laplace transform of e^at u(t) | find the laplace transform of e at u(t) and its roc

find the laplace transform of e at u(t) and its roc or laplace transform of e^at u(t) ?

Laplace Transform

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

Laplace Transform

Consider a LTI system with input x(t) = e^st and impulse response h(t). the output of the system

y(t) = H(s)e^st …………………(1)

H(s) = x(t)e^-st dt ……………..(2)

The function H(s) is referred to as laplace transform of h(t).

1. Lef x(t) is a general CT signal. the bilateral (two sided) laplace transform of x(t) is defined as

X(s) = x(t)e^-st dt ……………….(3)

where, the variable s is generally complex valued.

s = o + j_oo

2. The unilateral (one sided) laplace transform of the CT signal x(t) is

X₁(s) = x(t)e^-st dt ………………(4)

Symbolically, we can write laplace transform pair of the signal x(t) as

X(s) = L{x(t)}

X(t)  X(s)

3. The values of s for which the laplace transform. converges is called the region of Convergence (ROC).

Example 1. Consider the signal

x(t) = e^-at  u(t)

find the laplace transform of x(t).

Sol. The laplace transform of x(t) is

X(s) = x(t)e^-st dt

= e^-at  u(t) e^-st dt

= e^-at e^-st dt

= e^{-1 (a + s)}  dt

= -1/s + a [e^{-1(s + a)}

= 1/s + a

The integral in example 1 converges if e^{-1 (s + a)} is finite, the integral is from 0 to _oo, i.e., the time, t > 0. which means, s + a > 0 in order for the exponential function to decay

a + s > 0

a + 0 + j_oo > 0

a + 0 > 0

0 > – a

ROC : R {s} > – a

The ROC in example 1. represents a line in he complex plane as shown in figure.

Poles and Zeros of X(s)

Usually, X(s) is a rational function.

X(s) = a₀s^m + a₁s^m-1 + …………+ a_m/b_osⁿ +b₁s^n-1+…+b_n

= a_o(s – z₁) ………(s – z_m)/b_o(s – p₁)………….(s – p_n)

a_k and b_k are real constants.

n and m are positive integers.

if n > m, x(s) is proper.

if n < m, x(s) is improper.

the roots of the numerator polynomial are called the zeros z_k of x(s).

the roots of the denominator polynomial are called the poles p_k of x(s).

when s = p_k for k = 1, …………n, x(s) does not converge. hence, the ROC cannot contain any poles.

Example 2. let

X(s) = 2s + 4/s² + 4s + 3

find poles and zeros of the given system.

Sol. X(s) can be written as

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

z₁ = – 2, p₁ = – 1, p₂ = – 3.

Properties of the ROC

ROC consists of strips parallel to the j₀₀-axis in the s-plane.

1. ROC does not contain any poles.
2. If x(t) is a finite duration signal, x(t) = 0, t₁ < t < t₂ and is absolutely integrable, the ROC is the entire s plane.
3. If x(t) is a right sided signal, x(t) = 0, t > t_o, the ROC is of the form

R{s} > max {R {P_K}}

i.e., ROC is to the right of the rightmost pole, in this case the system is causal.

4. If x(t) is a left sided signal x(t) = 0, t > t_o the ROC is of the form

R{s} < min {R {P_K}}

i.e., ROC is to the left of the leftmost pole, in this case the system is anti-causal.

5. If x(t) is a double sided signal, the ROC is of the form p₁ < R{s} < p₂
6. If the ROC includes the j_oo-axis. fourier transform exists and the system is stable.

Eample 3.

let    x(s) = s + 3/(s + 1) (s – 2)

find ROC of the system and explain.

Sol.  poles at s = – 1, s = 2 three ROC’s

(1) R{s} < -1, property (4) applies, the signal is left sided, the ROC doesn’t include j_oo-axis hence, fourier transform doesn’t exist.

(2) -1 < R{s} <2, proprty (5) applies, the signal is double sided, the ROC includes the J_oo-axis, fourier transform exists.

(3) R{s} > 2, property (3) applies, the signal is right sided, the ROC doesn’t include j_oo-axis hence, fourier transform doesn’t exist

usually, laplace transform pairs are obtained using laplace, transform table.table of the most common laplace transform pairs is shown below.

