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Continuous-Time System Analysis Using The Laplace Transform

Dr. Mohamed BingabrUniversity of Central Oklahoma

Outline• Laplace Transform• Properties of Laplace Transform• Solution of Differential Equations• Analysis of Electrical Networks• Block Diagrams and System Realization• Frequency Response of an LTIC System• Filter Design by Placement of Poles and Zeros of H(s)

Why Laplace Transform?

Laplace transform is a mathematical tool to map signals and system behavior from the time-domain into the frequency domain.

𝑥𝑥 𝑡𝑡 = 𝑛𝑛=0

𝑥𝑥(𝑛𝑛𝑛𝑛)𝛿𝛿(𝑡𝑡 − 𝑛𝑛𝑛𝑛) 𝑥𝑥 𝑡𝑡 = 𝑖𝑖

𝑋𝑋(𝑠𝑠𝑖𝑖)𝑒𝑒𝑠𝑠𝑖𝑖𝑡𝑡

Time Domain Laplace Domain (frequency)

si : complex frequency = σi + jωi

σi : rate of decay ωi : rate of oscillation

x(t) = 3e-5t

ts

ii

iesXtx ∑∞

−∞=

≈ )()(

LT𝑋𝑋 𝑠𝑠 =

3𝑠𝑠 + 5 |𝑋𝑋 𝑠𝑠 | =

3

𝜎𝜎 + 5 2 + 𝜔𝜔2

x(t) can be expressed as a linear sum of exponential with different strength

|X(s)|

Sigma

Omegaσ

ω

Definition of Laplace Transform

Two-sided Laplace transform of x(t)

𝑋𝑋 𝑠𝑠 = −∞

∞𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑠𝑠𝑡𝑡𝑑𝑑𝑡𝑡

One-sided (unilateral) Laplace transform of causal signal x(t)

Two-sided (bilateral) inverse Laplace transform of X(s)

𝑥𝑥 𝑡𝑡 =1

2𝜋𝜋𝜋𝜋𝑐𝑐−𝑗𝑗∞

𝑐𝑐+𝑗𝑗∞𝑋𝑋(𝑠𝑠)𝑒𝑒𝑠𝑠𝑡𝑡𝑑𝑑𝑠𝑠

𝑋𝑋 𝑠𝑠 = 0

∞𝑥𝑥(𝑡𝑡)𝑒𝑒−𝑠𝑠𝑡𝑡𝑑𝑑𝑡𝑡

Examples of Popular Functions

Find the Laplace transform of the impulse δ(t).

ℒ 𝛿𝛿 𝑡𝑡 = −∞

∞𝛿𝛿 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑡𝑡𝑑𝑑𝑡𝑡 = 1 For all s

ℒ 𝑢𝑢 𝑡𝑡 = 0

∞𝑢𝑢 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑡𝑡𝑑𝑑𝑡𝑡

For Re[s]σ > 0

= −1𝑠𝑠𝑒𝑒−𝑠𝑠𝑡𝑡

0

∞

=1𝑠𝑠

Find the Laplace transform of the unit step u(t).

Examples of Popular Functions

Find the Laplace transform of 𝑒𝑒𝑎𝑎𝑡𝑡𝑢𝑢(𝑡𝑡).

ℒ 𝑒𝑒𝑎𝑎𝑡𝑡𝑢𝑢(𝑡𝑡) = 0

∞𝑒𝑒𝑎𝑎𝑡𝑡𝑒𝑒−𝑠𝑠𝑡𝑡𝑑𝑑𝑡𝑡 =

1𝑠𝑠 − 𝑎𝑎

ℒ cos𝜔𝜔0𝑡𝑡 𝑢𝑢(𝑡𝑡) = 0

∞12𝑒𝑒𝑗𝑗𝜔𝜔0𝑡𝑡 + 𝑒𝑒−𝑗𝑗𝜔𝜔0𝑡𝑡 𝑒𝑒−𝑠𝑠𝑡𝑡𝑑𝑑𝑡𝑡

=𝑠𝑠

𝑠𝑠2 + 𝜔𝜔02

For Re[s] > aσ > a

Find the Laplace transform of cos𝜔𝜔0𝑡𝑡 𝑢𝑢(𝑡𝑡).

