Engineering Mathematics Test 12

\(L\left( {1 - {e^t}} \right) = \frac{1}{s} - \frac{1}{{s - 1}}\)

\(L\left( {\frac{{\left( {1 - {e^t}} \right)}}{t}} \right) = \mathop \smallint \limits_s^\infty \left( {\frac{1}{s} - \frac{1}{{s - 1}}} \right)ds\)

\( = \left[ {\log \left( {\frac{s}{{s - 1}}} \right)} \right]_s^\infty = \log \left( {\frac{{s - 1}}{s}} \right)\)

Concept:

The definition of z-transform is given by,

\(X\left( z \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left( n \right){z^{ - n}}\) 

Let the signal, x(n) = an u(n)

The z-transform of the above-given signal is given by

\(X\left( z \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left( n \right){z^{ - n}}\)

\(X\left( z \right) = \mathop \sum \limits_{n = 0}^\infty {a^n}{z^{ - n}}\)

\(z\left( {{a^n}} \right) = \frac{z}{{z - a}}\)

Scaling property:

If x(n) ↔ X(z), then

an x(n) ↔ X(z/a)

Differentiation property:

If x(n) ↔ X(z), then

\(n\;x\left( n \right) \leftrightarrow - z\frac{d}{{dz}}\left( {X\left( z \right)} \right)\)

Calculation:

\(z\left( {{3^n}} \right) = \frac{z}{{z - 3}}\)

From the differentiation property of z-transform.

\(z\left( {n \cdot {3^n}} \right) = - z\frac{d}{{dz}}\left( {\frac{z}{{z - 3}}} \right)\)

\(= - z\left( {\frac{{\left( {z - 3} \right)\left( 1 \right) - z\left( 1 \right)}}{{{{\left( {z - 3} \right)}^2}}}} \right)\)

\(= \frac{{3z}}{{{{\left( {z - 3} \right)}^2}}}\)

Concept:

Final value theorem:

• A final value theorem allows the time domain behavior to be directly calculated by taking a limit of a frequency domain expression
• Final value theorem states that the final value of a system can be calculated by

\(f\left( \infty \right) = \mathop {\lim }\limits_{s \to 0} sF\left( s \right)\)

 Where F(s) is the Laplace transform of the function.

• For the final value theorem to be applicable system should be stable in steady-state and for that real part of poles should lie in the left side of s plane.

Initial value theorem:

\(C\left( 0 \right) = \mathop {\lim }\limits_{t \to 0} c\left( t \right) = \mathop {\lim }\limits_{s \to \infty } sC\left( s \right)\)

It is applicable only when the number of poles of C(s) is more than the number of zeros of C(s).

Calculation:

\(F\left( s \right) = \frac{b}{{s\left( {s + 1} \right)\left( {s + a} \right)}}\)

By using final value theorem, the steady state value of f(t) is,

\(f\left( \infty \right) = \mathop {\lim }\limits_{s \to 0} s \cdot \frac{b}{{s\left( {s + 1} \right)\left( {s + a} \right)}} = \frac{b}{a} = 0.5\) 

⇒ a = 2b

Concept:

The definition of unilateral Laplace transform is

\(X\left( s \right) = \mathop \smallint \nolimits_0^\infty x\left( t \right){e^{ - st}}dt\)

Laplace transform of function of f(t) is shown by L [f(t)] = F(s)

The effect of time-shifting in the frequency domain is represented as:

\(u\left( {t - t_0} \right) \leftrightarrow \;\frac{{{e^{ - st_0}}}}{s}\)

By using the first shifting rule

 If L [f(t)] = F (s), then

L [eat f(t)] = F (s – a)

Differentiation in the frequency domain can be represented as:

\(tf\left( t \right) \leftrightarrow - \frac{d}{{ds}}\left( {F\left( s \right)} \right)\)

Calculation:

f(t) = t² e-^2t cos (3t)

