what is definition of laplace transform of e^at meaning and find out the laplace transform of f(t) = e-at for t > 0 ?

**Laplace Transform**

**Laplace Transform** The beauty of the laplace transform method for solving differential equatio is that. it is a simple and systematic method which provides the total solution in one stroke by taking into account the initial condition in a natural way at beginning of the process itself.

**Laplace Transform**

If f(t) is the time domain function then its laplace transform is denoted by F(s) and is defined by the equation

L[F(t)] = F(s) = |f(t)_{e}^{-st} dt

s = o + j_{00} = complex frequency

0 = attenuation constant or damping factor

_{00} =angular frequency

To take into account the possibility that f(t) may be an impulse or one of its higher dervatives, the lower limit of integration is taken as 0. this includes the time just before the instant t = 0.

In the case when no impulse or higher order derivatives of impulses are involved, i.e., for continuous functions f (0-) = f(0^{+}), then for such functions, the integration can be effectively taken from zero to infinity.

L[f(t)] F (s) = | f(t)_{e}^{-st} dt

(when f(t) is continuous)

The given function f(t) possesses a laplace transform only if it satisfies the condition given by

-|f(t)_{e}^{-0t} dt <_{00}, for real, positive

inverse Laplace transform

f(t) = l^{-1} [f(s) = 1/2j |f(s)_{e}^{st} ds

**Example 1.** find out the laplace transform of f(t) = e^{-at} for t > 0.

**Sol.** As function existing for t > 0, its laplace transform [LT] can be obtained, as

F (s) = |f(t)_{e}^{-st} dt = |e^{-at} e^{-st} dt = |e^{-(s + a)t} dt

f(s) = 1/s + a

l[e^{-at}] = 1/s + a

**Properties of Laplace Transform**

**Linearity**

The transform of a finite sum of time functions is the sum of the laplace transforms of the individual functions

L[F_{1}(T) + F_{2} (T) + ….F_{N} (T)] = F_{1} (S) + F_{2} (S) +…..+ F_{N} (S)

**Scaling Theorem**

L[Kf (t)] = KF (s)

**Real Differentiation**

L{d/dt f(t)} = sf (s) – f (0^{–})

L [d^{n}/dt^{n} f(t)} = s^{n}F (s) – s^{n-1} f(0^{–}) – s^{n-2} f^{‘ }(0^{–}) – …….f’^{[n – (n – 1)] }(0^{–})

**Differentiation by s**

L (T) f(t) = – d/ds f(s)

**Complex Translation**

F (s – a) = L {e^{at} f(t) }

where, a is the complex number.

and L (e^{at} f (t)} = F (s + a)

**Real Translation (Shifting theorem)**

**L {f (t – T)} = e ^{-ts} f (s)**

**Initial value Theorem**

f (0^{+}) = lim f(t) = lim sf (s)

The only restrction is that f (t) must be continuous or at the most, a step discontinuity at t = 0.

**Final Value Theorem**

lim f (t) = lim sf (s)

The only restriction is that the roots of the denominator polynomial of f (s), i.e., poles of f (s) have negative or zero real parts.

**Example 2. ** Find the laplace transfom of sin _{oo}t.

**Sol.** sin _{oo}t = e^{j00t} – e^{-joot}/2j

L{sin_{ 00}t} = l {e^{joot} – e^{-joot}/2j}

= l {e^{j00t}/2j} – l {e^{-j00t}/2j}

l {e^{at}} = 1/s – a and l{e^{-at}} = 1/s + a

l {sin _{oot}} = 1/2j {1/s – j_{oo}} – 1/2j {1/s +j_{oo}}

= _{oo}/s^{2} + _{oo}^{2}

l {sni _{oo}t} = _{oo}/s^{2} + _{oo}^{2}

**Example 3.** Find the final value of f (t) = 3 + 2e^{-t} using final value theorem.

**Sol.** f(t) = 3 + 2e^{-t}

f(s) = l[3] + 2 l [e^{-t}]

f(s) = 3/s + 2/s + 1

f(s) = 5s + 3/s (s + 1)

by final value theorem,

f (_{oo}) = lim sF(s)

= lim s x [5s + 3/s(s + 1)]

f (_{oo}) = 3

**Standard Time Functions**

**Step Function**

f(t) = 0 for t < 0

= A for t > 0

**unit step function [u(t)]**

U(t) = 1 for t > 0

= 0 for t < 0

**Shifted unit step function**

u(t – a) = 0, t < a

= 1, t > a

**Example 4.** Express the waveform shown in figure in terms of step functions.

**Sol.** f_{1}(t) = 3u (t – 1)

f_{2} (t) = – 3u (t – 2)

f(t) = f_{1}(t) + f_{2}(t)

f(t) = 3u (t – 1) – 3u (t – 2)

