Transform Theory

Laplace transform of sin ht is

Laplace transform of cos (ωt) is

The Laplace transform of a function f(t) is $$frac{1}{{{{ ext{s}}^2}left( {{ ext{s}} + 1}
ight)}}.$$ xa0The function f(t) is

The inverse Laplace transform of $${ ext{H}}left( { ext{s}}
ight) = frac{{{ ext{s}} + 3}}{{{{ ext{s}}^2} + 2{ ext{s}} + 1}}$$ xa0 xa0for t ≥ 0 is

Let [{ ext{f}}left( { ext{x}}
ight) = left{ {x08egin{array}{*{20}{c}}
{ - pi ,,,{ ext{if}}}&{ - pi < { ext{x}} leqslant { ext{0}}} \
{pi ,,,{ ext{if}}}&{0 < { ext{x}} leqslant pi }
end{array}}
ight.] xa0 xa0 be a periodic function of period 2π. The coefficient of sin5x in the Fourier series expansion of f(x) in the interval [-π, π] is

Given the Fourier series in (-π, π) for f(x) = x cosx, the value of a 0 will be

Laplace transform of the function f(t) is given by $${ ext{F}}left( { ext{s}}
ight) = { ext{L}}left{ {{ ext{f}}left( { ext{t}}
ight)}
ight} = int_0^infty {{ ext{f}}left( { ext{t}}
ight){{ ext{e}}^{ - { ext{st}}}}{ ext{dt}}{ ext{.}}} $$ xa0 xa0 xa0 Laplace transform of the function shown below is given by

If L defines the Laplace Transform of a function, L [sin (at)] will be equal to

The Fourier series expansion of the saw-toothed waveform f(x) = x in (-π, π) of period 2π gives the series, $$1 - frac{1}{3} + frac{1}{5} - frac{1}{7} + ,....$$ The sum is equal to

Laplace transform of 8t 3

A solution for the differential equation [{
m{dot x}}left( {
m{t}}
ight) + 2{
m{x}}left( {
m{t}}
ight) = delta left( {
m{t}}
ight)] xa0 xa0with initial condition x(0 - ) = 0 is

Consider the differential equation $$frac{{{{ ext{d}}^2}{ ext{y}}left( { ext{t}}
ight)}}{{{ ext{d}}{{ ext{t}}^2}}} + 2frac{{{ ext{dy}}left( { ext{t}}
ight)}}{{{ ext{dt}}}} + { ext{y}}left( { ext{t}}
ight) = delta left( { ext{t}}
ight)$$ xa0 xa0 xa0with $${left. {{ ext{y}}left( { ext{t}}
ight)}
ight|_{{ ext{t}} = 0}} = - 2$$ xa0 and $${left. {frac{{{ ext{dy}}}}{{{ ext{dt}}}}}
ight|_{{ ext{t}} = 0}} = 0.$$ The numerical value of $${left. {frac{{{ ext{dy}}}}{{{ ext{dt}}}}}
ight|_{{ ext{t}} = 0}}$$ xa0 is

A delayed unit step function is defined as [{ ext{u}}left( {{ ext{t}} - { ext{a}}}
ight) = left{ {x08egin{array}{*{20}{c}}
{0,}&{{ ext{for t}} < { ext{a}}} \
{1,}&{{ ext{for t}} geqslant { ext{a}}}
end{array}}
ight..] xa0 xa0 xa0Its Laplace transform is

The Laplace Transform of f(t) = e 2t sin(5t) u(t) is

Let $${ ext{X}}left( { ext{s}}
ight) = frac{{3{ ext{s}} + 5}}{{{{ ext{s}}^2} + 10{ ext{s}} + 21}}$$ xa0 xa0be the Laplace Transform of a signal x(t). Then, x(0 + ) is

If f(t) is a function defined for all t ≥ 0, its Laplace transform F(s) is defined as

Laplace transform for the function f(x) = cosh(ax) is

If F(s) is the Laplace transform of function f(t), then Laplace transform of $$intlimits_0^{ ext{t}} {{ ext{f}}left( au
ight){ ext{d}} au } $$ xa0 is

The inverse Laplace transform of $$frac{1}{{left( {{{ ext{s}}^2} + { ext{s}}}
ight)}}$$ xa0is

The Fourier series of the function, [x08egin{array}{*{20}{c}}
{{ ext{f}}left( { ext{x}}
ight) = 0,}&{ - pi < { ext{x}} leqslant 0} \
{,,,,,,,,,,,,,,,,,,,,, = pi - { ext{x,}}}&{0 < { ext{x}} < pi }
end{array}] xa0 xa0 xa0in the interval $$left[ { - pi ,,pi }
ight]$$ xa0is $${ ext{f}}left( { ext{x}}
ight) = frac{pi }{4} + frac{2}{pi }left[ {frac{{cos { ext{x}}}}{{{1^2}}} + frac{{cos { ext{3x}}}}{{{3^3}}} + ,...}
ight] + left[ {frac{{sin { ext{x}}}}{1} + frac{{sin { ext{2x}}}}{2} + frac{{sin { ext{3x}}}}{3} + ,...}
ight]$$ The convergence of the above Fourier series at x = 0 gives

The inverse Laplace transform of the function $${ ext{F}}left( { ext{s}}
ight) = frac{1}{{{ ext{s}}left( {{ ext{s}} + 1}
ight)}}$$ xa0 is given by

Laplace transform analysis gives

Inverse Laplace transform of the function $$frac{s}{{{s^2} + 3s + 2}},$$ xa0 is

Which of the following is the advantage of using Laplace transform techniques?

