LaPlace Transforms in Design and Analysis of …...LaPlace Transforms in Design and Analysis of Circuits© Part 2 by Tom Bertenshaw Basic Circuit Analysis - Series Circuits Series

LaPlace Transforms in Design and Analysis of Circuits Part 2: Basic Series Circuit Analysis Course No: E03-010

Credit: 3 PDH

Thomas G. Bertenshaw, Ed.D., P.E.

Continuing Education and Development, Inc. 9 Greyridge Farm Court Stony Point, NY 10980 P: (877) 322-5800 F: (877) 322-4774 [email protected]

LaPlace Transforms in Design and Analysis of Circuits© Part 2

by Tom Bertenshaw

Basic Circuit Analysis - Series Circuits

Series RC Circuit A series RC circuit is a basic electrical building block. Frequently these circuits are configured to be either a low pass or a high pass filter. In later modules we will investigate the design of active filters, but an understanding of the underlying principles is fundamental. Analysis begins with understanding the role of the transfer function, how to develop the transfer function and its utility to predicting the time response of the circuit. Consider the following circuit:

1/sC

V(s)

I(s)

R

and suppose we wish to know the voltage drop across any of the components, and further suppose we wish to do so by using transfer functions. Using Ohm's Law we can write:

⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛ +=

RCs

sRsI

sCRsIsVin

1)(1)()(

further, RsIsVR )()( =

and

⎟⎠⎞

⎜⎝⎛=

sCsIsVC

1)()(

To identify the transfer function for either component we form the ratio of voltage across the component divided by the voltage applied. For example:

⎟⎠⎞

⎜⎝⎛ +

=

RCs

ssVsV

in

R

1)()( ← (Transfer function)

Revised: Aug 13, 2011 1

or ( )

⎟⎠⎞

⎜⎝⎛ +

=

RCs

ssVsV in

R 1)(

)(

that says the voltage across the resistor equals the derivative of the applied voltage divided by the denominator. It is clear that if the applied voltage is a DC, the voltage across the resistor is 0 as the derivative of a constant is 0; in other words all the voltage appears across the capacitor. That makes sense because a capacitor presents an open to DC. Also, please note that the factor is the "time constant". Further, this value is completely determined by the choice of values for both R & C.

RC

Assume that )sin()( tAsVin ω= , and since the LaPlace Transform of that sine function is

( )22 ωω+sA

then,

( )221)(

ω

ω

+⎟⎠⎞

⎜⎝⎛ +

=s

RCs

sAsVR ← Eq. 1

We have now positioned ourselves to predict the output as a function of the input. In the "s" domain, the transfer function times the driver equals the output (not so in the time domain, be careful with this, the time domain requires the use of the convolution integral). From Table 1 at the end of this discussion, it is apparent that the driver in Eq. 1 will invert to

)cos( tK ωω

where K is yet to be determined, and )cos( tωω is clearly the derivative of )sin( tω .

To completely invert to , it is first necessary to determine the appropriate numerators of a decomposed left side:

)(sVR )(tVR

( ) )(11)( 22

22 ωω

ω+

+⎟⎠⎞

⎜⎝⎛ +

=+⎟

⎠⎞

⎜⎝⎛ +

=s

C

RCs

B

sRC

s

sAsVR ←Eq. 1 (re-written)


where are determined by partial fraction expansion. The techniques of partial fraction expansion can be found in almost any Algebra text, or for convenience, there is a short tutorial included here as Appendix A.

CB &

As an example, from the above let

3155

=

==

RC

Aω

then,

( ) 2222 535)3(25)(

++

+=

++=

sB

sA

ssssVR

The poles ("poles" are the value(s) that cause the denominator to be zero) of the above equation are & , which requires we re-write as: 3− 5j±

553)(

jsC

jsB

sAsVR −

++

++

=

Multiplying both sides by , and then evaluating at 3+s 3−=s , we get:

A=−3475

and repeating at 5js −=

( ) Bjj

j=

−−−

10)53(125

and again at 5js =

Cjj

j=

+ )10)(53(125

584.11.1

584.11.1

321.2)(

jsj

jsj

ssVR −

−+

++

++

−= 2.eq←

The two complex denominators on the right hand side do not match any of the pairs in Table 1, so they must be rationalized.