**Ramp Function**

f(t) = At, t > 0

= 0, t < 0

**Unit ramp function**** [r(t)]**

r(t) = t, t > 0

= 0, t < 0

r(t) = tU(t) for t > 0

= 0 for t < 0

similarly, the ramp of magnitude A can be expressed in terms of unit step function as

x tU (t)

**Shifted unit ramp function**

f (t) = (t – a) for t > a

= 0 for t < a

Such a delayed ramp can be expressed using delayed unit step function, by the same amount,

f (t) = (t – a) u (t – a)

f (t) = A (t – a) U (t – a) for t > a

= 0 for t < 0

f (t) = – A x U (t)

f (t) = – A(t – a) U (t – a)

**Addition of two ramp functions**

f (t) = f_{1} (t) + f_{2} (t) = 0, constant for t > 0

**Example 5. **Express the waveform shown in figure in tems of the standard functions.

f_{1}(t) = r(t) = t U (t)

f_{2}(t) = – (t – 1) u (t – 1)

f_{3 }(t) = – (t – 1) u (t – 1)

f_{4}(t) = + (t – 2) u (t – 2)

hence, the given waveform can be expressed in terms of four ramps as

f(t) = f_{1}(t) + f_{2}(t) + f_{3}(t) + f_{4}(t)

f(t) = t U (t) – (t – 1) u (t – 1) – (t – 1) u (t – 1) + (t – 2) u (t – 2)

f (t) = t U (t) – 2 (t – 1) u (t – 1) + (t – 2) u (t – 2)

**Impulse Function**

The impulse function is also called delta function. the unit impulse function is denoted as (t).

area under the pulse is unity, i.e., (t) dt = 1

consider a reactangular pulse of unit area having width T_{1} and amplitude 1/t_{1}

A_{1} = Width x amplitude = t_{1} x 1/t_{1} = 1

A_{2} = t_{2} x 1/t_{2 } = 1, where t_{3} < t_{2} < t_{1}

Such a function i.e., pulse with infinite amplitude and zero width having unit area under the pulse is called unit impulse function or delta function.

(t) = lim f(t)

area = lim [t x 1/t] = 1

hence, the rectangular pulse can be expressed as

f (t) = u (t) – u (t – t)

this is a pulse of width t

(t) = lim 1/t [u (t) – u (t – t)]

**Delayed unit impulse function**

f (t) = (t – a)

**Important Properties of impulse function**

**Shifting property**

f (t) (t – t) dt = f(t)

**Replication property**

f (t) (t – t) d = f (t)

in other words, f(t)* (t) = f (t)

**Example 6.** Find the laplace transform of the function.

**Sol.** f_{1} (t) = u (t)

f_{2} (t) = u (t-1)

f_{3} (t) = -u (t-2)

f_{4} (t) = – u (t-3)

y(t)= f_{1}(t) + f_{2}(t) + f_{3}(t) + f_{4}(t)

y (t) = u (t) + u (t-1) – u (t – 2) – u(t – 3)

y(s) = 1/5 + 1/s e^{-1} – 1/s e^{-s} – 1/s e^{-2s}– 1/s e^{-3s}

y (s) = 1/s [1 + e^{-s} – e^{-2s} – e^{-3s}]

**Laplace Transform of a Periodic Function**

f (t + nt) = f (t), where n is positive or negative integer.

f (s) = 1/1 – e^{-st} f_{1} (s)

where, F_{1 }(s) is the laplace transform of the first cycle of the periodic function.

**Example 7.** Obtain the laplace transform of the square wave train shown in the figure

**Sol. **consider the first cycle of the function f(t) as shown in the figure.

Let t be the time period of f(t) which is 2t.

it can be noticed that f_{1}(t) starts with a step of 1 at t = 0

f_{a} (t) = u (t)

at t = t, there is instantaneous change in f_{1}(t) from 1 to -1, so there is step of -2 at t = 2t

f_{c}(t) = u (t – 2t)

f_{1} (t) = f_{a} (t) + f_{b} (t) + f_{c} (t)

f_{1} (t) = u (t) – 2u (t – t) + u (t – 2t)

f_{1}(s) = 1/s – 2/s e^{-ts} + 1/s e^{-2ts}

f (s) = f_{1} (s)/1 – e^{-ts} = 1/s [1 – 2e^{-ts} + e^{-2ts}]/1 – e^{-2ts}

f (s) = 1/s [1 – e^{-ts}]^{2}/s [1 + e^{-ts}] [1 – e^{-ts}]

f (s) = 1/s . 1 – e^{-ts}/1 + e^{-ts} = 1/s tanh (ts/2)

**Intro Exercise – 4**

- Laplace transform analysis gives

(a) time domain response only

(b) frequency domain response only

(c) both (a) and (b)

(d) none of the above

- Which of the following is an advantage of using laplace transform techniques?

