ay. A similar overlap-mode capacitor attached to the vehicle along the x-directioncan be used to measure ax. Through a similar analysis for the gap-closing mode capacitor shown in Fig TF12-2(a),we can arrive at a functional relationship that can be used to determine the vertical acceleration az by measuringcapacitance Cz.

If we designate the time when the ignition starts the engine as t = 0, we then can set the initial conditions on both thevelocity u of the vehicle and its acceleration a as zero at t = 0. That is, u(0) = a(0) = 0. The capacitor accelerometersmeasure continuous-time waveforms ax(t), ay(t), and az(t). Each waveform then can be used by an op-amp integratorcircuit to calculate the corresponding velocity waveform. For ux, for example,

ux(t) =t∫

0

ax(t) dt,

and similar expressions apply to uy and uz.Figure TF12-3 shows the Analog Devices ADXL202 accelerometer which uses the gap-closing mode to detect

accelerations on a tiny micromechanical capacitor structure that works on the same principle described above, althoughslightly more complicated geometrically. Commercial accelerometers, however, make use of negative feedback toprevent the plates from physically moving. When an acceleration force attempts to move the plate, an electric negative-feedback circuit applies a voltage across the plates to generate an electrical force between the plates that counteractsthe acceleration force exactly, thereby preventing any motion by the plate. The magnitude of the applied voltage


TECHNOLOGY BRIEF 12: MICROMECHANICAL SENSORS AND ACTUATORS 271

(a) The ADXL202 accelerometer employs many

gap-closing capacitor sensors to detect acceleration.

(Courtesy Analog Devices.)

(b) A silicon sensor that uses overlap mode fingers. The

white arrow shows the direction of motion of the moving

mass and its fingers in relation to the fixed anchors. Note

that the moving fingers move into and out of the fixed

fingers on either side of the mass during motion.

(Courtesy of the Adriatic Research Institute.)

Spring constant k

Mass mz

d

Fsp

Facc

Fsp = Facc

kz = maz

maz

kz =

Cz =maz

k( )

Spring

yw

d

FspFacc

Facc = Fsp

may = ky

may

ky =

Cy = =d

may

k( )d

Figure TF12-2: Adding a spring to a movable plate capacitor makes an accelerometer.

becomes a measure of the acceleration force that the capacitor plate is subjected to. Because of their small sizeand low power consumption, chip-based microfabricated silicon accelerometers are used in most modern cars toactivate the release mechanism of airbags. They also are used heavily in many toy applications to detect position,velocity and acceleration. The Nintendo Wii, for example, uses the newer ADXL330 accelerometer in each remoteto detect orientation and acceleration. Incidentally, a condenser microphone operates much like the device shown in


272 TECHNOLOGY BRIEF 12: MICROMECHANICAL SENSORS AND ACTUATORS

Figure TF12-3: The complete ADXL202 accelerometerchip. The center region holds the micromechanicalsensor; the majority of the chip space is used for theelectronic circuits that measure the capacitance change,provide feedback, convert the measurement into a digitalsignal, and perform self-tests. (Courtesy of AnalogDevices.)

Fig. TF12-2(a): as air pressure waves (sound) hit the spring-mounted plate, it moves and the change in capacitancecan be read and recorded.

A Capacitive Actuator: MEMS Electrostatic Resonators

Not surprisingly, we can drive the devices discussed previously in reverse to obtain actuators. Consider again theconfiguration in Fig. TF12-2(a). If the device is not experiencing any external forces and we apply a voltage V acrossthe two plates, an attractive force F will develop between the plates. This is because charges of opposite polarity onthe two plates give rise to an electrostatic force between them. This, in fact, is true for all capacitors. In the case of ouractuator, however, we replace the normally stiff, dielectric material with air (since air is itself a dielectric) and attach it toa spring as before. With this modification, an applied potential generates an electrostatic force that moves the plates.

This basic idea can be applied to a variety of applications. The most successful one to date is the Digital LightProjector (DLP) system (see Technology Brief 6 on page 106) that drives most digital projectors used today. In the DLP,hundreds of thousands of capacitor actuators are arranged in a 2-D array on a chip, with each actuator correspondingto a pixel on an image displayed by the projector. One capacitive plate of each pixel actuator (which is mirror smoothand can reflect light exceedingly well) is connected to the chip via a spring. In order to brighten or darken a pixel, avoltage is applied between the plates, causing the mirror to move into or out of the path of the projected light. Thesesame devices have been used for many other applications, including microfluidic valves and tiny force sensors usedto measure forces as small as a zeptonewton (1 zeptonewton = 10−21 newtons).


