Laplace Transform and Applications

EE/CE 3301-002 Handout Prof. Y. Chiu

1

Laplace Transform and Applications

We have seen the application of the phasor technique in solving dynamic circuits, consisting of R, L, C, independent

and controlled sources, for the sinusoidal steady-state response. The procedure shoots directly for the final, forced

(i.e., steady-state) response of the circuit while ignoring the initial transient (or natural) response part. When a

complete solution of the circuit starting at t = 0 is desired, or the excitation is not limited to sinewaves, other tech-

niques of solving are necessary―the Laplace transform is one of such techniques that will be covered in this note.

The most widely used bilateral Laplace transform of a general signal x(t) is given as (with the assumption that

the integral converges)

( ) ( ) ( ) ( ), ,stX s x t e dt X s x t+

−

−

⎯⎯→= L

(1)

where s = σ + jω is a complex variable. The bilateral Laplace transform degenerates to the Fourier transform (you

will learn this in ENGR 3302, Signals and Systems) once the real part of s is set to zero, i.e., Re{s} = σ = 0:

( ) ( ) ,j tX s j x t e dt

+−

−

= = (2)

which is useful when analyzing the convergence and inverse transform of the Laplace.

In analyzing linear time-invariant (LTI) circuits and systems with the input onset at t = 0 and the circuit or

system may have non-zero initial conditions or energy storage (for example, the step response of an RLC circuit),

the unilateral Laplace transform is often used instead:

( ) ( ) ( ) ( )0

, .stX s x t e dt X s x t−

+−

⎯⎯⎯→= UL

(3)

Comparing to the bilateral counterpart, the unilateral Laplace transform sets the lower limit of the integral to 0−,

which is to include any impulse or singularity functions at t = 0―a very useful resort to handle, for example, the

impulse response of the system or the simple action of throwing a switch at time zero. If no such impulse or singu-

larity occurs, setting the lower limit of integration to 0− or 0 yields no difference ( ) ( )0 0

0

0 0

.0stx t e dt x t e dt− −

−

= = If

x(t) does not contain any singularity at t = 0, the unilateral and the bilateral Laplace transforms can be related as

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0

, ,st stX s x t e dt x t u t e dt x t X s x t u t−

+ +− −

−

⎯⎯⎯→ ⎯⎯→= = UL L

(4)

where u(t) is the unit step function defined as (e.g., recall the step response of dynamic circuits)

( )( )

0, 0,

1, 0.

u t t

u t t

=

= (5)

Differentiating the step function gives us another useful function called the impulse or Dirac delta function:

( )( ) , 0,

0, o.w.

du t tt

dt

+ == = (6)

The impulse function is a signal of infinite amplitude and zero duration. It enables us to differentiate the step func-

tion u(t) at its discontinuity at t = 0 (see Fig. 1). Note that

( )( )

( ) 1.du t

t dt dt u tdt

+ +

+

−− −

= = = (7)


2

The output produced by a circuit or system when excited by δ(t) is called the impulse response, denoted as

h(t), which is of great importance in characterizing the circuit or system. The step and impulse functions are very

useful in circuit analysis. We note the sifting property of the delta function:

( ) ( ) ( ) ( ) ( ) ( ) ( ).x t d x t t d x t t d x t + + +

− − −

− = − = − = (8)

Examples of Laplace Transform

Exponential: ( ) ( ) ( ) ( )

00

1 1, 0 .

s a tat at stx t e u t a X s e e dt es a s a−

−

− +− − −= → = = − =

+ + (9)

In the last step of (9), we assumed that σ + a > 0 or σ > −a, so e−(s+a)∞ evaluates to 0―the constraint Re{s} = σ >

−a denotes the region of convergence, or the ROC, of the transform. In general, a Laplace transform must be

specified with its ROC.

Sinewave:

( ) ( ) ( )

( ) ( ) 2 20 0

sin

1 1 1sin .

2 2

j t j tst st

x t t u t

e eX s t e dt e dt

j j s j s j s

− −

−− −

= →

−= = = − =

− + + (10)

In (10), the ROC is Re{s} > 0. Fig. 2 shows the ROC’s of the above two transforms.

Step: ( ) ( ) ( )0

0

1 1.st stx t u t X s e dt e

s s−−

− −= → = = − = (11)

u(t)

t

1

0

(a)

δ(t)

t0

(b)

Fig. 1 (a) Unit step function, (b) im-pulse function, (c) approaching the step function by letting ε → 0, and (d) approaching the impulse func-tion by differentiating the waveform in (c) and letting ε → 0. Note the area underneath the curve is con-stant and equals 1.

u(t)

t

1

0 ε ε

(c)

δ(t)

t

1

2ε

0 ε ε

Area = 1

(d)

jω

σ0

ROC

s-plane

a

(a)

jω

σ0

ROC

s-plane

+jω

jω

(b)

Fig. 2 (a) The ROC of the Laplace trans-form of e−at, a > 0 and (b) the ROC of the transform of sin(ωt). The cross symbol marks the positions of the roots of the denominators, a.k.a. the poles of the transform X(s).