Unilateral Laplace Transform Properties

Bilateral laplace transform reduces to unilateral laplace transform for causal systems. in this section, we will discuss the unilateral laplace transform properties.

1. Linearity

X₁(t)  X₁(s)

x₂(t)  x₂(s)

ax₁(t) + bx₂(t) ax₁(s) + bx₂(s)

ROC = ROC₁  ROC₂

2. Time shifting

x(t)  x(s)

x(t – t_o)  x(s) e^-st0

for some positive fixed t_o, the ROC will be the same.

3. Shifting in the s-domain

x(t)   x(s)

e^s0t x(t)  x(s – s_o)

the ROC will be shifted by Re{s_o}

ROC = ROC₁ + Re{s_o}

4. Time scaling

x(t)  x(s)

x(at) 1/|a| x (s/a)

ROC = a ROC₁

Inverse Laplace Transform

It is the process of finding x(t) given x(s)

x(t) = L{x(s)}

there are two methods to obtain the inverse laplace transform.

Inversion Using Complex Line Integral

x(t) = 1/2j  x(s) e^st ds

the values of c in this integral depends on the ROC . we will not be using this line integral, rather we will use the inversion using laplace table.

Inversion Using Laplace Table

laplace transform can be written as

x(s) = num(s)/den(s)

= k (s – z₁)……(s – z_m)/(s – p₁) …………(s – p_n)

we always make sure that m < n. in this case, partial fraction expansion can be performed.

Partial Fraction Expansion

1. All poles have multiplicity of 1.

x(s) = c₁/s – p₁ + …….+ c_n /s – p_n

c_k = (s – p_k) x(s)|_{s = pk}

1. When one or more poles have multiplicity r.

in this case, x(s) has the term (s – p)^r.

x(s) = /s – p + (s – p)² +……..+(s – p)^r

the coefficient can be found as

1/(r – k) d^{r – k}/ds^{r – k} ((s – p_k)^r x(s))

Intro Exercise – 3

1. The laplace transform of signal u(t – 2) is

(a) -e^-2s/s

(b) e^-2s/s

(c) e^-2s/1 + s

(d) 0

2. The laplace transform of signal u(t + 2) is

(a) 1/s

(b) -1/s

(c) e^-2s/s

(d) -e^-2s/s

3. The laplace transform of signal e^-2t u(t + 1) is

(a) 1/s + 2

(b) e^-s/s + 2

(c) e^{-(s + 2)}/s + 2

(d) -e^-s/s + 2

4. The laplace transform of signal e^2t u(-t + 2) is

(a) e^{2(s – 2)} – 1/s – 2

(b) e^-2s/s + 2

(c) 1 – e^{-2(s – 2)}/s – 2

(d) e^-2s/s – 2

5. The laplace transform of signal sin 5t is

(a) 5/s² + 5

(b) s/s² + 5

(c) 5/s² + 25

(d) s/s² + 25

6. The laplace transform of signal u(t) -u(t – 2) is

(a) e^-2s – 1/s

(b) 1 – e^-25/s

(c) 2/s

(d) -2/s

7. The laplace transform of signal d/dt {te^-t u(t)} is

(a) 1/s(s + 1)²

(b) s/(s + 1)²

(c) e^-s/s + 1

(d) e^-s/(s + 1)²

8. The laplace transform of signal tu(t)^* cos 2t u(t) is

(a) 1/s(s² + 4²)

(b) 2/s²(s² + 4²)

(c) 1/s²(s² + 4²)

(d) s²/s² + 4²

9. The laplace transform of signal t³u(t) is

(a) 4/s⁴

(b) -3/s⁴

(c) 6/s⁴

(d) -6/s⁴

10. The laplace transform of signal u(t – 1)^* e^-2t u(t – 1) is

(a) e^{-2(s + 1)}/2s + 1

(b) e^{-2(s + 1)}/s + 1

(c) e^{-(s + 2)}/s + 2

(d) e^{-2(s + 1)}/s + 2

11. The laplace transform of signal e^-3 cos 2 d is

(a) -(s + 3)/s[(s + 3)² + 4)]

(b) (s + 3)/s[(s + 3)² + 4]

(c) s(s + 3)/(s + 3)² + 4

(d) -s (s + 3)/(s + 3)² + 4)

