Problem Find the exponential Fourier transform of the given \(f\left ( x\right ) \) and write \(f\relax (x) \) as a fourier integral.
\[ \ f(x)=\left \{ \begin {array} [c]{lll}\sin x & & \left \vert x\right \vert <\frac {\pi }{2}\\ \ & & \ \\ 0, & & \left \vert x\right \vert >\frac {\pi }{2}\end {array} \right . \]
Solution
Let \(\digamma \left (\alpha \right ) \) be the Fourier transform of \(f\left ( x\right ) \) defined as \(\digamma \left (\alpha \right ) =\frac {1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} f\relax (x) \ e^{-i\alpha x}\ dx\)
\begin {align*} \digamma \left (\alpha \right ) & =\frac {1}{2\pi }\ {\displaystyle \int \limits _{-\infty }^{\infty }} f\relax (x) \ e^{-i\alpha x}\ dx\\ 2\pi \ \digamma \left (\alpha \right ) & =\ {\displaystyle \int \limits _{-\frac {\pi }{2}}^{\frac {\pi }{2}}} \sin x\ \ e^{-i\alpha x}\ dx\ \ \ \end {align*}
Integration by parts, \(u=\sin x,du=\cos x,v=\frac {e^{-i\alpha x}}{-i\alpha }\), hence \(\int u\ dv\ =\) \(uv-\int du\ v\)
\begin {align*} I & =\ {\displaystyle \int \limits _{-\frac {\pi }{2}}^{\frac {\pi }{2}}} \sin x\ \ e^{-i\alpha x}\ dx\ \ \\ & =\left [ \sin x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}-{\displaystyle \int \limits _{-\frac {\pi }{2}}^{\frac {\pi }{2}}} \cos x\ \ \frac {e^{-i\alpha x}}{-i\alpha }\ dx\\ & =\left [ \sin x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}+\frac {1}{i\alpha }{\displaystyle \int \limits _{-\frac {\pi }{2}}^{\frac {\pi }{2}}} \cos x\ \ e^{-i\alpha x}\ dx \end {align*}
Integration by parts the second integral again. \(u=\cos x,du=-\sin x\)
\begin {align*} I & =\left [ \sin x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}+\frac {1}{i\alpha }\left \{ \left [ \cos x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}-{\displaystyle \int \limits _{-\frac {\pi }{2}}^{\frac {\pi }{2}}} \left (-\sin x\right ) \ \ \frac {e^{-i\alpha x}}{-i\alpha }\ dx\right \} \\ I & =\left [ \sin x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}+\frac {1}{i\alpha }\left \{ \left [ \cos x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}-\left ( \frac {1}{i\alpha }\right ) {\displaystyle \int \limits _{-\frac {\pi }{2}}^{\frac {\pi }{2}}} \sin x\ \ e^{-i\alpha x}\ dx\right \} \\ I & =\left [ \sin x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}+\frac {1}{i\alpha }\left [ \cos x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}-\frac {1}{i^{2}\alpha ^{2}}{\displaystyle \int \limits _{-\frac {\pi }{2}}^{\frac {\pi }{2}}} \sin x\ \ e^{-i\alpha x}\ dx\\ I & =\left [ \sin x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}+\frac {1}{i\alpha }\left [ \cos x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}+\frac {1}{\alpha ^{2}}{\displaystyle \int \limits _{-\frac {\pi }{2}}^{\frac {\pi }{2}}} \sin x\ \ e^{-i\alpha x}\ dx \end {align*}
But the last integral on the right above is the same as the integral we start with, so
\begin {align*} I & =\left [ \sin x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}+\frac {1}{i\alpha }\left [ \cos x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}+\frac {1}{\alpha ^{2}}I\\ I-\frac {1}{\alpha ^{2}}I & =\left [ \sin x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}-\frac {i}{\alpha }\left [ \cos x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\frac {\pi }{2}}^{\frac {\pi }{2}}\\ I\left (1-\frac {1}{\alpha ^{2}}\right ) & =\left [ \sin \left (\frac {\pi }{2}\right ) \ \frac {e^{-i\alpha \frac {\pi }{2}}}{-i\alpha }-\sin \left ( -\frac {\pi }{2}\right ) \ \frac {e^{-i\alpha \left (-\frac {\pi }{2}\right ) }}{-i\alpha }\right ] -\frac {i}{\alpha }\left [ \cos \left (\frac {\pi }{2}\right ) \ \frac {e^{-i\alpha \frac {\pi }{2}}}{-i\alpha }-\cos \left (-\frac {\pi }{2}\right ) \ \frac {e^{-i\alpha \left (-\frac {\pi }{2}\right ) }}{-i\alpha }\right ] \\ I\left (\frac {\alpha ^{2}-1}{\alpha ^{2}}\right ) & =\left [ \ \frac {e^{-i\alpha \frac {\pi }{2}}}{-i\alpha }+\ \frac {e^{i\alpha \left (\frac {\pi }{2}\right ) }}{-i\alpha }\right ] -\ \frac {i}{\alpha }\relax (0) \\ I & =\left (\frac {\alpha ^{2}}{\alpha ^{2}-1}\right ) \left [ \ \frac {-e^{-i\alpha \frac {\pi }{2}}}{i\alpha }-\ \frac {e^{i\alpha \frac {\pi }{2}}}{i\alpha }\right ] \\ I & =\left (\frac {\alpha }{\alpha ^{2}-1}\right ) \frac {-1}{i}\left [ \ e^{i\alpha \frac {\pi }{2}}+e^{-i\alpha \frac {\pi }{2}}\right ] \\ I & =\left (\frac {\alpha i}{\alpha ^{2}-1}\right ) \ \left [ \ e^{i\alpha \frac {\pi }{2}}+e^{-i\alpha \frac {\pi }{2}}\right ] \\ I & =\left (\frac {\alpha i}{\alpha ^{2}-1}\right ) 2\cos \left (\alpha \frac {\pi }{2}\right ) \end {align*}
Hence the Fourier transform of \(f\relax (x) \) is
\begin {align*} 2\pi \ \digamma \left (\alpha \right ) & =\ {\displaystyle \int \limits _{-\frac {\pi }{2}}^{\frac {\pi }{2}}} \sin x\ \ e^{-i\alpha x}\ dx\\ & =\left (\frac {\alpha i}{\alpha ^{2}-1}\right ) 2\cos \left (\alpha \frac {\pi }{2}\right ) \end {align*}
Therefore
\begin {align*} g\left (\alpha \right ) & =\frac {1}{\pi }\left (\frac {\alpha i}{\alpha ^{2}-1}\right ) \cos \left (\alpha \frac {\pi }{2}\right ) \\ & =\ \frac {\alpha \ i\ \cos \left (\alpha \frac {\pi }{2}\right ) }{\left ( \alpha ^{2}-1\right ) \pi } \end {align*}
To obtain \(f\relax (x) \) given its fourier transform \(\digamma \left ( \alpha \right ) ,\) then we apply the inverse fourier transform
\begin {align*} f\relax (x) & =\ {\displaystyle \int \limits _{-\infty }^{\infty }} \digamma \left (\alpha \right ) \ e^{i\alpha x}\ d\alpha \\ & =\ {\displaystyle \int \limits _{-\infty }^{\infty }} \ \frac {\alpha \ i\ \cos \left (\alpha \frac {\pi }{2}\right ) }{\left ( \alpha ^{2}-1\right ) \pi }\ \ e^{i\alpha x}\ d\alpha \\ & =\frac {1}{\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \ \frac {\alpha \ i\ \cos \left (\alpha \frac {\pi }{2}\right ) }{\alpha ^{2}-1}\ \ e^{i\alpha x}\ d\alpha \end {align*}
Problem Find the fourier sin transform of the given \(f\left ( x\right ) \) and write \(f\relax (x) \) as a fourier integral. Verify the answer is the same as the exponential fourier transform.
\[ \ f(x)=\left \{ \begin {array} [c]{lll}x & & \left \vert x\right \vert <1\\ \ & & \ \\ 0, & & \left \vert x\right \vert >1 \end {array} \right . \]
Solution
Let \(\mathcal {F}_{s}f\relax (x) \) be the Fourier sin transform of \(f\relax (x) \) defined as \(g_{s}\left (\alpha \right ) =\mathcal {F}_{s}f\relax (x) =\sqrt {\frac {2}{\pi }}{\displaystyle \int \limits _{0}^{\infty }} f\relax (x) \ \sin \left (\alpha x\right ) \ dx\). Hence, for the function above we get
\[ g_{s}\left (\alpha \right ) =\ \sqrt {\frac {2}{\pi }}{\displaystyle \int \limits _{0}^{1}} x\ \sin \left (\alpha x\right ) \ dx \]
Notice, we integrate from zero, not from -1, since the sin transform is defined only for positive x. Integrating by parts, \(u=x,v=\frac {-\cos \left ( \alpha x\right ) }{\alpha }\), hence \(\int u\ dv\ =\) \(uv-\int du\ v\)
\begin {align*} g_{s}\left (\alpha \right ) & =\ \sqrt {\frac {2}{\pi }}\left \{ \left [ x\left (\frac {-\cos \left (\alpha x\right ) }{\alpha }\right ) \right ] _{0}^{1}-\ {\displaystyle \int \limits _{0}^{1}} \ \frac {-\cos \left (\alpha x\right ) }{\alpha }\ dx\right \} \\ & =\sqrt {\frac {2}{\pi }}\left \{ \frac {-1}{\alpha }\left [ x\ \cos \left ( \alpha x\right ) \right ] _{0}^{1}+\frac {1}{\alpha }\ {\displaystyle \int \limits _{0}^{1}} \ \cos \left (\alpha x\right ) \ dx\right \} \\ & =\sqrt {\frac {2}{\pi }}\left \{ \frac {-1}{\alpha }\left [ x\ \cos \left ( \alpha x\right ) \right ] _{0}^{1}+\frac {1}{\alpha }\ \left [ \frac {\sin \left ( \alpha x\right ) }{\alpha }\right ] _{0}^{1}\right \} \\ & =\sqrt {\frac {2}{\pi }}\frac {1}{\alpha }\left \{ -\left [ \ \cos \left ( \alpha \right ) -0\right ] +\ \frac {1}{\alpha }\ \left [ \sin \left (\alpha x\right ) \right ] _{0}^{1}\right \} \\ & =\sqrt {\frac {2}{\pi }}\frac {1}{\alpha }\left \{ -\cos \left (\alpha \right ) \ +\ \frac {1}{\alpha }\ \left [ \sin \left (\alpha \right ) -0\right ] ^{\ }\right \} \\ & =\sqrt {\frac {2}{\pi }}\frac {1}{\alpha }\left (-\cos \left (\alpha \right ) +\ \frac {1}{\alpha }\ \sin \left (\alpha \right ) \right ) \\ & =\sqrt {\frac {2}{\pi }}\frac {1}{\alpha }\left (\frac {1}{\alpha }\ \sin \left ( \alpha \right ) -\cos \left (\alpha \right ) \ ^{\ }\right ) \end {align*}
Hence the Sin Fourier transform of \(f\relax (x) \) is
\[ g_{s}\left (\alpha \right ) =\sqrt {\frac {2}{\pi }}\frac {1}{\alpha }\left ( \frac {1}{\alpha }\ \sin \left (\alpha \right ) -\cos \left (\alpha \right ) \ ^{\ }\right ) \]
To obtain \(f\relax (x) \) given its sin fourier transform \(g_{s}\left ( \alpha \right ) ,\) then we apply the inverse sin fourier transform
\begin {align} f_{s}\relax (x) & =\sqrt {\frac {2}{\pi }}\ {\displaystyle \int \limits _{0}^{\infty }} g_{s}\left (\alpha \right ) \ \sin \alpha x\ d\alpha \nonumber \\ & =\ \sqrt {\frac {2}{\pi }}\ {\displaystyle \int \limits _{0}^{\infty }} \sqrt {\frac {2}{\pi }}\frac {1}{\alpha }\left (\frac {1}{\alpha }\ \sin \alpha -\cos \alpha ^{\ }\right ) \ \sin \alpha x\ d\alpha \nonumber \\ & =\frac {2}{\pi }\ {\displaystyle \int \limits _{0}^{\infty }} \ \left (\frac {1}{\alpha ^{2}}\ \sin \alpha -\frac {1}{\alpha }\cos \alpha \right ) \ \sin \alpha x\ d\alpha \nonumber \\ & =\frac {2}{\pi }\ {\displaystyle \int \limits _{0}^{\infty }} \ \left (\frac {\sin \alpha -\alpha \cos \alpha }{\alpha ^{2}}\right ) \ \sin \alpha x\ \ d\alpha \tag {A0} \end {align}