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The ROC is Re{s} > 0.

Impulse: ( ) ( ) ( ) ( ) 0

0

1.st sx t t X s t e dt e −

− −= → = = = (12)

The ROC is the entire s-plane.

Table 1 lists a few functions and their unilateral Laplace transforms commonly encountered in circuit analysis.

Properties of Laplace Transform

Linearity: ( ) ( ) ( ) ( )1 2 1 2.ax t bx t aX s bX s+ → + (13)

ROC: at least ROC1 ∩ ROC2, because the summation of X1(s) and X2(s) may result in pole cancellation.

Differentiation:

( )

( ) ( ) ( ) ( ) ( ) ( )0

0 0 0

0 .st st st st

dx t

dt

dx t e dt e dx t x t e s x t e dt sX s x

dt −− − −

− − − − −

→

= = + = − (14)

ROC: at least ROC ∩ {Re{s} > 0, so e−s∞ evaluates to 0}, considering possible pole cancellation.

Integration:

( ) ( ) ( )

( ) ( ) ( )

0 0 0 0 0

0 00

1

1 1.

t t tst st

tst st

x d x d e dt x d des

x d e x t e dt X ss s

− − − − −

− −−

− −

− −

→ = −

= − − =

(15)

Table 1 Common Unilateral Laplace Transform Pairs

x(t) X(s) ROC

( )t 1 Entire s-plane

( )u t 1

s Re{s} > 0

t 2

1

s Re{s} > 0

ate− 1

s a+ Re{s} > −a

( )sin t 2 2s

+ Re{s} > 0

( )cos t 2 2

s

s + Re{s} > 0

atte− ( )

2

1

s a+ Re{s} > −a

( )sinate t−

( )2 2s a

+ + Re{s} > −a

( )cosate t−

( )2 2

s a

s a

+

+ + Re{s} > −a


4

ROC: at least ROC ∩ {Re{s} > 0}, considering possible pole cancellation.

Convolution:

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

1 2 1 2 1 2

0 0

1 2 2 1 1 2,

st st

s s

x t x t x x t d e dt x x t e dt d

x e X s d X s x e d X s X s

− −

+ + − −

− −

+ +− −

− −

→ − = −

= = =

(16)

where we considered that x1(t) and x2(t) are identically zero for t < 0.

ROC: at least ROC1 ∩ ROC2, considering possible pole cancellation.

Initial-Value Theorem: ( )0

limt

x t+→

= ( )lims

sX s→

(17)

Final-Value Theorem: ( )limt

x t→

= ( )0

lims

sX s→

(18)

The proof of the initial- and final-value theorems can be approached by invoking (14) and is left as an exercise (see

§12.9 if you need help on this).

Table 2 lists some common properties of the unilateral La-

place transform and the initial- and final-value theorems.

Example 1: For the parallel RLC circuit shown in Fig. 3, find the

step response of vo(t) for t ≥ 0 using the Laplace transform

method. The circuit has no energy storage before t = 0.

Table 2 Properties of The Unilateral Laplace Transform

Property x(t) X(s) ROC

Linearity ( ) ( )1 1 2 2k x t k x t+ ( ) ( )1 1 2 2

k X s k X s+ At least ROC1 ∩ ROC2

s-domain shift ( )0s te x t ( )0

X s s− ROC shifted by Re{s0}

Time scaling ( ) , 0x kt k 1 s

Xk k

ROC scaled by k

Conjugation ( )x t ( )X s

ROC

Convolution† ( ) ( )1 2x t x t ( ) ( )1 2

X s X s At least ROC1 ∩ ROC2

t-domain derivative ( )d

x tdt

( ) ( )0sX s x −− At least ROC ∩ {Re{s} > 0}

2nd-order derivative ( )2

2

dx t

dt ( ) ( ) ( )2

0 0d

s X s sx xdt

− −− − At least ROC ∩ {Re{s} > 0}

s-domain derivative ( )tx t− ( )d

X sds

ROC

t-domain integral ( )0

t

x d −

( )1

X ss

At least ROC ∩ {Re{s} > 0}

Initial-value theorem ( )0

limt

x t+→

= ( )lims

sX s→

Final-value theorem ( )limt

x t→

= ( )0

lims

sX s→

† with the assumption that x1(t) and x2(t) are identically zero for t < 0.

