Chapter II Circuit Analysis Fundamentals - Texas...

ELEN-325. Introduction to Electronic Circuits: A Design-Oriented Approach Jose Silva-Martinez and Marvin Onabajo

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Chapter II

Circuit Analysis Fundamentals

From a design engineer’s perspective, it is more relevant to understand a circuit’s operation and limitations than

to find exact mathematical expressions or exact numerical solutions. Precise results can always be obtained

through proper circuit simulations and through numerical analysis with software programs such as MATLAB or

Maple. However, these tools cannot design the circuit for you, especially if you are dealing with analog circuits.

Even though design automation programs can synthesize complete digital circuits, and progress is being made

toward automating analog circuit design, engineers will still have to maintain knowledge about the functionality

of circuits and systems. Hence, this chapter will review and introduce some basic circuit analysis methodologies

that can be applied to gain insights into performance characteristics and design trade-offs. Generally, the

approach will be to identify and utilize appropriate simplifications that enable approximate analysis, which is an

essential skill when making initial component parameter selections for complex circuits and when performing

optimizations.

Circuit analysis techniques are of fundamental importance when determining the main parameters of an electronic

system. Even though software simulators are extremely useful as complementary tools, they cannot compensate for

a lack of theoretical understanding on the part of the designer. Therefore, an effective system design procedure

begins with selecting the most suitable active devices (transistors, diodes, etc.) and passive devices (e.g. resistors,

capacitors, inductors) along with power supplies, and signal sources. Next, the proper component interconnections

have to be defined, and the device parameters have to be chosen together with the operating point for each of the

active devices in order to realize the desired electronic operation. These steps require theoretical knowledge as well

as experience. For this reason, fundamentals of circuit analysis are revisited in this chapter before focusing more on

practical examples throughout the book.

Special attention is devoted to the physical interpretation of the results as well as the main properties of the basic

circuit components and topologies that reoccur frequently in complex circuits. It is important to investigate circuits

with the proper analysis techniques, but for an engineer it is even more significant to learn how to apply the results

during the design of efficient and reliable electronic systems. Finally, designs should be verified and completed by

optimizing component values through accurate simulations with more realistic and detailed models of the

components. Before we discuss the properties of basic electronic networks, it is necessary to introduce some

essential definitions for the analysis of signal components, system transfer functions, decibel notations, and

magnitude/phase responses.

II.i. DC and AC signals

A typical data acquisition system, such as the one shown in Figure 2.1, consists of a sensor (transducer), a

preamplifier, an analog filter, and the signal processor that typically includes an analog-to-digital converter. The

transducer detects the physical quantities to be measured and processed (temperature, pressure, glucose level,

frequency, wireless signal, etc.). Usually, the sensor’s output is a small signal (in the range of microvolts to

millivolts, i.e., in the range of 10-6 to 10-3 volts) and it must be amplified to fit within the linear input amplitude

range of the analog-to-digital converter. Since the desired signal is typically accompanied by undesired information

and random noise, the signal may be “cleaned” through a frequency-selective filter that removes most of the out-of-

band contents (at undesired frequencies) before the sensed signal is converted into a digital format and further

processed through dedicated software.

TransducerFilter (noise

removal)Preamp

Analog Processing Blocks

Electrical Signal

Digital

Signal

Processor

Digital Processing BlocksBiological

System

Analog-to-

Digital

Converter

Biological Signal

Fig. 2.1. Front-end of a typical bio-electronics system.


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The electrical signals at the output of the transducer are normally composed of two components: the DC (direct

current) and AC (alternating current) signals. As shown in Figure 2.2, the DC component is a time-invariant

quantity, while the AC component is a time-variant quantity and usually this AC component contains the relevant

information to be processed.

You will learn that the signals processed in the amplifiers also contain both DC and AC components. The general

expression for signals such as the one that appears at the output of the transducer in Figure 2.1 can be denoted as:

sAC(t) = SDD + sac(t) (2.0)

where SDD and sac denote the DC and AC signal components, respectively. Before we discuss the manipulation

techniques of these signals, let us define some of the nomenclature based on common conventions in electronics

literature. The following notations are used in this text to identify the nature of the signal components:

CAPITALCAPITAL labels represent only the DC component; e.g. SDD = 10V; IX = 2A.

lowercaselowercase stands for the AC component only; e.g. sac(t) = 2∙sin(t + ) V; isignal(t) = 2∙ej(t + ) A.

lowercaseCAPITAL stands for the combination of DC and AC components; e.g. sAC = SDD + sac(t); iSIGNAL = IX + isignal .

An example of individual signals and their combination is illustrated in Figure 2.2.

Time

DC component

AC component

Amplitude

Time

AmplitudeDC + AC

SDD

sac(t)

sAC(t) = SDD+sac(t)

0 0

(a) (b)

Fig. 2.2. Time domain signals: a) standalone DC and AC signals, b) combination of AC and DC signals.

Although signals found in most real world scenarios are not periodic, the analysis and design of electronic circuits is

conducted based on periodic signals (sinusoidal waveforms, pulse train, triangular waveforms, modulated

waveforms, etc.) because many waveforms can be approximated with summations of periodic signals. The case of

sinusoidal waveforms is especially interesting because periodic waveforms can be represented by a Fourier series

using sinusoids as a basis of functions. It is well known that many practical signals are continuous-time, periodic

with period T, and real functions f(t) are defined for all t; which allows representation with a simplified Fourier

series in the following form1:

T

tncosC)t(fn

n

200

, (2.1)

1 This is a simplified form of the Fourier series; the general form is

T;eC)t(f

n

tnj

n

2

00


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where 0 (2/T = 2f) is the fundamental frequency component in radians/sec used for series expansion. In practice,

the f(t) has to be expressed as a linear combination of sine and cosine waveforms, but to simplify our discussion let

us ignore the sine functions. Cn is the nth Fourier coefficient, and it is computed as follows:

dtetfT

C

Tt

t

tjn

n

0

0

01

. (2.2)

In this introduction, Equation 2.1 is a simplification of real signal representation with the intention to emphasize that

the signal’s spectrum often has sinusoidal components at frequencies which are multiples of 0. Furthermore, the

analytical expressions for the signals processed by the electronic circuits during the evaluation of relevant

performance characteristics typically have the same form as in Eq. 2.1. Examples on the use of Fourier series can be

found in a number of textbooks that deal with signal processing, circuit analysis, and circuit realizations.

Notice that C0 (n = 0) in 2.1 represents the DC component (average) of f(t) obtained with dttfT

C

Tt

t

0

0

10

.

Depending on the applications, this DC component may or may not contain desired information. The coefficient C1

(n = 1) is the fundamental (AC) component of the signal, which for practical purposes carries the most relevant

information to be processed in communication applications. The amplitude of the other terms, determined by Cn (n >

1) are known as the harmonic components and are rarely studied in introductory courses. However, they are relevant

and an important subject matter in advanced studies because they decide the quality of the signal. Specifically, the

high-order terms indicate the amount of signal distortion, and they generate undesired frequency components when

processed by nonlinear circuits, which limits performance.

Eqs 2.1 and 2.2 are beneficial because they allow one to avoid analyzing circuits for all possible input signals.

Instead, it is preferred to analyze them for the case of sinusoidal inputs and, from those results, infer system

behavior for any other kind of input signal. This is the so-called frequency domain analysis. In the frequency

domain, we usually extend the analysis of the signals from DC (= 0) up to very high frequencies. Even though the

frequency response of a system can be accurately computed and predicted, it is often not easy to find the exact

transfer function of a system, especially if the mathematical representation is complex. In other words, exact

mathematical derivations often imply too much effort for very little insight since a real electronic system may

contain millions of transistors! In these cases, it is important to have a good understanding of a circuit’s behavior

from a top level rather than being concerned with minor details that can be obtained from simulations. Few

observations about the properties of frequency responses and logarithmic functions lead to useful tools for

evaluating many complex analog circuits from a top-level perspective. Note, in some areas such as power

electronics, the time-domain analysis is more commonly used than the frequency domain analysis. Pulse and

impulse system responses are employed in those cases, and these topics will also be partially covered in the

following subsections.