Now we need to show that the above is the same as the inverse fourier transform found for problem 6. From back of the book, the IFT for problem 6 is given as
\[ f\relax (x) =\int _{-\infty }^{\infty }\frac {\sin \alpha -\alpha \cos \alpha }{i\pi \alpha ^{2}}e^{i\alpha x}d\alpha \]
Need to convert the above to \(f_{s}\relax (x) \). Since \(e^{i\alpha x}=\cos \alpha x+i\sin \alpha x\)
\begin {align} f\relax (x) & =\int _{-\infty }^{\infty }\frac {\sin \alpha -\alpha \cos \alpha }{i\pi \alpha ^{2}}\left (\cos \alpha x+i\sin \alpha x\right ) d\alpha \nonumber \\ & =\int _{-\infty }^{\infty }\frac {\left (\sin \alpha -\alpha \cos \alpha \right ) \cos \alpha x}{i\pi \alpha ^{2}}+\frac {\left (\sin \alpha -\alpha \cos \alpha \right ) i\sin \alpha x}{i\pi \alpha ^{2}}d\alpha \nonumber \\ & =\int _{-\infty }^{\infty }\frac {\left (\sin \alpha -\alpha \cos \alpha \right ) \cos \alpha x}{i\pi \alpha ^{2}}d\alpha +\int _{-\infty }^{\infty }\frac {\left ( \sin \alpha -\alpha \cos \alpha \right ) i\sin \alpha x}{i\pi \alpha ^{2}}d\alpha \tag {1} \end {align}
Looking at the first integral,
\begin {align*} & \int _{-\infty }^{\infty }\frac {\left (\overset {\text {odd}}{\overbrace {\sin \alpha }}-\overset {\text {odd}}{\overbrace {\alpha }}\overset {\text {even}}{\overbrace {\cos \alpha }}\right ) \overset {\text {even}}{\overbrace {\cos \alpha x}}}{i\pi \overset {\text {even}}{\overbrace {\alpha ^{2}}}}\ \ d\alpha \\ & =\int _{-\infty }^{\infty }\frac {\left (\text {odd}-\text {odd}\times \text {even}\right ) \times \text {even}}{\text {even}}\ \ d\alpha \\ & =\int _{-\infty }^{\infty }\frac {\left (\text {odd}-\text {odd}\ \right ) \times \text {even}}{\text {even}}\ \ d\alpha \\ & =\int _{-\infty }^{\infty }\frac {\text {odd\ }\times \text {even}}{\text {even}}\ \ d\alpha \\ & =\int _{-\infty }^{\infty }\frac {\text {odd}}{\text {even}}\ \ d\alpha \\ & =\int _{-\infty }^{\infty }\text {odd\ }\ d\alpha \end {align*}
Hence the integral vanishes. Hence (1) becomes
\begin {equation} f\relax (x) =\int _{-\infty }^{\infty }\frac {\left (\sin \alpha -\alpha \cos \alpha \right ) i\sin \alpha x}{i\pi \alpha ^{2}}d\alpha \tag {2} \end {equation}
Looking at the above
\begin {align*} \frac {\left (\overset {\text {odd}}{\overbrace {\sin \alpha }}-\overset {\text {odd}}{\overbrace {\alpha }}\overset {\text {even}}{\overbrace {\cos \alpha }}\right ) i\overset {\text {odd}}{\overbrace {\sin \alpha x}}}{i\pi \overset {\text {even}}{\overbrace {\alpha ^{2}}}} & =\frac {\left (\text {odd}-\text {odd}\times \text {even}\right ) \times \text {odd}}{\text {even}}\\ & =\frac {\text {odd}\ \ \times \text {odd}}{\text {even}}\\ & =\frac {\text {even}}{\text {even}}\\ & =\text {even} \end {align*}
Since the integrand is even, then \(\int _{-\infty }^{\infty }=2\int _{0}^{\infty }\) Hence (2) becomes
\begin {align} f\relax (x) & =2\int _{0}^{\infty }\frac {\left (\sin \alpha -\alpha \cos \alpha \right ) i\sin \alpha x}{i\pi \alpha ^{2}}\ \ d\alpha \tag {3}\\ & =\frac {2}{\pi }\int _{0}^{\infty }\frac {\left (\sin \alpha -\alpha \cos \alpha \right ) }{\alpha ^{2}}\sin \alpha x\ \ d\alpha \nonumber \end {align}
comparing this to equation (A1) above, we see that
\[ f_{s}\relax (x) =f\relax (x) \]
Which is what we are asked to show.
Problem Find the fourier sin transform of the given \(f\left ( x\right ) \) and write \(f\relax (x) \) as a fourier integral. Verify the answer is the same as the exponential fourier transform.
\[ \ f(x)=\left \{ \begin {array} [c]{lll}\sin x & & \left \vert x\right \vert <\frac {\pi }{2}\\ \ & & \ \\ 0, & & \left \vert x\right \vert >\frac {\pi }{2}\end {array} \right . \]
Solution
Let \(\mathcal {F}_{s}f\relax (x) \) be the Fourier sin transform of \(f\relax (x) \) defined as \(g_{s}\left (\alpha \right ) =\mathcal {F}_{s}f\relax (x) =\sqrt {\frac {2}{\pi }}{\displaystyle \int \limits _{0}^{\infty }} f\relax (x) \ \sin \left (\alpha x\right ) \ dx\). Hence, for the function above we get
\[ g_{s}\left (\alpha \right ) =\ \sqrt {\frac {2}{\pi }}{\displaystyle \int \limits _{0}^{1}} \sin x\ \sin \left (\alpha x\right ) \ dx \]
Notice, we integrate from zero, not from -1, since the sin transform is defined only for positive \(x\). Since \(\sin \beta \sin \gamma =\frac {1}{2}\cos \left (\beta -\gamma \right ) -\frac {1}{2}\cos \left (\beta +\gamma \right ) \) Then \(\sin x\ \sin \left (\alpha x\right ) =\frac {1}{2}\cos \left (x-\alpha x\right ) -\frac {1}{2}\cos \left (x+\alpha x\right ) \). Hence
\begin {align*} g_{s}\left (\alpha \right ) & =\ \sqrt {\frac {2}{\pi }}{\displaystyle \int \limits _{0}^{1}} \frac {1}{2}\cos \left (x-\alpha x\right ) -\frac {1}{2}\cos \left (x+\alpha x\right ) \ dx\\ & =\frac {1}{2}\sqrt {\frac {2}{\pi }}\left \{ \left [ \frac {\sin \left (x-\alpha x\right ) }{1-\alpha }\right ] _{0}^{1}-\left [ \frac {\sin \left (x+\alpha x\right ) }{1+\alpha }\right ] _{0}^{1}\right \} \\ & =\frac {1}{2}\sqrt {\frac {2}{\pi }}\left \{ \left [ \frac {\sin \left ( 1-\alpha \right ) }{1-\alpha }-0\right ] -\left [ \frac {\sin \left ( 1+\alpha \right ) }{1+\alpha }-0\right ] _{0}^{1}\right \} \\ & =\frac {1}{2}\sqrt {\frac {2}{\pi }}\left (\frac {\sin \left (1-\alpha \right ) }{1-\alpha }-\frac {\sin \left (1+\alpha \right ) }{1+\alpha }\right ) \end {align*}
Hence the Sin Fourier transform of \(f\relax (x) \) is
\[ g_{s}\left (\alpha \right ) =\frac {1}{2}\sqrt {\frac {2}{\pi }}\left (\frac {\sin \left (1-\alpha \right ) }{1-\alpha }-\frac {\sin \left (1+\alpha \right ) }{1+\alpha }\right ) \]
Therefore, to obtain \(f\relax (x) \) given its sin fourier transform \(g_{s}\left (\alpha \right ) \), we apply the inverse sin fourier transform
\begin {align} f_{s}\relax (x) & =\sqrt {\frac {2}{\pi }}\ {\displaystyle \int \limits _{0}^{\infty }} g_{s}\left (\alpha \right ) \ \sin \alpha x\ d\alpha \nonumber \\ & =\ \sqrt {\frac {2}{\pi }}\ {\displaystyle \int \limits _{0}^{\infty }} \frac {1}{2}\sqrt {\frac {2}{\pi }}\left (\frac {\sin \left (1-\alpha \right ) }{1-\alpha }-\frac {\sin \left (1+\alpha \right ) }{1+\alpha }\right ) \ \sin \alpha x\ d\alpha \nonumber \\ & =\frac {1}{\pi }\ {\displaystyle \int \limits _{0}^{\infty }} \ \left (\frac {\sin \left (1-\alpha \right ) }{1-\alpha }-\frac {\sin \left ( 1+\alpha \right ) }{1+\alpha }\right ) \ \sin \alpha x\ d\alpha \tag {1} \end {align}
Now we need to show that the above is the same as the exponential inverse fourier transform found for problem 12. The exponential IFT for problem 12 is
\begin {equation} f\relax (x) =\frac {1}{\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \ \frac {\alpha \ i\ \cos \left (\alpha \frac {\pi }{2}\right ) }{\alpha ^{2}-1}\ \ e^{i\alpha x}\ d\alpha \tag {2} \end {equation}
So Need to show that (1) and (2) are the same. Need to convert the above (1) to \(f_{s}\relax (x) \) in (2)\(.\)Since \(e^{i\alpha x}=\cos \alpha x+i\sin \alpha x\), (2) can be written as
\begin {align} f\relax (x) & =\frac {1}{\pi }\int _{-\infty }^{\infty }\frac {\alpha \ i\ \cos \left (\alpha \frac {\pi }{2}\right ) }{\alpha ^{2}-1}\ \left ( \cos \alpha x+i\sin \alpha x\right ) d\alpha \nonumber \\ & =\frac {1}{\pi }\left [ \int _{-\infty }^{\infty }\frac {\alpha \ i\ \cos \left ( \alpha \frac {\pi }{2}\right ) }{\alpha ^{2}-1}\cos \alpha x\ \ d\alpha +\int _{-\infty }^{\infty }\frac {\alpha \ i\ \cos \left (\alpha \frac {\pi }{2}\right ) }{\alpha ^{2}-1}\ i\sin \alpha x\ \ d\alpha \right ] \tag {3} \end {align}
Looking at the first integral,
\begin {align*} & \int _{-\infty }^{\infty }\frac {\overset {\text {odd}}{\overbrace {\alpha }}\ i\ \overset {\text {even}}{\overbrace {\cos \left (\alpha \frac {\pi }{2}\right ) }}}{\overset {\text {even}}{\overbrace {\alpha ^{2}}}-1}\overset {\text {even}}{\overbrace {\cos \alpha x}}\ \ d\alpha \\ & \\ & =\int _{-\infty }^{\infty }\frac {\left (\text {odd}\times \ \text {even}\right ) \times \text {even}}{\text {even}}\ \ d\alpha \\ & =\int _{-\infty }^{\infty }\frac {\text {odd}\ \times \text {even}}{\text {even}}\ \ d\alpha \\ & =\int _{-\infty }^{\infty }\frac {\text {odd}}{\text {even}}\ \ d\alpha \\ & =\int _{-\infty }^{\infty }\text {odd}\ \ d\alpha \end {align*}
Hence the integral vanishes. So (3) becomes
\begin {align*} f\relax (x) & =0+\frac {1}{\pi }\int _{-\infty }^{\infty }\frac {\alpha \ i\ \cos \left (\alpha \frac {\pi }{2}\right ) }{\alpha ^{2}-1}\ i\sin \alpha x\ \ d\alpha \\ & =\frac {1}{\pi }\int _{-\infty }^{\infty }\frac {\alpha \ i\ \cos \left ( \alpha \frac {\pi }{2}\right ) }{\alpha ^{2}-1}\ i\sin \alpha x\ \ d\alpha \end {align*}
Looking at the above integrand,