L

+

t = 0Idc CR vo(t)

Fig. 3 A parallel RLC circuit


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The integrodifferential equation describing the RLC circuit is

( )( )

( )( )

0

1.

to o

o dc

v t dv tv d C I u t

R L dt + + =

Performing Laplace transform on both sides of the above equation, we have

( ) ( )( )

1 1.

o oo dc

V s V sC sV s I

R L s s + + =

Solve, we have

( )( )2

.1 1

dco

I CV s

s RC s LC=

+ +

Lastly, we need to perform the inverse Laplace transform on Vo(s) to obtain vo(t):

( ) ( ) 1 .o ov t V s−=L

Poles and Zeros

Most of the Laplace transforms we have seen so far are rational functions, which can be expressed as the ratio of

two polynomials of variable s:

( )( )( )

( )( ) ( )

( )( ) ( )

11 21 1 0

11 1 0 1 2

,n n

nn nm m

m mm

s z s z s zN s a s a s a s aX s K

D s b s b s b s b s p s p s p

−−

−−

− − −+ + + += = =

+ + + + − − − (19)

where K = an/bm. if m > n, the ratio N(s)/D(s) is called a proper rational function; if m ≤ n, it is called an improper

rational function. An improper rational function can always be expressed as the sum of a proper one and a poly-

nomial of s by performing long division on N(s)/D(s)―see Example 3.

The roots of the numerator polynomial N(s), i.e., z1, z2, …, zn, are called the zeros of the function X(s), and the

roots of the denominator polynomial D(s), i.e., p1, p2, …, pm, are called the poles. We observe that on the s-plane

X(s) evaluates to zero for s = z1, z2, …, zn and to infinite for s = p1, p2, …, pm―thus the names zero and pole. In Fig.

2, examples are shown in the complex s-plane illustrating the poles s = −a and s = ±jω of the transforms of the

exponential and sinewave, respectively. These plots are termed the pole-zero plot. Since s = σ + jω, the real axis

(abscissa) of a pole-zero plot is often labeled σ and the imaginary axis (ordinate) is labeled jω. Note that the poles

are marked by × and zeros are marked by ○ in the pole-zero plot.

Note that equal number of poles and zeros exist in a rational function X(s)―if they are not at a finite location,

then they must be at infinity. For example, 1/(1 + s) has a pole at −1, but no obvious zeros. The zero is actually

located at ∞, as can be verified by the fact that 1/(1 + s) evaluates to 0 for s = ∞. Another example: s + 1 has a zero

at −1, but no obvious poles. The pole is actually at ∞, as can be seen when 1 + s evaluates to ∞ for s = ∞.

Inverse Laplace Transform―Partial Fraction Expansion

The inverse Laplace transform can be derived using the Fourier transform aided by the contour integral in the s-

plane. We give the expression of the inverse Laplace transform without diving into the detail of derivation (see

Oppenheim and Willsky, Signals & Systems, 2nd Ed., Prentice Hall):

( ) ( )1

,2

j

st

j

x t X s e dsj

+

−

= (20)

where the contour of integration is the straight line in the s-plane corresponding to all points satisfying Re{s} = σ.


6

For rational functions, a shortcut exists for evaluating the inverse Laplace transform, the partial fraction expan-

sion. The following example illustrates the exact procedures of this technique.

Example 2: Find the inverse Laplace transform of X(s) = 1/(s2 + 3s + 2).

( )( )( )2

1 1 1 1.

1 23 2 1 2X s

s ss s s s= = = −

+ ++ + + +

From Table 1, 1/(s + 1) corresponds to the signal e−tu(t) and 1/(s + 2) corresponds to the signal e−2tu(t). By super-

position, the inverse transform of X(s) is

( ) ( ) ( )2 .t tx t e e u t− −= +

The function X(s) in Example 2 is a proper rational function. For improper functions, we can always first

expand them into the sum of a polynomial and a proper rational function by performing long division on X(s), and

then perform the partial fraction expansion on the proper part to obtain the final result. The following example

illustrates this procedure.

Example 3: Find the inverse Laplace transform of X(s) = s3/(s2 + 3s + 2).

( ) ( )( )( )

( )3

2

7 6 1 83 3 .

1 23 2 1 2

s sX s s s

s ss s s s

+ −= = − + = − + +

+ ++ + + +

From Table 1, (s − 3) corresponds to the signal δ′(t) − 3δ(t). So, the inverse transform of X(s) is

( ) ( ) ( ) ( ) ( )23 8 .t tx t t t e e u t − −= − + − +

In general, for proper rational functions, we categorize four scenarios as follows.

A. X(s) with distinct real poles (using an example with three poles)

( )( )

( )( )( )31 2

1 2 31 2 3

,N s KK K

X ss p s p s ps p s p s p

= = + ++ + ++ + +

(21)

where K1, K2, and K3 can be solved by collecting all three terms, evaluating the ensuing numerator and equating it

to N(s). Alternatively, the coefficients can be directly found by evaluating (for K1)

( ) ( )( )

1

1

2 1

1 12

s p

s p

K s ps p X s K

s p=−

=−

++ = +

+

( )

1

3 1

3s p

K s p

s p=−

++

+ 1.K= (22)

K2 and K3 can be found in the same way. The inverse transform can then be looked up in Table 3.

B. X(s) with distinct complex poles

For circuits with real-valued components, the complex poles or zeros must appear in X(s) in conjugate pairs.

( )( )

( )( )

*

,N s K K

X ss j s js j s j

= = ++ − + ++ − + +

(23)1

where K can be found by evaluating

1 The coefficients K and K* appear in (23) in complex conjugate because the polynomial coefficients {an} and {bm} are real.