II.2. Decibels and Bode Plots

i) Magnitude response. In electronics, the magnitude response is usually plotted in decibels (logarithmic scale),

making it easier to compare strong and weak signals. In general, it is often very convenient to use logarithmic scales

in cases where it is hard to visualize signal differences using linear scales. The magnitude in decibels (dB) of a

complex function f(x) is defined as:

)(log20)(log10 10

2

10 xfxfxfdB

, (2.3)

where |f(x)| stands for the magnitude of f(x). The expression 10∙log10 (|f(x)|2) is commonly used in communication

systems where the power of the signal (|f(x)|2) is employed, while the second definition 20∙log10 (|f(x)|) is more

popular in baseband electronic applications (e.g. biomedical, audio, and video frequencies) where the main

quantities of interest are either voltage or current. In any case, the use of decibel notation (logarithmic scale) is very

convenient when dealing with intricate transfer functions. Some relevant properties of the logarithmic function are

listed in Table 2.1. As an engineering student, you most likely already have a good background in logarithm

operations, but be sure to master these properties as they will be used extensively throughout your study of

electronics.


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Table 2.1 Fundamental properties of the logarithmic function.

Notice that:

10∙log10(1) = 0 dB shows that 0dB with logarithmic scale corresponds to unity gain with linear scale.

The magnitude of 2-1/2 (=0.7071) corresponds to - 3dB. The frequency at which the gain of a circuit is reduced

by 3dB is often referred to as 3dB-freuncy (f3dB), cutoff frequency, or corner frequency.

The multiplication of functions in linear scale is equivalent to addition of logarithmic functions. In many cases,

it is a lot easier to manipulate logarithmic equations as well as to get a better intuition about the characteristics

of a system through plots of the transfer function in logarithmic scale.

The ratio of two quantities maps into the subtraction of two logarithms.

Exponents in linear expressions correspond to scaling factors in front of logarithms.

To appreciate the benefits of the logarithmic scale, let us consider the following complex function:

)/(1

1

22

1

2

1

AxjA

A

jxA

Axf

, (2.4)

where A1 and A2 are real numbers, and jx is the imaginary part of the denominator. Notice that j2 = -1 and that x is a

real variable which can take values in the range of {0 < x < }. A2 in the above equation defines the frequency of the

pole, which is a significant parameter because it corresponds to the frequency at which the gain (A1/A2) of this first-

order transfer function is reduced by 3dB. The squared magnitude (power) of this function can be expressed as:

22

2

2

12)(*)()(

xA

Axfxfxf

, (2.5)

where f*(x) is the complex conjugate of f(x). The magnitude response (plot of the magnitude of f(x) versus x) for

this function with A1 = A2 = 10 is depicted in Figure 2.3a using linear scale. You can observe that it is very difficult

to extract information from this plot. When x is small, the value of f(x) is close to unity, and it decreases rapidly as x

approaches 10. For example, it is very difficult to accurately predict the value of f(x) for x >> A2 = 10 based on this

plot. Of course, you can see that the function has a tendency to go to zero, but it would be hard to quantitatively

estimate how small the values are when x is large. The same transfer function is plotted with dB-scale on the y-axis

in Figure 2.3b, where the x-axis is in log10 scale. To obtain this dB-log plot, Eq. 2.5 is substituted into Eq. 2.3 prior

to plotting. Clearly, we can extract significantly more information by visual inspection of this plot, especially under

consideration of the properties in Table 2.1. For instance, the function is approximately flat from very small x-values

up to x = 5 [note: f(0) = 1 = 0dB], and the -3dB value occurs at x = A2 = 10 [note: f(10) = 0.707]. The gain of the

Function Properties

log10(1) 0

log10(10) 1

log10(xN) N∙log10(x)

log10(1/xN) -N∙log10(x)

10∙log10(1/2) = 20∙log10(2-1/2) -3.01

10∙log10(2) = 20∙log10(21/2) +3.01

log10(f∙g) log10(f) + log10(g)

log10(f/g) log10(f) - log10(g)

log10(10N) N

log10(AfX/BgX) log10(AfX) - log10(BgX)

= log10(A) + X∙log10(f) - log10(B) - X∙log10(g)


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transfer function at x = 1000 is -40 dB, which corresponds to a magnitude equal to 0.01. You can check this value

using your calculator and Equation 2.3.

0E+0 2E+2 4E+2 6E+2 8E+2 1E+3

x

0.00

0.20

0.40

0.60

0.80

1.00

Mag

nit

ude

.1E+0 1E+1 1E+2 1E+3

x

-40

-30

-20

-10

0

Mag

nit

ud

e (d

B)

(a) (b)

Figure 2.3. Magnitude response of the function 10/(10 + jx) using: a) linear-linear scale, b) dB-log scale.

To get more insight, let us more fully examine the previous transfer function. Taking advantage of the properties of

the logarithmic function, the magnitude of the transfer function given in Eq. 2.5 can be expressed in dB:

)(log*10)(log10 22

210

2

110 xAAxfdB

(2.6)

Notice that:

For x << A2 this equation can be approximated as 20∙log10(A1) - 20∙log10(A2) = 20∙log10(A1/A2), which is equal

to 0dB for the example case with A1 = A2 = 10.

For x = A2 = 10, the magnitude of f(x) becomes

|f(x)| = 10∙log10(A12) - 10∙log10(A2

2 ∙ 2)

= 20∙log10(10) - 10∙[log10(102) + log10(2)]

= 20∙log10(10) - 20∙log10(10) - 10∙log10(2)

= -10∙log10(2) = -3.01dB.

For x >> A2 equation 2.6 can be approximated as 10∙log10(A12) - 10∙log10(x2) = 20∙log10(10) - 20∙log10(x) dB.

Notice that the expression consists of a constant term and a negative term that decreases proportional to the -

20∙log10 function of x. If we use a log10 scale on the x-axis, this corresponds to a straight line with a slope of -

20, which agrees well with the plot shown in Figure 2.3b for x >> A2 = 10. If x = 1000, then |f(1000)| ≈ 20 -

20∙log10(103) = -40 dB, which again fits well with the visual inspection that can be made in Figure 2.3b.

The simple analysis above can be performed for any x value.

The transfer functions can be easily plotted in log10-dB scale after some additional observations. For frequencies

beyond the pole frequency (A2 in the previous example) |f(x)| is approximately equal to 20∙log10(A1) - 20∙log10(x)

dB; some values for the previous case are provided in Table 2.2. Beyond the frequency of the pole, a one-decade

increase on the x-axis corresponds to a gain change of -20 dB (equivalently, this can be referred to as 20dB

attenuation). Hence, the slope of the function with log10-scale on the X-axis is -20dB/decade. It can also be shown

that an increase by a factor of 2 (1 octave) on the x-axis corresponds to an attenuation of 6dB for |f(x)| with x >>

A2, leading to a slope of –6dB/octave. Scales in octaves are frequently used in music and medical equipment, and

they provide more resolution (change by a factor of 2 per division on the x-axis) compared to log10-scale on the x-

axis. In both cases, the scales of the axes make it easier to plot complex transfer functions and to identify relevant

properties through visual inspection of their magnitude responses.


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Table 2.2 Example values for the magnitude of the 20∙log10(x) (x < 1) function in dB.

x

10-1 -20

10-2 -40

10-3 -60

10-4 -80

10-5 -100

ii) The phase response of the first-order transfer function (solid line) displayed in Figure 2.4 requires a bit more

detailed analysis than for the magnitude response. A brief analysis is presented in Appendix A (to be added). It is

important to recognize that a complex expression can always be written in polar form: f(x) = |f(x)|ej, where is the

phase of f(x) defined as = tan-1[Imaginary(f)/Real(f)]. Due to the properties of the exponential function, it is

evident that the phase properties listed in the Table 2.2 hold; where R and Im represent the real and imaginary parts,

respectively.

Table 2.2 Relevant characteristics of complex numbers.

Function Evaluation

ImjRxf

RIm/tan

ImRxf

exfxf

f

j f

1

22

xgxf gfjexgxf

xg

xf

gfje

xg

xf

xgxgxg

xfxfxf

m

n

...

...

21

21

m

k

gk

n

i

fij

m

ne

xgxgxg

xfxfxf11

...