TECHNOLOGY BRIEF 13: TOUCHSCREENS AND ACTIVE DIGITIZERS 289

Technology Brief 13: Touchscreens and Active Digitizers

Touchscreen is the common name given to a wide variety of technologies that allow computer displays to directlysense information from the user. In older systems, this usually meant the display could detect and pinpoint wherea user touched the screen surface; newer systems can detect touch location as well as the associated pressure atmultiple locations simultaneously, with very high resolution. This has led to a surge of applications in mobile computing,cell phones, personal digital assistants (PDA), and consumer appliances. One of the most common modern interactivescreens is the tablet-style notebook computer, which uses a pen or stylo to input pressure-sensitive information, therebyallowing a user to write realistically and directly onto the screen.

Numerous technologies have been developed since the invention of the electronic touch interface in 1971 bySamuel C. Hurst. Some of the earlier technologies were susceptible to dust, damage from repeat use, and poortransparency. These issues largely have been resolved over the years (even for older technologies) as experienceand advanced material selection have led to improved devices. With the explosion of consumer interest in portable,interactive electronics, newer technologies have emerged that are more suitable for these applications. Figure TF13-1summarizes the general categories of touchscreens in use today. Historically, touchscreens were manufacturedseparately from displays and added as an extra layer of the display. More recently, display companies have begun tomanufacture sensing technology directly into the displays; some of the newer technologies reflect this.

Resistive

Resistive touchscreens are perhaps the simplest to understand. A thin, flexible membrane is separated from a plasticbase by insulating spacers. Both the thin membrane and the plastic base are coated on the inside with a transparentconductive film (indium tin oxide (ITO) often is used). When the membrane is touched, the two conductive surfacescome into contact. Detector circuits at the edges of the screen can detect this change in resistance between the twomembranes and pinpoint the location on the X–Y plane. Older designs of this type were susceptible to membranedamage (from repeated flexing) and suffered from poor transparency.

Capacitive

Capacitive touchscreens employ a single thin, transparent conductive film (usually ITO) on a plastic or glass base. Theconductive film is coated with another thin, transparent insulator for protection. Since the human body stores charge,a finger tip moved close to the surface of the film effectively forms a capacitor with the film as one of the plates and thefinger as the other. The protective coating and the air form the intervening dielectric insulator. This capacitive couplingchanges how a current flowing across the film surface is distributed; by placing electrodes at the screen corners andapplying an ac electric signal, the location of the finger capacitance can be calculated precisely. A variant of this ideais to divide the sensing area into many smaller squares (just like pixels on the display) and to sense the change incapacitance across each of them continuously and independently. Capacitive technologies are much more resistantto wear and tear (since they are not flexed) than resistive touchscreen and are somewhat more transparent (80 to 85percent transparency) since they have fewer films. These types of screens can be used to detect metal objects as well,so pens with conductive tips can be used on writing interfaces.

Pressure

Touch also can be detected mechanically. Pressure sensors can be placed at the corners of the display screen or eventhe entire display assembly, so whenever the screen is depressed, the four corners will experience different stressesdepending on the (X,Y) position of the pressure point. Pressure screens benefit from high resistance to wear and tearand no losses in transparency (since there is no need to add layers over the display screen).


290 TECHNOLOGY BRIEF 13: TOUCHSCREENS AND ACTIVE DIGITIZERS

(a) Resistive (b) Capacitive

(c) Pressure (d) Acoustic

(e) Infrared (f) Active digitizer

Spacer

MembraneConductive film

Plastic

Membrane

Cfinger

Conductive film

Plastic

Force

Stress Stress

Strainsensor

Strainsensor

Acousticsensor

Acousticemitter

5 MHz Acoustic wave

Acousticdampening

Screen

IR beam

Screen

DetectorLED

Electromagneticradiation

Pen

Wires

LCR

Figure TF13-1: Touchscreen technologies: (a) resistive, (b) capacitive, (c) pressure/strain sensor, (d) acoustic, (e) infrared,and (f) active digitizer.