\(L\left\{ {\cos 3t} \right\} = \frac{s}{{{s^2} + 9}}\)

By using differentiation property,

\(L\left\{ {t\cos 3t} \right\} = \frac{{ - d}}{{ds}}\left( {\frac{s}{{{s^2} + 9}}} \right)\)

\( = - \left[ {\frac{{\left( {{s^2} + 9} \right)\left( 1 \right) - s\left( {25} \right)}}{{{{\left( {{s^2} + 9} \right)}^2}}}} \right]\)

\(= \frac{{{s^2} - 9}}{{{{\left( {{s^2} + 9} \right)}^2}}}\)

By using differentiation property,

\(L\left\{ {{t^2}\cos 3t} \right\} = \frac{{ - d}}{{ds}}\left( {\frac{{{s^2} - 9}}{{{{\left( {{s^2} + 9} \right)}^2}}}} \right)\)

\(= - \left[ {\frac{{{{\left( {{s^2} + 9} \right)}^2}\left( {25} \right) - \left( {{s^2} - 9} \right)\left( {2\left( {{s^2} + 9} \right)} \right)\left( {25} \right)}}{{{{\left( {{s^2} + 9} \right)}^4}}}} \right]\)

\(= - \left[ {\frac{{\left( {{s^2} + 9} \right)\left( {25} \right) - \left( {2{s^2} - 18} \right)\left( {25} \right)}}{{{{\left( {{s^2} + 9} \right)}^3}}}} \right]\)

\(= \frac{{25\left( {{{25}^2} - 18 - {s^2} - 9} \right)}}{{{{\left( {{s^2} + 9} \right)}^3}}}\)

\(= \frac{{25\left( {{s^2} - 27} \right)}}{{{{\left( {{s^2} + 9} \right)}^3}}}\)

By using the shifting property,

\(L\left\{ {{e^{ - 2t}}{t^2}\cos t} \right\} = \frac{{2\left( {s + 2} \right)\left( {{{\left( {s + 2} \right)}^2} - 27} \right)}}{{{{\left( {{{\left( {s + 2} \right)}^2} + 9} \right)}^3}}}\)

\(= \frac{{2\left( {s + 2} \right)\left( {{s^2} + 4s - 23} \right)}}{{{{\left( {{s^2} + 4s + 13} \right)}^3}}}\)

\(F\left( s \right) = \frac{{3\left( {s + 3} \right)}}{{{s^2} + 65 + 8}}\) 

By applying partial fraction method,

\(F\left( s \right) = \frac{{3\left( {s + 3} \right)}}{{\left( {s + 2} \right)\left( {s + 4} \right)}} = \frac{A}{{\left( {s + 2} \right)}} + \frac{B}{{\left( {s + 4} \right)}}\) 

⇒ 35 + 9 = A (s + 4) + B (s + 2)

By comparing both the sides,

A + B = 3and 4A + 2B = 9

⇒ A = B = 1.5

\(F\left( s \right) = \frac{{1.5}}{{\left( {s + 2} \right)}} + \frac{{1.5}}{{\left( {s + 4} \right)}}\) 

By applying inverse Laplace transform, we get

f(t) = 1.5 e^-2t + 1.5 e^-4t

At t = 0.5,

\(f\left( {0.5} \right) = 1.5\left( {\frac{1}{e} + \frac{1}{{{e^2}}}} \right) = 0.7548\) 

\(\frac{{dt}}{{dt}} + ay = {e^{ - bt}}\)

Integrating factor \(= {e^{\smallint pdt}} = {e^{at}}\)

Solution of y is,

\(y\left( {{e^{at}}} \right) = \smallint {e^{at}}.{e^{ - bt}}dt + c\)

\(\Rightarrow y{e^{at}} = \frac{{{e^{\left( {a - b} \right)t}}}}{{a - b}} + c\)

Given that, y(0) = 0

\(\Rightarrow 0 = \frac{1}{{a - b}} + c \Rightarrow c = - \frac{1}{{\left( {a - b} \right)}}\)