( )2222 5

568.35

2.2321.2)(

++

++

+−

=ss

ss

sVR

and that yields (note that ω was factored from the numerator of the third in anticipation of inversion as a sine function in the time domain):


tetttf 321.2)5sin(68.3)5cos(2.2)( −−+= 3.eq← or re-writing:

to ettf 321.2)315sin(29.4)( −−+≅

Eq 3

-5-4-3-2-1012345

1 17 33 49 65 81 97 113 129 145 161 177 193 209 225 241 257 273 289 305 321 337 353 369 385 401

Time in .01sec

Am

plitu

de

Note that , satisfying the assertion that will contain a function of the derivative of the driver. A short discussion on obtaining phase angles is included in Appendix B.

)595cos()315sin( 00 −=+ tt )(tf

The above procedure is long and tedious and open to error, so let us see if we can find a "short cut" method for dealing with solutions of polynomials that contain complex roots (sinusoids). First, merely factor the denominator;

35)25)(3(25)( 222 +

++

=++

=s

Cs

BssssVR

Second, multiply both sides by the complex factor, in this case ( )22 5+s ; Third, set ωjs += , and importantly, factor and isolate ω from the numerator result:

( ) oo

jjB 31529.43144.21

53125

∠≅∠≈+

= and 21.2−=C


or ( )

321.2

553129.4)( 22 +−

+∠

=ss

sVo

R

then inverting: to

R ettV 321.2)315sin(29.4)( −−+=

The answers are consistent, and the "short cut" is easier. A purist might reject the short cut as being intellectually lazy, but the author is a charter member of that group so we will adopt that method as our modus operendii. Continuing with uncovering the voltage drops across each of the components, we will now pursue the drop at the capacitor, of course using the same driver of . )5sin(5 t The transfer function at the capacitor is developed as follows;

⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛ +=

=

RCs

sRsi

SCRsisV

sCsisV

in

C

1)(1)()(

)()(

Forming the transfer function, then manipulating these expressions to obtain as a function of the driver, we get:

)(sVC

( )⎟⎟⎠⎞

⎜⎜⎝

⎛+

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛ +

= 22 525

1

1

)(s

RCs

RCsVC

Letting 31=

RC, as before:

( )( ) 2222 535375)(

++

+=

++=

sB

sA

sssVC

Using the usual method,

21.23475

==A

( ) )5957.2)(5(

53515 o

jB −∠=

+= ω factored )5( ω=

so then,

22 5)5957.2)(5(

321.2)(

+−∠

++

=ss

sVo

C

and


)595sin(57.221.2)( 3 otC tetV −+= − 4.eq←

Another quick check is that are out of phase, as they should be. As proficiency is developed with this form of circuit analysis, speed and accuracy increase rapidly, and obtaining quantitative answers for the outputs in the time domain becomes far less tedious.

)(&)( tVtV CR090

Eq 4

-3

-2

-1

0

1

2

3

4

1 51 101 151 201 251 301 351 401

Time in .01sec

Am

plitu

de

Vr+Vc

-6

-4

-2

0

2

4

6

1 50 99 148 197

Time in .01sec

Am

plitu

de

Series1


The above figure is the sum of the steady state sum of Equations 3 & 4; the voltages across the resistor and the capacitor. Its gratifying that it mirrors the driver, verifying that our computations are correct.