TECHNOLOGY BRIEF 13: TOUCHSCREENS AND ACTIVE DIGITIZERS 291

Acoustic

A completely different way to detect touch relies on the transmission of high-frequency acoustic energy across thesurface of the display material. Bursts of 5 MHz tones are launched by acoustic actuators from two corners of thescreen. Acoustic reflectors all along the edges of the screen re-direct the incoming waves to the sensors. Any timean object comes into contact with the screen, it dampens or absorbs some fraction of the energy traveling across thematerial. The exact (X,Y) position can be calculated from the energy hitting the acoustic sensors. The contact forcecan be calculated as well, because the acoustic energy is dampened more or less depending on how hard the screen ispressed. Acoustic touchscreens also can detect multiple touches simultaneously, whereas resistive, capacitive (exceptthose that use sensing “pixels”), and pressure touchscreens cannot do so easily.

Infrared

One of the oldest and least used technologies is the infrared touchscreen. This technology relies on infrared emitters(usually IR diodes; see Technology Brief 5 on page 96) aligned along two adjoining edges of the screen and infrareddetectors aligned across from the emitters at the other two edges. The position of a touch event can be determinedthrough a process based on which light paths are interrupted. The detection of multiple simultaneous touch eventsis possible. Infrared screens are somewhat bulky, prone to damage or interference from dust and debris, and needspecial modifications to work in daylight. They largely have been displaced by newer technologies.

Electromagnetic Resonance

The newest technology in widespread use is the electromagnetic resonance detection scheme used by most moderntablet PCs, developed by The Wacom Company. Strictly speaking, tablet PC screens are not touchscreens; they arecalled active digitizers because they can detect the presence and location of the tablet pen as it approaches thescreen (even without contact). In this scheme, a very thin wire grid is integrated within the display screen (whichusually is a flat-profile LCD display, see Technology Brief 6: Displays on page 106). The pen itself contains a simpleRLC resonator (see Section 6-1) with no power supply. The wire grid alternates between two modes (transmit andreceive) every ∼ 20 milliseconds. The grid essentially acts as an antenna. During the transmit mode, an ac signal isapplied to the grid and part of that signal is emitted into the air around the display. As the pen approaches the grid,some energy from the grid travels across to the pen’s resonator which begins to oscillate. In receive mode, the grid isused to “listen” for ac signals at the resonator frequency; if those signals are present, the grid can pinpoint where theyare across the screen. A tuning fork provides a good analogy. Imagine a surface vibrating at a musical note; if a tuningfork designed to vibrate at that note comes very close to that surface, it will begin to oscillate at the same frequency.Even if we were to stop the surface vibrations, the tuning fork will continue to make a sound for a little while longer (asthe resonance dies down). In a similar way, the laptop screen continuously transmits a signal and listens for the pen’selectromagnetic resonance. Functions (such as buttons and pressure information) can be added to the pen by havingthe buttons change the capacitance value of the LCR when pressed; in this way, the resonance frequency will shift(see Section 6-2), and the shift can be detected by the grid and interpreted as a button press.


7-10 APPLICATION NOTE: POWER-SUPPLY CIRCUITS 351

Example 7-18: Power-Supply Design

A power supply with the circuit configuration shown inFig. 7-36 has the following specifications: the input voltageis 60 Hz with an rms amplitude Vrms = 110 V whereVrms = Vs/

√2 (the rms value of a sinusoidal function is

discussed in Chapter 8), N1/N2 = 5, C = 2 mF, Rs = 50 �,RL = 1 k�, Vz = 24 V, and Rz = 20 �. Determine vout, theripple voltage, and the ripple fraction relative to vout.

Solution:At the secondary side of the transformer,

vs1(t) =(

N2

N1

)(Vs cos 377t)

= 1

5× 110

√2 cos 377t = 31.11 cos 377t V.

Hence, Vs1 = 31.11V, which is greater than the zener voltageVz = 24 V.

The output voltage is then

vout = Vz = 24 V.

In Example 7-17, we established that Trect = 8.33 ms. Also,

Rz ‖ RL = 20 × 1000

20 + 1000= 19.6 �.

Application of Eq. (7.143) gives

Vr = [(Vs1 − 1.4) − Vz]Trect

RsC× (Rz ‖ RL)

Rs + (Rz ‖ RL)

= [(31.11 − 1.4) − 24]50 × 2 × 10−3 (8.33 × 10−3) × 19.6

50 + 19.6

= 0.13 V (peak-to-peak).

Hence,

Ripple fraction = (Vr/2)

Vz= 0.13/2

24= 0.0027,

which represents a relative variation of less than±0.3 percent.