...

21

21

To plot the phase of a transfer function by hand, you must first evaluate the phase of each individual term and add

the phases of the functions in the numerator while subtracting the phase of the functions in the denominator. For

instance, the phase response of Eq. 2.4 is depicted (solid line) in Figure 2.4 (A1 = A2 = 10). Correspondingly, the

phase shift of the complex function can be assessed from the following expression:

)/(tan0][][

][][)]([

2

1

22 AxjxAPhaseAPhase

rDenominatoPhaseNumeratorPhasexfPhaseradians

. (2.7a)

This equation is giving the phase in radians. To convert the phase to degrees, you must multiply the value in radians

by 360/2≈ 57.3, yielding

radiansdegrees)]([3.57)]([ xfPhasexfPhase . (2.7b)

The phase plot shown in Figure 2.4 is a nonlinear function. Some observations will allow us to approximate this

function with simpler but quite useful piecewise linear functions:

The phase at x = 0 is 0 degrees because the function is real and positive for this value of x. Negative real

functions have an initial phase of radians (or 180 degrees) at x = 0.


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The phase shift of f(x) at the location of the pole x = A2 is -45 degrees.

The phase at x = is - radians (or -90 degrees).

If we evaluate the derivative of the phase response at the pole’s location (x = A2), it takes on the larger value of

more than -45 degrees/decade.

Figure 2.4. Phase response for the transfer function in Eq. 2.4 with A1 = A2 = 10.

As a specific example, let us approximate the phase response (Eq. 2.7a) of the transfer function by a piecewise linear

function such that below x = 0.1∙A2 = 1 where the phase is approximated as 0 degrees, and above x = 10∙A2 = 100, it

is approximated by -90 degrees. The phase around x = A2 = 10 (the frequency of the pole) is approximated to have a

slope of -45 degrees per decade. Both the exact phase response (solid trace) and the piecewise linear approximation

(dashed trace) are shown in Figure 2.4. The largest error introduced by the approximation is +/-5.7 degrees, which is

acceptable for most hand calculations.

The previously discussed properties are also valid for the general case. For instance, consider the following complex

function:

xjAA

Axf

32

1)(

. (2.8)

This equation can be rearranged to:

jxA

A

A

A

xf

3

2

3

1

)( . (2.8b)

Therefore, this equation becomes similar to Eq. 2.5., and can be expressed in dB as:

2

2

3

2

2102

3

2

110 log10log10)( x

A

A

A

Axf

dB. (2.9)

Now, the transfer function can be easily sketched by examining each part separately for the different ranges that

variable x can be in while noting that:


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3

210

3

110

2

102

3

2

110

3

22

210

2

110

3

22

210

2

1102

3

2

2102

3

2

110

log20log20loglog10

3loglog10

log20log20log10log10

)(

A

Axifx

A

Ax

A

A

A

AxifAA

A

AxifAA

A

A

A

A

xfdB

(2.10)

It is straightforward to plot the approximate transfer function, as demonstrated in Figure 2.5. Using a piecewise

linear approximation is a popular method that is used to get an idea of a circuit’s frequency response and its

important characteristics. In this example, the magnitude of the transfer function can be approximated as constant

[=10∙log10(A12) - 10∙log10(A2

2) = 20∙log10(A1/A2)] up to x = A2/A3. After this frequency, the magnitude of the transfer

function decreases with a slope of –20 dB/decade. A maximum error of -3 dB occurs at x = A2/A3 with the linear

approximation. For most of our practical applications, we will use the piecewise linear approximation unless it is

critical to make more accurate calculations. It should be evident that the parameter A2/A3 plays an important role in

shaping the transfer function, since this parameter is in fact the location of the pole for which further implications

are discussed in the following sections.

3 dB

exact

Appoximation

A2/A3

x(log)

20·log10(A1/A2)

-20 dB/decade

Fig. 2.5. Plot of the transfer function described by Eq. 2.9 with its piecewise linear approximation. (The maximum

error of -3 dB occurs at the frequency (A2/A3) of the pole.)

Since f(x) is complex, we have to consider its phase response as well. Considering f(x) given by Eq. 2.8, the phase

response is

x

A

A

AA

xAPhasexfPhase

2

31

32

1

1 tantan][)]([. (2.11)

This function is similar to the one plotted in Figure 2.4. For x = 0, the phase shift presented by the circuit is 0

degrees. It is evident from Eq. 2.11 that the phase shift is -0.785 radians (= -45 degrees) for x = A2/A3; the phase

shift when x approaches infinity is -1.57 radians (-90 degrees). Notice that the phase shift is close to the limits of the

piecewise linear approximation (around 0º when x ≈ 0.1∙A2/A3, and around -90º when x ≈ 10∙A2/A3). The transition

from 0 to -90 degrees takes place within approximately two decades around the system’s pole x = A2/A3. The slope

of the phase around a pole is approximately –45 degrees/decade, and the maximum error is +/-5.7 degrees.

Additionally, the phase shift is -45 degrees at the pole (x = A2/A3).

iii) Magnitude and phase of a general function. The complex transfer function might consist of a combination of

poles and zeros; hence, another benefit of the logarithmic scale is that it helps to distinguish between the two cases.

In the previous examples, we found that a pole reduces the magnitude of the transfer function with a roll-off of -20


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dB/decade beyond the pole’s location. Furthermore, the pole introduces negative phase shift. A zero, on the other

hand, increases the transfer function with a slope of +20 dB/decade and introduces positive phase shift.

Let us consider a first-order transfer function with a pole and a zero:

jxA

A

jxA

A

A

A

xjAA

xjAAxf

4

3

2

1

4

2

43

21)(. (2.12)

A1/A2 and A3/A4 are termed the zero and pole of the complex function, respectively. The zero corresponds to the

positive value of x that solves Xz + A1/A2 = 0 in the numerator, where Xz = jxz. This yields Xz = -A1/A2. Similarly,

the pole corresponds to the value of Xp such that Xp + A3/A4=0 in the denominator; hence, Xp = - A3/A4.

The magnitude (in dB) of f(x) can be expressed as follows:

2

2

4

310

2

2

2

110

2

4

210 log10log10log10)( x

A

Ax

A

A

A

Axf

(dB) . (2.13)

Similar to the case of a single pole function, the three terms of this expression can be approximated for different

ranges of x. As an example, let us consider A1/A2 < A3/A4. In this case, the frequency of the zero is lower than that of

the pole, resulting in a zero location on the x-axis that is to the left of the pole location. From Eq. 2.13 it follows that

.log10

,3log10log10log10

,log10log10log10

,3log10

,log10log10log10log10

)(

4

3

2

4

210

4

3

2

4

310

2

2

2

110

2

4

210

4

3

2

1

2

4

310

2

2

2

110

2

4

210

4

3

2

1

2

3

110

4

3

2

1

2

3

110

2

4

310

2

2

110

2

4

210

)(

A

Axif

A

A

A

Axif

A

Ax

A

A

A

A

A

Ax

A

Aif

A

Ax

A

A

A

A

A

A

A

Axif

A

A

A

A

A

Axif

A

A

A

A

A

A

A

A

xfdB

.

(2.14)

These results can be plotted by using the piecewise linear approximation shown in Figure 2.6.

3 dB

A2/A1

x(log)20·log10(A1/A3)

20 dB/decade

A3/A4

3 dB

20·log10(A2/A4)

Fig. 2.6. Magnitude response of a function with a pole-zero pair where zero location < pole location.


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To plot phase response of transfer functions, we can use the aforementioned properties. For the transfer function

from the previous example, the phase in radians can be obtained through the following rearrangements:

.tantan

)(

3

41

1

21

4

3

2

1

4

3

2

1

A

xA

A

xA

jxA

APhasejx

A

APhase

jxA

A

jxA

A

PhasexfPhase

(2.15)

As evident above, the phase contributions of the zero and pole can be observed independently, and the overall phase

response is the superposition of the individual responses, which is plotted in Figure 2.7 for the case in which the

zero is located at a significantly lower frequency than the pole (A1/A2 << A3/A4). Since the zero is located to the left

of the pole, the phase starts at 0 degrees and increases due to the impact of the zero, reaching a phase shift

approximately equal to +45 degrees at the location of the zero. The phase goes up to +90 degrees at x = 10∙A1/A2 <<

A3/A4, and remains constant until the pole begins to affect the phase response by introducing negative excess phase

that is noticeable at x-values around the pole location A3/A4. At the pole value (x = A3/A4) the phase shifts due to the

zero is +90 degrees while the pole contributes with -45 degrees, leading to an overall phase equal to +45 degrees. At

very high frequencies. The phase shift is zero because the +90 degrees introduced by the zero is cancelled by the -90

degrees from the pole. In the next subsection, we will consider the effects that occur when poles and zeros are

located close to each other.