For Re[s] > 0σ > 0

Inverse Laplace Transform

The inverse Laplace transform, finding x(t)from X(s), involves integration in the complex plane that is not covered in this class.

Partial fractions expansion, Laplace properties, and the table will be used to find the inverse Laplace transform.

Example 1 of Inverse Laplace Transform

Use partial fraction expansion to find the inverse Laplace transform of

𝑋𝑋 𝑠𝑠 =7𝑠𝑠 + 6

𝑠𝑠2 + 5𝑠𝑠 + 6

𝑥𝑥 𝑡𝑡 = −8𝑒𝑒−2𝑡𝑡 + 15𝑒𝑒−3𝑡𝑡 𝑢𝑢(𝑡𝑡)Answer:



𝐻𝐻 𝑠𝑠 =2𝑠𝑠2 + 5

𝑠𝑠2 + 3𝑠𝑠 + 2

ℎ 𝑡𝑡 = 2𝛿𝛿 𝑡𝑡 + 7𝑒𝑒−𝑡𝑡 − 13𝑒𝑒−2𝑡𝑡 𝑢𝑢(𝑡𝑡)Answer:



𝑋𝑋 𝑠𝑠 =6(𝑠𝑠 + 34)

𝑠𝑠(𝑠𝑠2 + 10𝑠𝑠 + 34)

𝑥𝑥 𝑡𝑡 = 6 + 10𝑒𝑒−5𝑡𝑡cos(3𝑡𝑡 + 126.9o) 𝑢𝑢(𝑡𝑡)Answer:

𝑋𝑋 𝑠𝑠 =𝑘𝑘1𝑠𝑠

+𝐴𝐴𝑠𝑠 + 𝐵𝐵

𝑠𝑠2 + 10𝑠𝑠 + 34

𝑋𝑋 𝑠𝑠 =𝑘𝑘1𝑠𝑠

+𝑘𝑘2

𝑠𝑠 + 5 − 𝜋𝜋3+

𝑘𝑘3𝑠𝑠 + 5 + 𝜋𝜋3



𝐻𝐻 𝑠𝑠 =8𝑠𝑠 + 10

(𝑠𝑠 + 1) 𝑠𝑠 + 2 2

ℎ 𝑡𝑡 = 2𝑒𝑒−𝑡𝑡 − 2𝑒𝑒−2𝑡𝑡 + 6𝑡𝑡𝑒𝑒−2𝑡𝑡 𝑢𝑢(𝑡𝑡)Answer:

𝐻𝐻 𝑠𝑠 =𝑘𝑘1

(𝑠𝑠 + 1)+

𝑘𝑘2(𝑠𝑠 + 2)

+𝑘𝑘3

(𝑠𝑠 + 2)2

Properties of The Laplace Transform

Time Shifting Property

𝑥𝑥 𝑡𝑡 𝑋𝑋 𝑠𝑠

𝑥𝑥 𝑡𝑡 − 𝑡𝑡0 𝑋𝑋 𝑠𝑠 𝑒𝑒−𝑡𝑡0𝑠𝑠

Time delaying x(t) by to is equivalent to multiplying X(s) by 𝑒𝑒−𝑡𝑡0𝑠𝑠(phase shifting).

For causal signal

𝑥𝑥 𝑡𝑡 − 𝑡𝑡0 𝑢𝑢(t − 𝑡𝑡0) 𝑋𝑋 𝑠𝑠 𝑒𝑒−𝑡𝑡0𝑠𝑠

𝑥𝑥 𝑡𝑡 𝑢𝑢(𝑡𝑡) 𝑋𝑋 𝑠𝑠

Example: Time Shifting Property

Find the Laplace transform of the signal x(t)

Frequency Shifting Property


𝑥𝑥 𝑡𝑡 𝑒𝑒𝑠𝑠𝑜𝑜𝑡𝑡 𝑋𝑋 𝑠𝑠 − 𝑠𝑠𝑜𝑜

Multiplying the signal x(t) by 𝑒𝑒𝑠𝑠0𝑡𝑡 is equivalent to frequency shifting by so in the Laplace domain.