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

Find the inverse Laplace transform of $$Fleft( s
ight) = frac{{{s^2} + 2s - 2}}{{sleft( {s + 4}
ight)left( {s - 5}
ight)}}$$

Give transfer function $$Hleft( s
ight) = frac{{s + 2}}{{{s^2} + s + 4}},$$ xa0 xa0under steady state condition, the sinusoidal input and output are, respectively x(t) = cos 2t, y(t) = cos(2t + $$phi $$), then angle $$phi $$ will be

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

The Laplace transform of $$Ileft( t
ight)$$ xa0is given by $$Ileft( s
ight) = frac{5}{{sleft( {{s^2} + 2}
ight)}}.$$ xa0 xa0As $$t o infty $$ xa0the value of $$Ileft( t
ight)$$ xa0tends to

Given that $$Fleft( s
ight)$$ xa0is the one-side Laplace transform of $$fleft( t
ight),$$ xa0the Laplace transform of $$int_0^t {fleft( au
ight)d au } $$ xa0 is

The Laplace transform of $$ileft( t
ight)$$ xa0is given by $$Ileft( s
ight) = frac{2}{{sleft( {1 + s}
ight)}}.$$ xa0 xa0As $$t o infty $$ xa0the value of $$ileft( t
ight)$$ xa0tends to

The bilateral Laplace transform of $${e^{ - 1}}uleft( {t + 2}
ight)$$ xa0 is

The bilateral Laplace transform of $${e^t}cos 2tuleft( { - t}
ight) + {e^{ - t}}uleft( t
ight) + {e^{frac{t}{2}}}uleft( t
ight)$$ xa0 xa0xa0 is

The bilateral Laplace transform of u(-t + 3) is

The Laplace transform of $$xleft( t
ight)$$ xa0is $$Xleft( s
ight) = {e^{ - 2s}}frac{{6{s^2} + s}}{{{s^2} + 2s - 2}}$$ The initial value of $$xleft( t
ight)$$ xa0is

The bilateral Laplace transform of $${e^{left( {3t + 6}
ight)}}uleft( {t + 3}
ight)$$ xa0 is

The Laplace transform of $$xleft( t
ight)$$ xa0is $$Xleft( s
ight) = frac{{{e^{ - 3s}}left( {2{s^2} + 1}
ight)}}{{sleft( {s + 1}
ight)left( {s + 4}
ight)}}$$ The final value of $$xleft( t
ight)$$ xa0is

The Laplace transform of $$xleft( t
ight)$$ xa0is $$Xleft( s
ight) = frac{{2{s^2} + 3}}{{{s^2} + 5s + 1}}$$ The initial value of $$xleft( t
ight)$$ xa0is

The trigonometric Fourier series of an even function does not have the

The inverse Laplace transform of the function $$frac{{s + 5}}{{left( {s + 1}
ight)left( {s + 3}
ight)}}$$ xa0 is

The Laplace transform F(s) of the exponential function. f(t) = e at when t ≥ 0, where a is a constant and (s - a) > 0, is

Evaluate $$intlimits_0^infty {frac{{sin { ext{t}}}}{{ ext{t}}}{ ext{dt}}} $$

If the Laplace transform of $${{ ext{e}}^{omega { ext{t}}}}$$ xa0is $$frac{1}{{{ ext{s}} - omega }},$$ xa0the Laplace transform of tcosh t is

The function f(t) satisfies the differential equation $$frac{{{{ ext{d}}^2}{ ext{f}}}}{{{ ext{d}}{{ ext{t}}^2}}} + { ext{f}} = 0$$ xa0 and the auxiliary conditions, f(0) = 0, $$frac{{{ ext{df}}}}{{{ ext{dt}}}}left( 0
ight) = 4.$$ xa0The Laplace transform of f(t) is given by

The Laplace transform of e i5t where $${ ext{i}} = sqrt { - 1} ,$$ xa0 is

Laplace transform of the function sin ωt is

For the function [{ ext{f}}left( { ext{x}}
ight) = left{ {x08egin{array}{*{20}{c}}
{ - 2,}&{ - pi < { ext{x}} < 0} \
{2,}&{0 < { ext{x}} < pi }
end{array}}
ight.] The value of a n in the Fourier series expansion of f(x) is

Laplace transform of cos (ωt) is $$frac{{ ext{s}}}{{{{ ext{s}}^2} + {omega ^2}}}.$$ xa0The Laplace transform of e -2t cos(4t) is

The Laplace transform of sinh(at) is

The Fourier cosine series for an even function f(x) is given by $${ ext{f}}left( { ext{x}}
ight) = {{ ext{a}}_0} + sumlimits_{{ ext{n}} = 1}^infty {{{ ext{a}}_{ ext{n}}}cos left( {{ ext{nx}}}
ight)} $$ The value of the coefficient a 2 for the function f(x) = cos 2 (x) in [0, π] is