7-11 Multisim Analysis of ac Circuits

Even though we usually treat the wires in a circuit as idealshort circuits, in reality a wire has a small but non-zeroresistance. Also, as noted earlier in Section 5-7.1, whentwo wires are in close proximity to one another, they forma non-zero capacitor. A pair of parallel wires on a circuitboard is modeled as a distributed transmission line with eachsmall length segment � represented by a series resistance R

and a shunt capacitance C, as depicted by the circuit modelshown in Fig. 7-37. For a parallel-wire segment of length �,R and C are given by

R = 2�

πa2σ

(low-frequency approximation)(a

√f σ ≤ 500),

(7.144a)

R =√

πf μ

σ

(�

πa

)(high-frequency approximation)(a

√f σ ≥ 1250),

(7.144b)

and

C = πε�

ln(d/a)for (d/2a)2 � 1, (7.144c)

where a is the wire radius, d is the separation betweenthe wires, f is the frequency of the signal propagatingalong the wires, μ and σ are respectively the magneticpermeability and conductivity of the wire material, and ε

is the permittivity of the material between the two wires.Note that R represents the resistance of both wires. There isactually a third distributed element to consider in the generalcase of a transmission line: the distributed inductance. Thisinductance is placed in series with the resistance R of eachsegment. It arises because current flowing through thetransmission-line wires gives rise to a magnetic field aroundthe wires and, hence, an inductance (as discussed in Section5-3). However, modeling the behavior of a transmissionline with all three components is rather complex. So, forthe purposes of this section, we will ignore the inductancealtogether so that we may illustrate the performance of an RCtransmission line using Multisim. Keeping this in mind, thedistributed model shown in Fig. 7-37 allows us to representthe wires by a series of cascaded RC circuits. For themodel to faithfully represent the behavior of the real two-wire configuration, each RC stage should represent a physicallength � that is no longer than a fraction (≈ 10 percent) ofthe distance that the signal travels during one period of thesignal frequency. Thus, � should be on the order of


7-10 APPLICATION NOTE: POWER-SUPPLY CIRCUITS 355

Example 7-20: Measuring Phase Shift

Run a TransientAnalysis on the Multisim circuit in Fig. 7-40after replacing the pulse generator with a 1-V amplitude,10-MHz ac source. The goal is to determine the phase ofnode 2, relative to the phase of node 1 (the voltage source).Select a Start Time of 2.7 μs and an End Time (TSTOP) of3.0 μs, and set TSTEP and TMAX to 1e-10 seconds so asto generate smooth-looking curves. [We did not choose aStart Time of 0 s simply because it takes the circuit a fewmicroseconds to reach its steady-state solution.]

Solution: Figure 7-42 shows the traces of selected nodesV(1), V(2), and V(6) on Grapher View. Clicking onthe Show/Hide Cursors button enables the measurementcursor, which can be used to quantify the amplitude (verticalaxis) and time (horizontal axis) for each curve. To measurethe phase shift between nodes V(2) and V(1), two cursorsare needed.

Step 1: Place cursor 1 slightly to the left of a maximum ofthe V(1) trace.

Step 2: Click on the trace for V(1) to select it. Whitetriangles will appear on the V(1) trace.

Step 3: Right-click the cursor itself and select Go to nextY Max=>. On row x1, at column V(1), the value inthe table should be 2.7250 μs.

Step 4: Repeat the process using cursor 2 to select the nearbymaximum of the V(2) trace. The entry in row x2, atcolumn V(2), should be 2.7312 μs.

The time difference between the two values is

�t = 2.7312 μs − 2.7250 μs = 0.0062 μs.

Given that f = 10 MHz, the period is

T = 1

f= 1

107 = 10−7 = 0.1 μs.

Application of Eq. (7.11) gives

φ = 2π

(�t

T

)= 360◦ ×

(0.0062

0.1

)= 22.3◦.

V(1)

V(2)

V(6)

Figure 7-42: Multisim Grapher Plot of voltage nodes V(1), V(2), and V(6) in the circuit of Fig. 7-38.


356 CHAPTER 7 AC ANALYSIS

Figure 7-43: Using Measurement Probes to determine phase and amplitude of signal at various points on transmission line.

We also can determine the ratio of the amplitude of V(2) tothat of V(1). The ratio of y2 in column V(2) to y1 in columnV(1) gives

V(2)

V(1)= 0.656

1� 66 percent.