\begin {align*} \frac {\overset {\text {odd}}{\overbrace {\alpha }}\ i\ \overset {\text {even}}{\overbrace {\cos \left (\alpha \frac {\pi }{2}\right ) }}}{\overset {\text {even}}{\overbrace {\alpha ^{2}}}-1}\ i\overset {\text {odd}}{\overbrace {\sin \alpha x}} & =\frac {\left (\text {odd}\ \times \text {even}\right ) \times \text {odd}}{\text {even}}\\ & =\frac {\text {odd}\ \ \times \text {odd}}{\text {even}}\\ & =\frac {\text {even}}{\text {even}}\\ & =\text {even} \end {align*}
Since the integrand is even, then \(\int _{-\infty }^{\infty }=2\int _{0}^{\infty }\). Hence (2) becomes
\begin {align*} f\relax (x) & =\frac {2}{\pi }\int _{0}^{\infty }\frac {\alpha \ i\ \cos \left (\alpha \frac {\pi }{2}\right ) }{\alpha ^{2}-1}\ i\sin \alpha x\ \ d\alpha \\ & =\frac {-2}{\pi }\int _{0}^{\infty }\frac {\alpha \ \ \cos \left (\alpha \frac {\pi }{2}\right ) }{\alpha ^{2}-1}\ \sin \alpha x\ \ d\alpha \\ & =\frac {2}{\pi }\int _{0}^{\infty }\frac {\alpha \ \ \cos \left (\alpha \frac {\pi }{2}\right ) }{1-\alpha ^{2}}\ \sin \alpha x\ \ d\alpha \end {align*}
But \(1-\alpha ^{2}=\left (1+\alpha \right ) \left (1-\alpha \right ) \), therefore
\[ f\relax (x) =\frac {2}{\pi }\int _{0}^{\infty }\frac {\alpha \ \ \cos \left ( \alpha \frac {\pi }{2}\right ) }{\left (1+\alpha \right ) \left (1-\alpha \right ) }\ \sin \alpha x\ \ d\alpha \]
But \(\cos \left (\alpha \frac {\pi }{2}\right ) \sin \alpha x=\frac {1}{2}\left ( -\sin \left (\alpha \frac {\pi }{2}-\alpha x\right ) +\sin \left (\alpha \frac {\pi }{2}+\alpha x\right ) \right ) \), hence the above becomes
\begin {align*} f\relax (x) & =\frac {2}{\pi }\int _{0}^{\infty }\frac {\alpha \ \ \frac {1}{2}\left (-\sin \left (\alpha \frac {\pi }{2}-\alpha x\right ) +\sin \left ( \alpha \frac {\pi }{2}+\alpha x\right ) \right ) }{\left (1+\alpha \right ) \left (1-\alpha \right ) }\ \ \ d\alpha \\ f\relax (x) & =\frac {1}{\pi }\int _{0}^{\infty }\frac {\alpha \ \ \left ( -\sin \left (\alpha \frac {\pi }{2}-\alpha x\right ) +\sin \left (\alpha \frac {\pi }{2}+\alpha x\right ) \right ) }{\left (1+\alpha \right ) \left ( 1-\alpha \right ) }\ \ \ d\alpha \\ f\relax (x) & =\frac {1}{\pi }\int _{0}^{\infty }\frac {\alpha \sin \left ( \alpha \frac {\pi }{2}+\alpha x\right ) }{\left (1+\alpha \right ) \left ( 1-\alpha \right ) }-\frac {\ \ \alpha \sin \left (\alpha \frac {\pi }{2}-\alpha x\right ) }{\left (1+\alpha \right ) \left (1-\alpha \right ) }\ \ \ d\alpha \end {align*}
Problem Find the fourier transform of the given \(f\relax (x) \) \(=e^{\frac {-x^{2}}{2\sigma ^{2}}}\)
Solution
Let \(\digamma \left (\alpha \right ) \) be the Fourier sin transform of \(f\relax (x) \) defined as
\[ \digamma \left (\alpha \right ) =\frac {1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} f\relax (x) \ e^{-i\alpha x}\ dx \]
So, for the function above we get
\begin {align} \digamma \left (\alpha \right ) & =\ \frac {1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} e^{\frac {-x^{2}}{2\sigma ^{2}}}\ e^{-i\alpha x}\ dx\nonumber \\ & =\ \frac {1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} e^{\frac {-x^{2}}{2\sigma ^{2}}-i\alpha x}\ dx\nonumber \\ & =\ \frac {1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} e^{\ \frac {-x^{2}-i\alpha x\left (2\sigma ^{2}\right ) }{2\sigma ^{2}}}\ dx\nonumber \\ & =\ \frac {1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} e^{\ \frac {-x^{2}-2i\alpha \sigma ^{2}x\ }{2\sigma ^{2}}}\ dx \tag {1} \end {align}
looking at the exponent \(\frac {-x^{2}-2i\alpha \sigma ^{2}x\ }{2\sigma ^{2}}\). completing the square in \(x\) gives
\[ x^{2}+2i\alpha \sigma ^{2}x\ =\left (x+Z\ \right ) ^{2}-Y \]
Solving for \(Z,Y\) gives
\[ x^{2}+2i\alpha \sigma ^{2}x\ =x^{2}+2xZ+Z^{2}-Y\ \]
Therefore \(Z=i\alpha \sigma ^{2},Z^{2}-Y=0\), and \(Y=-\alpha ^{2}\sigma ^{4}\). Hence Exponent can be written as
\begin {align*} \frac {x^{2}+2i\alpha \sigma ^{2}x\ }{-2\sigma ^{2}} & =\frac {\left ( x+i\alpha \sigma ^{2}\ \right ) ^{2}-\left (-\alpha ^{2}\sigma ^{4}\right ) }{-2\sigma ^{2}}\\ & =\frac {\left (x+i\alpha \sigma ^{2}\ \right ) ^{2}+\alpha ^{2}\sigma ^{4}}{-2\sigma ^{2}}\\ & =\frac {\left (x+i\alpha \sigma ^{2}\ \right ) ^{2}}{-2\sigma ^{2}}-\frac {\alpha ^{2}\sigma ^{4}}{2\sigma ^{2}}\\ & =\frac {\left (x+i\alpha \sigma ^{2}\ \right ) ^{2}}{-2\sigma ^{2}}-\frac {\alpha ^{2}\sigma ^{2}}{2} \end {align*}
The integral (1) becomes
\begin {align*} \digamma \left (\alpha \right ) & =\frac {1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} e^{\ \frac {\left (x+i\alpha \sigma ^{2}\ \right ) ^{2}}{-2\sigma ^{2}}-\frac {\alpha ^{2}\sigma ^{2}}{2}}\ dx\\ & =\frac {1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} e^{\ \frac {\left (x+i\alpha \sigma ^{2}\ \right ) ^{2}}{-2\sigma ^{2}}}\ e^{-\frac {\alpha ^{2}\sigma ^{2}}{2}}\ dx \end {align*}
Moving \(e^{-\frac {\alpha ^{2}\sigma ^{2}}{2}}\) outside the integral because it does not depend on \(x\) gives
\[ \digamma \left (\alpha \right ) =\frac {e^{-\frac {\alpha ^{2}\sigma ^{2}}{2}}}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} e^{\ \frac {\left (x+i\alpha \sigma ^{2}\ \right ) ^{2}}{-2\sigma ^{2}}}\ \ dx \]
Let \(y=x+i\alpha \sigma ^{2}\), \(dy=dx\) and the limits do not change. Hence we get
\[ \digamma \left (\alpha \right ) =\frac {e^{-\frac {\alpha ^{2}\sigma ^{2}}{2}}}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} e^{\ \frac {y\ ^{2}}{-2\sigma ^{2}}}\ \ dy \]
Since the exponential function is raised to a square power, then we can write \({\displaystyle \int \limits _{-\infty }^{\infty }} e^{\ y^{2}}=2{\displaystyle \int \limits _{0}^{\infty }} e^{\ y^{2}}\ \) (since even function). Hence above integral becomes
\[ \digamma \left (\alpha \right ) =\frac {e^{-\frac {\alpha ^{2}\sigma ^{2}}{2}}}{\pi }{\displaystyle \int \limits _{0}^{\infty }} e^{\ \frac {y\ ^{2}}{-2\sigma ^{2}}}\ \ dy \]
Let \(\zeta =\frac {y}{\sqrt {2}\sigma }\) \(,\) then \(y=\sqrt {2}\sigma \zeta \), and \(y^{2}=2\sigma ^{2}\zeta ^{2}\). Hence \(d\zeta =\frac {1}{\sqrt {2}\sigma }dy\) and the above integral becomes
\begin {align} \digamma \left (\alpha \right ) & =\frac {e^{-\frac {\alpha ^{2}\sigma ^{2}}{2}}}{\pi }{\displaystyle \int \limits _{0}^{\infty }} e^{\ -\zeta ^{2}}\ \ \sqrt {2}\sigma \ d\zeta \nonumber \\ \digamma \left (\alpha \right ) & =\sqrt {2}\sigma \frac {e^{-\frac {\alpha ^{2}\sigma ^{2}}{2}}}{\pi }{\displaystyle \int \limits _{0}^{\infty }} e^{\ -\zeta ^{2}}\ \ d\zeta \tag {2} \end {align}
Now from equation 9.5 on page 468
\begin {align*} \frac {2}{\sqrt {\pi }}{\displaystyle \int \limits _{0}^{\infty }} e^{\ -\zeta ^{2}}\ \ d\zeta & =1\\{\displaystyle \int \limits _{0}^{\infty }} e^{\ -\zeta ^{2}}\ \ d\zeta & =\frac {\sqrt {\pi }}{2} \end {align*}
Hence (2) becomes
\begin {align*} \digamma \left (\alpha \right ) & =\sqrt {2}\sigma \frac {e^{-\frac {\alpha ^{2}\sigma ^{2}}{2}}}{\pi }\left (\frac {\sqrt {\pi }}{2}\right ) \\ \digamma \left (\alpha \right ) & =\ \frac {\sigma }{\sqrt {2\pi }}\ e^{-\frac {\alpha ^{2}\sigma ^{2}}{2}} \end {align*}
Which is what we are asked to show.
Problem Show that
\(\int _{0}^{\infty }\frac {1-\cos \pi \alpha }{\alpha }\sin \alpha \ d\alpha =\frac {\pi }{2}\) and \(\int _{0}^{\infty }\frac {1-\cos \pi \alpha }{\alpha }\sin \pi \alpha \ d\alpha =\frac {\pi }{4}\)
Solution
From problem 17, the Fourier sin transform for \(f\relax (x) \) shown in problem 3 is
\[ g_{s}\left (\alpha \right ) =\frac {\sqrt {2}\left (1-\cos \pi \alpha \right ) }{\sqrt {\pi }\alpha }\]
From equation 4.14 page 651, \(f\relax (x) \) can be obtained from inverse sin transform is
\begin {align} f_{s}\relax (x) & =\sqrt {\frac {2}{\pi }}\int _{0}^{\infty }g_{s}\left ( \alpha \right ) \sin \alpha x\ d\alpha \nonumber \\ & =\sqrt {\frac {2}{\pi }}\int _{0}^{\infty }\left (\frac {\sqrt {2}\left ( 1-\cos \pi \alpha \right ) }{\sqrt {\pi }\alpha }\right ) \sin \alpha x\ d\alpha \nonumber \\ & =\frac {2}{\pi }\int _{0}^{\infty }\frac {\left (1-\cos \pi \alpha \right ) }{\alpha }\sin \alpha x\ d\alpha \tag {1} \end {align}
Now, from the definition of \(f\relax (x) \), which is
\[ \ f(x)=\left \{ \begin {array} [c]{lll}-1 & & -\pi <x<0\\ \ 1 & & \ 0<x<\pi \\ 0, & & \left \vert x\right \vert >\pi \end {array} \right . \]
We see that for \(x=1\), \(f\relax (x) =1\), hence substitute in (1) we get
\begin {align*} 1 & =-\frac {2}{\pi }\int _{0}^{\infty }\frac {\left (1-\cos \pi \alpha \right ) }{\alpha }\sin \alpha \ d\alpha \\ \frac {\pi }{2} & =\int _{0}^{\infty }\frac {\left (1-\cos \pi \alpha \right ) }{\alpha }\sin \alpha \ d\alpha \end {align*}
Which is the first result we required to show. For the second result, let \(x=\pi \) hence \(f\left (\pi \right ) =\ \)average value of \(f\relax (x) \) at \(x=\pi \). Which is given by \(\frac {f\left (\pi _{-}\right ) +f\left ( \pi _{+}\right ) }{2}=\frac {1+0}{2}=\frac {1}{2}\). Hence substitute in (1) we get
\begin {align*} f_{s}\left (x=\pi \right ) & =\frac {1}{2}=\frac {2}{\pi }\int _{0}^{\infty }\frac {\left (1-\cos \pi \alpha \right ) }{\alpha }\sin \alpha \pi \ d\alpha \\ \frac {\pi }{4} & =\int _{0}^{\infty }\frac {\left (1-\cos \pi \alpha \right ) }{\alpha }\sin \alpha \pi \ d\alpha \end {align*}
Which is the second result we are asked to show.