(a) permits use of simple algebra

(b) converts fuctions in the t-domain into s-domain

(c) initial conditions are automatically taken care of

(d) all of the above

- Which of the following correctly defines laplace transform of a function in the time domain?

(a) L {f(t)] |f (t) e^{-st} dt

(b) L {f(t) e^{+st} dt

(c) L {f(t) = |f(t)^{-st} e-^{st} dt

(d) L {f(t) ] = | f(s) e^{-st} dt

- The initial value theorem does not hold good for which of the following functions?

(a) ramp function

(b) delta function

(c) step function

(d) hyperbolic function

- The laplace transform of l(t) is given by l(s) = 5/s(s
^{2}+ 2) . as t –_{ 00}the value of l(t) tends to

(a) 0

(b) 1

(c) 5/2

(d) _{00}

- Consider the signal e
^{-t}U(t). the laplace transform of the derivative of the signal is

(a) s + 1

(b) s/(s + 1)

(c) s/s + 2

(d) 1/ (s + 1)

- Find the value of Z(is) the figure given below

(a) 3s^{2} + 8s + 7/s(5s + 6)

(b) s(5s + 6)/3s^{2} + 8s + 7

(c) 3s^{2} + 7s + 6/s(5s + 6)

(d) 2s (8s + 3)/8s^{2} + 10s + 3

- Inverse laplace transform of the function s/s
^{2}+ 3s + 2, is

(a) -e^{-t} + 2e^{-2t}

(b) e^{-t} – 2e^{-2t}

(c) e^{-t} + 2e^{-2t}

(d) 2e^{-t} + e^{-2t}

- Current I(t) = 2e
^{-2(1 – 3)}U(t – 3) A for t > 0 and voltage V(t) is given as 2 (t – 3), then the circuit components are

(a) R and C

(b) R and L

(c) L and C

(d) R only

- Find the value of Z (s) in the figure given below

(a) s^{2} + 1.5s + 1

(b) s^{2} + 3s + 1/s(s + 1

(c) 2s^{2} + 3s + 2/s(s + 1)

(d) 2s^{2} + 3s + 1/2s (s + 1)

- Let I(t) = 1/8 [e
^{-3t}+ 3 1e^{-5t}]. the initial and final values of currents are, respectively

(a) Zero, 2 A

(b) zero, 4 A

(c) 4 A, zero,

(d) 2 A, zero

- Find the inverse laplace transform of F(s) = s
^{2}+ 2s – 2/s(s + 4) (s – 5)

(a) (1/10 + 1/6 e^{-4t} + 11/15e^{5t}) u(t)

(b) (1/10 + 1/6 e^{6t} + 10/15 e^{5t}) u (t)

(c) (1 + 1/6 e^{-4t} + 10/15 e^{5t}) u(t)

(d) (1/10 – 1/6 e^{4t} + 10/15 e^{-5t}) u(t)

- The circuit is shown in figure below the current ratio transfer function I
_{0}/I_{S}is

(a) s(s + 4)/s^{2} + 3s + 4

(b) s (2s + 2)/(s + 1) (2s + 3)

(c) s^{2} + 3s + 4/s (s + 4)

(d) (s + 1) (s + 3)/s(s + 4)

- The voltage of the source, i.e., V
_{S}if I(t) = – 20e^{-2t}

(a) 10e^{-2t}

(b) 10e^{2t}

(c) 20e^{-2t}

(d) 20e^{-2t}

- A unit step current of 1 A is applied to a network whose driving point impedance is

Z(s) = V(s) /I (s) = (s + 2)/(s + 1)^{2}

The steady state and initial values of the voltage developed across the source would be, respectively

(a) 3/4 V, 1 V

(b) 1/4 V, 3/4 V

(c) 2 V, zero

(d) 1 V, 3/4 V

- A voltage V(T) = 6e
^{-2t}is applied at t = 0 to a serise RL circuit with L = 1 H. if I(t) = 6 [e^{-2t}– e^{-3t}], then R will have a value of