7

( ) ( )( )*

s j

s j

K s js j X s K

s j

=− +

=− +

+ −+ − = +

+ +.K= (24)

The inverse transform can then be looked up in Table 3.

C. X(s) with repeated real poles

( )( )

( ) ( ) ( )31 2

3 3 21

1 1 1

,N s KK K

X ss ps p s p s p

= = + +++ + +

(25)

where K1, K2, and K3 can be found by evaluating

( ) ( )( )

( )1

1

3

3 2 1

1 1 2

1s p

s p

K s ps p X s K

s p=−

=−

++ = +

+

( )

1

3

3 1

1s p

K s p

s p=−

++

+ 1.K=

( ) ( ) ( ) ( )1 1

3 2

1 1 2 1 3 1 2.

s p s p

d ds p X s K K s p K s p K

ds ds=− =−

+ = + + + + = (26)

( ) ( ) ( ) ( )1 1

2 23 2

1 1 2 1 3 1 32 22 .

s p s p

d ds p X s K K s p K s p K

ds ds=− =−

+ = + + + + =


D. X(s) with repeated complex poles

For circuits with real-valued components, the complex poles or zeros must appear in X(s) in conjugate pairs.

( )( )

( ) ( ) ( ) ( )

* *1 1 2 2

2 2 2 2,

N s K K K KX s

s j s js j s j s j s j = = + + +

+ − + ++ − + + + − + + (27)

where K1 and K2 can be found by evaluating

( ) ( ) ( ) ( )2 2

1 2, .

s j s j

ds j X s K s j X s K

ds

=− + =− +

+ − = + − = (28)

Table 3 Four Useful Laplace Transform Pairs

Nature of Poles X(s) x(t)

Distinct real K

s a+ ( )atKe u t−

Repeated real ( )

2

K

s a+ ( )atKte u t−

Distinct complex conjugate

*K K

s j s j +

+ − + + ( ) ( )2 cos

tK e t u t − +

Repeated complex conjugate ( ) ( )

*

2 2

K K

s j s j +

+ − + + ( ) ( )2 cos

tt K e t u t − +


8


Circuit Elements in the s-Domain

In the following narration, we will use i(t), v(t) or simply i, v to denote the time-domain current, voltage, respec-

tively; and use I(s), V(s) or I, V to denote the s-domain current, voltage, respectively.

Resistor in the s-domain:

( ) ( ) ( ) ( ).v t R i t V s R I s⎯⎯→= = L

(29)

Inductor in the s-domain:

( )

( )( ) ( ) ( ) ( )

( )( ) ( )

0

0 0

0 ,

or .

di tv t L V s L sI s i sL I s LI

dt

V s LI V s II s

sL sL s

− ⎯⎯→

= = − = −

+= = +

L

(30)

The two representations (essentially in the Thévenin and Norton forms) are illustrated in Fig. 4.

Capacitor in the s-domain:

( )

( )( ) ( ) ( ) ( )

( )( ) ( )

0

0 0

0 ,

or .

dv ti t C I s C sV s v sC V s CV

dt

I s CV I s VV s

sC sC s

− ⎯⎯→

= = − = −

+= = +

L

(31)

The two representations of (31) are also illustrated in Fig. 4.

Applications of Laplace Transform in Circuit Analysis

First, we note that the Ohm’s law, KCL, KVL, nodal and mesh analyses, and source transformations (Thévenin and

Norton) are all applicable to the s-domain representation of a circuit. Thus, the treatments of R, L, and C are unified

under the concept of impedance (Z)―the impedances of R, L, and C are R, sL, 1/sC, respectively, all with a unit of

I

sL

LI0

+V

(a)

I

sL

+ V

I0/s

(b) Fig. 4 The s-domain equivalent circuits of the inductor and capacitor in Thévenin and Norton forms

I

+V

V0/s

1/sC

(c)

I + V

1/sC

CV0

(d)


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Ω. In addition, the initial conditions (i.e.,

energy storage in L’s and C’s) are auto-

matically accounted for during the La-

place transform (Fig. 4). The s-domain

analysis does not differentiate the transi-

ent (natural) response from the steady-

state (forced) response of a circuit and

treats them in a unified way. It does not

require the input excitation of the circuit

to be sinusoidal either. Thus, it is a general

technique for circuit analysis that yields

complete solutions from t = 0 to ∞. We

will illustrate these points through the fol-

lowing examples.

Example 4: The step response of an RLC

circuit (Example 1 revisited with initial

conditions). For the parallel RLC circuit

shown in Fig. 5, find the step response of v(t) for t ≥ 0 using the Laplace transform method. The initial conditions

are iL(0−) = 29 mA and vo(0−) = 50 V.

Since all elements are in parallel in the circuit, we choose the Norton representation for the L and C with initial

conditions shown in Fig. 5. The R, L, and C are translated into the impedances R, sL, and 1/sC, respectively. The

input DC current source and the switch are replaced by a step function which is represented as 1/s in the s-domain.