A2/A

1

x(log)0 degrees

A3/A

4

90 degrees

45 degrees

Fig. 2.7. Phase response of the first-order transfer function under examination.

iv) Bode plots: General case. A Bode plot is a well-known graphical representation of magnitude and phase plots

vs. frequency in logarithmic scale. In most practical cases, the characteristic equation of an electronic system

consists of several terms that model individual sections connected in a cascade forming a multi-stage system.

Although the sections may need a more complex representation, let us assume that each stage can be described by a

first-order equation so that the electronic system can be mathematically represented as:

M

i i

ii

N

k k

kk

M

i

ii

N

k

kk

jxD

CD

jxB

AB

xjDC

xjBA

xf

1

1

1

1)(,

(2.16)


- - 11

where the symbol denotes the multiplication of the factors. As will be evident soon, this factorization is very

convenient, especially for low-pass transfer functions. Using the properties of the logarithmic function, the

magnitude (in dB) of the generalized transfer function yields:

M

i

i

N

k

k

dBx

C

A

xf

1

2

1

2

10)(0log10)( , (2.17a)

M

i i

iN

k k

k

M

i

i

N

k

k

dBx

D

Cx

B

A

D

B

xf1

2

2

10

1

2

2

10

1

2

1

2

10)(log10log10log10)( . (2.17b)

Note that 2.17a is only valid at DC (x = 0), while 2.17b holds at any frequency. Generally, the complete expression

must be considered because we do not have information about the location of the poles and/or zeros. However,

simplifications can be made when relative pole/zero locations are known.

Also notice that we normalized the functions with respect to the zeros and poles to make an analysis by inspection

easier. Some insightful characteristics involving the above equations follow:

1. By setting x = 0 in Eq. 2.17b, the DC gain of the function (magnitude of the transfer function at x = 0) can

be obtained. This is the starting point of the overall magnitude response on the y-axis, and it depends on the

coefficients Ak and Ci since other terms will cancel each other.

2. The zeros and poles must be identified. After frequency x = 0, each zero of the transfer function will

increase the magnitude at a rate of 20 dB/decade. Similarly, each pole will decrease the magnitude of the

function by -20dB/decade after the pole’s location on the x-axis. Since a dB-scale is used on the y-axis

together with logarithmic scale on the x-axis, the aforementioned effects of poles and zeros on the

magnitude can be added algebraically.

3. At a given x-value, the roll-off (slope) of the magnitude can be easily obtained by the following equation:

dB20xbelowpoles of #xbelowzeros of #xx f of slope 000 (2.18)

4. Finally, you can evaluate the function for x → to see that the final value depends on several factors:

a. If the number of zeros is greater than the number of poles, the transfer function goes to infinity

(positive infinite in dB) as x → .

b. If the number of poles is greater than the zeros, which is the typical case in electronics, the final

value is zero (negative infinite in dB) when x → .

c. In case the number of poles and zeros are equal, then the magnitude’s final value is a function of

the coefficients Bk and Di.

Similar rules can be established for the phase response. The general phase equation can be written as follows:

M

i i

iN

k k

k

M

i i

iN

k k

k

C

xD

A

xB

jxD

CPhasejx

B

APhasexfPhase

1

1

1

1

11

tantan

)]([

.

(2.19)


- - 12

Each zero introduces +45 degrees at the location of the zero and roughly +90 degrees after one decade beyond the

zero’s location, while each pole introduces -45 degrees at its location and -90 degrees after about a decade.

According to Eq. 2.19, the phase contribution of the poles and zeros should be algebraically added.

Example: For a better understanding of the use of these rules, let us consider the following exemplary equation with

two poles and two zeros:

xj

xj

xj

xjxf

20020000

501000

1001000

10500)( (2.20)

Basic algebra allows us to put this equation in a more convenient form:

jx

jx

jx

jxxf

100

20

10

50

20000

500)( . (2.21)

The DC gain can be identified as |f(0)| = 0.025 (-32dB). Further inspection reveals that the zeroes of the system are

located at xZ1 = 20 and xZ2 = 50, while the poles are placed at x P1 = 10 and x P2 = 100. In decibels, this function yields

22

10

22

10

22

10

22

10)(100log1010log1050log1020log1032)( xxxxdBxf

dB . (2.22)

Figure 2.8 illustrates the magnitude response that is sketched using a piecewise linear approximation based on the

following observations of the above expression: For x < 10, the magnitude is roughly constant and equal to -32 dB

(since 0 zeros and 0 poles below x = 10 give us a slope of 0 according to Eq. 2.18). Within the interval 10 < x < 20,

the gain decreases with a slope of -20 dB/decade due to the pole located at x = 10 (1 pole below x = 20). At x = 20,

we have the effect of the pole (-20 dB/decade) and the effect of the zero at x = 20, which contributes to a slope of

+20 dB/decade after this frequency. Thus, in the interval 20 > x > 50, the effect of the pole and zero cancel each

other and the gain remains constant, since 1 zero and 1 pole lead to a slope of 0 in this interval. The zero located at x

= 50 comes into the picture for x > 50, adding an extra increment of +20 dB/decade. Finally at x > 100, all poles and

zeros determine the shape of the overall function. Since the number of poles (2) and zeros (2) is the same in this

case, the slope of the transfer function for x > 100 is zero. The final value of the transfer function is obtained by

evaluating the original function at = , which for this example results in 10∙50/(100∙200) = 0.025 (-32 dB).

x(log)-10

|f(x)| (dB)

10 102

-20

-40

-32

Magnitude

(dB)

Fig. 2.8. Magnitude response of the complex transfer function given by Eq. 2.21.

Let us now consider the phase response of the previous example (Eq. 2.21). You are strongly encouraged to check

your work so as to make sure that the phase of this function can be written as


- - 13

100tan

10tan

50tan

20tan 1111 xxxx

xfPhase . (2.23)

The piecewise linear approximation for this phase response is shown in Figure 2.9. The DC phase shift is obtained

by evaluating it from the original transfer function at x = 0, yielding 0 degrees in this case. After this frequency, we

have to consider the phase contributions of the poles and zeros and add them together. The pole located at x = 10

introduces phase shift that goes from zero degrees at x-values below 1, to -45 degrees at x = 10, and reaching -90

degrees at x > 100. Its effect can be approximated by a linear phase variation with a roll-off of -45 degrees/decade

within the range 1 < x < 100. Similarly, the zero located at x = 20 introduces 0 degrees for x < 2, +45 degrees at x =

20, and +90 degrees at x > 200. The zero located at x = 50 raises the phase at a rate of +45 degrees/decade starting at

x = 5 and ending at x = 500. Finally, the last pole adds a roll-off of -45 degrees/decade in the range 10 < x < 1000.

Figure 2.9 also includes the superposition of the individual phase contributions that form the overall phase response.

Since piecewise linear approximations of phase responses are sometimes more obscure than those of magnitude

responses, you will have to plot several example in order to master bode plots.

1x(log)

0

10 10220 50

-45

-90

45

90

Phase (f(x))

1st pole 2nd pole

1st zero 2nd zero

Fig. 2.9. Phase response approximation for the complex function given by Eq. 2.21.

II.3. Frequency Response of First-Order Systems.

Steady-state analysis: resistive, capacitive, and inductive impedances. When applying fundamental circuit

theory, the voltage-current relationship for a resistor is dictated by Ohm’s law. Fortunately, most of the resistors in

micro-electronic circuits can be approximated as linear devices even though they are not perfectly linear. Hence, it is

easy to describe their electrical characteristics with simple algebra:

RR iRv (2.24)

where vR is the voltage drop across a resistor while current iR flows through it. The voltage polarity and current

direction are defined in Figure 2.10a. The impedance of a device is defined as the ratio of the voltage applied to its

terminals divided by the resulting current flow, which in case of a resistor gives:

Ri

vZ

R

RR . (2.25)

A resistor’s impedance is real and positive; therefore, the voltage and current are in phase with each other. Since

resistance and impedance are equivalent for resistors, the terms are commonly used interchangeably. But when

dealing with capacitors or inductors, be aware that the impedance is frequency-dependent and not equal to the

resistance.

Current and voltage variables associated with a capacitor (Figure 2.10b) are related by the following fundamental

relationship:


- - 14

+

VR

-

iR

+

VC

-

iC

+

VL

-

iL

(a) (b) (c)

Fig. 2.10. Passive component models: a) resistor, b) capacitor, and c) inductor.

dt

dvCi c

c . (2.26)

In this expression, C is the capacitance, usually measured in Farads (F). Some capacitors exhibit nonlinear

capacitance characteristics; however in this course, we will assume that capacitors remain constant (time, voltage,

and current invariant) under all operating conditions. When expressing the AC voltage applied at the capacitor

terminals in the general complex form vc = VC∙ejt, the expression for the current flowing through it can be derived

from Eq. 2.26 as

cc vCji . (2.27)

Thus, contrary to a resistor in which voltage and current are in phase, the voltage across the capacitor’s plates and

the current flowing through the capacitor are 90 degrees out of phase. Another remarkable property evident from

2.27 is that the current flowing between the capacitor plates increases proportionally with the frequency of operation

(signal frequency), leading to very large values at high frequencies even if the voltage variations are small. Based on

Eq. 2.27, it is obvious that the impedance of an ideal capacitor C is a function of frequency, and it is given by

sCCji

vZ

c

cc

11

(2.28)

where (=2f) is the signal frequency in radians/sec. It follows that the impedance of a capacitor is negative and

imaginary: ZC = 1/(jC) = -j∙(1/C). To simplify the algebra, usually the frequency variable is defined as s = j.

Notice that s is an imaginary variable.

At = 0, the magnitude of the capacitive impedance is infinite (open circuit). Consequentially, any DC voltage

applied at the terminals of the capacitor does not generate any current in steady state. For this reason, capacitors are

often utilized to “couple” electronic circuits together while maintaining different DC levels at the interconnected

inputs and outputs. In such practical applications, the capacitors are often referred to as “DC blocking” capacitors.

Some transient current is generated if a capacitor’s initial condition is different than the applied voltage, but this

current vanishes as soon as the capacitor is charged to the DC voltage across its terminals. It is also important to

recognize that the capacitive impedance decreases with frequency, and this low impedance affects the behavior of

electronic systems at very high frequencies. As an example, let us consider a 100pF capacitor (10-10 Farads). The

magnitude of its impedance is 107 at 1Krad/sec but drops down to 100 at 100Mrad/secand the capacitor

only has an equivalent impedance of 10 at 1Grad/sec. In most practical linear circuits, with the exception of some

special techniques such as switched-capacitor circuits, the effects of the capacitors in the range of a few picofarads

can be neglected at low frequencies (< 1MHz) because their impedance is usually very high compared to resistors.

However, you should check any assumptions before simplifying your circuit analysis by omitting capacitors in

equations. Before closing this section, it is important to note that the voltage and current related to the same

capacitor are phase-shifted because the phase of a purely capacitive impedance is -90 degrees.

Following a similar procedure as for the capacitor, the impedance of the inductor under steady state conditions can

also be found easily. The relationships between the quantities annotated in the ideal inductor model of Figure 2.10c

are given below.


- - 15

dt

diLv L

L (2.29)

where L is the inductance in henry (H). If the inductor’s current has the general complex form iL = IL∙ejt, then the

voltage across its terminals is

LL iLjv . (2.30)

This voltage generated across the inductor and the current flowing through it are +90 degrees out of phase. The

impedance of the inductor can be derived from Eq. 2.30 as

sLLji

vZ

L

LL . (2.31)

Contrary to the case of the capacitor, the inductor presents very small impedance across its terminals at low

frequency (ZL = 0 at DC) and very high impedance (open circuit) at very high frequencies.

Voltage Divider. The voltage divider consists of two impedances connected in series. Its output voltage is defined

as the voltage drop across one of the elements, which is a fraction of the input voltage. A typical voltage divider is

shown in Figure 2.11, in which impedances Z1 and Z2 are employed to reduce the input voltage vi to a lower output

voltage vo. By using basic circuit theory (KVL) it can be easily shown that the voltage across the grounded

impedance Z2 is given by the following expression:

21

20

ZZ

Z

v

v

i . (2.32)

If the nonzero impedances Z1 and Z2 are of the same type (either real or imaginary), the amplitude of the output

voltage v0 is smaller than the amplitude of the applied signal vi. Therefore, this topology causes signal loss, and it

can be referred to as an attenuator. In the most general case, Z1 and Z2 can be composed by several elements, leading

to a variety of possible transfer functions with different properties. The frequency response of some of these

functions is analyzed next.

Z1

Z2

vi v0

vi v0

R2

R1

(a) (b)

Fig. 2.11. Voltage divider: a) general scheme and b) resistive voltage divider.

Let us first consider the case of a simple resistive divider shown in Figure 2.11b. The voltage gain in this case is

21

20

RR

R

v

v

i . (2.33)

Since the resistors are ideally voltage and frequency independent elements, the voltage gain of the resistive voltage

divider is a real number with a magnitude less than unity and a phase shift of zero degrees for all frequencies. Thus,

the output signal is an attenuated replica of the incoming signal. Figure 2.12 shows the magnitude response of the

resistive voltage divider for different example resistor values. In theory the circuit does not have any frequency-


- - 16

dependent limitations, but parasitic capacitive and inductive elements are unavoidable in practice, which will affect

the magnitude and phase responses. For these shown simulation results, R1 = 1k and R2 = {100, 1.1 k, and

2.1k}. The voltage attenuation factors are 100/1100 (-20.8 dB), 1100/2100 (-5.6 dB), and 2100/3100 (-3.4 dB) for

the respective values of R2.

1E+1 1E+2 1E+3 1E+4 1E+5

Frequency (Hz)

-25

-20

-15

-10

-5

0

Volt

age

Gai

n (

dB

)

Fig. 2.12. Magnitude response of the resistive voltage divider.

It is important to emphasize that the attenuation factor reduces if the series resistance R1 is smaller than the

grounded resistance R2 because the input voltage is split between the two impedances, and the larger one absorbs

most of the signal. In fact, as R1 approaches zero, the voltage gain becomes close to unity (i.e., the output signal is

almost equal to the input signal). The output voltage (across Z2) is the difference between the input voltage and the

voltage drop across the series impedance Z1. Hence, the larger the series resistance Z2 is, the larger the voltage

drop across it, and the smaller the output voltage will be. Notice that if a voltage divider is formed by resistors, then

2

10

1R

R

vv i

. (2.33b)

First-order low-pass transfer function. To obtain more insights into the properties of electronic networks

containing capacitors, we will analyze the voltage divider with a capacitor and resistor. If Z2 is replaced by a

capacitor (C2) in the voltage divider of Figure 2.11 while keeping Z2 = R2, then the circuit becomes a low-pass filter

(Figure 2.13). Low-pass filters are frequently used to suppress undesired high-frequency signal components and

noise. In audio applications, for example, the signal bandwidth approximately is 20kHz, and received signals above

that frequency should be suppressed before the information is converted into digital format for further processing

and recording.

vi v0

R1+

vC2

-

+ vR1 -

C2

Fig. 2.13. First-order low-pass filter.