(a) 2/3

(b) 1

(c) 3

(d) 1/3

- Consider laplace transform of two signals as 1/s + 2 and 1/s + 1 the resultant signal pbtained after frequency convolution of these two signals along the constant multiplier 1/2j is

(a) e^{-2t} + e^{-t}

(b) e^{-3t}

(c) e^{-2t}

(d) e^{-t}

- Find I(t) in the figure given below

(a) 10 – 8e^{2000t}

(b) 10 + 8e^{-2000t}

(c) 10 – 8e^{-2000t}

(d) none of these

- The impedance of following network consists of I(t) = e
^{-1}– e^{-(t – 2)}

(a) resistor only

(b) resistor and capacitor

(c) capacitor, inductor and resistor

(d) resistro and inductor

- Given transfer function H(s) = s + 2/s
^{2}+ s + 4, unber steady state condition, the sinusoidal input and output are, respectively x(t) = cos 2t y(t) = cos (2t + 0), then angle 0 will be

(a) 45^{0}

(b) 0^{0}

(c) -45^{0}

(d) – 90^{0}

**Answers with Solutions**

- (c)

we know that laplace transform changes the time domain function f(t) to the frequency domain function f(s). similarly, inverse laplace transformation converts frequency domain function f(s) to the time domain function f(t).

- (d)

all the given conditions are the advantages of using laplace transform techniques.

- (a)

laplace transform is defined by

l{f(t)} = f(t)e^{-st} dt

- (b)

laplace transform of delta function

l{(t)} = 1

initial value theorem lim f(t) = lim sF(s)

f(t) = s(t)

f(s) = 1

so, delta function does not hold good.

- (c)

by final value theorem, lim I(s) .s

lim (5s/s^{2} + 2) 1/s = lim 5/s^{2} + 2 = 5/2

- (c)

f(t) = e^{-1}U(t)

derivation of function f(t) is

f'(t) = e^{-1} (t) + u(t) (-e^{-1})

f'(t) = -e^{-t} u(t) + e^{-t} (t)

taking laplace transform,

f'(s) = – 1/s + 1 + 1

f’ (s) = s /s + 1

- (d)

Z(s) = 2s ||[1 + 2(2 + 1/s)/2 + 2 + 1/s]

Z(s) = 2s[8s + 3/8s^{2} + 10s + 3]

- (a)

given, f(s) = s/s^{2} + 3s + 2 = s/(s + 1) (s + 2)

applying partial fraction method.

s/(s + 1)(s + 2) = A/(s + 1) + B/(s + 2)

A = s/s + 2|_{s = – 1}

A = – 1

B = s/s + 1|_{s = – 2} = 2

f(s) = s/(s + 1)(s + 2) = -1/(s + 1) + 2/(s + 2)

taking inverse laplace transform,

f(t) = – e^{-t} + 2e^{-2t}

- (b)

given, i(t) = 2e^{-2(t – 3)}u(t – 3)

taking laplace transform of i(t),

I(s) = 2e^{-3s}/s + 2

sI(s) + 2I (s) = 2e^{-3s}

taking inverse laplace transform we have

2i(t) = di(t)/dt = 2 (-3)

given V(t) = 2 (t – 3)

2i (t) + di (t)/dt = V (t)

V(t) = Ri (t) + l di(t)/dt

R = 2

L = 1 H

So, the circuit components are R and L

- (a)

Z(s) = 1/s + (1) (1 + 2s)/1 + 1 + 2s

Z(s) = s^{2} + 1.5s + 1/s(1 + s)

- (c)

given i(t) = 1/8 [e^{-3t} + 31e^{-5t}]

taking laplace transform of I (t)

I(s) = 1/8 [1/s + 3] + 31/8 [1/s + 5]

by initial value therorm,

lim f(t) = lim sF (s)

i(0^{+}) = lim sF (s) = lim [1/8(s/s + 3) + 31/8 (s/s + 5)]

i(0^{+}) = 4 A

by final value theorem lim I(t) = lim sF (s)

i(_{00}) = lim sI (s) = lim [1/8 (s/s + 3) + 31/8 (s/s + 5)

i(_{00}) = 0 A

- (a)

given, F (s) = s^{2} + 2s – 2/s(s + 4) (s – 5)

first expanding into partial fractions,

f(s) = a/s + b/s + 4 + c/s – 5

A = s^{2} + 2s – 2/(s + 4)(s – 5)|_{s = 0} = 1/10

B = s^{2 }+ 2s – 2/s(s – 5)|_{s = – 4} = 1/6

C = s^{2} + 2s – 2/s (s + 4)|_{s = 5} = 11/15

hence, f(s) = 1/10s + 1/6(s + 4) + 11/15(s – 5)

taking inverse laplace transform

f(t) = (1/10 + 1/6 e^{-4t} + 11/15e^{5t}) u(t)