Exercise 7-18: Determine the amplitude and phase ofV(6) in the circuit of Example 7-20, relative to those ofV(1).

Answer: (See )

Additional Method to Measure Amplitude and Phase

Let us continue working with the transmission-line circuit ofthe previous two examples. Place a Measurement Probe(of the type we introduced in Chapters 2 and 3) at each ofthe appropriate nodes in the circuit. Double-click on theProbe, and under the Parameters tab, select the appropriateparameters so that only V(p-p), Vgain(ac), and Phase are

printed in the Probe output. Additionally, with the exceptionof Probe 1 (located right above V1), at the top of the ProbeProperties window, check Use reference probe, and selectProbe 1. Note that “phase” here refers to the phase differencebetween the voltage at the specific probe and the referenceprobe. So if a particular signal is leading the reference node,then the phase will appear negative, and if a particular signalis lagging the reference node, then the phase will appearpositive. This is the opposite of how we are taught to thinkof phase, so keep this at the front of your mind when usingthis approach.

Run the Interactive Simulation by pressing F5 (or anyof the appropriate buttons or toggles, which you shouldknow by now) and the result should resemble that shownin Fig. 7-43. We can see that the Phase at Node 2 is 22.6◦,which of course is opposite to what we see in Fig. 7-42,where the signal at V(2) is behind V(1) by 22.3◦. However,we must remember that the phase values are flipped in theMeasurement Probe readings, so the values actually arein agreement. Additionally, we see in Fig. 7-43 that theVgain(ac) at Node 2 is “654m” (which corresponds to 65.4percent), which is very nearly in agreement with the value of66 percent obtained in Example 7-20.


444 CHAPTER 9 FREQUENCY RESPONSE OF CIRCUITS AND FILTERS

Example 9-12: Bandreject Filter

Design a bandreject filter with the specifications:(a) Gain = −50, (b) bandstop extends from 20 kHz to40 kHz, and (c) gain roll-off rate = −40 dB/decade along bothboundaries of the bandstop.

Solution: The specified roll-off rate requiresthe use of two identical lowpass filters with

(b) Circuit diagram

(a) Block diagram

(c) Filter frequency response

LP filter

Vs Vout

Summingamplifier

GLP = −1ωLP = 4π 104 rad/s

LP filterGLP = −1

ωLP = 4π 104 rad/s

G = −50HP filterGHP = −1

ωHP = 8π 104 rad/s

HP filterGHP = −1

ωHP = 8π 104 rad/s

Vout

+_

+_

R

RR

R

CLP

Rf

CLP

CHP +_

+_

+_

R

RR

RR

R

Vs

CHP

M [dB]

ω (rad/s)103 104 105 106 1070

10

20

3034

40

2-stage LP filterωLP = 1.26 105 rad/s

2-stage HP filterωHP = 2.5 105 rad/s

Figure 9-28: Bandreject filter of Example 9-12.

ωLP = 2π × 2 × 104 = 4π × 104 rad/s, and two identicalhighpass filters with ωHP = 8π × 104 rad/s. To minimizeperformance variations among identical pairs, identicalresistors will be used in all four units (Fig. 9-28(b)), whichmeans that they all will have unity gain. The overall gain of−50 will be provided by the summing amplifier.


9-8 APPLICATION NOTE: MODULATION AND THE SUPERHETERODYNE RECEIVER 445

Somewhat arbitrarily, we select R = 1 k�. From

ωLP = 4π × 104 = 1

RCLPCLP = 7.96 nF � 8 nF,

ωHP = 8π × 104 = 1

RCHPCHP � 4 nF.

The value of Rf is specified by the gain of the summing amplifieras

G = −50 = −Rf

RRf = 50 k�.

The transfer function of the bandreject filter is given by

H(ω) = G[H2LP + H2

HP]

= −50

[(1

1 + jωRCLP

)2

+(

jωRCHP

1 + jωRCHP

)2]

= −50

[(1

1 + jω/4π × 104

)2

+(

jω/8π × 104

1 + jω/8π × 104

)2]

.