Using the expression for the capacitor’s impedance (Eq. 2.28) and the voltage divider relationship (Eq. 2.32), the

voltage gain of the resulting first-order system becomes

R1 = 1k, R2 = 2.1k

R1 = 1k, R2 = 1.1k

R1 = 1k, R2 = 100


- - 17

21

21

21

21

2

1

2

21

20

1

1

1

1

1

1

CRs

CR

CRj

CR

CjR

Cj

ZZ

Z

v

v

i

. (2.34)

It should be evident that the voltage gain has become a complex function of the frequency variable due to the

combination of the resistor (real impedance) and the capacitor (imaginary impedance). For very low frequencies

(≈ 0), the transfer function is positive and real (unity in this case). At the frequency of the pole (= p =

1/[R1C2]), the transfer function becomes equal to p/(p + jp) =1/(1 + j), and the magnitude of the transfer function

is equal to 0.7071 (-3 dB). On the other hand, the phase shift between output and input is close to zero at very low

frequencies and -45 degrees at = p. For frequencies beyond p = 1/(R1C2), the transfer function is dominated by

the factor 1/jR1C2), causing the magnitude of the transfer function to decrease as the signal frequency increases.

The magnitude of the transfer function reduces at high frequencies because the capacitive impedance is lower at

high frequency. This is explained by the fact that the value of the resistive impedance is greater than that of the

capacitive impedance beyond p. Consequently, most of the input signal appears across the larger impedance, which

is the resistor when >> p. The regions of operation for this example filter can be summarized as follows:

.1

log201

log20

,1

3

,1

0

log10

21

10

21

10

21

212

010

CRif

CR

CRifdB

CRifdB

v

v

i

(2.35)

The two regions of operation (pass-band with flat response and stop-band with roll-off response) can be identified

by observing the log-log plot depicted in Figure 2.14a for two example cases: p = 1/(R1C2) = 2∙300 rad/sec (fp =

300Hz) and p = 1/(R1C2) = 2∙10000 rad/sec (fp = 10kHz). Recall that the pole’s frequency p is determined by the

R1C2 product.

It is interesting to revisit the physical explanation for the behavior of this first-order low-pass circuit. For low

frequencies (<< 1/[R1C2]), the series resistance R1 is much smaller than the magnitude of the grounded capacitor

impedance given by 1/C2). Since the input voltage is split between the two elements (vin = vR1 + vC2) and one of

the impedances is much greater than the other one, most of the voltage will appear across the higher impedance.

Most of the input voltage is absorbed by the capacitor at low frequencies, leading to vC2 = vin at DC (unity gain and

zero phase shift). At = 1/(R1C2), the magnitudes of both impedances are equal, and the input voltage is equally

split between the two elements, resulting in |vR1| = |vC2| = 0.7vin since the current flowing through both elements is

the same. But, the resistance is real and the capacitive impedance is imaginary. At high frequencies, the impedance

of the capacitor reduces further, and most of the input signal is then absorbed by the series resistor. The higher the

frequency is, the smaller the capacitive impedance and the larger the attenuation factor will be. At very high

frequencies, the capacitor’s impedance becomes extremely small (practically zero) and effectively shortens the

output node to ground (i.e. the capacitor forms a short circuit at = ).


- - 18

1E+1 1E+2 1E+3 1E+4 1E+5

Frequency (Hz)

-50

-40

-30

-20

-10

0

Volt

age

Gai

n (

dB

)

Fig. 2.14a. Magnitude response of the first-order low-pass filter for two cases:

1/(R1C2) = 2∙300 and 1/(R1C2) = 2∙10000.

1E+1 1E+2 1E+3 1E+4 1E+5

Frequency (Hz)

-90.0

-67.5

-45.0

-22.5

0.0

Phas

e (D

egre

es)

Fig. 2.14b. Phase response of the first-order low-pass filter for the two example cases.

The phase response (in radians) of the low-pass filter circuit’s transfer function is obtained from 2.34 as

21

1

21

211 tan01

1

tan CR

CRj

CR

v

vPhase

i

o

. (2.35)

The corresponding phase plot is depicted in Figure 2.14b. As discussed in the previous subsection, the phase

variation is mainly allocated within two decades around the pole’s frequency (0.1P < <P). For the case P =

1/(R1C2) = 2∙10000 rad/sec (fP = 10kHz), the phase shift transitions from 0 to -90 degrees within the frequency

range of 1kHz - 100kHz, being exactly -45 degrees at fP = 10kHz.

Of course, the magnitude and phase plots are correlated with the time response of the circuit. If a sinusoidal signal is

applied at the input of the low-pass filter, which is a linear system, then the output will also be a sinusoidal function

but with a different amplitude and different phase. In case the frequency of the input signal is below the frequency of

the pole, the amplitude of the output signal is very close to the amplitude of the input signal, and the phase


- - 19

difference is small as well, as shown in Figure 2.15a. At the pole’s frequency, the amplitude of the output is 0.7∙vin

and the phase lag is -45 degrees (Figure 2.15b). For frequencies beyond 10 times the pole’s frequency, the phase lag

is close to -90 degrees and the magnitude of the output signal is very small as visualized in the Figure 2.15c.

Input signal Output signal

(a) (b) (c)

Fig. 2.15. Signals present at the input (continuous curve) and output (dashed curve) of the first-order low-pass

filter: a) < p, b) = p (the magnitude of the output signal is 0.7∙Vin and the phase shift is –45

degrees), and c) > p (the output signal is small and the phase shift is close to –90 degrees)

Before we continue plotting the transfer functions with Bode plot approximations, it is desirable to introduce another

useful parameter: the circuit’s time constant. The first-order low-pass filter is described by Eq. 2.34, and if the s

variable is used instead of j, the pole is located at

21

1

CRS p . (2.36)

Sp is a left-hand plane pole in the complex s-plane. It will be evident soon that it is convenient to use p = 1/(R1C2).

The circuit’s time constant is defined as the product R1C2, and the unit of this RC product is indeed a time-domain

quantity expressed in seconds. In general, if the circuits are composed by active elements, capacitors, and resistors,

then the frequency of both poles and zeros can be determined by corresponding RC products (time constants).

High-pass transfer function. Another interesting circuit is the so-called first-order high-pass filter. It can be

obtained by connecting a capacitor in series with a resistor as shown in Figure 2.16. This configuration is found at

the input of many AC-coupled amplifiers because it blocks the DC level from the previous stage (at the vi terminal).

Using basic circuit analysis, the transfer function can be obtained as:

121

2

20

11

CRs

s

sCR

R

v

v

i

, (2.37)

where s = j. A zero is located at z = 0 and a pole occurs at p = 1/R2C1. As mentioned before, a zero increases the

magnitude at a rate of +20dB/decade, while a pole decreases it at the same rate. As a result, three regions can be

identified when plotting the magnitude response of the first-order high-pass filter: a) the behavior for << p; b) the

magnitude at =p; and c) at the trend for >> p. The fundamental equations for the square of the magnitude in

those regions are as follows:


- - 20

vi v0

C1

R2

Fig. 2.16. First-order high-pass filter.

;1

1

,1

5.0

,1

)(

12

12

12

2

12

2

0

CRif

CRif

CRifCR

v

v

i

(2.38a)

or, in decibels:

.1

0

,1

3

,1

log20log20

12

12

12

101210

0

CRif

CRif

CRifCR

v

v

dBi

(2.38b)

The transfer function starts at zero (- in dB) and increases at a rate of +20dB/decade when the frequency is swept

up to the pole frequency, at which the gain is -3dB. At higher frequencies, the output voltage is almost in phase with

the input signal and the amplitudes of both signals become nearly identical.

It is important to notice that the capacitive impedance is extremely large at DC and low frequencies, hence most of

the input signal is absorbed by this element when it is in the direct signal path from the input to the output.

Therefore, the voltage drop across the resistor is very small at low frequencies. At very high frequencies, the

capacitive impedance decreases and can even be considered as zero for quick estimates at frequencies much larger

than the pole frequency. In this case, the input signal is mostly absorbed by the resistor, which leads to a voltage

gain that is close to unity (0dB). The magnitude and phase responses of the filter depicted in Figure 2.17 are for two

cases: p = 1/R1C2 = 2∙300 rad/sec and p = 1/R1C2 = 2∙10000 rad/sec, where the -3dB frequencies occur when the

plots cross the horizontal dashed lines.


- - 21

1E+1 1E+2 1E+3 1E+4 1E+5

Frequency (Hz)

-60

-50

-40

-30

-20

-10

0

Volt

age

Gai

n (

dB

)

Fig. 2.17a. Magnitude response of the first-order high-pass filter for two cases: fp = 300Hz and fp = 10kHz.