The Impulse Response

What is the native response of a circuit? That question interests us because the native response is the transient portion of the output (the driver forms the steady state portion of the output - )(tδ or a pulse being an exception as there is no steady state output). Since the transform of 1)( ↔tδ we will use the impulse as a driver, the output is then the native, natural or impulse response i.e.,

)1(1

1

)(⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+=

RCs

RCsVC ← transfer function with impulse driver

so

RCt

eRC

tf−

=1)(

again for the sake of illustration only, letting 31=

RC

Impulse Response

0

0.5

1

1.5

2

2.5

3

3.5

1 50 99 148 197

Time in Hundredths of a Second

Am

plitu

de


The Step Response

The step response isolates the response of the transfer function to DC. Like the Impulse Response, the output is expressed as the transfer function times the input, in this case a unit step, i.e.

RCsss

RCs

RCsVC 1111

1

1

)(+

−=⎟⎠⎞

⎜⎝⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+=

RCt

C etV−

−= 1)(

Step Response

0

0.2

0.4

0.6

0.8

1

1.2

1 50 99 148

Hundredths of a Second

Am

plitu

de

It is interesting to note the relationship between the impulse and unit step response, both mathematically and graphically. But the real significance is that the impulse response is the native time response of the circuit, and the step response is the response to a DC driver (or DC component of a driver).


The Series RLC Circuit

The series RLC circuit is a fundamental building block in circuitry, even though the desired circuit response can often be obtained using active circuits. To understand RLC like behavior, as well as to analyze and/or design a circuit to obtain a specific response, it is very desirable that a thorough grounding in the fundamentals is well understood. One of basic parameters of electrical design, the circuit resonant frequency, is uncovered by understanding RLC behavior. Series RLC circuits are sometimes referred to as "series tank circuits", because they do possess an inherent resonant frequency. At resonance ( ) the impedance of the network is at the minimum. 0=− CL jXjX As usual our approach will be via the transfer function. This approach not only provides for a prediction of output in the time domain, but also positions us for analysis and design work in the frequency domain, as we shall eventually see in later modules.

R sL

1/sC

V(in)

I(s)

Circuit voltage is (Kirchhoff's voltage loop):

)1)(()(sC

sLRsIsV ++=

We need to re-arrange the voltage loop expression as it will become the denominator of any transfer function and it MUST be in a form compatible with Table 1 for inversion.

⎟⎠⎞

⎜⎝⎛ ++⎟⎠⎞

⎜⎝⎛=

LCs

LRs

sLsIsV 1)()( 2

A nice quadratic! That is important as a quadratic has either a pair of real roots or a pair of complex roots; in either case it is invertible after PFE. And as we will see as time goes on, that this quadratic plays a major role, often the dominant role, in any circuit that the solution to the quadratic is complex.

The voltage across the capacitor is (Ohm's law):

sCsIsVc 1*)()( =

and across the inductor: sLsIsVL *)()( =

and across the resistor:


RsIsV R *)()( =

As we have discussed, the ratio of the voltage across any component divided by the source voltage, by definition, is the transfer function:

LCs

LRs

LCsVsVC

1

1

)()(

2 ++=

and

LCs

LRs

ssVsVL

1)()(

2

2

++=

finally,

LCs

LRs

LRs

sVsVR

1)()(

2 ++=

for the sake of illustration and to generate examples, assume

291

10

=

=

LC

LR

Lets investigate the impulse response across each of the components. In general;

)1(2910

)()( 2 ⎟⎠⎞

⎜⎝⎛

++=

sssNsVX

The roots of this denominator are 25 j±− , so re-writing:

( ))1(

2510)(

22 ⎟⎟⎠

⎞⎜⎜⎝

⎛

++=

sssVR

( ))1(

2529)(

22 ⎟⎟⎠

⎞⎜⎜⎝

⎛

++=

ssVC

( ))1(

25)(

22

2

⎟⎟⎠

⎞⎜⎜⎝

⎛

++=

sssVL


You are invited to verify that: ( ) 452910 22 ++=++ sss

and that the roots are truly 25 js ±−=

There a couple of important things to be aware of regarding the denominator. First if