Now, the solution becomes,

\(y{e^{at}} = \frac{{{e^{\left( {a - b} \right)t}}}}{{a - b}} - \frac{1}{{a - b}}\)

\(\Rightarrow y\left( t \right) = \frac{{{e^{ - bt}}}}{{a - b}} - \frac{{{e^{ - at}}}}{{a - b}}\)

Apply Laplace transform

\(\Rightarrow y\left( s \right) = \frac{1}{{a - b}}\left[ {\frac{1}{{s + b}} - \frac{1}{{s + a}}} \right]\)

Explanation:

Let F(t) = e^-at – e^-bt

{L{F(t)} = L{e^-at – e^-bt}

\(= \frac{1}{{s + a}} - \frac{1}{{s - b}}\)

Now \(L\left\{ {\frac{{{e^{ - at}} - {e^{ - bt}}}}{t}} \right\} = \mathop \smallint \limits_s^\infty \left( {\frac{1}{{\left( {s + a} \right)}} - \frac{1}{{\left( {s + b} \right)}}} \right)ds\)

\( = \left[ {\log \left( {s - a} \right) - \log \left( {s + b} \right)} \right]_s^\infty \)

\(= \left[ {\log \left( {\frac{{s + a}}{{s + b}}} \right)} \right]_s^\infty \)

\( = \mathop {\lim }\limits_{s \to \infty } \log \left( {\frac{{s + a}}{{s + b}}} \right) - \log \left( {\frac{{s + a}}{{s + b}}} \right)\)

\( = \mathop {\lim }\limits_{s \to \infty } \log \left( {\frac{{1 + a/s}}{{1 + b/s}}} \right) - \log \left[ {\frac{{s + a}}{{s + b}}} \right]\)

\( = \log \left[ {\frac{{1 + 0}}{{1 + 0}}} \right] + \log {\left( {\frac{{s + a}}{{s + b}}} \right)^{ - 1}} = \log \left( {\frac{{s + b}}{{s + a}}} \right)\)

∵ By definition

\(L\left\{ {F\left( t \right)} \right\} = \mathop \smallint \limits_o^\infty {e^{ - st}}.F\left( t \right)dt\)

\(L\left[ {\frac{{{e^{ - at}} - {e^{ - bt}}}}{t}} \right] = \mathop \smallint \limits_0^\infty {e^{ - st}}\left[ {\frac{{{e^{ - at}} - {e^{ - bt}}}}{t}} \right]dt\)

Put s = 0

\(\mathop \smallint \limits_0^\infty \frac{{{e^{ - at}} - {e^{ - bt}}}}{t}dt = \log \frac{b}{a}\)

Laplace transform of \(\mathop \smallint \limits_0^\infty \frac{{{e^{ - at}} - {e^{ - bt}}}}{t}dt\) = Laplace transform of \(\log \frac{b}{a}\)

\( = \frac{1}{s}\log \frac{b}{a}\)

Concept:

By initial value theorem,

\({u_o} = \mathop {{\rm{lt}}}\limits_{z \to \infty } U\left( z \right)\)

Calculation:

Given:

\(U\left( z \right) = \frac{{2{z^2} - 3z}}{{3{z^2} + 4}}\)

\(= \mathop {{\rm{lt}}}\limits_{z \to \infty } \;\frac{{2{z^2} - 3z}}{{3{z^2} + 4}}\)

\(= \mathop {{\rm{lt}}}\limits_{z \to \infty } \frac{{2{z^2}\left( {1 - \frac{3}{z}} \right)}}{{3{z^2}\left( {1 + \frac{4}{{{z^2}}}} \right)}} = \frac{2}{3}\)

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{1 - \left| x \right|,\;\; - 1 \le x < 1}\\{0,\;\;\;\;\;otherwise}\end{array}} \right.\) 