12. The laplace transform of signal t d/dt {e^-t cos t u(t)} is

(a) -(s² + 4s + 2)/(s² + 2s + 2)²

(b) (s² + 4s + 2)/(s² + 2s + 2)²

(c) (s² + 2s + 2)/(s² + 4s + 2)²

(d) -(s² + 2s + 2)/(s² + 4s + 2)²

13. The time signal cprresponding to s + 3/s² + 3s + 2 is

(a) 2e^-2t + e^-1) u(t)

(b) (2e^-t – e^-2t) u(t)

(c) (2e^-2t – e^-t) u(t)

(d) (2e^-t + e^-2t) u(t)

14. The time signal corresponding to 2s² + 10s + 11/s² + 5s + 6

(a) 2(t) + (e^-3t – e^-2t) u(t)

(b) 2(t) + (e^-2t – e^-3t) u(t)

(c) 2(t) + (e^-2t + e^-3t) u(t)

(d) 2(t) – (e^-2t + e^-3t) u(t)

15. The time signal corresponding to 2s – 1/s² + 2s + 1 is

(a) (3e^-t – 2te^-1) u(t)

(b) (3e^-t + 2te^-t) u(t)

(c) (2e^-t – 3te^-t) u(t)

(d) (2e^-t + 3te^-t) u(t)

16. The time signal corresponding to 5s + 4/s³ + 3s² + 2s is

(a) (2 + e^-t + 3e^-2t) u(t)

(b) (2 + e^-t – 3e^-2t) u(t)

(c) (3 + e^-t – 3e^-2t) u(t)

(d) (3 + e^-t + 3e^-2t) u(t)

17. The time signal corresponding s² – 3/(s + 2) (s² + 2s + 1) is

(a) (e^-2t – 2te^-t) u(t)

(b) (e^2t + 2te^-t) u(t)

(c) (e^-t – 2te^-2t) u(t)

(d) (e^-t + 2te^-2t) u(t)

18. The time signal corresponding to 3s + 2/s² + 2s + 10 is

(a) [3e^-t cos 3t – 1e^-t/3 sin 3t] u(t)

(b) 3e^-t sin 3t – 1e^-t/3 cos 3t] u(t)

(c) (3e^-t cos 3t – e^-t sin 3t) u(t)

(d) (3e^-t sin 3t + 3e^-t cos 3t) u(t)

19. The time signal corresponding to 4s² + 8s + 10/(s + 2) (s² + 2s + 5) is

(a) (2e^-2t + 2e^-t sin 2t – 2e^-t cos 2t) u(t)

(b) (2e^-2t  + 2e^-t cos 2t – 2e^-t sin 2t) u(t)

(c) (2e^-2t + 2e^-t cos 2t – e^-t sin 2t) u(t)

(d) (2e^-2t + 2e^-t sin 2t – e^-1 cos 2t) u(t)

20. The time signal corresponsing to 3s² + 10s + 10/(s + 2) (s² + 6s + 10) is

(a) (e^-2t + 2e^-3t cos t 2e^-3t sin t) u(t)

(b) (e^-2t + 2e^-3t cos t – 6e^-3t sin t) u(t)

(c) (e^-2t + 2e^-3t cos t – 2e^-3t sin t) u(t)

(d) (9e^-2t  – 6e^-3t cos t + 3e^-3t sin t) u(t)

Answers with Solutions

1. (b)

x(s) = x(t) e^-st dt

= e^-st dt = e^-2s/s

2. (a)

x(s) = x(t) e^-3t dt = u(t + 2)^-3t dt

= e^-3t dt = 1/s

3. (a)

x(s) = e^-2t e^-st dt = 1/s + 2

4. (c)

x(s) = x(t) e^-st dt = e^2t u(-t + 2) e^-st dt

= e^{t(2 – s)} dt = e^{2(2 – s)} – 1/2 – s = 1 – e^{-2(s – 2)}/s – 2

5. (c)

x(s) = (e^j5t – e^-j5t)/2j e^-st dt = 5/s² + 25

6. (b)

x(s) = e^st dt = 1 – e^-2s/s

7. (b)

q(t) = te^-t u(t)   q(s) = 1/(s + 1)²

x(t) = d/dt p(t)   x(s) = s/(s + 1)²

8. (a)

p(t) = tu(t)   p(s) = 1/s²

q(t) = cos 2t u(t)   Q(s) = s/s² + 4²

x(t) = p(t) * q(t)    x(s) = p(s) Q(s)

x(s) = 1/s(s² + 4²)