Note that an s-domain current bears a dimension of A⸱s due to the integration of the transform (and for the same

reason a voltage bears a dimension of V⸱s).

The total current is 0 0

1 1,dcI I CV

s s+ + and the total impedance is

11 1

.sCR sL

−

+ + Therefore, the output volt-

age is

( )0

0 0 0

22

1 150 2 5

.1 1 1 1 8 4 1.6 9

dcdc

I II I CV sV

s es s CV s I Zs e s esC s s

R sL RC LC

++ + +

−= = = =

+ ++ + + +

Comparing this result to that of Example 1, ( )( )2

,1 1

dcI CV s

s RC s LC=

+ + we see that the initial conditions of L and

C are automatically and properly handled using the Laplace transform method2.

We need to perform the inverse Laplace transform on V(s) to obtain v(t). But first we need to perform the partial

fraction expansion on V(s):

( )( ) ( )

2 2 2

50 2 5 50 2 5 2.2 6 50.

4 48 4 1.6 9 4 4 4 4

s e s e eV s

s es e s e s e s e

− − −= = = +

++ + + +

And by looking up in Table 3, we have

( ) ( ) ( ) ( )1 4 4 4 42.2 6 50 V.e t e tv t V s e te e u t− − −= = − +L

2 It is a good practice to double check the dimensions of all the quantities in the final expression of V(s): in the numerator, two

terms bear the dimension of V/sec, whereas in the denominator all three terms bear the dimension of 1/sec2; so, the ratio has a

dimension of V⸱sec, as expected for an s-domain voltage.

25 mH

+

t = 024 mA 25 nF500 Ω v(t)

iL(t)

s/40

Ω

+

24/s

mA-s

40/s

MΩ

500

Ωv(t)

1.25

μA-s

29/s

mA-s

( ) 0V s I

sL s+ ( ) 0sC V s CV −

Fig. 5 A parallel RLC circuit with initial energy storage


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Example 5: The series RLC circuit shown in Fig. 6 is driven by

a sinusoidal input ( ) 15cos4 4 V.v t e t= Find i(t) for t ≥ 0 using

the Laplace transform method. No initial energy storage.

First, the R, L, and C are replaced by impedances. The sin-

ewave and the switch together are represented in the s-domain

as

( ) ( ) 2 2

1515cos4 4 V s.

4 4

sV s e t u t

s e= =

+L

The total impedance is 1

.R sLsC

+ + Therefore, the current is

( )

( )( )

2 2

2

2 2 2

15

4 41 8

4 316

240.

4 4 6.4 4 1.6 9

sV s eI s

s eZe

s

s

s e s e s e

+= =+ +

=+ + +

As an extra step, we can also apply the initial- and final-value theorems to check the above expression:

From the initial-value theorem: ( ) ( )( )( )

3

2 2 20

240lim lim 0,

4 4 6.4 4 1.6 9st

si t sI s

s e s e s e+ →→

= = =+ + +

From the final-value theorem: ( ) ( )( )( )

3

0 2 2 2

240lim lim 0.

4 4 6.4 4 1.6 9t s

si t sI s

s e s e s e→ →= = =

+ + +

The theorems correctly predicted i(0+) = 0 as the inductor does not have initial energy storage. But the i(∞) = 0 is

questionable as we know that, driven by a sinusoidal input, the current will not settle to a fixed value when t→∞.

So, what is wrong? It turns out that the final-value theorem does not apply in this case as I(s) has two poles on the

jω-axis! The theorem works only when the transform has all its poles, except one single pole at origin, reside in the

LHP of the s-plane (related to the stability of the system, which is beyond the scope of this note).

Lastly, in order to obtain i(t), we perform the partial fraction expansion on I(s):

( )( )( )

* *21 1 2 2

2 2 2

240.

4 4 4 4 3.2 4 2.4 4 3.2 4 2.4 44 4 6.4 4 1.6 9

K K K KsI s

s j e s j e s e j e s e j es e s e s e= = + + +

− + + − + ++ + +

Solve for K1 and K2, we have K1 = 1.875 3,e− K2 = 3.125 3 126.87 .e− − And finally we have

( )

( ) ( ) ( )

1

3.2 4

1.875 3 1.875 3 3.125 3 126.87 3.125 3 126.87

4 4 4 4 3.2 4 2.4 4 3.2 4 2.4 4

3.75cos4 4 6.25 cos 2.4 4 126.87

steady-state transient

e t

e e e ei t

s j e s j e s e j e s e j e

e t u t e e t u t

−

−

− − − − − = + + +

− + + − + +

= + −

L

mA

We observe that the first term represents the steady-state response of the circuit. It exhibits the same frequency as

the input sinewave and persists for all time. We can also use the phasor technique to solve for this term. The second

term, however, is the transient response of the circuit due to the sudden onset of the sinewave at t = 0 (even the

circuit has zero initial energy storage) and perishes when t→∞. Its characteristics are set by the tank (i.e., R, L and

62.5 mH

t = 0

v(t)

10 nF

4 kΩ i(t)

s/16 Ω

t = 0

1e8/s MΩ

4 kΩ I(s)

2 2

15V s

4 4

s

s e

+

Fig. 6 A series RLC circuit with sinusoidal input


11

C values) and independent of the input. Note that the Laplace transform technique properly accounts for all the

initial conditions of the circuit and yields the complete solution for t ≥ 0. In contrast, the phasor technique cannot

solve for the transient part of the circuit response.

