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2. Definition of the Laplace Transform

The Laplace transform provides a useful method of solving certain types of differential equations when certain initial conditions are given, especially when the initial values are zero.The Laplace transform is also very useful in the area of circuit analysis (which we see later in the Applications section) . It is often easier to analyse the circuit in its Laplace form, than to form differential equations. The techniques of Laplace transform are not only used in circuit analysis, but also in Proportional-Integral-Derivative (PID) controllers DC motor speed control systems DC motor position control systems Second order systems of differential equations (underdamped, overdamped and critically damped)

Definition of Laplace Transform of f(t)The Laplace transform of a function f(t) for t > 0 is defined by the following integral defined over 0 to :{ f(t)} = The resulting expression is a function of s, which we write as F(s). In words we say "The Laplace Transform of f(t) equals function F of s"and write:{f(t)} = F(s)Similarly, the Laplace transform of a function g(t) would be written:{g(t)} = G(s)

The Good NewsIn practice, we do not need to actually find this infinite integral for each function f(t) that we have to find the Laplace Transform for. There is a table of Laplace Transforms which we can use.Go to the Table of Laplace Transformations. Table of Laplace TransformationsThe following Table of Laplace Transforms is very useful when solving problems in science and engineering that require Laplace transform.Each expression in the right hand column (the Laplace Transforms) comes from finding the infinite integral that we saw in the Definition of a Laplace Transform section.Time Function f(t) f(t) = -1{F(s)} Laplace Transform of f(t)F(s) = { f(t)}

1s > 0

t (unit-ramp function)s > 0

tn (n, a positive integer)s > 0

eats > a

sin ts > 0

cos ts > 0

tng(t), for n = 1, 2, ...

t sin ts > ||

t cos ts > ||

g(at) Scale property

eatg(t)G(s a) Shift property

eattn, for n = 1, 2, ...s > a

te-ts > -1

1 e-t/Ts > -1/T

eatsin ts > a

eatcos ts > a

u(t)s > 0

u(t a)s > 0

u(t a)g(t a)e-asG(s) Time-displacement theorem

g'(t)sG(s) g(0)

g''(t)s2 G(s) s g(0) g'(0)

g(n)(t) sn G(s) sn-1 g(0) sn-2 g'(0) ... g(n-1)(0)

In the following sections we see how to use the Table of Laplace Transformations to solve problems.6. Laplace Transforms of IntegralsWe first saw the following properties in the Table of Laplace Transforms.

1. If G(s) = {g(t)}, then

2. For the general integral, if

is the value of the integral when t = 0, then:

EXAMPLESUse the above information and the Table of Laplace Transforms to find the Laplace transforms of the following integrals:(a) AnswerIn this example, g(t) = cos at and from the Table of Laplace Transforms, we have:G(s) = cos at = s / (s2 + a2) Now, applying the first rule above, we have:

(b)

AnswerThis is similar to example (b). We find the transform of the function g(t) = eatcos bt, then divide by s, since we are finding the Laplace transform of the integral of g(t) evaluated from 0 to t.

(c)

AnswerThis follows the same process as examples (a) and (b). Find the Laplace transform of the function g(t) = te-3t then divide by s.

(d)

AnswerRecall from the Double Agle Formula thatsin 2 = 2 sin cos We can use this to re-express our integrand (the part we are integrating):

So the Laplace Transform of the integral becomes:

3. Some Properties of Laplace TransformsWe saw some of the following properties in the Table of Laplace Transforms.Property 1. Constant Multiple If a is a constant and f(t) is a function of t, then{a f(t)} = a{f(t)}Example{7 sin t} = 7{sin t}[This is not surprising, since the Laplace Transform is an integral and the same property applies for integrals.] Property 2. Linearity PropertyIf a and b are constants while f(t) and g(t) are functions of t, then{a f(t) + b g(t)} = a{f(t)} + b{g(t)}Example{3t + 6t2 } = 3 {t} + 6{t2}Property 3. Change of Scale PropertyIf {f(t)} = F(s) then Example

Property 4. Shifting Property (Shift Theorem){eatf(t)} = F(s a)Example{e3tf(t)} = F(s 3)Property 5.

Property 6.The Laplace transforms of the real (or imaginary) part of a complex function is equal to the real (or imaginary) part of the transform of the complex function.Let Re denote the real part of a complex function C(t) and Im denote the imaginary part of C(t), then{Re[C(t)]} = Re {C(t)}and{Im[C(t)]} = Im {C(t)}If you need some background, go to Complex Numbers. EXAMPLESObtain the Laplace transforms of the following functions, using the Table of Laplace Transforms and the properties given above.(We can, of course, use Scientific Notebook to find each of these. Sometimes it needs some more steps to get it in the same form as the Table).(a) f(t) = 4t2AnswerWe use:

and the Constant Multiple property from above, to obtain:

(b) v(t) = 5 sin 4tAnswerWe use: Clearly, = 4.

(c) g(t) = t cos 7tAnswerWe use It is simple to obtain

DEMONSTRATION of PROPERTY 5:{t f(t)}For example (c), we could have also used Property 5:

with f(t) = cos 7t.Now So

So

This is the same result that we obtained using the formula.For a reminder on derivatives of a fraction, see Derivatives of Products and Quotients.