Problem Show that
(a) represent as an exponential fourier transform the function
\[ \ f(x)=\left \{ \begin {array} [c]{lll}\sin x & & 0<x<\pi \\ \ \ & & \ \ \\ 0, & & \text {otherwise}\end {array} \right . \]
(b) Show that the result can be written as
\[ f\relax (x) =\frac {1}{\pi }\int _{0}^{\infty }\frac {\cos \alpha x+\cos \alpha \left (x-\pi \right ) }{1-\alpha ^{2}}d\alpha \]
Solution
The exponential Fourier transform is defined as
\[ g\left (\alpha \right ) =\frac {1}{2\pi }\int _{-\infty }^{\infty }f\left ( x\right ) e^{-i\alpha x}\ dx \]
Applying the function \(f\relax (x) \) gives
\[ g\left (\alpha \right ) =\frac {1}{2\pi }\int _{0}^{\pi }\sin x\ e^{-i\alpha x}\ dx \]
But
\[ \sin x=\frac {e^{ix}-e^{-ix}}{2i}\]
Hence the transform can be written as
\begin {align*} g\left (\alpha \right ) & =\frac {1}{2\pi }\int _{0}^{\pi }\frac {e^{ix}-e^{-ix}}{2i}\ e^{-i\alpha x}\ dx\\ & =\frac {1}{4i\pi }\int _{0}^{\pi }e^{ix-i\alpha x}-e^{-ix-i\alpha x}\ \ dx\\ & =\frac {1}{4i\pi }\int _{0}^{\pi }e^{x\left (i-i\alpha \right ) }-e^{x\left ( -i-i\alpha \right ) }\ \ dx\\ & =\frac {1}{4i\pi }\left (\left [ \ \frac {e^{x\left (i-i\alpha \right ) }}{i-i\alpha }\right ] _{0}^{\pi }\ -\left [ \ \frac {e^{x\left (-i-i\alpha \right ) }}{-i-i\alpha }\right ] _{0}^{\pi }\right ) \\ & =\frac {1}{4i\pi }\left (\left [ \ \frac {e^{\pi \left (i-i\alpha \right ) }}{i-i\alpha }-\frac {1}{i-i\alpha }\right ] \ -\left [ \ \frac {e^{\pi \left ( -i-i\alpha \right ) }}{-i-i\alpha }-\ \frac {1}{-i-i\alpha }\right ] \right ) \\ & =\frac {1}{4i\pi }\left (\frac {1}{i-i\alpha }\left [ \ e^{\pi \left ( i-i\alpha \right ) }-1\right ] \ -\frac {1}{-i-i\alpha }\left [ \ e^{\pi \left ( -i-i\alpha \right ) }-\ 1\right ] \right ) \\ & =\frac {1}{4i\pi }\left (\frac {1}{i-i\alpha }\left [ \ e^{\pi \left ( i-i\alpha \right ) }-1\right ] \ +\frac {1}{i+i\alpha }\left [ \ e^{\pi \left ( -i-i\alpha \right ) }-\ 1\right ] \right ) \ \\ & =\frac {1}{i}\frac {1}{4i\pi }\left (\frac {1}{1-\alpha }\left [ \ e^{\pi \left (i-i\alpha \right ) }-1\right ] \ +\frac {1}{1+\alpha }\left [ \ e^{\pi \left (-i-i\alpha \right ) }-\ 1\right ] \right ) \ \\ & =\ \frac {-1}{4\pi }\left (\frac {e^{\pi \left (i-i\alpha \right ) }}{1-\alpha }-\frac {1}{1-\alpha }\ +\frac {e^{\pi \left (-i-i\alpha \right ) }}{1+\alpha }-\ \frac {1}{1+\alpha }\right ) \\ & =\ \frac {-1}{4\pi }\left (\frac {\left (1+\alpha \right ) e^{\pi \left ( i-i\alpha \right ) }+\left (1-\alpha \right ) e^{\pi \left (-i-i\alpha \right ) }}{\left (1-\alpha \right ) \left (1+\alpha \right ) }\ -\frac {\left ( 1+\alpha \right ) +\left (1-\alpha \right ) }{\left (1-\alpha \right ) \left ( 1+\alpha \right ) }\ \right ) \\ & =\ \frac {-1}{4\pi }\left (\frac {\left (1+\alpha \right ) e^{\pi \left ( i-i\alpha \right ) }+\left (1-\alpha \right ) e^{\pi \left (-i-i\alpha \right ) }}{\left (1-\alpha ^{2}\right ) }\ -\frac {2}{\left (1-\alpha ^{2}\right ) }\ \right ) \\ & =\ \frac {-1}{\left (1-\alpha ^{2}\right ) 4\pi }\left (e^{\pi \left ( i-i\alpha \right ) }+\alpha e^{\pi \left (i-i\alpha \right ) }+e^{\pi \left ( -i-i\alpha \right ) }-\alpha e^{\pi \left (-i-i\alpha \right ) }\ \ -2\right ) \\ & =\ \frac {-1}{\left (1-\alpha ^{2}\right ) 4\pi }\left (e^{\pi i\ }e^{-i\pi \alpha }+\alpha e^{\pi i\ }e^{-i\pi \alpha }+e^{-\pi i\ }e^{-i\pi \alpha }-\alpha e^{-\pi i\ }e^{-i\pi \alpha }\ \ -2\right ) \end {align*}
But \(e^{\pi i\ }=-1\) and \(e^{-\pi i\ }=-1\)
\begin {align*} g\left (\alpha \right ) & =\ \frac {-1}{\left (1-\alpha ^{2}\right ) 4\pi }\left (-e^{-i\pi \alpha }-\overbrace {\alpha e^{-i\pi \alpha }}-e^{-i\pi \alpha }+\overbrace {\alpha e^{-i\pi \alpha }}\ \ -2\right ) \\ g\left (\alpha \right ) & =\ \frac {-1}{\left (1-\alpha ^{2}\right ) 4\pi }\left (-e^{-i\pi \alpha }\ -e^{-i\pi \alpha }-2\right ) \\ g\left (\alpha \right ) & =\ \frac {1}{\left (1-\alpha ^{2}\right ) 4\pi }\left (e^{-i\pi \alpha }\ +e^{-i\pi \alpha }+2\right ) \\ g\left (\alpha \right ) & =\ \frac {2e^{-i\pi \alpha }\ +2}{\left ( 1-\alpha ^{2}\right ) 4\pi } \end {align*}
Hence the exponential fourier transform is
\[ g\left (\alpha \right ) =\ \frac {e^{-i\pi \alpha }\ +1}{\left (1-\alpha ^{2}\right ) 2\pi }\]
Therefore \(f\relax (x) \) can be rewritten as
\begin {align} f\relax (x) & =\ \int _{-\infty }^{\infty }g\left (\alpha \right ) \ e^{i\alpha x}\ d\alpha \nonumber \\ & =\ \int _{-\infty }^{\infty }\frac {e^{-i\pi \alpha }\ +1}{\left (1-\alpha ^{2}\right ) 2\pi }\ e^{i\alpha x}\ d\alpha \nonumber \\ & =\ \frac {1}{2\pi }\int _{-\infty }^{\infty }\frac {1+e^{-i\pi \alpha }\ }{\left ( 1-\alpha ^{2}\right ) }\ e^{i\alpha x}\ d\alpha \tag {1} \end {align}
Which is the answer required to show.
Part(b)
Now need to show that the above can be written as \[ f\relax (x) =\frac {1}{\pi }\int _{0}^{\infty }\frac {\cos \alpha x+\cos \alpha \left (x-\pi \right ) }{1-\alpha ^{2}}d\alpha \]
From (1)
\begin {align*} f\relax (x) & =\frac {1}{2\pi }\int _{-\infty }^{\infty }\frac {1+e^{-i\pi \alpha }\ }{\left (1-\alpha ^{2}\right ) }\ e^{i\alpha x}\ d\alpha \\ & =\frac {1}{2\pi }\int _{-\infty }^{\infty }\frac {e^{i\alpha x}+e^{i\alpha x}e^{-i\pi \alpha }\ }{\left (1-\alpha ^{2}\right ) }\ d\alpha \\ & =\frac {1}{2\pi }\int _{-\infty }^{\infty }\frac {e^{i\alpha x}+e^{i\alpha x-i\pi \alpha }\ }{\left (1-\alpha ^{2}\right ) }\ d\alpha \\ & =\frac {1}{2\pi }\int _{-\infty }^{\infty }\frac {e^{i\alpha x}+e^{i\alpha \left ( x-\pi \right ) }\ }{\left (1-\alpha ^{2}\right ) }\ d\alpha \\ & =\frac {1}{2\pi }\int _{-\infty }^{\infty }\frac {2\left (\cos \alpha x+\cos \alpha \left (x-\pi \right ) \right ) \ }{\left (1-\alpha ^{2}\right ) }\ d\alpha \\ & =\frac {1}{\pi }\int _{-\infty }^{\infty }\frac {\cos \alpha x+\cos \alpha \left ( x-\pi \right ) \ }{\left (1-\alpha ^{2}\right ) }\ d\alpha \end {align*}
Which is what is required to show.
Problem Find the exponential fourier transform of the given \(f\left ( x\right ) \) and write \(f\relax (x) \) as a fourier integral.\[ \ f(x)=\left \{ \begin {array} [c]{lll}-1, & & -\pi <x<0\\ 1, & & 0<x<\pi \\ 0, & & \left \vert x\right \vert >\pi \end {array} \right . \] Solution
Let \(\digamma \left (\alpha \right ) \) be the Fourier transform of \(f\left ( x\right ) \) defined as \(\digamma \left (\alpha \right ) =\frac {1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} f\relax (x) \ e^{-i\alpha x}\ dx\), hence\begin {align*} \ \digamma \left (\alpha \right ) & =\frac {1}{2\pi }\ {\displaystyle \int \limits _{-\infty }^{\infty }} f\relax (x) \ e^{-i\alpha x}\ dx\\ 2\pi \ \digamma \left (\alpha \right ) & =\ {\displaystyle \int \limits _{-\pi }^{0}} -\ e^{-i\alpha x}\ dx+{\displaystyle \int \limits _{0}^{\pi }} \ e^{-i\alpha x}\ dx\\ & =\ -\left [ \ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-\pi }^{0}\ +\left [ \ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{0}^{\pi }\\ & =\ \frac {1}{i\alpha }\left [ \ e^{-i\alpha x}\right ] _{-\pi }^{0}-\frac {1}{i\alpha }\left [ \ e^{-i\alpha x}\right ] _{0}^{\pi }\\ & =\frac {1}{i\alpha }\left [ \ e^{0}-e^{i\alpha \pi }\right ] -\frac {1}{i\alpha }\left [ \ e^{-i\alpha \pi }-e^{0}\right ] _{0}^{\pi }\\ & =\frac {1}{i\alpha }\left [ \ 1-e^{i\alpha \pi }\right ] -\frac {1}{i\alpha }\left [ \ e^{-i\alpha \pi }-1\right ] _{0}^{\pi }\\ & =\frac {1}{i\alpha }-\frac {e^{i\alpha \pi }}{i\alpha }-\frac {e^{-i\alpha \pi }}{i\alpha }\ +\frac {1}{i\alpha }\\ & =\frac {2}{i\alpha }-\frac {1}{i\alpha }\left (e^{i\alpha \pi }+e^{-i\alpha \pi }\right ) \end {align*}
But \(e^{i\alpha \pi }+e^{-i\alpha \pi }=2\cos \alpha \pi \). Hence\begin {align*} 2\pi \ \digamma \left (\alpha \right ) & =\frac {2}{i\alpha }-\frac {1}{i\alpha }\left (2\cos \alpha \pi \right ) \\ & =\ \frac {2}{i\alpha }\left (1-\cos \alpha \pi \right ) \end {align*}
Hence the Fourier transform of \(f\relax (x) \) is\begin {align*} \digamma \left (\alpha \right ) & =\frac {1}{2\pi }\left [ \ \frac {2}{i\alpha }\left (1-\cos \alpha \pi \right ) \right ] \\ & =\ \frac {1}{\pi \alpha i}\ \left (1-\cos \alpha \pi \right ) \end {align*}
To obtain \(f\relax (x) \) given its fourier transform \(\digamma \left ( \alpha \right ) ,\) then we apply the inverse fourier transform\begin {align*} f\relax (x) & =\ {\displaystyle \int \limits _{-\infty }^{\infty }} \digamma \left (\alpha \right ) \ e^{i\alpha x}\ d\alpha \\ & =\ {\displaystyle \int \limits _{-\infty }^{\infty }} \frac {1}{\pi \alpha i}\ \left (1-\cos \alpha \pi \right ) \ e^{i\alpha x}\ d\alpha \\ & =\frac {1}{\pi i}{\displaystyle \int \limits _{-\infty }^{\infty }} \ \frac {1}{\alpha }\left (1-\ \cos \alpha \pi \right ) \ e^{i\alpha x}\ d\alpha \end {align*}
Problem Find the exponential fourier transform of the given \(f\left ( x\right ) \) and write \(f\relax (x) \) as a fourier integral.\[ \ f(x)=\left \{ \begin {array} [c]{lll}1, & & 0<x<1\\ \ & & \ \\ 0, & & \text {otherwise}\end {array} \right . \] Solution
Let \(\digamma \relax (s) \) be the Fourier transform of \(f\left ( x\right ) \) defined as \(\digamma \relax (s) =\frac {1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} f\relax (x) \ e^{-i\alpha x}\ dx\)\begin {align*} \digamma \relax (s) & =\frac {1}{2\pi }\ {\displaystyle \int \limits _{-\infty }^{\infty }} f\relax (x) \ e^{-isx}\ dx\\ 2\pi \ \digamma \relax (s) & =\ {\displaystyle \int \limits _{0}^{1}} e^{-isx}\ dx\ \\ & =\ \ \left [ \ \frac {e^{-isx}}{-is}\right ] _{0}^{1}\ \ \\ & =\ \frac {-1}{is}\left [ \ e^{-isx}\right ] _{0}^{1}\ \\ & =\frac {-1}{is}\left [ \ e^{-is}-e^{0}\right ] \ \\ & =\frac {-1}{is}\left [ \ e^{-is}-1\right ] \ \end {align*}
Hence the Fourier transform of \(f\relax (x) \) is\begin {align*} \digamma \relax (s) & =\frac {1}{2\pi }\left [ \ \frac {-1}{is}\left [ \ e^{-is}-1\right ] \right ] \\ & =\ \frac {i}{2\pi s}\ \left (e^{-is}-1\right ) \end {align*}
To obtain \(f\relax (x) \) given its fourier transform \(\digamma \left ( s\right ) ,\) then we apply the inverse fourier transform\begin {align*} f\relax (x) & =\ {\displaystyle \int \limits _{-\infty }^{\infty }} \digamma \relax (s) \ e^{isx}\ ds\\ & =\ {\displaystyle \int \limits _{-\infty }^{\infty }} \ \frac {i}{2\pi s}\ \left (e^{-is}-1\right ) \ e^{isx}\ ds\\ & =\frac {i}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \ \frac {1}{s}\left (e^{-is}-1\right ) \ e^{isx}\ ds \end {align*}
Problem Find the exponential fourier transform of the given \(f\left ( x\right ) \) and write \(f\relax (x) \) as a fourier integral.