1E+1 1E+2 1E+3 1E+4 1E+5

Frequency (Hz)

0.0

22.5

45.0

67.5

90.0

Phas

e (D

egre

es)

Fig. 2.17b. Phase response of the first-order high-pass filter for two cases: fp = 300Hz and fp = 10kHz.

General first-order transfer function. The most general first-order filter shown in Figure 2.18 realizes a zero and a

pole. Regardless of where the zero and pole are located, the circuit can be analyzed by inspection:

1. At very low frequencies the capacitors can be considered as open circuits, leading to a low-frequency

gain given by R2/(R1+R2) and a phase shift of zero degrees.

2. For very high frequencies, the capacitors dominate the voltage division because their impedances

become significantly smaller than the resistances. For this reason, the high-frequency gain is

determined by C1/(C1+C2). Under this condition, the phase shift is also 0 degrees.

3. If the low-frequency gain is smaller than the high-frequency gain, then the zero is at a lower frequency

than the pole, allowing the magnitude response to increase.

4. If the low-frequency gain is greater than the high-frequency gain, then the frequency of the pole is

lower than the zero location to cause a decrease of the magnitude.


- - 22

vi v0

C1

R2 C2

R1

Fig. 2.18. General first-order filter.

To obtain the circuit’s transfer function, the equation should be written as:

21

21

21

11

21

10

1

1

CCRR

RRs

CRs

CC

C

v

v

i

. (2.39)

The DC gain and approximate low-frequency gain can be obtained removing the capacitors under the open circuit

assumption, and solving the resulting circuit with ω = 0; which leads to

21

2

0

0

RR

R

v

v

i

. (2.40)

At intermediate frequencies, the gain depends on the locations of the zero and pole, which are given by

.

1

,1

21

21

21

11

CCRR

RR

CR

p

z

(2.41)

Finally, the magnitude of the transfer function at high frequencies can be obtained by evaluating the transfer

function for = , which results in

21

10

CC

C

v

v

i

. (2.42)

Piecewise linear approximations can be easily plotted for both the magnitude and the phase by following the rules

discussed in previous sections. The magnitude responses in Figure 2.19 and the phase responses in Figure 2.20

represent the following example cases:

i) C1 = 0.159F, C2 = 1.59F, R1 = R2 = 100,

ii) C1 = 0.159F, C2 = 1.59F, R1 = R2 = 4.1k.


- - 23

In the first case, the DC gain is 1/2 (-6dB) and the high-frequency gain is 0.0909 (-20.8dB). The frequency of the

zero is z = 62.8 krad/sec (fz =10kHz) and the pole is located at p = 11.435 krad/sec (fp = 1.82kHz). Similarly, the

student can easily verify the plots that correspond to the second case.

1E+1 1E+2 1E+3 1E+4 1E+5

Frequency (Hz)

-25

-20

-15

-10

-5

0

Volt

age

Gai

n (

dB

)

Fig. 2.19. Magnitude responses of the generalized first-order filter for cases i) and ii).

1E+1 1E+2 1E+3 1E+4 1E+5

Frequency (Hz)

-50

-40

-30

-20

-10

0

Phas

e (D

egre

es)

Fig. 2.20. Phase responses of the generalized first-order filter for cases i) and ii).

The current divider. Many important integrated circuit realizations are based on the relationships between input

and output currents because the signals at the device terminals are currents or are easily expressed as such (e.g., at

the collector of a bipolar junction transistor). The basic structure of the current divider is visualized in Figure 2.21.

Z2ii i0Z1 i1

Fig. 2.21. Current divider.

Piece-wise linear

approximation


- - 24

Here, the problem is to find the output current i0 as function of the input current signal ii and the associated

impedances. Certainly, the output current depends on the relationship between the impedances Z1 and Z2.

Conventional circuit analysis techniques allow us to find the current gain:

21

1

ZZ

Z

i

i

i

o

. (2.43a)

The current flowing through Z2 increases with increment of Z1 or decrement of Z2. Here, it is important to note that

most of the current flows through the smaller impedance. Remember that the relation ii = i1 + i2 must hold at the

summing node; hence, if more current flows through one of the impedances, then less current flows through the

other one such that the addition remains equal to the input current. A similarity with the voltage divider is that the

current gain is a function of the relative values of the impedances rather than the absolute values of the impedances,

as clarified in the following rearranged expression:

121

1

ZZi

i

i

o

. (2.43b)

For a resistive current divider (Z1 = R1, Z2 = R2), it can be observed that the output current is significantly reduced if

R2 >> R1, since most of the incoming current under this condition flows through the smaller impedance (R1), and

very little current is collected at the output. On the other hand, most of the current flows through R2 if R2 << R1.

Similar to the voltage divider, the current transfer function of the current divider becomes a frequency-dependent

complex function in the case where its impedance elements are combinations of resistors and capacitors. A low-pass

transfer function is obtained if the configuration shown in Figure 2.23 is constructed. Using Eq. 2.43, it follows that

12

12

2

1

1

1

1

1

1

CRs

CR

RCj

Cj

i

i

i

o

. (2.44)

At low frequencies, the capacitive impedance is much higher than R2, and most of the current flows through the

resistor. Consequently, the current gain is close to unity. At = 1/R2C1 (when R2 = |1/jC1|) the magnitude of the

current gain is equal to 1/21/2, and for higher frequencies the current gain reduces inversely proportional to . This

structure is useful for the analysis of multistage current-mode amplifiers. The magnitude and phase plots of the

current gain are similar to the ones obtained for the voltage divider and are not shown here, but we encourage you to

sketch them with piecewise linear approximations as well as plot them with a software program.

iii0R2

C1

Fig. 2.23. A simple resistor-capacitor (RC) current divider.

It is interesting to note that the current iC1 flowing through the capacitor C1 exhibits high-pass behavior, but this can

be easily understood by observing that the addition of iC1 and io must be equal to ii. Since io reduces at high

frequencies, the current flowing through C1 must increase to maintain the sum of the currents equal to the input

current. The current gain relating iC1 to the input current is given by the following expression:


- - 25

12

12

2

1

201

11 CRj

CRj

RCj

R

i

ii

i

i

i

i

i

C

. (2.45)

The time constant concept. From the previous examples, it is clear that the resistor-capacitor (RC) products define

the frequencies of poles and zeros of the transfer function. The RC product is also termed as time constant , and

frequencies of poles (or zeros) (in radians/second) are usually defined by 1/. Furthermore, finding the time

constants lumped to the nodes in a circuit allows us to identify its poles and zeros. You will learn more about circuit

analysis through such inspections in the following section.


- - 26

II.4. First order system: Time response.

If the frequency response of a circuit is known, the time response can also be obtained by using the properties of the

Laplace transform which relates the time and frequency domains. Time domain analysis will be considered in the

following chapters, but it is sufficient to say now that frequency domain analysis is simpler to apply, especially for

complex systems under steady state conditions. Both techniques allow us to get insights into the behavior of the

circuit. Sinusoidal signals are often assumed in frequency analysis of linear time-invariant circuits because they

make it easy to extrapolate the results to cases in which the input signal is smooth (no jumps, spikes, or square

waves).

If f(t) is a function defined for t > 0, the Laplace transform is denoted as F(s) and defined as

0

dttfetfsF st , (2.46)

where s = j The Laplace transform of f(t) exists if and only if Eq. 2.46 converges to a finite value for all values of

s. Some important properties of the Laplace transform are listed in Table 2.3.

Property Expression

Linearity sFCsFCtfCtfCtfCtfC 221122112211

Frequency

shifting asFtfeat

Scaling

a

sF

aatf

1

Differentiation 0' FssFtf

Second

differentiation 0'0'' 2 FsFsFstf

Integration s

sFdxxf

t

0

Frequency

differentiation sf

ds

dtft

n

nnn 1

Table 2.3. Properties of the Laplace transform.

Among other properties of Laplace transforms that are not discussed in this book, we have to consider the initial and

final value theorems, impulse and step transformations, and system response for periodic functions. Please refer to a

text book on linear systems to learn more about these topics.