9-8 Application Note: Modulation andthe Superheterodyne Receiver

9-8.1 Modulation

In the language of electronic communication, the term signalrefers to the information to be communicated between twodifferent locations or between two different circuits, and theterm carrier refers to the sinusoidal waveform that carriesthe information. The latter is of the form

vc(t) = A cos 2πfct, (9.95)

where A is its amplitude and fc is its carrier frequency.The sinusoid can be used to carry information by varyingits amplitude—in which case A becomes A(t)—whilekeeping fc constant. In the example shown in Fig. 9-29(a),multiplication of the signal waveform by the sinusoidalcarrier generates an amplitude-modulated (AM) carrierwhose envelope is identical with the signal waveform.Alternatively, we can apply frequency modulation (FM)by keeping A constant and varying fc in a fashion thatmimics the variation of the signal waveform, as illustrated byFig. 9-29(b). FM usually requires more bandwidth than AM,but it is also more immune to noise and interference, therebydelivering a higher-quality sound thanAM. Many other typesof modulation techniques also are available, including phasemodulation and pulse-code modulation.

9-8.2 The Superheterodyne Receiver

Let us assume the signal vs(t) in Fig. 9-29(a) is an audiosignal and the carrier frequency fc = 1 MHz. Let us alsoassume that the signal was used to generate an amplitude-modulated waveform, which was then fed into a transmitantenna. After propagating through the air along manydifferent directions (as dictated by the antenna radiationpattern), part of the AM waveform was intercepted bya receive antenna connected to an AM receiver. Priorto 1918, the receiver would have been a tuned-radiofrequency receiver or a regenerative receiver, both of whichsuffered from poor frequency selectivity and low immunityto noise. In either case, the receiver would have demodulatedthe AM signal by suppressing the carrier and preservingthe envelope, thereby retrieving the original signal vs(t)

(or more realistically, some distorted version of vs(t)).To overcome the shortcomings of such receivers, EdwinArmstrong introduced the heterodyne receiver in 1918 byproposing the addition of a receiver stage to convert thecarrier frequency of the AM signal fc to a fixed lowerfrequency (now called the intermediate frequency fIF)before detection (demodulation). [Armstrong also inventedfrequency modulation in 1935.] The superheterodyneconcept proved to be one of the foundational enablers of20th-century radio transmission. It is still in use in mostAM and FM analog receivers, although it slowly is gettingsupplanted by the concept of software radio (Section 9-8.4).

The spectrum of M [dB] = 20 log |H| is displayed inFig. 9-28(c).

Exercise 9-15: The bandreject filter of Example 9-12 usestwo lowpass-filter stages and two highpass-filter stages.If three stages of each were used instead, what would theexpression for H(ω) be in that case?

Answer:

H(ω) = 50

[(1

1 + jω/4π × 104

)3

+(

jω/8π × 104

1 + jω/8π × 104

)3]

.

(See )


PROBLEMS 517

is(t)

t

6 A

6e−2t

00

Figure P10.34: Waveform for Problem 10.34.

∗10.35 The current source shown in the circuit of Fig. P10.35 isgiven by the displayed waveform. Determine vout(t) for t ≥ 0,given that R1 = 1 �, R2 = 0.5 �, and L = 0.5 H.

is(t)

t

1.5 A

0.5 A1 A ω = 4 rad/s

+

_

L

R2R1is(t) vout(t)

Figure P10.35: Circuit for Problem 10.35 and current waveform forProblems 10.35 and 10.36.

10.36 If the circuit shown in Fig. P10.36 is excited by thecurrent waveform is(t) shown in Fig. P10.35, determine i(t)

for t ≥ 0, given that R1 = 10 �, R2 = 5 �, and C = 0.02 F.

is(t)

i(t)

R1 R2C

Figure P10.36: Circuit for Problems 10.36 to 10.38.

10.37 If the circuit shown in Fig. P10.36 is excited by currentwaveform is(t) = 36te−6t u(t) mA, determine i(t) for t ≥ 0,given that R1 = 2 �, R2 = 4 �, and C = (1/8) F.

10.38 If the circuit shown in Fig. P10.36 is excited by a currentwaveform given by is(t) = 9te−3t u(t) mA, determine i(t) fort ≥ 0, given that R1 = 1 �, R2 = 3 �, and C = 1/3 F.

10.39 The circuit shown in Fig. P10.39 first was introducedin Problem 5.62. Then, a time-domain solution was sought forvout1(t) and vout2(t) for t ≥ 0, given that vi(t) = 10u(t) mV,VCC = 10 V for both op amps, and the two capacitors had nochange prior to t = 0. Analyze the circuit and plot vout1(t) andvout2(t) using the Laplace transform technique.

vi vout1 vout2

+

_+_

Vcc = 10 VVcc = 10 V

5 kΩ

4 μF5 μF

1 MΩ

Figure P10.39: Circuit for Problems 10.39 and 10.40.