\[ \ f(x)=\left \{ \begin {array} [c]{lll}\left \vert x\right \vert & & \left \vert x\right \vert <1\\ \ & & \ \\ 0, & & \left \vert x\right \vert >1 \end {array} \right . \]
Solution
Let \(\digamma \left (\alpha \right ) \) be the Fourier transform of \(f\left ( x\right ) \) defined as \(\digamma \left (\alpha \right ) =\frac {1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} f\relax (x) \ e^{-i\alpha x}\ dx\)
\begin {align*} \ \digamma \left (\alpha \right ) & =\frac {1}{2\pi }\ {\displaystyle \int \limits _{-\infty }^{\infty }} f\relax (x) \ e^{-i\alpha x}\ dx\\ 2\pi \ \digamma \left (\alpha \right ) & =\ {\displaystyle \int \limits _{-1}^{0}} -x\ \ e^{-i\alpha x}\ dx\ +{\displaystyle \int \limits _{0}^{1}} x\ \ e^{-i\alpha x}\ dx\ \end {align*}
Integrating by parts, \(u=x,v=\frac {e^{-i\alpha x}}{-i\alpha }\), hence \(\int u\ dv\ =\) \(uv-\int du\ v\). The first integral is
\begin {align*} {\displaystyle \int \limits _{-1}^{0}} -x\ \ e^{-i\alpha x}\ dx & =\ \left [ \left (-x\right ) \ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{-1}^{0}+{\displaystyle \int \limits _{-1}^{0}} \ \ \frac {e^{-i\alpha x}}{-i\alpha }\ dx\\ & =\ \left [ \relax (x) \ \frac {e^{-i\alpha x}}{i\alpha }\right ] _{-1}^{0}+{\displaystyle \int \limits _{-1}^{0}} \ \ \frac {e^{-i\alpha x}}{-i\alpha }\ dx\\ & =\frac {1}{i\alpha }\ \ \left [ 0-(-1)\times e^{i\alpha }\right ] -\frac {1}{i\alpha }{\displaystyle \int \limits _{-1}^{0}} \ e^{-i\alpha x}\ dx\ \\ & =\frac {1}{i\alpha }\ \ \left [ e^{i\alpha }\right ] -\frac {1}{i\alpha }\left [ \ \frac {e^{-i\alpha x}\ }{-i\alpha }\right ] _{-1}^{0}\\ & =\frac {1}{i\alpha }\ \ \left [ e^{i\alpha }\right ] +\frac {1}{i^{2}\alpha ^{2}}\left [ \ e^{-i\alpha x}\ \right ] _{-1}^{0}\\ & =\frac {1}{i\alpha }\ \ \left [ e^{i\alpha }\right ] -\frac {1}{\alpha ^{2}}\left [ \ 1-\ e^{i\alpha }\right ] \\ & =\frac {e^{i\alpha }}{i\alpha }\ -\frac {1}{\alpha ^{2}}+\ \frac {e^{i\alpha }}{\alpha ^{2}} \end {align*}
And the second integral
\begin {align*} {\displaystyle \int \limits _{0}^{1}} x\ \ e^{-i\alpha x}\ dx & =\ \left [ x\ \frac {e^{-i\alpha x}}{-i\alpha }\right ] _{0}^{1}-{\displaystyle \int \limits _{0}^{1}} \ \ \frac {e^{-i\alpha x}}{-i\alpha }\ dx\\ & =\frac {1}{-i\alpha }\ \ \left [ 1\times e^{-i\alpha }-0\right ] +\frac {1}{i\alpha }{\displaystyle \int \limits _{0}^{1}} \ e^{-i\alpha x}\ dx\ \\ & =\frac {1}{-i\alpha }\ \ \left [ e^{-i\alpha }\right ] +\frac {1}{i\alpha }\left [ \ \frac {e^{-i\alpha x}\ }{-i\alpha }\right ] _{0}^{1}\\ & =\frac {1}{-i\alpha }\ \ \left [ e^{-i\alpha }\right ] -\frac {1}{i^{2}\alpha ^{2}}\left [ \ e^{-i\alpha x}\ \right ] _{0}^{1}\\ & =\frac {1}{-i\alpha }\ \ \left [ e^{-i\alpha }\right ] +\frac {1}{\alpha ^{2}}\left [ \ e^{-i\alpha }-1\ \right ] \\ & =\frac {e^{-i\alpha }}{-i\alpha }\ \ +\frac {\ e^{-i\alpha }}{\alpha ^{2}}-\frac {1}{\alpha ^{2}}\ \end {align*}
Hence
\begin {align*} 2\pi \ \ \digamma \left (\alpha \right ) & =\left (\frac {e^{i\alpha }}{i\alpha }\ -\frac {1}{\alpha ^{2}}+\ \frac {e^{i\alpha }}{\alpha ^{2}}\right ) +\left (\frac {e^{-i\alpha }}{-i\alpha }\ \ +\frac {\ e^{-i\alpha }}{\alpha ^{2}}-\frac {1}{\alpha ^{2}}\right ) \ \\ 2\pi \ \ \digamma \left (\alpha \right ) & =\frac {e^{i\alpha }}{i\alpha }\ -\frac {1}{\alpha ^{2}}+\frac {\ e^{i\alpha }}{\alpha ^{2}}-\frac {e^{-i\alpha }}{i\alpha }\ \ +\frac {\ e^{-i\alpha }}{\alpha ^{2}}-\frac {1}{\alpha ^{2}}\\ & =\frac {e^{i\alpha }}{i\alpha }\ +\ \frac {e^{i\alpha }}{\alpha ^{2}}-\frac {e^{-i\alpha }}{i\alpha }\ \ +\frac {\ e^{-i\alpha }}{\alpha ^{2}}\ \ -\frac {2}{\alpha ^{2}}\\ & =\frac {1}{\alpha }\left (\frac {e^{i\alpha }}{i}-\frac {e^{-i\alpha }}{i}\right ) +\frac {1}{\alpha ^{2}}\left (e^{i\alpha }+e^{-i\alpha }\ \right ) -\frac {2}{\alpha ^{2}}\\ & =\frac {1}{\alpha }\left (\frac {e^{i\alpha }-e^{-i\alpha }}{i}\right ) +\frac {1}{\alpha ^{2}}\left (e^{i\alpha }+e^{-i\alpha }\ \right ) -\frac {2}{\alpha ^{2}} \end {align*}
But \(e^{i\alpha }+e^{-i\alpha }=2\cos \alpha \) and \(\frac {e^{i\alpha }-e^{-i\alpha }}{i}=2\sin \alpha \), Hence the above becomes
\begin {align*} 2\pi \ \ \digamma \left (\alpha \right ) & =\frac {1}{\alpha }\left ( 2\sin \alpha \right ) +\frac {1}{\alpha ^{2}}\left (2\cos \alpha \ \right ) -\frac {2}{\alpha ^{2}}\\ & =\frac {2}{\alpha ^{2}}\left [ \alpha \sin \alpha +\cos \alpha \ -1\right ] \\ \ \digamma \left (\alpha \right ) & =\frac {1}{\pi \alpha ^{2}}\left [ \alpha \sin \alpha +\cos \alpha \ -1\right ] \end {align*}
Hence the Fourier transform of \(f\relax (x) \) is
\[ \digamma \left (\alpha \right ) =\frac {1}{\pi \alpha ^{2}}\left [ \alpha \sin \alpha +\cos \alpha \ -1\right ] \]
To obtain \(f\relax (x) \) given its fourier transform \(\digamma \left ( \alpha \right ) ,\) then we apply the inverse fourier transform
\begin {align*} f\relax (x) & =\ {\displaystyle \int \limits _{-\infty }^{\infty }} \digamma \left (\alpha \right ) \ e^{i\alpha x}\ d\alpha \\ & =\ {\displaystyle \int \limits _{-\infty }^{\infty }} \ \frac {1}{\pi \alpha ^{2}}\left [ \alpha \sin \alpha +\cos \alpha \ -1\right ] \ \ e^{i\alpha x}\ d\alpha \\ & =\frac {1}{\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \ \frac {1}{\alpha ^{2}}\left [ \alpha \sin \alpha +\cos \alpha \ -1\right ] \ \ e^{i\alpha x}\ d\alpha \end {align*}
Problem Show that \(\ g\relax (t) \circledast h\left ( t\right ) =\ h\relax (t) \circledast g\relax (t) \ \)
Solution
By definition,
\begin {equation} g\relax (t) \circledast h\relax (t) =\int _{0}^{t}g\left ( t-\tau \right ) \ h\left (\tau \right ) \ d\tau \tag {1} \end {equation}
Let \(u=t-\tau \), \(du=-d\tau \), when \(\tau =0\), \(u=t\), when \(\tau =t\), \(u=0\), Hence The RHS becomes
\begin {align*} \int _{0}^{t}g\left (t-\tau \right ) \ h\left (\tau \right ) \ d\tau & =\int _{u=t}^{u=0}g\relax (u) \ h\left (t-u\right ) \ (-du)\\ & =-\int _{u=t}^{u=0}g\relax (u) \ h\left (t-u\right ) \ du\\ & =\int _{u=0}^{u=t}g\relax (u) \ h\left (t-u\right ) \ du \end {align*}
Since \(u\) is a dummy variable of integration, call it anything we want, say \(\tau \) so above integral becomes
\begin {align} \int _{0}^{t}g\left (t-\tau \right ) \ h\left (\tau \right ) \ d\tau & =\int _{0}^{t}g\left (\tau \right ) \ h\left (t-\tau \right ) \ d\tau \nonumber \\ \int _{0}^{t}g\left (t-\tau \right ) \ h\left (\tau \right ) \ d\tau & =\int _{0}^{t}h\left (t-\tau \right ) \ g\left (\tau \right ) \ \ d\tau \tag {2} \end {align}
Hence from (2) \(g\relax (t) \circledast h\relax (t) =\ h\left ( t\right ) \circledast g\relax (t) \ \)
Problem
Use convolution integral to find the inverse transform of \(\frac {1}{p\left ( p^{2}+a^{2}\right ) ^{2}}\)
Solution
\[ \frac {1}{p\left (p^{2}+a^{2}\right ) ^{2}}=\frac {1}{p}\frac {1}{\left ( p^{2}+a^{2}\right ) ^{2}}=GH \]
From Tables using L1 and L17 \(g\relax (t) =1\,\) and \(h\left ( t\right ) =\frac {\sin at-at\ \cos at}{2a^{3}}\). Hence the inverse transform of \(GH=\) \(\ g\relax (t) \circledast h\relax (t) \). Using L34
\begin {equation} \ g\relax (t) \circledast h\relax (t) =\int _{0}^{t}g\left ( t-\tau \right ) \ h\left (\tau \right ) \ d\tau \tag {L34} \end {equation}
Hence
\begin {align} \ g\relax (t) \circledast h\relax (t) & =\int _{0}^{t}1\times \frac {\sin a\left (t-\tau \right ) -a\left (t-\tau \right ) \ \cos a\left (t-\tau \right ) }{2a^{3}}\ \ \ d\tau \nonumber \\ & =\frac {1}{2a^{3}}\int _{0}^{t}\sin a\left (t-\tau \right ) -a\left ( t-\tau \right ) \ \cos a\left (t-\tau \right ) \ d\tau \nonumber \\ & =\frac {1}{2a^{3}}\left [ \int _{0}^{t}\sin a\left (t-\tau \right ) \ d\tau -at\int _{0}^{t}\ \cos a\left (t-\tau \right ) d\tau +a\int _{0}^{t}\tau \cos a\left (t-\tau \right ) \ d\tau \right ] \tag {1} \end {align}
The last integral can be integrated by parts. \(u=\tau \), \(v=\frac {\sin a\left ( t-\tau \right ) }{-a}\)