Example: Frequency Shifting Property

Find the Laplace transform of 𝑒𝑒−𝑎𝑎𝑡𝑡𝑐𝑐𝑐𝑐𝑠𝑠𝑐𝑐𝑡𝑡 𝑢𝑢(𝑡𝑡)

Time-Differentiation Property

𝑥𝑥 𝑡𝑡 𝑋𝑋 𝑠𝑠𝑑𝑑𝑥𝑥𝑑𝑑𝑡𝑡 𝑠𝑠𝑋𝑋 𝑠𝑠 − 𝑥𝑥(0−)

Frequency-differentiation property

𝑑𝑑2𝑥𝑥𝑑𝑑𝑡𝑡2 𝑠𝑠2𝑋𝑋 𝑠𝑠 − 𝑠𝑠𝑥𝑥 0− − 𝑥(0−)

𝑑𝑑3𝑥𝑥𝑑𝑑𝑡𝑡3 𝑠𝑠3𝑋𝑋 𝑠𝑠 − 𝑠𝑠2𝑥𝑥 0− − 𝑠𝑠𝑥 0− − 𝑥 (0−)

𝑑𝑑𝑛𝑛𝑥𝑥𝑑𝑑𝑡𝑡𝑛𝑛 𝑠𝑠𝑛𝑛𝑋𝑋 𝑠𝑠 −

𝑘𝑘=1

𝑛𝑛

𝑠𝑠𝑛𝑛−𝑘𝑘𝑥𝑘𝑘−1(0−)

𝑡𝑡𝑥𝑥 𝑡𝑡 −𝑑𝑑𝑑𝑑𝑠𝑠𝑋𝑋 𝑠𝑠

Example: Time Differentiation Property

Find the Laplace transform of the signal x(t)

Answer: 𝑋𝑋 𝑠𝑠 =1𝑠𝑠2

1 − 3𝑒𝑒−2𝑠𝑠 + 2𝑒𝑒−3𝑠𝑠

Time-Integration Property


𝑋𝑋(𝑠𝑠)𝑠𝑠

The dual property of time-integration property is the frequency-integration.

0−

𝑡𝑡

𝑥𝑥 𝜏𝜏 𝑑𝑑𝜏𝜏

𝑥𝑥(𝑡𝑡)𝑡𝑡

𝑠𝑠

∞

𝑋𝑋 𝑧𝑧 𝑑𝑑𝑧𝑧

Scaling Property


1𝑎𝑎𝑋𝑋

𝑠𝑠𝑎𝑎

for a > 0

Expansion in time domain result in compression in the s(frequency) domain, and the inverse is true.

𝑥𝑥 𝑎𝑎𝑡𝑡

Convolution Property


𝑋𝑋 𝑠𝑠 𝑌𝑌(𝑠𝑠)

Convolution in the time domain is equivalent to multiplication in the Laplace (frequency) domain.

𝑥𝑥 𝑡𝑡 ∗ 𝑦𝑦(𝑡𝑡)

y 𝑡𝑡 𝑌𝑌 𝑠𝑠

12𝜋𝜋𝜋𝜋

[𝑋𝑋 𝑠𝑠 ∗ 𝑌𝑌 𝑠𝑠 ]𝑥𝑥 𝑡𝑡 𝑦𝑦(𝑡𝑡)

Multiplication in the time domain is equivalent to multiplication in the Laplace (frequency) domain.

Laplace and Time Domains analysis of Systems

The convolution property is very useful in simplifying the system analysis in the Laplace domain.

Impulse Responseh(t)

Transfer FunctionH(s)

x(t) y(t)

X(s) Y(s)

𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 ∗ ℎ(𝑡𝑡)

𝑌𝑌 𝑠𝑠 = 𝑋𝑋 𝑠𝑠 𝐻𝐻 𝑠𝑠

Zero-state response. All initial conditions are zeros

𝐻𝐻 𝑠𝑠 =𝑌𝑌(𝑠𝑠)𝑋𝑋(𝑠𝑠)

Example: Convolution Property

Find y(t) where 𝑦𝑦 𝑡𝑡 = 𝑒𝑒𝑎𝑎𝑡𝑡𝑢𝑢(𝑡𝑡) ∗ 𝑒𝑒𝑏𝑏𝑡𝑡𝑢𝑢(𝑡𝑡)

𝑦𝑦 𝑡𝑡 = 1𝑎𝑎−𝑏𝑏

[𝑒𝑒𝑎𝑎𝑡𝑡 − 𝑒𝑒𝑏𝑏𝑡𝑡]𝑢𝑢(𝑡𝑡)Answer:

Summary of Laplace

Transform Properties

Using Laplace to Solve Differential Equations and System Analysis

Solving differential equation of third order and higher in the time domain to find the output of the system y(t) is challenging.