\(X\left( \omega \right) = \mathop \smallint \limits_{ - 1}^1 \left( {1 - \left| x \right|} \right){e^{ - j\omega x}}dx\) 

\( = \mathop \smallint \limits_{ - 1}^0 \left( {1 + x} \right){e^{ - j\omega x}}dx + \mathop \smallint \limits_0^1 \left( {1 - x} \right){e^{ - j\omega x}}dx\) 

\(\mathop \smallint \limits_{ - 1}^0 \left( {1 + x} \right){e^{ - j\omega x}}dx = \left( {1 + x} \right)\left[ {\frac{{{e^{ - j\omega x}}}}{{ - j\omega }}} \right] - \smallint \frac{{{e^{ - j\omega x}}}}{{ - j\omega }}dx\) 

\( = \left[ {\left( {1 + x} \right)\left[ {\frac{{{1^{ - j\omega x}}}}{{ - j\omega }}} \right] - \left[ {\frac{{{e^{ - j\omega x}}}}{{{{\left( {j\omega } \right)}^2}}}} \right]} \right]_{ - 1}^0\) 

\( = \left( {\frac{1}{{ - j\omega }} - \frac{1}{{{{\left( {j\omega } \right)}^2}}}} \right) - \left( {0 - \frac{{{e^{j\omega }}}}{{{{\left( {j\omega } \right)}^2}}}} \right)\) 

\(= \frac{1}{{ - j\omega }} - \frac{1}{{{{\left( {j\omega } \right)}^2}}} + \frac{{{e^{j\omega }}}}{{{{\left( {j\omega } \right)}^2}}}\) 

\(\mathop \smallint \limits_0^1 \left( {1 - x} \right){e^{ - j\omega x}}dx\) 

\(= \left( {1 - x} \right)\left[ {\frac{{{e^{ - j\omega x}}}}{{ - j\omega }}} \right] - \smallint \frac{{{e^{ - j\omega x}}}}{{ - j\omega }}\left( { - 1} \right)dx\) 

\( = \left[ {\left( {1 - x} \right)\left[ {\frac{{{e^{ - j\omega x}}}}{{ - j\omega }}} \right] + \left[ {\frac{{{e^{ - j\omega x}}}}{{{{\left( {j\omega } \right)}^2}}}} \right]} \right]_0^1\) 

\( = \left[ {0 + \frac{{{e^{ - j\omega }}}}{{{{\left( {j\omega } \right)}^2}}}} \right] - \left[ {\frac{1}{{ - j\omega }} + \frac{1}{{{{\left( {j\omega } \right)}^2}}}} \right]\) 

\(= \frac{{{e^{ - j\omega }}}}{{{{\left( {j\omega } \right)}^2}}} + \frac{1}{{j\omega }} - \frac{1}{{{{\left( {j\omega } \right)}^2}}}\) 

\(X\left( \omega \right) = \frac{{ - 1}}{{j\omega }} - \frac{1}{{{{\left( {j\omega } \right)}^2}}} + \frac{{{e^{j\omega }}}}{{{{\left( {j\omega } \right)}^2}}} + \frac{{{e^{ - j\omega }}}}{{{{\left( {j\omega } \right)}^2}}} + \frac{1}{{j\omega }} - \frac{1}{{{{\left( {j\omega } \right)}^2}}}\) 

\(= \frac{{ - 2}}{{{{\left( {j\omega } \right)}^2}}} + \frac{{{e^{j\omega }} + {e^{ - j\omega }}}}{{{{\left( {j\omega } \right)}^2}}}\) 

\(= \frac{2}{{{{\left( {j\omega } \right)}^2}}}\left[ { - 1 + \frac{{\left( {{e^{j\omega }} - {e^{ - j\omega }}} \right)2}}{2}} \right] = \frac{2}{{{{\left( {j\omega } \right)}^2}}}\left[ {1 - 2\cos j\omega } \right]\) 

\(= \frac{2}{{{s^2}}}\left[ {1 - 2\cos s} \right]\)