\begin {align*} \int _{0}^{t}\tau \cos a\left (t-\tau \right ) \ d\tau & =\ \left [ \tau \frac {\sin a\left (t-\tau \right ) }{-a}\right ] _{0}^{t}-\int _{0}^{t}\frac {\sin a\left (t-\tau \right ) }{-a}\ d\tau \\ & =\frac {-1}{a}\left [ \tau \sin a\left (t-\tau \right ) \right ] _{0}^{t}+\frac {1}{a}\int _{0}^{t}\sin a\left (t-\tau \right ) \ d\tau \\ \ & =\frac {-1}{a}\left [ \tau \sin a\left (t-\tau \right ) \right ] _{0}^{t}+\frac {1}{a}\left [ \ \frac {-\cos a\left (t-\tau \right ) }{-a}\right ] _{0}^{t}\\ \ \ & =\frac {-1}{a}\left [ \tau \sin a\left (t-\tau \right ) \right ] _{0}^{t}+\frac {1}{a^{2}}\left [ \ \cos a\left (t-\tau \right ) \right ] _{0}^{t}\\ & =\frac {-1}{a}\left [ t\sin a\left (t-t\right ) -0\right ] +\frac {1}{a^{2}}\left [ \ \cos a\left (t-t\right ) -\cos a\left (t-0\right ) \right ] \\ \ & =\frac {-1}{a}\left [ 0\right ] +\frac {1}{a^{2}}\left [ \ \cos a\left ( 0\right ) -\cos a\relax (t) \right ] \\ \ & =\ \ \frac {1}{a^{2}}\left [ \ 1-\cos at\right ] \end {align*}
Hence (1) becomes
\begin {align*} g\relax (t) \circledast h\relax (t) & =\frac {1}{2a^{3}}\left [ \int _{0}^{t}\sin a\left (t-\tau \right ) \ d\tau -at\int _{0}^{t}\ \cos a\left ( t-\tau \right ) d\tau +a\frac {1}{a^{2}}\left [ \ 1-\cos at\right ] \right ] \\ g\relax (t) \circledast h\relax (t) & =\frac {1}{2a^{3}}\left [ \left [ \frac {-\cos a\left (t-\tau \right ) }{-a\tau }\right ] _{0}^{t}-at\ \left [ \frac {\ \sin a\left (t-\tau \right ) }{-a\tau }\right ] +\ \frac {1}{a}\left [ \ 1-\cos at\right ] \right ] \\ & =\frac {1}{2a^{3}}\left [ \frac {1}{a}\left [ \cos a\left (t-\tau \right ) \right ] _{0}^{t}+t\ \left [ \ \sin a\left (t-\tau \right ) \right ] _{0}^{t}+\ \frac {1}{a}\left [ \ 1-\cos at\right ] \right ] \\ & =\frac {1}{2a^{3}}\left [ \frac {1}{a}\left [ \cos a\left (t-t\right ) -\cos a\left (t-0\right ) \right ] +t\ \left [ \ \sin a\left (t-t\right ) -\sin a\left (t-0\right ) \right ] _{0}^{t}+\ \frac {1}{a}\left [ \ 1-\cos at\right ] \right ] \\ & =\frac {1}{2a^{3}}\left [ \frac {1}{a}\left [ 1-\cos at\right ] +t\ \left [ \ -\sin at\right ] +\ \frac {1}{a}\left [ \ 1-\cos at\right ] \right ] \\ & =\frac {1}{2a^{3}}\left [ \frac {1}{a}\left [ 1-\cos at\right ] -t\ \ \sin at+\ \frac {1}{a}\left [ \ 1-\cos at\right ] \right ] \\ & =\frac {1}{2a^{4}}\left (2-2\cos at-at\ \ \sin at\right ) \end {align*}
So the inverse Laplace transform of \(\frac {1}{p\left (p^{2}+a^{2}\right ) ^{2}}\) is \[ \frac {1}{2a^{4}}\left (2-2\cos at-at\ \ \sin at\right ) \]
Problem Use L34 and L2 to find the inverse transform of \(G\left ( p\right ) H\relax (p) \) when \(G\relax (p) =\frac {1}{\left ( p+a\right ) }\) and \(H\relax (p) =\frac {1}{\left (p+b\right ) }\) your result should be L7
Solution
\begin {equation} \mathcal {L}\left (e^{-at}\right ) =\frac {1}{p+a} \tag {L2} \end {equation}
\begin {equation} g\relax (t) \circledast h\relax (t) =\int _{0}^{t}g\left ( t-\tau \right ) \ h\left (\tau \right ) \ d\tau \tag {L34} \end {equation}
Using L2, \(g(t)=\mathcal {L}^{-1}\frac {1}{\left (p+a\right ) }=e^{-at}\), and \(h\relax (t) =\mathcal {L}^{-1}\frac {1}{\left (p+b\right ) }=e^{-bt}\). Now Let \(Y(p)=G\relax (p) H\relax (p) \), But
\[ G\relax (p) H\relax (p) =\mathcal {L}\left \{ \ g\relax (t) \circledast h\relax (t) \ \right \} \]
Then
\begin {align*} y\relax (t) & =g\relax (t) \circledast h\relax (t) =\int _{0}^{t}g\left (t-\tau \right ) \ h\left (\tau \right ) \ d\tau \\ & =\int _{0}^{t}e^{-a\left (t-\tau \right ) }\ \ e^{-b\tau }\ d\tau \\ & =\int _{0}^{t}e^{-at+a\tau }\ \ e^{-b\tau }\ d\tau \\ & =\int _{0}^{t}e^{-at\ }e^{a\tau }\ \ e^{-b\tau }\ d\tau \end {align*}
\(e^{-at}\) can be moved outside the integral
\begin {align*} y\relax (t) & =e^{-at\ }\int _{0}^{t}e^{a\tau }\ \ e^{-b\tau }\ d\tau \\ y\relax (t) & =e^{-at\ }\int _{0}^{t}e^{a\tau -b\tau }\ \ \ \ d\tau \\ y\relax (t) & =e^{-at\ }\int _{0}^{t}e^{\tau \left (a-b\right ) }\ \ \ \ d\tau \\ y\relax (t) & =e^{-at\ }\ \left [ \frac {e^{\tau \left (a-b\right ) }}{a-b}\right ] _{0}^{t}\\ y\relax (t) & =\frac {e^{-at\ }}{a-b}\ \left [ e^{t\left (a-b\right ) }-1\right ] \\ y\relax (t) & =\frac {e^{-at+t\left (a-b\right ) \ }-e^{-at\ }}{a-b}\ \ \\ y\relax (t) & =\frac {e^{-bt\ }-e^{-at\ }}{a-b}\\ y\relax (t) & =\frac {e^{-at\ }-e^{-bt\ }}{b-a} \end {align*}
Which is L7 as required to show.
Problem
Verify Parseval’s theorem for \(f\relax (x) =e^{-\left \vert x\right \vert }\) and \(g\left (\alpha \right ) =\)Fourier transform of \(f\relax (x) \)
Solution
Parseval theorem says that total energy in a signal equal to the sum of the energies in the harmonics that make up the signal. i.e.
\[ \int _{-\infty }^{\infty }\left \vert g\left (\alpha \right ) \right \vert ^{2}\ d\alpha =\frac {1}{2\pi }\int _{-\infty }^{\infty }\left \vert f\left ( x\right ) \right \vert ^{2}\ dx \]
\begin {align} \frac {1}{2\pi }\int _{-\infty }^{\infty }\left \vert f\relax (x) \right \vert ^{2}\ dx & =\frac {1}{2\pi }\int _{-\infty }^{\infty }\left \vert e^{-\left \vert x\right \vert }\right \vert ^{2}\ dx\nonumber \\ & =\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{-2\left \vert x\right \vert }\ dx\nonumber \\ & =\frac {1}{2\pi }\left \{ \int _{-\infty }^{0}e^{2x}\ dx+\int _{0}^{\infty }e^{-2x}\ dx\right \} \nonumber \\ & =\frac {1}{2\pi }\left \{ \left [ \ \frac {e^{2x}}{2}\right ] _{-\infty }^{0}+\ \left [ \frac {e^{-2x}}{-2}\right ] _{0}^{\infty }\right \} \nonumber \\ & =\frac {1}{4\pi }\left \{ \left [ \ e^{2x}\right ] _{-\infty }^{0}-\ \left [ e^{-2x}\right ] _{0}^{\infty }\right \} \nonumber \\ & =\frac {1}{4\pi }\left \{ \left [ \ e^{0}-0\right ] -\ \left [ 0-e^{0}\right ] \right \} \nonumber \\ & =\frac {1}{4\pi }\left \{ 1+1\right \} \nonumber \\ & =\frac {1}{2\pi }\ \tag {1} \end {align}
Now we find the Fourier transform for \(f\relax (x) \)
\begin {align*} g\left (\alpha \right ) & =\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{-\left \vert x\right \vert }e^{-ix\alpha }dx\\ & =\frac {1}{2\pi }\left [ \int _{-\infty }^{0}e^{x}e^{-ix\alpha }dx+\int _{0}^{\infty }e^{-x}e^{-ix\alpha }dx\right ] \\ & =\frac {1}{2\pi }\left [ \int _{-\infty }^{0}e^{x\left (1-i\alpha \right ) }\ dx+\int _{0}^{\infty }e^{\ x\left (-1-i\alpha \right ) }\ dx\right ] \\ & =\frac {1}{2\pi }\ \left (\left [ \frac {e^{x\left (1-i\alpha \right ) }}{1-i\alpha }\right ] _{-\infty }^{0}+\left [ \frac {e^{x\left (-1-i\alpha \right ) }}{-1-i\alpha }\right ] _{0}^{\infty }\right ) \\ & =\frac {1}{2\pi }\ \left (\frac {1}{1-i\alpha }\left [ e^{x\left ( 1-i\alpha \right ) }\right ] _{-\infty }^{0}-\frac {1}{1+i\alpha }\left [ e^{x\left (-1-i\alpha \right ) }\right ] _{0}^{\infty }\right ) \\ & =\frac {1}{2\pi }\ \left (\frac {1}{1-i\alpha }\left [ e^{x\left ( 1-i\alpha \right ) }\right ] _{-\infty }^{0}-\frac {1}{1+i\alpha }\left [ e^{x\left (-1-i\alpha \right ) }\right ] _{0}^{\infty }\right ) \\ & =\frac {1}{2\pi }\ \left (\frac {1}{1-i\alpha }\left [ 1-e^{-\infty \left ( 1-i\alpha \right ) }\right ] -\frac {1}{1+i\alpha }\left [ e^{\infty \left ( -1-i\alpha \right ) }-1\right ] \right ) \\ & =\frac {1}{2\pi }\ \left (\frac {1}{1-i\alpha }\left [ 1\right ] -\frac {1}{1+i\alpha }\left [ -1\right ] \right ) \\ & =\frac {1}{2\pi }\ \left (\frac {1}{1-i\alpha }\ +\frac {1}{1+i\alpha }\ \right ) \\ & =\frac {1}{2\pi }\ \left (\ \frac {1+i\alpha +1-i\alpha }{\left ( 1-i\alpha \right ) \left (1+i\alpha \right ) }\ \right ) \\ & =\frac {1}{2\pi }\ \left (\frac {2\ }{1+\alpha ^{2}}\right ) \\ & =\frac {1\ }{\pi \left (1+\alpha ^{2}\right ) }\ \end {align*}
So
\begin {align*} \int _{-\infty }^{\infty }\left \vert g\left (\alpha \right ) \right \vert ^{2}\ d\alpha & =\int _{-\infty }^{\infty }\left \vert \frac {1\ }{\pi \left ( 1+\alpha ^{2}\right ) }\right \vert ^{2}\ d\alpha \\ & =\frac {1}{\pi ^{2}}\int _{-\infty }^{\infty }\frac {1\ }{\left (1+\alpha ^{2}\right ) ^{2}}d\alpha \end {align*}
But \(\int _{-\infty }^{\infty }\frac {1\ }{\left (1+\alpha ^{2}\right ) ^{2}}d\alpha =\frac {\pi }{2}\), Hence
\begin {align} \int _{-\infty }^{\infty }\left \vert g\left (\alpha \right ) \right \vert ^{2}\ d\alpha & =\frac {1}{\pi ^{2}}\ \left (\frac {\pi }{2}\right ) \ \nonumber \\ & =\frac {1}{2\pi }\ \tag {2} \end {align}
Comparing (1) and (2). They are the same. Hence\(\ \)\[ \int _{-\infty }^{\infty }\left \vert g\left (\alpha \right ) \right \vert ^{2}\ d\alpha =\frac {1}{2\pi }\int _{-\infty }^{\infty }\left \vert f\left ( x\right ) \right \vert ^{2}\ dx \] was verified for this problem as required.