10.40 Repeat Problem 10.39 retaining all element val-ues and conditions but changing the input voltage tovi(t) = 0.4te−2t u(t).

Section 10-8: Transfer Function and Impulse Response

10.41 A system is characterized by a transfer function givenby

H(s) = 18s + 10

s2 + 6s + 5.

Determine the output response y(t), if the input excitation isgiven by the following functions.

(a) x1(t) = u(t)

∗(b) x2(t) = 2t u(t)

(c) x3(t) = 2e−4t u(t)

(d) x4(t) = [4 cos 4t] u(t)

10.42 When excited by a unit step function at t = 0, a systemgenerates the output response

y(t) = [5 − 10t + 20 sin 2t] u(t).

Determine (a) the system transfer function and (b) the impulseresponse.


11-2 FOURIER SERIES REPRESENTATION 531

(a) M waveform

f(t)

t−1−2−3−4−5 10

03 4 5

1

2

0.5

(b) Amplitude spectrum

ωω0 2ω0 3ω0 4ω0 5ω0

0.2

0.43

0.375

0.1

0.23

0.12

0

0.50.4

0.3

0.1

An

A1

A2A3 A4 = 0 A5

a0 (d) 5 terms

(f) 1000 terms

Gibbsphenomenon

Amplitude

1000

10

5

(e) 10 terms

−0.2

0.5

1

0

−0.2−5 −4 −3 −2 −1 0 1 2 3 4 5

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5 −4 −3 −2 −1 0 1 2 3 4 5

0.5

1

0

0.5

1

0−0.2

Figure 11-6: Plots for Example 11-3.

over the positive half period is:

f (t) ={

12 (1 + t) 0 ≤ t ≤ 1 s,

0 1 ≤ t ≤ 2 s.

Application of Eq. (11.31) yields:

a0 = 2

T

T/2∫0

f (t) dt = 2

4

1∫0

1

2(1 + t) dt = 0.375,

an = 4

T

T/2∫0

f (t) cos nω0t dt = 4

4

1∫0

1

2(1 + t) cos nω0t dt

= 2

nπsin

nπ

2+ 2

n2π2

(cos

nπ

2− 1

),

and

bn = 0.

Since bn = 0,

An = |an| φn ={

0 if an > 0,

180◦ if an < 0.


11-5 FOURIER TRANSFORM 543

−4 −2 00

1

2

1.5

0.5

2 4nω0

|cn|Waveform Spectrum

T = 5 s

0

1

2

1.5

0.5

−4 −2 0 2 4nω0

|cn|

T = 10 s

τ = 1 s

T = 10 s

0

1

2

1.5

0.5

−4 −2 0 2 4nω0

|cn| T = 20 s

τ = 1 s

T = 20 s

τ = 1 s

T = 5 st (s)

t (s)

t (s)

Figure 11-12: Line spectra for pulse trains with T/τ = 5, 10, and 20.

(a) When its argument is zero, the sinc function is equal to 1,

sinc(0) = sin(x)

x

∣∣∣∣x=0

= 1. (11.55)

Verification of this property can be established by applyingl’Hopital’s rule to Eq. (11.54) and then setting x = 0.

(b) Since sin(mπ) = 0 for any integer value of m, the same istrue for the sinc function,

sinc(mπ) = 0. (11.56)

(c) Because both sin x and x are odd functions, their ratio isan even function. Hence, the sinc function possesses evensymmetry relative to the vertical axis. Consequently,

cn = c−n. (11.57)

Evaluation of Eq. (11.53) with A = 10 leads to the line spectradisplayed in Fig. 11-12. The general shape of the envelopeis dictated by the sinc function, exhibiting a symmetricalpattern with a peak at n = 0, a major lobe extending betweenn = −T/τ and n = T/τ , and progressively smaller amplitudelobes on both sides. The density of spectral lines depends onthe ratio of T/τ , so in the limit as T → ∞, the line spectrumbecomes a continuum.

11-5.2 Nonperiodic Waveforms

In Example 11-8, we noted that as the period T → ∞ theperiodic function becomes nonperiodic and the associated linespectrum evolves from one containing discrete lines into acontinuum. We now will explore this evolution in mathematicalterms, culminating in a definition for the Fourier transform of


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