Solving differentials of any order in the Laplace domain is very easy since Laplace transform transfers differential equation into algebraic equation that can be easily solved to find Y(s).

If you set all initial conditions to zero then you will obtain only the zero-state response of the output y(t).

𝑑𝑑𝑛𝑛𝑦𝑦𝑑𝑑𝑡𝑡𝑛𝑛

=𝑑𝑑𝑚𝑚𝑥𝑥𝑑𝑑𝑡𝑡𝑚𝑚

𝑠𝑠𝑛𝑛𝑌𝑌 𝑠𝑠 −𝑘𝑘=1

𝑛𝑛

𝑠𝑠𝑛𝑛−𝑘𝑘𝑦𝑘𝑘−1 0− = 𝑠𝑠𝑚𝑚𝑋𝑋 𝑠𝑠 −𝑘𝑘=1

𝑚𝑚

𝑠𝑠𝑚𝑚−𝑘𝑘𝑥𝑘𝑘−1(0−)

Initial Conditions𝑠𝑠𝑛𝑛𝑌𝑌 𝑠𝑠 = 𝑠𝑠𝑚𝑚𝑋𝑋 𝑠𝑠 𝑌𝑌 𝑠𝑠 =𝑠𝑠𝑚𝑚

𝑠𝑠𝑛𝑛𝑋𝑋 𝑠𝑠

ExampleFind the transfer function H(s) of an LTIC system described by the equation

and the system response y(t) if the input x(t) = 3e-5tu(t) and all the initial conditions are zero; that is the system is in the zero state (relaxed).

Answer :

)()32()( 325 tueeety ttt −−− +−−=

)()()(6)(5)(2

2

txdt

tdxtydt

tdydt

tyd+=++

Example: System analysis in Laplace domain

Solve the differential equation to find y(t)

Given that 𝑦𝑦 0− = 2 and 𝑦 0− = 1 and the input 𝑥𝑥 𝑡𝑡 = 𝑒𝑒−4𝑡𝑡𝑢𝑢(𝑡𝑡)

Make sure to show the zero-input response, the zero-state response, and the transfer function of the system H(s).

𝑑𝑑2𝑦𝑦𝑑𝑑𝑡𝑡2

+ 5𝑑𝑑𝑦𝑦𝑑𝑑𝑡𝑡

+ 6𝑦𝑦 𝑡𝑡 =𝑑𝑑𝑥𝑥𝑑𝑑𝑡𝑡

+ 𝑥𝑥(𝑡𝑡)

Answer:

𝑦𝑦 𝑡𝑡 = 7𝑒𝑒−2𝑡𝑡 − 5𝑒𝑒−3𝑡𝑡 𝑢𝑢 𝑡𝑡 + −0.5𝑒𝑒−2𝑡𝑡 + 2𝑒𝑒−3𝑡𝑡 − 1.5𝑒𝑒−4𝑡𝑡 𝑢𝑢 𝑡𝑡

zero-input response zero-state response

ExampleIn the circuit, the switch is in the closed position for a long time before t = 0, when it is opened instantaneously. Find the inductor current y(t) for t ≥ 0.