Question: The interesting question here is that the circuit has no energy storage at t = 0−, so why it still produces a

transient response for t > 0? The answer lies in the fact that the sinewave is discontinuous at t = 0 when the switch

closes, i.e., had the sinewave been present at all time, the L and C would have some energy stored in them at t = 0.

But we know this is not true as the circuit is initially at rest. So, we postulate that the circuit will produce no transient

response if the L and C have just the right amount of energy stored in them; and then at the onset of the sinewave

at t = 0, it will just produce the steady-state response only, as if it had been driven by the sinewave for a long time

before t = 0.

Exercise: Calculate the initial conditions for the L and C in the above example so no transient response is produced.

(Answer: evaluate the steady-state part of the response at t = 0, we have IL(0−) = 3.75 mA, VC(0−) = 0. If you plug

these values into the circuit and rederive I(s), you should obtain K1 = 1.875e-3, K2 = 0.)

Convolution Integral of LTI System

Suppose the impulse response of a linear time-invariant (LTI)3 system is known, i.e., the system outputs h(t) when

driven by the impulse function δ(t), we want to derive the system response y(t) to a general input x(t). First, we note

the sifting property of the delta function, ( ) ( ) ( ) ,x t x t d +

−

= − which can be viewed as a bunch of time-shifted

delta functions δ(t−τ) weighted by x(τ) to sum up to x(t), i.e., x(t) can be viewed as a “linear combination” of many

individual weighted functions of the form δ(t−τ). Since the system response to δ(t) is known as h(t), by time invar-

iance, the response to δ(t−τ) must be h(t−τ). Then by superposition (or linearity), the LTI system output y(t) in

response to x(t) must be

( ) ( ) ( ) ( ) ( ) ( ), also denoted as .y t x h t d y t x t h t +

−

= − = (32)

The Eq. (32) is called the convolution integral of the LTI system. It reveals that once the impulse response h(t) of

a system is known, its response to an arbitrary input x(t) is also known by virtue of the convolution integral. It is

easy to show that (32) can be equivalently expressed as

( ) ( ) ( ) ( ) ( ) ( ), also denoted as .y t h x t d y t h t x t +

−

= − = (33)

In practice, the convolution integral provides a means to calculate a system’s output directly in the time domain,

without going through the s-domain transforms. It is useful to numerical analysis such as in experiments, wherein

an analytical form of h(t) is unknown.

Transfer Function

We see that, in analyzing circuits using the Laplace transform technique, we

always arrive at an expression of X(s) that has a part of the denominator (as a

polynomial in s) independent of the input signal. For example, in the parallel

RLC circuit, this polynomial is ( )2 1 1 ;s RC s LC+ + and in the series RLC cir-

cuit, it is ( )2 1 .s R L s LC+ + In addition, this part of the denominator seem-

ingly determines the characteristics of the transient or natural response of the

3 A system is linear if the superposition rule applies; it is time-invariant if a time-shifted input will result in a time-shifted

output and nothing else, i.e., if an input x(t) produces an output y(t), then the input x(t−τ) must produce the output y(t−τ).

H(s)

h(t)

X(s)

x(t)

Y(s)

y(t)

Fig. 7 The transfer function of an LTI system


12

circuit (recall the characteristic equation in the RLC circuits). To examine this further, we consider the LTI system

shown in Fig. 7, where the system produces an output y(t) when excited by an input x(t). We express the ratio of

the Laplace transforms of the output and input as a new term, the transfer function H(s) of the system:

( )( )( )

.Y s

H sX s

= (34)

In doing so, we also note that we need to ensure that the system is initially at rest, i.e., no energy storage at t = 0.

Otherwise, x(t) will no longer be the sole excitation and the H(s) derived is then ambiguous. The H(s) derived in

(34) is independent of the input X(s). A circuit can produce many transfer functions, depending on where exactly

the input and the output are.

Since the Laplace transform of the delta function is 1, the output of the system in Fig. 7 will be H(s)―with a

time-domain counterpart h(t)―if the input is replaced by δ(t), i.e., H(s) is essentially the Laplace transform of

the impulse response h(t) of the system. Recall we stated before that, once the impulse response of a system is

known, its response to any signal is also known by virtue of the convolution integral. We thus relate the convolution

integral and the transfer function as

( ) ( ) ( ) ( ) ( ) ( ).y t h t x t Y s H s X s⎯⎯⎯→⎯⎯⎯= = L

(35)

The Eq. (35) is of no surprise―we recognize it as the convolution property of the Laplace transform. The impulse

response and the transfer function both characterize a system or circuit, one in the time domain and the other in the

s-domain. The impulse response is also a form of natural response of the system (see Example 6 for the equiva-

lence).