, then the roots of the denominator become 0=R 29j± . That is apparent as the denominator in that case becomes:

( ) ( )20

222 129 ω+=⎟⎠⎞

⎜⎝⎛ +=+ s

LCss

This frequency is commonly denoted as 0ω and it is the highest frequency the circuit is naturally capable of; i.e., it is the circuit's resonant frequency. The circuit may be driven to a higher frequency by an excitation source, but recalling the physics of a coil, the higher the frequency is that it is driven by, the more it begins to behave as an open. You can always uncover the resonant frequency by setting the factor on . Clearly then 01 =s

LC1 is the resonant frequency. All these remarks apply only to the case where the

roots of the quadratic are complex; if they are real there is no resonant frequency. You are invited to verify that if the roots are real, the inverted form becomes

tt BeAetf βα ±± ±=)( . In the example above, the frequency is 2 and is commonly denoted as dω , the damped frequency ( dω is always < 0ω , an inspection of the general quadratic will disclose why). Notice that in the above denominator once we choose the value for one of the

components, the values of the others are set by the relationships of LCL

R 1& . In other

words, we want to design a circuit in which 290 =ω , 2=dω and 52

=L

R . Choosing a

capacitor with a value of 1 farad, requires we choose a coil of 186 mili-henrys and a resistor of 1.86 ohms. Not realistic circuit values for small signal circuits, but they serve our purposes as an illustration (for ease of math reasons we have chosen a circuit with an

0ω of 5.38 rads/s). Note that L

R2

.assumes the function of the multiplier on t in the

expression LRt

e 2−

which, in turn, identifies RL2 as the circuit time constant.

As an exercise, why is the time constant RL2 instead of

RL ? As cursory inspection of the

way in which the denominator must be arranged to guarantee inversion from the domain t, the time domain provides the answer.


Taking the inverse transforms of the impulse response one at a time,

)2sin(5.14)( 5 tetV tC

−=

)(sVR cannot be inverted as it stands, so we will add and subtract 50 in the numerator to yield an invertible pair;

22222 2)5()2(25

2)5()5(10

2)5(50)5(10)(

++−

+++

=++−+

=ss

ss

ssVR

or )2.1582sin(9.26))(2(sin(25)2(cos(10)( 55 ott

R tettetV +=−= −−

See Appendix B for a Word About Phase Angles. A quick review of Table 1 shows that cannot be directly inverted as it stands, and addition/subtraction as in the case of will not resolve the issue. So, a new rule is called for, and it is this:

)(sVL

)(sVR

When the numerator is the same order of as the denominator, long division MUST be applied as necessary to obtain a remainder in which the denominator is at least one order higher than the numerator.

s

In the above example of , then: )(sVL

1 22 2910 sss ++

29102 ++ ss 2910 −− s

so

291029101)( 2 ++

+−=

ssssVL

this will further devolve to (using addition/subtraction of 21± ):

( )( )

( )( ) 45

25.1045

5101)( 22 ++−

+++

−=ss

ssVL

and inverts to: ( ) ( )( )ttettf t 2sin05.12cos10)()( 5 +−= −δ

( )( )ot tet 6.432sin5.14)( 5 +−= −δ


Impulse Response Inductor Voltage

-10

-5

0

5

10

15

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49

Hundredths of a second

Ampl

itude

The above figure fairly well captures the effect that )(tδ has on the impulse response.

To amplify the effect of LRt

e 2−

, dω is arbitrarily increased below, simply to illustrate how the exponential decay can dominate the impulse response.