10 V

2 Ω

t=0

5 Ω

1 H

0.2 F

y(t)

x(t)𝐿𝐿𝑑𝑑𝑦𝑦𝑑𝑑𝑡𝑡

+ 𝑅𝑅𝑦𝑦(𝑡𝑡) +1𝐶𝐶−∞

𝑡𝑡

𝑦𝑦(𝜏𝜏)𝑑𝑑𝜏𝜏 = 𝑥𝑥(𝑡𝑡)

ss

dy

ssYsYyssY 10

)(5)(5)(2)0()(

0

=+++−∫−

∞−−

ττ

Ay 25

10)0( ==− 2)0()(0

=== −

∞−∫−

CVqdy cττ

ssssYsYyssY 1010)(5)(2)0()( =+++− −

)()6.262cos(5)( tutety ot += −

Frequency Response of a LTI System

522)( 2 ++

=ssssH

|H(s)|

𝐻𝐻 𝜋𝜋𝜔𝜔 =2𝜋𝜋𝜔𝜔

−𝜔𝜔2 + 2𝜋𝜋𝜔𝜔 + 5

𝑠𝑠 = 𝜎𝜎 + 𝜋𝜋𝜔𝜔

If σ = 0

Transfer Function Frequency Response

Frequency Response of a LTI System

𝜔𝜔

𝜎𝜎

Laplace and Frequency Response of a LTI System

𝑥𝑥 𝑡𝑡 = Acos(𝜔𝜔𝑜𝑜𝑡𝑡 + 45)𝑦𝑦 𝑡𝑡 = 𝐻𝐻(𝜋𝜋𝜔𝜔𝑜𝑜) Acos 𝜔𝜔𝑜𝑜𝑡𝑡 + 45 + ∠𝐻𝐻(𝜋𝜋𝜔𝜔𝑜𝑜)

𝑥𝑥 𝑡𝑡 = 𝑖𝑖

𝑋𝑋(𝑠𝑠𝑖𝑖)𝑒𝑒𝑠𝑠𝑖𝑖𝑡𝑡 𝑦𝑦 𝑡𝑡 = 𝑖𝑖

𝐻𝐻(𝑠𝑠𝑖𝑖)𝑋𝑋(𝑠𝑠𝑖𝑖)𝑒𝑒𝑠𝑠𝑖𝑖𝑡𝑡

X(s) H(s) Y(s)

𝑠𝑠 = 𝜎𝜎 + 𝜋𝜋𝜔𝜔If σ = 0 in the complex frequency s, so s = jω then system analysis is simplified further and it becomes frequency response.

X( jω ) H( jω) Y( jω)

𝑥𝑥 𝑡𝑡 = 𝑖𝑖

𝑋𝑋(𝜋𝜋𝜔𝜔𝑖𝑖)𝑒𝑒𝑗𝑗𝜔𝜔𝑖𝑖𝑡𝑡 𝑦𝑦 𝑡𝑡 = 𝑖𝑖

𝐻𝐻(𝜋𝜋𝜔𝜔𝑖𝑖)𝑋𝑋(𝜋𝜋𝜔𝜔𝑖𝑖)𝑒𝑒𝑗𝑗𝜔𝜔𝑖𝑖𝑡𝑡

Amplitude Response Phase Response

Example: Frequency Response

For the system described by the transfer function

𝐻𝐻 𝑠𝑠 = 𝑠𝑠 + 0.1𝑠𝑠 + 5

Find the frequency response H(jω), and the system response y(t) for input x(t) = cos2t and x(t) = cos(10t-50° )

Replace s by jω

𝐻𝐻(𝜋𝜋𝜔𝜔) =𝜋𝜋𝜔𝜔 + 0.1𝜋𝜋𝜔𝜔 + 5

Magnitude and Phase of the Frequency Response

2

2

0.01( )25

H j ωωω+

=+

1 1( ) ( ) tan tan0.1 5

j H j ω ωω ω − − Φ = ∠ = −


For input x(t)=cos2t, we have:

Therefore

2

2

2 0.01( 2) 0.3722 25

H j += =

+1 12 2( 2) tan tan 65.3

0.1 5j − − Φ = − =

( ) 0.372cos(2 65.3 )y t t= +


For input x(t)= cos(10t - 50°), we will use the amplitude and phase response curves directly:

Therefore

( 10) 0.894H j =

( 10) ( 10) 26j H jΦ = ∠ =

( ) 0.894cos(10 50 26 ) 0.894cos(10 24 )y t t t= − + = +

Frequency Response of Ideal Delay

H(s) of an ideal T sec delay is:

𝐻𝐻 𝑠𝑠 = 𝑒𝑒−𝑠𝑠𝑠𝑠 𝐻𝐻 𝜋𝜋𝜔𝜔 = 𝑒𝑒−𝑗𝑗𝜔𝜔𝑠𝑠

The magnitude 𝐻𝐻 𝜋𝜋𝜔𝜔 = 1The phase 𝜃𝜃 𝜋𝜋𝜔𝜔 = −𝜔𝜔𝑛𝑛An ideal delay system does not effect the amplitude of the input but phase shift the input with –T gradient.