For systems and circuits that can be described by linear, constant-coefficient ordinary differential equations,

H(s) must be rational, i.e.,

( )( )( )

( )( ) ( )

( )( ) ( )

11 21 1 0

11 1 0 1 2

.

n nnn n

m mm mm

s z s z s zN s a s a s a s aH s K

D s b s b s b s b s p s p s p

−−

−−

− − −+ + + += = =

+ + + + − − − (36)

In addition, for circuits with real, positive component values (i.e., pas-

sive components such as R, L, and C), the coefficients {an} and {bm}

of the numerator and denominator polynomials, respectively, must be

real and positive too; thus, the poles and zeros of H(s) either must be

real or appear in complex conjugate pairs. If the system is BIBO

(bounded input, bounded output) stable, all poles of H(s) must reside

in the LHP of the s-plane, whereas the zeros can reside anywhere in

the s-plane. The poles set the neper and resonant frequencies (or the

damping factor and undamped natural frequency) of the circuit. They

are the roots of the denominator polynomial of H(s) (or the roots of

the characteristic equation of the system).

Example 6: Impulse function in circuit analysis.

The circuit shown in Fig. 8(a) is driven by δ(t). Calculate how much

energy is instantaneously deposited into the L and C at t = 0. In Fig.

8(b), the delta function is removed and the 1-μF capacitor now has an

initial voltage of 1000 V. Show that the circuit will produce an identi-

cal output as that of Fig. 8(a).

As δ(t) ≠ 0 only at t = 0, if vL(t) and iC(t) are known for the L and

C, respectively, we can integrate them from 0− to 0+ to determine the

50 mH

δ(t) 1 μF250 Ω

1 kΩ

+

vC(t)

+vL(t)

iC(t)

(a)

50 mH1 μF

250 Ω

1 kΩ t = 0

+

vC(0 )

= 1000 V

(b)

Fig. 8 Equivalence between (a) the impulse function and (b) the initial condition in pro-ducing transient response in circuits


13

energy deposited into the storage elements at t = 0. We have

1

2

1,

.

C

LC

C LL

IVsC I

R sL

I VV R

sC sL

−= +

= +

→

1 61

,1 3 50 31 6 250

.50 3

C

LC

C LL

e IVs I

e s ee I V

Vs s e

−= +

−

= +−

Solve, we have

2

2

2

5 3mA s,

6 3 2.5 71 3

V s.6 3 2.5 7

C

L

s e sI

s e s ee s

Vs e s e

+=

+ +

= + +

Note that the rational expression of IC is improper while it is proper for VL. Factoring the expressions,

*1 1

*2 2

1 mA s,3 3 4 3 3 3 4 3

V s.3 3 4 3 3 3 4 3

C

L

K KI

s e j e s e j e

K KV

s e j e s e j e

= + + + − + +

= + + − + +

Since we are integrating from 0− to 0+, we need an impulse function to yield a non-zero result. In the above expres-

sions, only IC will produce a delta function when an inverse Laplace transform is performed. So, we have

( ) ( )

( ) ( ) ( ) ( )

0

0

0 0 0

0 0 0

10 0,

1 1 10 other terms 1 3 V.

L L

C C

i v t dtL

v i t dt t dt dt eC C C

+

−

+ + +

− − −

+

+

= =

= = + =

At this point, we realize that the circuit in Fig. 8(b) essentially bears the initial condition just obtained above. Thus,

the two circuits must produce identical outputs for t > 0.

There is another method to derive the equivalence between the two circuits by inspecting the s-domain equations

directly. We now show how this can be done. Suppose both L and C have initial energy storage, the system equations

can be modified as

01

2 0.

0,C L

C

LLC

V VI

R sL

VV V R

CV

LIsL

−= +

= −+

−

Comparing these equations to those obtained before, we realize that the two sets of equations will be identical if

R1CV0 = 1 and LI0 = 0 hold, which yield V0 = 1e3 V and I0 = 0―this is exactly the initial conditions given in the

circuit of Fig. 8(b).

Transfer Function and Steady-State Sinusoidal Response

We first establish the fact that the exponential function is an eigenfunction of the LTI system. Consider an LTI

system H(s) excited by an exponential input est, the output is given by the convolution integral as


14

( ) ( ) ( ) ( ) ( ) ( ) ,s tst st s sty t h t e h e d e h e d H s e

+ +

− −

− −

= = = = (37)

where we used the bilateral Laplace transform instead of the unilateral one. Eq. (37) reveals that convolving h(t)

with est is equivalent to multiplying its Laplace transform H(s) with est, i.e., the function est is an eigenfunction

of the system H(s).