Inductor Impulse V @ many rads

-15

-10

-5

0

5

10

15

20

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96

Hundredths of second

Ampl

itude

To repeat, note that if the factor on is 0 (means 1s )0=R then the circuit oscillates at

29=ω . Whereas when the frequency is lower (in this case 2) - and that will be true with all oscillating circuits. When

0≠R0=R the circuit will theoretically oscillate


undiminished forever. But even if we left the resistor out, there is residual resistance in the leads (at any temperature above absolute 0) such that there is an RI 2 loss, and no matter how slight the loss, oscillations will eventually decay to 0. In a word; in the real world no perpetual motion is allowed. Arguments concerning absolute zero and quantum electrodynamics almost never apply in practice, i.e., Newtonian physics is just fine for common use. Moving on, suppose the above RLC circuit is driven by a )200sin(10 tVin = , then,

( ) ⎟⎠⎞

⎜⎝⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛

++= 2222

2

2002000

25)(

ssssVL

Division is no longer required as the denominator is now 2 orders higher than the numerator. Nevertheless, a PFE is required in order to invert : )(sVL

2222234

2

2002)5(58002000229102000

++

++=

++++ sB

sA

sssss

Using the techniques of Appendix A (PFE), and solving for A & B:

( )( ) ( )( )22

2

25 200252252000++−

+−=+−=↓ j

jA js

oA 6.437. −∠=

( )

( ) ( )( )420052002002000

2

2

200 ++==↓ j

jB js

086.210∠≈B

and )86.2200sin(10)6.432sin(72.)( 5 oot ttetf ++−= −

)(tf is our prediction of the output given an input of . )200sin(10 t

It is important to note that elements of both the impulse response and the driver are present in the output. The impulse response element is the transient response and the driver element is the steady state response. In general, that statement is true across the board, i.e., there will always be elements of both the impulse and the driver in the time domain output. In the case above, the transient portion is:


ot te 6.432sin(72. 5 −− )

corresponding to the contribution of the native or impulse response of the circuit to excitation. The steady-state portion of the output is

)86.2200sin(10 ot + Again, the steady state response is of the same form as the driver, e.g., )200sin( φ±tA and will remain until excitation is terminated.

Transient & Steady State Respons

-15

-10

-5

0

5

10

15

1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 86 91 96 101

Time Hundredths of a second

Ampl

itude

In the above graph, the amplitude of the transient portion has been distorted purposefully to better identify its contribution to the output. It should be clear that by the end of the second cycle it has fully disappeared. As an exercise, you are invited to solve for the impulse response for each of the components of an RLC series circuit, using the component values as shown.


.5

V(in)

I(s)

.0488H

.5fV(in)

Kirchhoff's voltage loop

∫++= dttiCdt

diLRtitVin )(1)()(

transforming directly

⎟⎠⎞

⎜⎝⎛ ++=

sCsLRsisVin

1)()(

rearranging

⎟⎠⎞

⎜⎝⎛ ++⎟⎠⎞

⎜⎝⎛=

LCs

LRs

sLsisVin

1)()( 2

plugging in values

( )4110)()( 2 ++⎟⎠⎞

⎜⎝⎛≅ ss

sLsisVin

so (here is where you are invited to do the work to find the impulse response);

)4sin(25.10)( 5 tetV tC

−≅ ( ))4sin(5)4cos(10)( 5 ttetV t

R −≅ − ( ))4sin(225.)4cos(10)()( 5 ttettV t

L +−+≅ −δ Let's convert to a sine function: )(tVL

)774sin(25.10)()( 5 ot

L tettV −+≅ −δ

If we assume a driver of and then solve for , we get: )5sin(2 t )(tVL

)75.1075sin(7.4)284sin(8.7)( 5 oot

L ttetV −+−≅ −


Inductor Voltage

-10

-8

-6

-4

-2

0

2

4

6

1 51 101 151 201 251 301 351 401 451

Time in .01sec

Am

plitu

de

Because series tank circuits present minimum impedance at resonance, they are useful as notch filters to trap for unwanted frequencies. Of course active filters now perform this same function, and those will be discussed in later modules.