Problem
Use convolution integral to find the inverse transform of \(\frac {1}{\left ( p+a\right ) \left (p+b\right ) ^{2}}\)
Solution
\[ \frac {1}{\left (p+a\right ) \left (p+b\right ) ^{2}}=\frac {1}{\left ( p+a\right ) }\frac {1}{\left (p+b\right ) ^{2}}\]
From but from L6
\begin {equation} \mathcal {L}\left (t^{k}e^{-at}\right ) =\frac {k!}{\left (p+a\right ) ^{k+1}} \tag {L2} \end {equation}
Hence \(\frac {1}{\left (p+a\right ) }=\mathcal {L}\left (e^{-at}\right ) \) and \(\frac {1}{\left (p+b\right ) ^{2}}=\mathcal {L}\left (e^{-bt}\right ) \). Hence the inverse transform of \(\frac {1}{\left ( p+a\right ) }\frac {1}{\left (p+b\right ) ^{2}}=e^{-at}\circledast te^{-bt}\). Using L34
\begin {equation} \ g\relax (t) \circledast h\relax (t) =\int _{0}^{t}g\left ( t-\tau \right ) \ h\left (\tau \right ) \ d\tau \tag {L34} \end {equation}
Hence
\begin {align*} e^{-at}\circledast te^{-bt} & =\int _{0}^{t}e^{-a\left (t-\tau \right ) }\ \tau e^{-b\tau }\ d\tau \\ & =\int _{0}^{t}e^{-at+a\tau }\ \tau e^{-b\tau }\ d\tau \\ & =\int _{0}^{t}e^{-at}e^{a\tau }\ \tau e^{-b\tau }\ d\tau \\ & =e^{-at}\int _{0}^{t}e^{a\tau }\ \tau e^{-b\tau }\ d\tau \\ & =e^{-at}\int _{0}^{t}\tau e^{\tau \left (a-b\right ) }\ \ d\tau \end {align*}
Integrate by parts. \(u=\tau \), \(v=\frac {e^{\tau \left (a-b\right ) }}{a-b}\)
\begin {align*} e^{-at}\circledast te^{-bt} & =e^{-at}\left \{ \ \left [ \tau \frac {e^{\tau \left (a-b\right ) }}{a-b}\right ] _{0}^{t}-\int _{0}^{t}\frac {e^{\tau \left (a-b\right ) }}{a-b}\ \ d\tau \right \} \\ & =e^{-at}\left \{ \ \left [ \tau \frac {e^{\tau \left (a-b\right ) }}{a-b}\right ] _{0}^{t}-\frac {1}{a-b}\left [ \frac {e^{\tau \left (a-b\right ) }}{a-b}\right ] _{0}^{t}\ \ \right \} \\ & =e^{-at}\ \left (\frac {1}{a-b}\left [ \tau e^{\tau \left (a-b\right ) }\right ] _{0}^{t}-\frac {1}{\left (a-b\right ) ^{2}}\ \left [ e^{\tau \left ( a-b\right ) }\right ] _{0}^{t}\ \right ) \ \\ & =e^{-at}\ \left (\frac {1}{a-b}\left [ te^{t\left (a-b\right ) }\right ] -\frac {1}{\left (a-b\right ) ^{2}}\ \left [ e^{t\left (a-b\right ) }-1\right ] \ \right ) \\ & =e^{-at}\ \left (\frac {te^{ta-tb}}{a-b}-\frac {e^{ta-tb}-1}{\left ( a-b\right ) ^{2}}\right ) \\ & =\ \frac {te^{-tb}}{a-b}-\frac {e^{-tb}-e^{-at}}{\left (a-b\right ) ^{2}}\\ & =\ \frac {\left (a-b\right ) te^{-tb}-e^{-tb}+e^{-at}}{\left (a-b\right ) ^{2}}\ \\ & =\ \frac {\left (\left (a-b\right ) t-1\right ) e^{-tb}+e^{-at}}{\left ( a-b\right ) ^{2}}\ \end {align*}
So the inverse laplace transform of \(\frac {1}{\left (p+a\right ) \left ( p+b\right ) ^{2}}\) is \[ \frac {\left (\left (a-b\right ) t-1\right ) \ e^{-tb}+e^{-at}}{\left ( a-b\right ) ^{2}}\]
Problem
Find the inverse laplace transform using 6.6 of the function \(\frac {1}{p^{4}-1}\)
Solution
6.6 states that \(f\relax (t) \) = sum of all residues of \(F\left ( z\right ) e^{zt}\) at all poles. Poles of \(F\relax (z) =\frac {1}{z^{4}-1}\)are at \(\pm 1,\pm i\). Hence
\[ F\relax (z) =\frac {e^{zt}}{\left (z-1\right ) \left (z+1\right ) \left (z-i\right ) \left (z+i\right ) \ }\]
Since each pole is of order 1, we use equation 6.1 page 599 which says
\[ \text {Residue of }F\relax (z) \ \text {at\ }z=z_{0}\ \ \text {is\ \ }\lim _{z\rightarrow z_{0}}\ \ \left (z-z_{0}\right ) F\relax (z) \]
Hence sum of residue is
\begin {align*} R & =\lim _{z\rightarrow +1}\frac {e^{zt}}{\ \left (z+1\right ) \left ( z-i\right ) \left (z+i\right ) \ }+\lim _{z\rightarrow -1}\frac {e^{zt}}{\left ( z-1\right ) \left (z-i\right ) \left (z+i\right ) \ }\\ & +\lim _{z\rightarrow +i}\frac {e^{zt}}{\left (z-1\right ) \left (z+1\right ) \left (z+i\right ) \ }+\lim _{z\rightarrow -i}\frac {e^{zt}}{\left (z-1\right ) \left (z+1\right ) \left (z-i\right ) \ }\\ & \\ & =\ \frac {e^{t}}{\ \left (1+1\right ) \left (1-i\right ) \left ( 1+i\right ) \ }+\ \frac {e^{-t}}{\left (-1-1\right ) \left (-1-i\right ) \left (-1+i\right ) \ }\\ & +\ \frac {e^{it}}{\left (i-1\right ) \left (i+1\right ) \left ( i+i\right ) \ }+\ \frac {e^{-it}}{\left (-i-1\right ) \left (-i+1\right ) \left (-i-i\right ) \ }\\ & \\ & =\frac {e^{t}}{\ 4\ \ }+\ \frac {e^{-t}}{-4}+\frac {e^{it}}{-4i\ }+\ \frac {e^{-it}}{4i\ }\\ & =\ \left (\frac {e^{t}-e^{-t}}{4}\right ) -\frac {1}{2}\left (\frac {e^{it}-e^{-it}\ }{2i}\right ) \\ & =\left (\frac {e^{t}-e^{-t}}{4}\right ) -\frac {1}{2}\left (\sin t\right ) \end {align*}
Problem
Find the inverse laplace transform using 6.6 of the function \(\frac {p^{3}}{p^{4}-16}\)
Solution
6.6 states that \(f\relax (t) \) = sum of all residues of \(F\left ( z\right ) e^{zt}\) at all poles. Poles of \(F\relax (z) =\frac {z^{3}}{z^{4}-16}\)are at \(\pm 2,\pm 2i\), Hence
\[ F\relax (z) =\frac {z^{3}e^{zt}}{\left (z-2\right ) \left (z+2\right ) \left (z-2i\right ) \left (z+2i\right ) \ }\]
Since each pole is of order 1, we use equation 6.1 page 599 which says
\[ \text {Residue of }F\relax (z) \ \text {at\ }z=z_{0}\ \ \text {is\ \ }\lim _{z\rightarrow z_{0}}\ \ \left (z-z_{0}\right ) F\relax (z) \]
Hence sum of residue is
\begin {align*} R & =\lim _{z\rightarrow +2}\frac {z^{3}e^{zt}}{\ \left (z+2\right ) \left ( z-2i\right ) \left (z+2i\right ) \ }+\lim _{z\rightarrow -2}\frac {z^{3}e^{zt}}{\left (z-2\right ) \left (z-2i\right ) \left (z+2i\right ) \ }\\ & +\lim _{z\rightarrow +2i}\frac {z^{3}e^{zt}}{\left (z-2\right ) \left ( z+2\right ) \left (z+2i\right ) \ }+\lim _{z\rightarrow -2i}\frac {z^{3}e^{zt}}{\left (z-2\right ) \left (z+2\right ) \left (z-2i\right ) \ }\\ & \\ & \\ & =\ \frac {8e^{2t}}{\ \left (2+2\right ) \left (2-2i\right ) \left ( 2+2i\right ) \ }+\frac {-8e^{-2t}}{\left (-2-2\right ) \left (-2-2i\right ) \left (-2+2i\right ) \ }\\ & +\ \frac {\left (2i\right ) ^{3}e^{2it}}{\left (2i-2\right ) \left ( 2i+2\right ) \left (2i+2i\right ) \ }+\frac {\left (-2i\right ) ^{3}e^{-2it}}{\left (-2i-2\right ) \left (-2i+2\right ) \left (-2i-2i\right ) \ }\\ & \\ & \\ & =\frac {8e^{2t}}{\ \relax (4) \ 8\ }+\frac {-8e^{-2t}}{\left ( -4\right ) 8\ }+\frac {\left (-8i\right ) e^{2it}}{\ -8\left (4i\right ) \ }+\frac {\left (8i\right ) e^{-2it}}{\ -8\left (-4i\right ) \ }\\ & =\frac {e^{2t}}{\ 4\ \ }+\frac {e^{-2t}}{4}+\frac {e^{2it}}{\ 4\ }+\frac {e^{-2it}}{\ 4\ }\\ & =\ \frac {e^{2t}+e^{-2t}}{4}+\frac {1}{2}\left (\cos 2t\right ) \end {align*}
So inverse Laplace transform of \(\frac {p^{3}}{p^{4}-16}\) is \[ \frac {e^{2t}+e^{-2t}}{4}+\frac {1}{2}\left (\cos 2t\right ) \]
Problem
Find the inverse laplace transform using 6.6 of the function \(\frac {3p^{2}}{p^{3}+8}\)
Solution
6.6 states that \(f\relax (t) \) = sum of all residues of \(F\left ( z\right ) e^{zt}\) at all poles. To find poles, look at \(p^{3}+8=0,\) hence \(p^{3}=-8\), \(p=-8^{\frac {1}{3}}\). Let
\[ 8^{\frac {1}{3}}=re^{i\theta }\]
Then the roots are
\begin {align*} & =8^{^{\frac {1}{3}}}e^{\frac {i0}{3}},8^{^{\frac {1}{3}}}e^{\frac {i\left ( 0+2\pi \right ) }{3}},8^{^{\frac {1}{3}}}e^{\frac {i\left (0+4\pi \right ) }{3}}\\ & =8^{^{\frac {1}{3}}}\ ,8^{^{\frac {1}{3}}}(\cos \frac {2\pi }{3}+i\sin \frac {2\pi }{3}),8^{^{\frac {1}{3}}}(\cos \frac {4\pi }{3}+i\sin \frac {4\pi }{3})\\ & =2\ ,2(\cos 120^{0}+i\sin 120^{0}),2(\cos 240^{0}+i\sin 240^{0})\\ & =2\ ,2(-\frac {1}{2}+i\frac {\sqrt {3}}{2}),2(-\frac {1}{2}+i\frac {-\sqrt {3}}{2})\\ & =2,-1+i\sqrt {3},-1-i\sqrt {3} \end {align*}
Hence
\[ p=-2,1-i\sqrt {3},1+i\sqrt {3}\]
And
\begin {align*} F\relax (z) & =\frac {3z^{2}}{z^{3}+8}e^{zt}\\ & =\frac {3z^{2}}{\left (z+2\right ) \left (z-\left (1-i\sqrt {3}\right ) \right ) \left (z-\left (1+i\sqrt {3}\right ) \right ) }e^{zt} \end {align*}
Since each pole is of order 1, we use equation 6.1 page 599 which says
\[ \text {Residue of }F\relax (z) \ \text {at\ }z=z_{0}\ \ \text {is\ \ }\lim _{z\rightarrow z_{0}}\ \ \left (z-z_{0}\right ) F\relax (z) \]
Hence sum of residue is