- - 27

The Laplace transform of some important functions are given in Table 2.4 as follows:

Table 2.4. Laplace transforms of some relevant functions.

Function F(t) Laplace transform F(s)

(t) 1

(t-a) -ase

u(t-a) s

e-as

1 , t ≥ 0 (unit step) s

1

tn-1/(n-1)! ns

1

tn-1∙eat/(n-1)! nas

1

ab

eeatbt

bsas

1

, a b

ab

eaeb atbt

bsas

s

, a b

To illustrate the relationship between the time domain analysis and the s-domain transfer function, let us consider

the following s-domain transfer function:

01

1

1

01

1

1

....

....

bsbsbs

asasas

sv

svn

n

n

m

m

m

i

o

. (2.47)

As illustrated in Figure 2.24 for a first-order system, any system can be characterized in the frequency domain by

noting the magnitude and phase of the input and output signals at different frequencies to obtain its transfer function

H(). Alternatively, the characterization can be performed by determining the transient impulse response of the

circuit; i.e., by applying vi(t) = (t) and finding vo(t) = h(t)∙(t), where h(t) = ℓ-1H(). In general, the system response

to any input signal can be obtained by calculating the convolution of the impulse response h(t) of the circuit with the

input signal. Another reason why the impulse response is frequently used for the time-domain characterization is

that it can be easily obtained from the s-domain transfer function, especially because there are plenty of tables

available in the literature.


- - 28

Impulse input

First-Order

System Impulse response

Sinusoidal input Output signal

Figure 2. 24. System characterization with sinusoidal functions for frequency domain analysis and the impulse

response for time domain analysis.

The impulse input signal in the time domain corresponds to vi(s) = ((t)) = 1 according to the first property in

Table 2.4. Hence, the time domain impulse response of 2.47 is computed as follows:

01

1

1

01

1

11

....

....

bsbsbs

asasastv

n

n

n

m

m

m

o . (2.48)

It can be seen clearly from the above that the Laplace transform of the transfer function corresponds to the system’s

impulse response in the time domain.

Example: Impulse response. Let us find the impulse and pulse (or step) response of a first-order single-pole

transfer function. If the s-domain transfer function to be considered is

svbs

asv io

0

0

, (2.49)

then the impulse response is obtained by finding the inverse Laplace transform of this equation. From Table 2.4 and

the linearity property, it follows that the impulse response (with vi(s) = 1) of the system is

tb

o eabs

atv 0

0

0

01

. (2.50)

This expression corresponds to the typical exponential function that results from the analytical solution of a first-

order systems, as in the case of a single resistor-capacitor combination for example. The magnitude response of the

first-order filter is displayed in Figure 2.25a. Its DC gain is given by a0/b0, while the -3dB frequency is determined

by b0. The corresponding unit impulse response is shown in Figure 2.25b, where = 1/b0 is the time constant of the

first-order system.

3 dB

b0

(log)

20·log10(a0/b0)

Time (Secs)

~0.018

v0 (t)

~0.37~0.13

~0.05

~0.007

1

(a)

(b)

Fig. 2.25 A first-order low-pass system: a) Magnitude response, b) unit impulse response with = 1/b0.


- - 29

In the previous analysis, it has been assumed that the initial conditions of the system are zero such that v0(0) = 0.

When an impulse is applied at the input, the steady state of the system is disturbed, and the output voltage jumps to

the amplitude of the input impulse, which is 1 in this example. The system returns back to its steady state by

exhibiting an exponentially decaying behavior. The system’s time constant time (= 1/b0) is dictated by the pole

frequency b0. After one time constant (at t = the output response has decayed 63%, and it will continue to decay

to 87% of its original value at t = 2= 2/b0 seconds. After t = 4 the output is very close to its final value and the

deviation from the steady state value is less than 2%. Settling errors below 1% are acceptable for many applications.

Hence, it takes at least five time constants after the impulse has been applied before the system returns back to its

steady state based on the 1% settling criterion. In the particular case, the steady state was defined with zero input

voltage, but you should be aware that systems can also be characterized when the DC input and steady state output

are non-zero prior to applying the impulse response.

Example: Step response. The step response of a linear system can easily be derived from the impulse response by

using the integration property in Table 2.3. Since the step function in the s-domain corresponds to 1/s in the

frequency response, the time domain output response of the system described by Eq. 2.49 to a unit input step yields

)0()1()( 0

0

0

0

ostep

tb

t

impulseostepo vVeb

adttvtv

(2.51)

where vo(t)|impulse is given by Eq. 2.50. Vstep is the amplitude of the applied step, and vo(0) is the initial value of the

output voltage at t = 0. The step response of the circuit also exhibits an exponential behavior. If the initial

conditions are zero, then the final output voltage (value at t = ) is given by (a0/b0)∙Vstep. After one time constant,

the output voltage is around 63% of its final value. As shown in Figure 2.26, the output reaches around 99.3% of the

final value after five time constants. The time to reach 99% of the final value is often called 1% (error) settling time,

and the time required to reach this condition is approximately t = 5= 5/b0. Thus, fast settling behavior implies

small time constants (RC products) or, equivalently, poles at high frequencies. You should remember that the

smaller the RC products are, the faster a system’s response is to disturbances.

Time (Secs)

~0.98

v0 (t)

~0.63

~0.86~0.95

~0.993

Vstep(a0/b0)

Fig. 2.26 Step response of the first-order system defined by Eq. 2.49.

General case. The general transfer function in Eq. 2.47 can be rearranged to

01

1

101

1

1 ........ asasassvbsbsbssv m

m

m

i

n

n

n

o

. (2.52)

Applying the Laplace transform to the above expression and assuming zero initial conditions, we can obtain the time

domain input-output relationship:


- - 30

tva

dt

tvda

dt

tvda

dt

tvd

tvbdt

tvdb

dt

tvdb

dt

tvd

ii

m

i

m

mm

i

m

oo

n

o

n

nn

o

n

01

1

11

1

1

01

1

11

1

1

...

...

. (2.53)

The time domain output can be obtained by solving this differential equation for a given input. Usually, the solution

of this equation is not trivial. It is a lot easier to solve the circuit in the frequency domain to find the input-output

voltage relationship, and then utilize the inverse Laplace transform to obtain the system’s time domain response.

Example: Transformation of a frequency domain transfer function to determine the impulse response. Find

the impulse response of the system described by the following second-order s-domain transfer function:

svjsjs

sv io)1010()1010(

10

. (2.54)

The poles of this function are complex conjugates given by

10102,1

jjsp . (2.55)

After substituting vi(s) = 1 from Eq. 2.54, the impulse response can be obtained by using the transform in Table 2.4

for the second-order function with two different poles. The result is

1221

112

))(( pp

tsts

pp

oss

eek

ssss

ktv

pp

(2.56)

where the constant k is equal to 10. If the poles are complex conjugates as in Eq. 2.55, the system’s output voltage

given by 2.56 becomes

)sin(22

tkej

eeek

j

eektv t

tjtjt

tjtj

o

. (2.57)

After substituting the numerical values, we finally get

)10sin(10 10 tetv t

o . (2.58)

This impulse response corresponds to a decaying sinusoidal function with an oscillating frequency (in radians per

second) determined by the imaginary part of the complex conjugate poles. The amplitude of the sinusoidal function

is modulated by an exponential function, for which the factor in the exponent depends on the real part of the poles.

A general representation of this impulse response is depicted in Figure 2.27.


- - 31

Time (Secs)

v0 (t)

Fig. 2.27. Typical pulse response of a second-order transfer function. Notice that <0.

Notice that the impulse response of a second-order system with poles located on the left hand side of the s-plane

(< 0) eventually returns back to its steady state (zero output voltage). The impulse response of systems with poles

located on the right hand side of the s-plane leads to a positive exponent, causing the response to diverge to +/-

and thereby making the system unstable. A system is unstable if its output is unbounded as a result of a bounded

input. A necessary condition for system stability is to have poles located on the left hand side of the complex s-

plane.

Concluding Remarks.

Appendix A. Complex Algebra

Complex number representation

Basic operations

Polar representation: Magnitude and Phase

tKe

tKe t sin

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