The group delay is the time delay of all frequencies 𝑑𝑑𝜃𝜃

𝑑𝑑𝜔𝜔= −𝑛𝑛

x(t) DelayT

y(t)=x(t-T)X(s) Y(s)= X(s) e-sT

Frequency Response of an Ideal Differentiator

H(s) of an ideal differentiator is:

𝐻𝐻 𝑠𝑠 = 𝑠𝑠 𝐻𝐻 𝜋𝜋𝜔𝜔 = 𝜋𝜋𝜔𝜔 = 𝜔𝜔𝑒𝑒𝑗𝑗𝑗𝑗/2

The magnitude 𝐻𝐻 𝜋𝜋𝜔𝜔 = 𝜔𝜔The phase is π/2If x(t) = cosωt what is the output

Differentiators amplify inputs with high frequencies such as noise and for that it is avoided in system design.

(cos ) sin cos( / 2)d t t tdt

ω ω ω ω ω π= − = +

x(t) 𝑑𝑑𝑑𝑑𝑡𝑡

𝑦𝑦 𝑡𝑡 =𝑑𝑑𝑥𝑥𝑑𝑑𝑡𝑡X(s)

Y(s)= sX(s)

Frequency Response of an ideal Integrator

H(s) of an ideal integrator is:

𝐻𝐻 𝑠𝑠 =1𝑠𝑠

𝐻𝐻 𝜋𝜋𝜔𝜔 =1𝜋𝜋𝜔𝜔

=1𝜔𝜔𝑒𝑒−𝑗𝑗𝑗𝑗/2

The magnitude 𝐻𝐻 𝜋𝜋𝜔𝜔 = 1/𝜔𝜔The phase is -π/2If x(t) = cosωt what is the output

Integrator supress inputs with high frequencies such as noise and for that it is used in system design.

x(t)𝑥𝑥 𝑡𝑡 𝑑𝑑𝑡𝑡

X(s)𝑌𝑌 𝑠𝑠 = 1

𝑠𝑠X(s)𝑦𝑦 𝑡𝑡 = 𝑥𝑥 𝑡𝑡 𝑑𝑑𝑡𝑡

1 1cos sin cos( / 2)t dt t tω ω ω πω ω

= = −∫

Internal Stability

• Internal Stability (Asymptotic)– If and only if all the poles are in the LHP– Unstable if, and only if, one or both of the

following conditions exist:• At least one pole is in the RHP• There are repeated poles on the imaginary axis

– Marginally stable if, and only if, there are no poles in the RHP, and there are some unrepeated poles on the imaginary axis.

External Stability BIBOThe transfer function H(s) can only indicate the external stability of the system BIBO.

NNN

MMM

asasbsbsbsH

++++++

= −

−

......)( 1

1

110

ExampleIs the system below BIBO and asymptotically (internally) stable?

11−S 1

1+−

SS

x(t) y(t)

BIBO stable if M ≤ N and all poles are in the LHP

1. Find the Laplace transform X(s) for signal x(t) using the integral2. Find the Laplace transform X(s) for signal x(t) using the Laplace properties and

table3. Find the signal x(t) by the inverse Laplace transform of X(s) using the partial

fraction expansion, Laplace properties, and the table.4. Given the differential equation of a system, the input x(t), and the initial conditions

of the system, you should be able to find the transfer function H(s), the zero-input response, the zero-state response, the forced response, the natural response, or the total response for the given input x(t).

5. Given the system differential equation or the transfer function H(s), you should be able to find the frequency response H(jω), its magnitude and phase. If a dc or sinusoidal input is given then you should be able to find the output y(t) and the steady state response.

6. Given the system differential equation or the transfer function H(s), you should be able to find the roots and zeros of the system and determine its external and internal stability.

Topics for Test 2

continuous-time system analysis using the laplace transformmbingabr/signals_systems/sigsys... ·...

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