\begin {align*} R & =\lim _{z\rightarrow -2}\frac {3z^{2}}{\ \left (z-\left (1-i\sqrt {3}\right ) \right ) \left (z-\left (1+i\sqrt {3}\right ) \right ) }e^{zt}+\lim _{z\rightarrow \left (1-i\sqrt {3}\right ) }\ \frac {3z^{2}}{\left ( z+2\right ) \left (z-\left (1+i\sqrt {3}\right ) \right ) }e^{zt}\\ & +\lim _{z\rightarrow \left (1+i\sqrt {3}\right ) }\frac {3z^{2}}{\left ( z+2\right ) \left (z-\left (1-i\sqrt {3}\right ) \right ) }e^{zt}\\ & =\ \frac {3\left (-2\right ) ^{2}}{\ \left (-2-\left (1-i\sqrt {3}\right ) \right ) \left (-2-\left (1+i\sqrt {3}\right ) \right ) }e^{-2t}+\ \frac {3\left (1-i\sqrt {3}\right ) ^{2}}{\left (\left (1-i\sqrt {3}\right ) +2\right ) \left (\left (1-i\sqrt {3}\right ) -\left ( 1+i\sqrt {3}\right ) \right ) }e^{\left (1-i\sqrt {3}\right ) t}\ \\ & +\ \frac {3\left (1+i\sqrt {3}\right ) ^{2}}{\left (\left (1+i\sqrt {3}\right ) +2\right ) \left (\left (1+i\sqrt {3}\right ) -\left ( 1-i\sqrt {3}\right ) \right ) }e^{\left (1+i\sqrt {3}\right ) t}\\ & =\frac {12}{\ \left (-3+i\sqrt {3}\right ) \left (-3-i\sqrt {3}\right ) }e^{-2t}+\ \frac {3\left (-2-2i\sqrt {3}\right ) }{\left (3-i\sqrt {3}\right ) \left (-2i\sqrt {3}\right ) }e^{\left (1-i\sqrt {3}\right ) t}+\ \frac {3\left (-2+2i\sqrt {3}\right ) }{\left (3+i\sqrt {3}\right ) \left ( 2i\sqrt {3}\right ) }e^{\left (1+i\sqrt {3}\right ) t}\\ & =\ \frac {12}{\ 12}e^{-2t}+\ \frac {\ -6-6i\sqrt {3}}{-6i\sqrt {3}-6}e^{\left ( 1-i\sqrt {3}\right ) t}+\ \frac {-6+6i\sqrt {3}}{6i\sqrt {3}-6}e^{\left ( 1+i\sqrt {3}\right ) t}\\ & =\ e^{-2t}+\ e^{\left (1-i\sqrt {3}\right ) t}+e^{\left (1+i\sqrt {3}\right ) t}\\ & =\ e^{-2t}+\ e^{t}e^{-i\sqrt {3}t}+e^{t}e^{i\sqrt {3}t}\\ & =\ e^{-2t}+\ e^{t}\left (e^{i\sqrt {3}t}+e^{-i\sqrt {3}t}\right ) \\ & =\ e^{-2t}+\ e^{t}\left (2\cos \sqrt {3}t\right ) \end {align*}
So inverse Laplace transform of \(\frac {3p^{2}}{p^{3}+8}\) is \[ e^{-2t}+\ e^{t}\left (2\cos \sqrt {3}t\right ) \]
Problem
Find the inverse laplace transform using 6.6 of the function \(\frac {p}{p^{4}-1}\)
Solution
6.6 states that \(f\relax (t) \) = sum of all residues of \(F\left ( z\right ) e^{zt}\) at all poles. To find poles, look at \(p^{4}-1=0,\) hence \(p^{4}=1\), \(p=1^{\frac {1}{4}}\). Let
\[ 1^{\frac {1}{4}}=re^{i\theta }\]
Then roots are
\begin {align*} & =1^{^{\frac {1}{4}}}e^{\frac {i0}{4}},1^{^{\frac {1}{4}}}e^{\frac {i\left ( 0+2\pi \right ) }{4}},1^{^{\frac {1}{4}}}e^{\frac {i\left (0+4\pi \right ) }{4}},^{^{\frac {1}{4}}}e^{\frac {i\left (0+6\pi \right ) }{4}}\\ & =1\ ,1\left (\cos \frac {2\pi }{4}+i\sin \frac {2\pi }{4}\right ) ,1\left ( \cos \frac {4\pi }{4}+i\sin \frac {4\pi }{4}\right ) ,1\left (\cos \frac {6\pi }{4}+i\sin \frac {6\pi }{4}\right ) \\ & =1\ ,\left (0+i\right ) ,\left (-1+i\ 0\right ) ,\left (0-i\ 1\right ) \\ & =\ 1,i,-1,-i \end {align*}
Therefore
\[ p=\ 1,i,-1,-i \]
Hence
\begin {align*} F\relax (z) & =\frac {z}{z^{4}-1}e^{zt}\\ & =\frac {z}{\left (z-1\right ) \left (z-i\right ) \left (z+1\right ) \left (z+i\right ) }e^{zt} \end {align*}
Since each pole is of order 1, we use equation 6.1 page 599 which says
\[ \text {Residue of }F\relax (z) \ \text {at\ }z=z_{0}\ \ \text {is\ \ }\lim _{z\rightarrow z_{0}}\ \ \left (z-z_{0}\right ) F\relax (z) \]
Hence sum of residue is
\begin {align*} R & =\lim _{z\rightarrow 1}\frac {z}{\ \left (z-i\right ) \left (z+1\right ) \left (z+i\right ) }e^{zt}+\lim _{z\rightarrow i}\ \frac {z}{\left ( z-1\right ) \left (z+1\right ) \left (z+i\right ) }e^{zt}\\ & +\lim _{z\rightarrow -1}\frac {z}{\left (z-1\right ) \left (z-i\right ) \left (z+i\right ) }e^{zt}+\lim _{z\rightarrow -i}\frac {z}{\left (z-1\right ) \left (z-i\right ) \left (z+1\right ) }e^{zt}\\ & \\ & =\ \frac {1}{\ \left (1-i\right ) \left (1+1\right ) \left (1+i\right ) }e^{t}+\ \frac {i}{\left (i-1\right ) \left (i+1\right ) \left (i+i\right ) }e^{it}\ \\ & +\ \frac {-1}{\left (-1-1\right ) \left (-1-i\right ) \left (-1+i\right ) }e^{-t}+\frac {-i}{\left (-i-1\right ) \left (-i-i\right ) \left ( -i+1\right ) }e^{-it}\\ & \\ & =\frac {1}{\ 4}e^{t}+\ \frac {i}{-4i}e^{it}\ +\frac {1}{4}e^{-t}+\frac {-i}{4i}e^{-it}\\ & =\frac {1}{\ 4}\left (e^{t}+e^{-t}\right ) -\frac {1}{2}\left (\cos t\right ) \end {align*}
So inverse Laplace transform of \(\frac {p}{p^{4}-1}\) is \[ \frac {1}{\ 4}\left (e^{t}+e^{-t}\right ) -\frac {1}{2}\left (\cos t\right ) \]
Problem
Using the \(\delta \) function method, Find the response of the following system to a unit impulse. \(\frac {d^{4}y}{dy^{4}}-y=\delta \left (t-t_{0}\right ) \)
Solution
Taking the laplace transform of each side gives (assuming initial conditions for the system are at rest)
\begin {align*} Yp^{4}-Y & =e^{-pt_{0}}\\ Y\ & =\frac {e^{-pt_{0}}}{p^{4}-1}\\ Y\ & =\frac {e^{-pt_{0}}}{\left (p^{2}-1\right ) \left (p^{2}+1\right ) } \end {align*}
Finding the inverse laplace of \(\frac {1}{\left (p-1\right ) \left ( p+1\right ) \left (p^{2}+1\right ) }=\frac {1}{\left (p-1\right ) \left ( p+1\right ) }\frac {1}{\left (p^{2}+1\right ) }=GH\). Then \(g\relax (t) =\frac {e^{t}-e^{-t}}{2}\,\)using L7 and\(,\) \(h\relax (t) =\sin t\) using L3. Hence the inverse transform is \(\ \ \ \ \ \)
\begin {align*} g\relax (t) \circledast h\relax (t) & =\int _{0}^{t}\frac {e^{\tau }-e^{-\tau }}{2}\sin \left (t-\tau \right ) \ d\tau \\ & =\frac {1}{2}\left (\sinh t-\sin t\right ) \end {align*}
Using L28 with the result above we get
\[ y\relax (t) =\left \{ \begin {array} [c]{lll}\frac {1}{2}\left (\sinh \left (t-t_{0}\right ) -\sin \left (t-t_{0}\right ) \right ) & & t>t_{0}\\ 0 & & t<t_{0}\end {array} \right . \]
Or by expressing \(\sinh \) using \(\exp \), the above becomes
\[ y\relax (t) =\left \{ \begin {array} [c]{lll}\frac {1}{4}\left (e^{t-t_{0}}-e^{-t+t_{0}}-2\sin \left (t-t_{0}\right ) \right ) & & t>t_{0}\\ 0 & & t<t_{0}\end {array} \right . \]
Problem
Using the \(\delta \) function method, Find the response of the following system to a unit impulse. \(y^{\prime \prime }+2y^{\prime }+y=\delta \left ( t-t_{0}\right ) \)
Solution
Take the laplace transform of each side we get (assume initial conditions for a system at rest)
\begin {align*} Yp^{2}+2Yp+Y & =e^{-pt_{0}}\\ Y\ & =\frac {e^{-pt_{0}}}{p^{2}+2p+1}\\ Y\ & =\frac {e^{-pt_{0}}}{\left (p+1\right ) ^{2}} \end {align*}
Using L28 and L6 (for \(k=1\))
\[ y\relax (t) =\left \{ \begin {array} [c]{lll}\left (t-t_{0}\right ) e^{-\left (t-t_{0}\right ) } & & t>t_{0}\\ 0 & & t<t_{0}\end {array} \right . \]
Problem
Using the \(\delta \) function method, Find the response of the following system to a unit impulse. \(y^{\prime \prime }+2y^{\prime }+10y=\delta \left ( t-t_{0}\right ) \)
Solution
Taking the laplace transform of each side we get (assume initial conditions for a system at rest)\begin {align*} Yp^{2}+2Yp+10Y & =e^{-pt_{0}}\\ Y\ & =\frac {e^{-pt_{0}}}{p^{2}+2p+10}\\ Y\ & =\frac {e^{-pt_{0}}}{\left (p-a\right ) \left (p-b\right ) } \end {align*}
Where \(a=-1+3i,\ b=-1-3i\) the roots of \(p^{2}+2p+10\). Using L28 and L7\[ y\relax (t) =\left \{ \begin {array} [c]{lll}\frac {e^{a\left (t-t_{0}\right ) }-e^{b\left (t-t_{0}\right ) }}{\left ( -b\right ) -\left (-a\right ) } & & t>t_{0}\\ 0 & & t<t_{0}\end {array} \right . \] Replacing values for \(a,b\) gives\begin {align*} y\relax (t) & =\frac {e^{a\left (t-t_{0}\right ) }-e^{b\left ( t-t_{0}\right ) }}{a-b}\\ & =\frac {e^{\left (-1+3i\right ) \left (t-t_{0}\right ) }-e^{\left ( -1-3i\right ) \left (t-t_{0}\right ) }}{\left (-1+3i\right ) -\left ( -1-3i\right ) }\\ & =\frac {e^{\left (-1+3i\right ) \left (t-t_{0}\right ) }-e^{\left ( -1-3i\right ) \left (t-t_{0}\right ) }}{6i}\\ & =\frac {\ e^{-t+t_{0}+3it-3it_{0}}-e^{-t+t_{0}-3it+3it_{0}}}{6i}\\ & =e^{-t+t_{0}}\frac {\ e^{3i\left (t-t_{0}\right ) }-e^{-3i\left ( t-t_{0}\right ) }}{6i}\\ & =e^{-t+t_{0}}\left (\frac {1}{3}\right ) \left (\frac {\ e^{3i\left ( t-t_{0}\right ) }-e^{-3i\left (t-t_{0}\right ) }}{2i}\right ) \\ & =\frac {e^{-t+t_{0}}}{3}\sin 3\left (t-t_{0}\right ) \end {align*}
Therefore\[ y\relax (t) =\left \{ \begin {array} [c]{lll}\frac {e^{-t+t_{0}}}{3}\sin 3\left (t-t_{0}\right ) & & t>t_{0}\\ 0 & & t<t_{0}\end {array} \right . \]