- (b)

using current divider rule,

I_{0} = (2s + 5/2s + 5 + 3/5) x I_{S}

I_{0}/I_{S} = s(2s + 5)/(s + 1) (2s + 3)

I_{0}/I_{S} = s(2s + 5)/(s + 1) (2s + 3)

- (a)

given, I(t) = – 20e^{-2t}

I_{C} = C dV(t)/dt

I_{C} = 2 d/dt (-20e^{-2t}) or I_{C} = 80e^{-2t}

I = I(t) + I_{E}

I = – 20e^{-2t} + 80e^{-2t}

I = 60 e^{-2t}

V_{S} = V_{R} + V_{L} + V(t)

V_{S} = 1 x 60e^{-2t} + L di/dt + V (t)

V_{S} = 1 x 60 e^{-2t} + 1/4 d/dt (60 e^{-2t}) – 20e^{-1}

V_{S} = 60 e^{-2t} – 15 x 2 e^{-2t} – 20 e^{-2t}

V_{S} = 60 e^{-2t} – 30e^{-2t} – 20e^{-2t}

V_{S} = 10e^{-2t}

- (c)

given, I(t) = u(t)

taking laplace transform,

I(s) = 1/s, V (s) = (s + 2)/s(s + 1)^{2}

by initial value theorem,

V (0) = lim sV (s) = 0

by final value theorem,

V (_{00}) = lim sV (s) = 2

- (c)

writing KVL,

Ri(t) + l di(t)/dt = v(t)

taking laplace transform,

I(s) = 6 [1/s + 2 – 1/s + 3]

v (s) = 6/s + 2

RI (s) + LsI (s) = V (s)

6R (1/s + 2 – 1/s + 3) + 6s (1/s + 2 – 1/s + 3) = 6/s + 2

6R (1/(s + 2) (s + 3) + 6s (1/(s + 2) (s + 3) = 6/s + 2

1/(s + 2) (s + 3) (R + s) = 1/(s + 2)

R + s = (s + 3)

R = 3

- (b)

according to the frequency convolution property of laplace transform,

I_{3} (t) = I_{1} (t) . I_{2}(t) – 1/2 i [f_{1}(s)* f_{2}(s)]

f_{1}(s) = 1/s + 2

f_{1}(t) = e^{-2t} u(t)

f_{2} (s) = 1/s + 1

i_{2} (t) = e^{-1} u(t)

i_{3} (t) = e^{-3t} u(t)

- (c)

v = RI + L di/dt

using laplace transfrm,

v (s) = RI (s) + l [sI (s) – I (0^{+})]

50/s = 5I(s) + 2.5 x 10^{-3} [sI (s) – 2]

I (s) = 10/s + (-8)/s + 2000

I (t) = 10 – 8e^{-2000t}

- (b)

given, v(t) = 10u (t) – 10u (t – 2)

taking laplace transform,

V (s) = 10/s – 10/s e^{-2t}

v (s) = 10/s (1 – e^{-2t})

given, i(t) = e^{-1} – e^{-(t – 2)}

taking laplace transform

I (s) = 1/s + 1 – 1/s + 1 e^{-2s}

I (s) = 1 – e^{-2s}/s + 1

Z (s) = V(s)/I(s) = 10/s (1 – e^{-2s})/1/s + 1 (1 – e^{-2s})

Z (s) = 10. (s + 1)/s = 10 + 1/(1/10)s

Z (s) = R + 1/sc

R = 10 and C = 1/10 F

- (c)

given, H (s) = Y(S)/X(S) = (s + 2)/(s^{2} + s + 4) ……..(i)

x(t) = cos 2t

y(t) = cos (2t + 0)

from x(t) = cos 2t, we couclude that _{00} = 2

on putting s = j_{oo} = j2 in eq. (i), we get

H (2 j) = 2 j + 2/j^{2}2^{2} + 2 j + 4 = 2(j + 1)/-4 + 2 j + 4 = 2 (1 + j)/2j

= 2 <45^{0}/<90^{0}

or y(t)/x(t) = <(45^{0} – 90^{0}) = < – 45^{0}