Transform )(tf )(sF

1 K sK

2 tKe σ− σ+s

K

3 )sin( tK ω 22 ω

ω+s

K

4 )cos( tK ω 22 ω+s

Ks

5 )sin( tKe t ωσ−

( ) 22 ωσω++s

K

6 )cos( tKe t ωσ− ( )( )( )22 ωσ

σ++

+s

sK

7 )(tδ 1

7a* )(tKδ K

8 )( atKu − s

Ke as−

9 )(' tf )0()( fssF − 10 ∫ dttf )(

sf

ssF )0()(+

11 )()( tbgtaf + )()( sbGsaF +12 t

2

1s

13 atte − ( )2

1as +

Table 1

* K is preserved for practical circuit reasons, not for theoretical reasons as ∞∗K is approximately equal to ∞

Table 1 is not all inclusive and other pairs will be examined and added when needed. But for beginning analysis purposes Table 1 is adequate.

It is very important to understand that to be able to transform any to an ,

must be reduced to one of the forms so far developed. If it is not in one of these forms it cannot be operated on until it is. Study the right hand side forms, they identify the left hand side.

)(sF )(tf)(sF

Transforms 12 and 13 are found as follows:


ttf =)(

20

1)(s

dttesF st == ∫∞ −

For transform 13, assume

attetf −=)( ( )

( )20

1)(as

dttesF tas

+== ∫

∞ +−

Finding transform 12 and 13 is a straightforward exercise in integration by parts.


Appendix A

Partial Fraction Expansion

Partial Fraction Expansion is a technique to decompose a ratio of polynomials into a sum of factors; for example:

5420952

2 ++

+=

+++

sB

sA

sss ← Ex. 1

where are to be determined. In general, then, BA &

npsN

psB

psA

sDsN

+++

++

+= ........

)()(

21

←Eq. 1

)(&)( sDsN are the numerator and denominator polynomials. through are the

roots of the factors ( . is used because these are also the poles, as will be shown in the module on Root-Locus - note that there is no restriction on to be real, it can be and often is, complex).

1p np

ip

ip

To solve for any of factor numerators, A for example, a) multiply both sides of Eq. 1 by

, b) cancel from both sides and from the denominator of ( 1ps + ) )( 1ps + A , then c) set . The result looks like this: 1ps −=

( ) ( ) Apppp

pN

n

=+−+−

−

121

1

.......)(

@ 1ps −=

Let's try it on Ex. 1:

( )( )( )( )

( ) ( )54

44

54524

++

+++

=++++

ssB

ssA

ssss

( )

54

552

++

+=++

ssBA

ss

now set 4−=s

( ) 054

542+=

+−+− A

so 3−=A

Following the same procedure for B


( )( )( )( )

( ) ( )55

45

54525

++

+++

=++++

ssB

ssA

ssss

( ) BssA

ss

+++

=++

45

452

setting , we get: 5−=s

( ) B+=+−+− 045

552

or 5=B

Let's check it:

43

55

20952

2 +−

+=

+++

sssss

Comparing the rightside with Table 1 we can see that the solution can now be inverted to the time domain (recall, a factor must be in one of the forms in Table 1 to be inverted). In fact this solution inverts to:

tt eetf 45 35)( −− −= Both values of the time constant are a direct function of circuit component values. It is very important to note that multiplication in the 's' domain is not multiplication in the time domain.

Checking the solution by cross multiplying the denominators to form the LCD, we have:

209153205

20952

22 ++−−+

=++

+ss

ssss

s

209

522 ++

+=

sss Q.E.D.