9. (c)

p(t) = tu(t)   p(s) = 1/s²

q(t) = tp(t)   Q(s) = d/ds p(s) = -2/s³

x(t) = – tq(t)   x(s) = d/ds Q(s) = 6/s⁴

tⁿu(t)   n’/s^{n + 1}

10. (d)

p(t) = u(t)   p(s) = 1/s

q(t) = u(t – 1)  Q(s) = e^-s/s²

r(t) = e^-2t u(t)   R(s) = 1/s + 2

v(t) = e^-2t u(t – 1) v(s) e^{-(s + 2)}/s²

x(t) = q(t) * v(t)  x(s) = Q(s) V(s)

x(s) e^{-2(s + 1)}/s + 2

11. (b)

e^-at cos t u(t)   s + a/(s + a)² + 1

p(t) = e^-3t cos 2tu(t)  p(s) = s + 3/(s + 3)² + 4

p d  1/s   p d + p(s)/s

x(s) = (s + 3)/s[(s + 3)² + 4]

12. (a)

p(t) = e^-t cos t u(t)   p(s) = s + 1/(s + 1)² + 1

q(t) = d/dt p(t)  Q(s) = s (s + 1)/(s + 1)² + 1

x(t) = tq(t)   x(s) = – d/ds Q(s)

x(s) = -(s² + 4s + 2)/(s² + 2s + 2)²

13. (b)

x(s) = s + 3/(s² + 3s + 2) = A/s + 1 + B/s + 2

A = s + 3/s + 2 = 2, B = s + 3/s + 1  = – 1

x(t) = [2e^-t -e^-2t ] u(t)

14. (a)

x(s) = 2 – 1/(s + 2) (s + 3)

= 2 – 1/(s + 2) + 1/(s + 3) e^at u(t)  1/s – a

x(t) = 2(t) + (e^-3t – e^-2t) u(t)

15. (c)

x(s) = 2s – 1/s² + 2s + 1 = A/(s + 1) + B/(s + 1)²

B = (2s – 2) = – 3, A = 2

[e^-at + u(t)   1/(s + a)²

x(t) = x(t) = (2e^-t – 3te^-t) u(t)

16. (b)

x(s) = 5s + 4/s³ + 3s² + 2s = A/s + B/s + 1 + C/s + 2

A = sx(s)|_{s = 0} = 2, B = (s + 1) x(s)|_{s = – 1} = 1

C = (s + 2) x(s)|_{s = – 2} = – 3

x(t) = (2 + e^-t – 3e^-2t) u(t)

17. (a)

x(s) = s² – 3/(s + 2) (s² + 2s + 1)

= A/(s + 2) + B/(s + 1) + C/(s + 1)²

A = (s + 2) x(s)|_{s = – 2} = 1

C = (s + 1)² x(s)|_{s = – 1} = – 2

A + B = 1   B = 0

x(t) = [e^-2t – 2te^-t] u(t)

18. (a)

x(s) = 3s + 2/s² + 2s + 10

= 3(s + 1)/(s + 1)² + 3² – 1/(s + 1)² + 3²

x(t) = [3e^-t cos 3t – 1/3 e^-t sin 3t] u(t)

[e^-at cos t u(t)   (s + a)/(s + a)²

e^-at sin u(t)   /(s + a)²

19. (c)

x(s) = 4s² + 8s + 10/(s + 2) (s² + 2s + 5)

= A/(s + 2) + B(s + 1)/(s + 1)² + 2² + C/(s + 1)² + 2²

A = (s + 2) x(s)|_{s = – 2} = 2

A + B  = 4   B = 2

5A + 2B + 2C = 10

C = – 2

x(t) = (2e^-2t + 2e^-t cos  2t – e^-t sin 2t) u(t)

20. (b)

x(s)  = 3x² + 10s + 10/(s + 2) (s² + 6s + 10)

= A/(s + 2) + B(s + 3)/(s + 3)² + 1 + C/(s + 3)² + 1

A = (s + 2) x(s)|_{s = – 2} = 1

10A + 6B + 2C = 10    C = – 6

x(t) = (e^-2t + 2e^-3t cos t – 6e^3t sin t) u(t)