Now suppose the system is driven by a sinusoidal input and we want to find the steady-state response. Using

the Euler identity, we have

( ) ( )2 2

cos .2

j t j t j t j tA Ae e

AA t e e − −

+= = + (38)

Thus, the sinusoidal functions are also eigenfunctions of the LTI system for the special case of s = jω. The steady-

state sinusoidal response of the system can then be expressed as

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( )( ) ( )

( ) ( )

cos2 2

2 2

2 2

2

cos .

j t j t

s j s j

j t j t

j j j jj t j t

j t j j t j

A Ah t A t H s e H s e

A AH j e H j e

A AH j e e H j e e

AH j e e

H j A t j

−

= =−

−

− −

+ − +

= +

= + −

= +

= +

= +

(39)4

Eq. (39) states that the steady-state sinusoidal response of a

system can be obtained by simply weighting the input sin-

ewave by the magnitude of H(s = jω), evaluated at the sin-

ewave frequency ω, and phase-shifting it by the phase of H(s

= jω), also evaluated at the sinewave frequency ω. Thus, si-

nusoidal signals can be used to characterize an LTI system

by sweeping the signal frequency and evaluating the ensuing

magnitude and phase of the output. This technique is widely

used in experiments.

The transfer function H(s) evaluated for the special case

s = jω is called the frequency response of the system (see

Fig. 9),

( ) ( ) ( ) ( ) ( ) ( ).j j

s jH s H j H j e H j j

== = = (40)

The plot of the magnitude and phase responses of ( ) ( ) ( )H j H j j = vs. the frequency ω is called the Bode

plot, an important tool for evaluating LTI systems’ characteristics. To process a wide range of frequencies and

magnitudes, the magnitude plot is often rendered on a log-log scale while the phase plot on a semi-log scale.

4 In (39), we used the fact that H(s*) = H*(s)⸺can be shown as ( ) ( ) ( ) ( )*

*

* *s t stH s h t e dt h t e dt H s+ +

− −

− −

= = = ⸺if h(t) is

real. So, ( ) ( ) ( ) ( )* .j j

H j H j H j e

−

− ==

jω

σ0

s-planeH(s = jω)

Fig. 9 Frequency response of an LTI system can be obtained by evaluating H(s) on the jω-axis.


15

Example 7: Plot the following transfer

function in MATLAB.

( )1

.1

H ss

=+

The transfer function has one pole

at s = −1 rad/s and a zero at ∞. The fol-

lowing MATLAB commands generate

the Bode plot in Fig. 10.

>> H = tf(1, [1 1])

>> bode(H)

The magnitude response ( )H j

displays a lowpass characteristic, i.e.,

it rolls off at high frequencies. The

low-frequency gain is 1 (or 0 dB), and

the gain asymptotically approaches 0

(or −∞ dB) with a roll-off slope of −20

dB/decade (or −6 dB/octave) at high

frequencies.

The phase ( )j starts off at 0°

and decreases asymptotically to −90°

at high frequencies.

We also note here the behavior of

the magnitude and phase responses at

the proximity of the pole frequency,

i.e., ( )H j and ( )j evaluated

at ω = 1 rad/s:

( )1 rad/s

1 rad/s

1

1

1 or 3.01 dB

2

H jj

=

=

=+

= −

and

( )1 rad/s

1

1

45 or .4

jj

= =

+

= − −

The angular frequency 1 rad /s is termed the −3-dB bandwidth (ω–3-dB) of the system H(s).

Example 8: Plot the following transfer function in MATLAB.

( ) 2

1.

2 1H s

s s=

+ +

Fig. 10 Bode plot of the first-order lowpass transfer function in Example 7

Fig. 11 Bode plot of the second-order lowpass transfer function in Example 8


16

The transfer function has two poles at s = −1 rad/s and two zeros at ∞. The following MATLAB commands

generate the Bode plot in Fig. 11.

>> H = tf(1, [1 2 1])

>> bode(H)

The magnitude response ( )H j still displays a lowpass characteristic. In this case, the low-frequency gain is

still 1 (or 0 dB), but the gain rolls off now with a slope of −40 dB/decade (or −12 dB/octave) at high frequencies

(due to the double poles).

The phase ( )j starts off at 0° and decreases gradually to −180° at high frequencies.

At the pole frequency, i.e., at ω = 1 rad/s, we have

( )( )

21 rad/s

1 rad/s

1 1 or 6.02 dB

21H j

j

=

=

= = −+

and

( )( )

21 rad/s

190 or .

21j

j

= = = − −

+

The −3-dB bandwidth (ω–3-dB) of the system H(s) is therefore no longer at ω = 1 rad/s⸺it is instead at a lower

frequency.

The transfer function in Example 7 is a first-order lowpass function and that of Example 8 is a second-order

lowpass function. The order of a transfer function H(s) is determined by its number of poles.

Exercise: calculate the −3-dB bandwidth of the transfer function H(s) in Example 8.

(Answer: equate ( )H j to −3-dB and solve, we obtain ω−3-dB = 0.644 rad/s.)

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Laplace Transform and Applications