This general procedure is repeated for each non-repeating root regardless of the number. When the roots are repeated a slight modification is called for. For example:

( ) ( ) ( ) 1221243

22 ++

++

+=

++−

sC

sB

sA

sss

The denominator must be assumed to be part of the solution because it is a factor of the LCD (although

( 12+s )B can be 0). Finding are straightforward as in the example

above, so: CA &


00143

++=+− A

ss ( )

)(2)( 2

sDssN +

= ←Eq. 2

at we have: ,2−=s10=A

finding C

( )C

ss

=+−

2243

at , we have: 1−=s7−=C

So far, so good - but what about B ? The typical procedure is to take the derivative of Eq. 2 and evaluate the result at the root value. For example:

( ) (( )

)21

4313143

+−−+

=⎟⎠⎞

⎜⎝⎛

+−

sss

dsssd

evaluating that at 2−=s7=B

therefore,

( ) ( ) ( ) 17

27

210

1243

22 +−

++

+=

++−

ssssss

Again, we compare the denominators to the forms required in Table 1, and find that:

=)(tf ttt eetetf −−− −+= 7710)( 22

To check the answer we cross multiply the numerators as appropriate and form the LCD.

( ) ( )12

432 ++−

=ss

s

Then there is the case of complex roots, for example:

( )( ) 32321134143

2 jsC

jsB

sA

ssss

−++

+++

+=

++++

Again, using the procedure we have established, evaluated at 1− :

Ass

s=

+++

13443

2


or 1.=A

Now, we will find B by letting 32 js −−=

( )( )( ) B

jjjj

=−+−−+−−

+−−3232132

4323

or 48.05. jB +−=

We need not solve for C as the other component of a complex conjugate pair with the complex conjugate of B.

( )( ) 3248.05.

3248.05.

11.

134143

2 jsj

jsj

sssss

−+−−

++++−

++

=+++

+

However, the solution above does not match a form in Table 1, so we must find a denominator for the complex fractions that matches a transform pair from Table 1. The LCD of the complex pair is:

( ) 9)2(134)32(32 22 ++=++=−+++ sssjsjs

Cross multiplying the numerators of the complex fraction and adjusting to extract the necessary numerator for the cosine and sine factors, we end up with:

( )( )

( )( )

( ) 222222 323)967(.

3221.

11.

327.21.

11.)(

++−

+++

++

=++

−+

+=

sss

sss

ssF

)(sF is now easily capable of being inverted:

)3sin(967.)3cos(1.1.)( 22 teteetf ttt −−− −+=

Again, note that although the factors in s domain are multiplicative, they are not in the time domain (very important point, particularly for design work - the time domain requires the use of the convolution integral; multiplication in the s domain ↔ convolution in the time domain). To check, again after cross multiplying the numerators of to re-form the LCD, we get:

)(sF

= ( )( ))134143

2 ++++

ssss

as it must be, or A, B & C are not correct.


Some examples, obtain answers to match those below:

51

31

11

1523923183

23

2

++

++

+=

+++++

ssssssss

( ) 25542.2.

25553

22 +++

+−

=++

−ss

sssss

s

( ) ( ) )1(11

18

211

2)1(53

22 ++

+−

+−

=++

−sssss

s


Appendix B

A Word About Phase Angles Define a 2 space with coordinate axes as shown below:

cosine axis

sine axis

Now consider a statement such as:

)sin(5.1)cos( tt ωω +

We will interpret that as 1 unit coincident with the cosine axis and 1.5 units coincident with the sine axis. In other words, it represents two sides of a triangle with an adjacent of 1.5 units and an opposite of 1 unit.

1 unit

+1.5 units

That being the case, the enclosed angle at origin is about , and will re-write the above sum as:

o7.33

)7.33sin(8.1 ot +ω

Consider the sum in the LRC circuit in the body of Part 2:


))(2(sin(5.2)2(cos(10)( 5 ttetV tR −= −

Applying the same rationale:

1 unit

-2.5 units Then as measured from the +sine axis (defined to be , the resultant is: )0o

)2.1582sin(9.26 5 ot te +−


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LaPlace Transforms in Design and Analysis of …...LaPlace Transforms in Design and Analysis of Circuits© Part 2 by Tom Bertenshaw Basic Circuit Analysis - Series Circuits Series