(d) f(t) = e2t sin 3tAnswerWe use

and simple substitution yields:

DEMONSTRATION OF No 4: SHIFTING PROPERTYFor example (d) we could have used:{eatg(t)} = G(s a)Let g(t) = sin 3t

So

This is the same result we obtained before for example (d).

(e) f(t) = t4e-jt

AnswerFrom the Laplace table, we have: So

(f) f(t) = te-t cos 4t

AnswerWe will use: Let Then Now So

(g) f(t) = t2 sin 5tAnswerWe will use , with . Now

Now for the second derivative:

For the formula, we need:

So

(h) f(t) = t3 cos t = t2(t cos t)

AnswerSo we are letting . This time we need the 2nd derivative of

(From the Table of Laplace Transforms.)First derivative:

Second derivative:

Now .So

(i) f(t) = cos23t, given that

AnswerFor this one, we need to apply the Scale Property:

Here, a = 3 so

10. Applications of Laplace TransformsCircuit EquationsThere are two (related) approaches:1. Derive the circuit (differential) equations in the time domain, then transform these ODEs to the s-domain; 2. Transform the circuit to the s-domain, then derive the circuit equations in the s-domain (using the concept of "impedance"). We will use the first approach. We will derive the system equations(s) in the t-plane, then transform the equations to the s-plane. We will usually then transform back to the t-plane.EXAMPLE 1Consider the circuit when the switch is closed at t = 0 with VC(0) = 1.0 V. Solve for the current i(t) in the circuit.

Answer

Multiplying throughout by :

Taking Laplace transform:

Now in this example, we are told .

So

That is: Therefore: NOTE:

Collecting I terms and subtracting from both sides:

Finding the inverse Laplace transform gives us

Note: Throughout this page these problems are also solved using Scientific Notebook. They are TEX files and you need Scientific Notebook or similar, to view them.Alternative answer using Scientific Notebook. (.tex file)

EXAMPLE 2Solve for i(t) for the circuit, given that V(t) = 10 sin5t V, R = 4 W and L = 2 H.

Answer

So

Equating coefficients of :gives Equating coefficients of :gives So

So we haveA

Alternative answer using Scientific Notebook. (.tex file) EXAMPLE 3In the circuit shown below, the capacitor is uncharged at time t = 0. If the switch is then closed, find the currents i1 and i2, and the charge on C at time t greater than zero.

AnswerNOTE: We could either: Set up the equations, take Laplace of each, then solve simultaneously Set up the equations, solve simultaneously, then take Laplace.It is easier in this example to do the second method. In many examples, it is easier to do the first method.For the first loop, we have:

For the second loop, we have:

Substituting (2) into (1) gives:

Next we take the Laplace Transform of both sides.Note:

In this example, . So

Now taking Inverse Laplace:

And using result (2) from above, we have:

For charge on the capacitor, we first need voltage across the capacitor:

So, since , we have:

Graph of q(t):

Alternative answer using Scientific Notebook. (.tex file)

EXAMPLE 4In the circuit shown, the capacitor has an initial charge of 1 mC and the switch is in position 1 long enough to establish the steady state. The switch is moved from position 1 to 2 at t = 0. Obtain the transient current i(t) for t > 0.

AnswerPosition 1, after a `long time': APosition 2: ()We apply emf, and consider the sum of the potential difference across elements.In position 2, there is no emf.

Finding Laplace Transform:

Multiplying by :

Solving for I and completing the square on the denominator gives us:

So the transient current is:

We could transform the trigonometric part of this to a single expression:2 cos 222.2t 0.45 sin 222.2t = R cos(222.2t + )

So

Alternative answer using Scientific Notebook. (.tex file) EXAMPLE 5The system is quiescent. Find the loop current i2(t).

AnswerQuiescent implies i1, i2 and their derivatives are zero for t = 0, ie i1(0) = i2(0) = i1'(0) = i2'(0) = 0.For loop 1:

For loop 2:

Substituting our result from (1) gives:

Taking Laplace transform:

Let So

So Taking Inverse Laplace:So

Alternative answer using Scientific Notebook. (.tex file) EXAMPLE 6Consider a series RLC circuit where R = 20 W, L = 0.05 H and C = 10-4 F and is driven by an alternating emf given by E = 100 cos 200t. Given that both the circuit current i and the capacitor charge q are zero at time t = 0, find an expression for i(t) in the region t > 0.AnswerWe use the following:

and obtain:

After multiplying throughout by 20, we have:

Taking Laplace transform and using the fact that i(0) = 0:

Using Scientific Notebook to find the partial fractions:

So

So+ cos200t 2 sin 200t

NOTE: Scientific Notebook can do all this for us very easily. In one step, we have:

+ cos200t 2 sin 200t

Transient part:

Steady state part:

Alternative answer using Scientific Notebook. (.tex file)

EXAMPLE 7A rectangular pulse vR(t) is applied to the RC circuit shown. Find the response, v(t).Graph of vR(t):

Note: v(t) = 0 V for all t < 0 s implies v(0-) = 0 V.AnswerNow To solve this, we need to work in voltages, not current.We start with .The voltage across a capacitor is given by .It follows that .So for this example we have:

Substituting known values:

Then

Taking Laplace:Since , we have:

So, taking inverse Laplace

NOTE: For the part: , we use:

So we have:

Solution Using Scientific Notebook1. To find the Inverse Laplace:

2. To solve the original DE:

Exact solution for v(t):

To see what this means, we could write it as follows:

To see what our expression for v(t) means, we graph it as follows: