Find Fourier transform of (a) \(e^{-2\left ( t-1\right ) }u\left ( t-1\right ) \)
Solution\begin{align*} X\left ( \omega \right ) & =\int _{-\infty }^{\infty }x\left ( t\right ) e^{-j\omega t}dt\\ & =\int _{1}^{\infty }e^{-2\left ( t-1\right ) }e^{-i\omega t}dt\\ & =\int _{1}^{\infty }e^{-2t}e^{2}e^{-i\omega t}dt\\ & =e^{2}\int _{1}^{\infty }e^{-t\left ( 2+i\omega \right ) }dt\\ & =\frac{e^{2}}{-\left ( 2+i\omega \right ) }\left [ e^{-t\left ( 2+i\omega \right ) }\right ] _{1}^{\infty } \end{align*}
Assuming \(\operatorname{Im}\left ( \omega \right ) <2\) then\begin{align*} X\left ( \omega \right ) & =\frac{e^{2}}{-\left ( 2+i\omega \right ) }\left [ 0-e^{-2}e^{-i\omega }\right ] \\ & =\frac{e^{2}}{\left ( 2+i\omega \right ) }\left [ e^{-2}e^{-i\omega }\right ] \\ & =\frac{e^{-i\omega }}{\left ( 2+i\omega \right ) } \end{align*}
Determine the Fourier transform of each of the following periodic signals (a) \(\sin \left ( 2\pi t+\frac{\pi }{4}\right ) \) (b) \(1+\cos \left ( 6\pi t+\frac{\pi }{8}\right ) \)
Solution
Since this is periodic signal, then we can not use \(X\left ( \omega \right ) =\int _{-\infty }^{\infty }x\left ( t\right ) e^{-j\omega t}dt\) which is for aperiodic signal. Instead we need to use 4.22 in the textbook which is \[ X\left ( \omega \right ) =\sum _{k=-\infty }^{\infty }2\pi a_{k}\delta \left ( \omega -k\omega _{0}\right ) \] From \(\sin \left ( 2\pi t+\frac{\pi }{4}\right ) \) we see that \(\omega _{0}=2\pi \), hence\[ X\left ( \omega \right ) =\sum _{k=-\infty }^{\infty }2\pi a_{k}\delta \left ( \omega -2k\pi \right ) \] Writing \(\sin \left ( \omega _{0}t+\frac{\pi }{4}\right ) =\frac{1}{2j}e^{j\left ( \omega _{0}t+\frac{\pi }{4}\right ) }-\frac{1}{2j}e^{-j\left ( \omega _{0}t+\frac{\pi }{4}\right ) }=\left ( \frac{1}{2j}e^{j\frac{\pi }{4}}\right ) e^{j\omega _{0}t}-\left ( \frac{1}{2j}e^{-j\frac{\pi }{4}}\right ) e^{-j\omega _{0}t}\) shows that \(a_{1}=\frac{1}{2j}e^{j\frac{\pi }{4}}\) and \(a_{-1}=-\frac{1}{2j}e^{-j\frac{\pi }{4}}\) and \(a_{k}=0\) for all other \(k\). Hence above simplifies to\begin{align*} X\left ( \omega \right ) & =2\pi \left ( \frac{1}{2j}e^{j\frac{\pi }{4}}\right ) \delta \left ( \omega -2\pi \right ) +2\pi \left ( -\frac{1}{2j}e^{-j\frac{\pi }{4}}\right ) \delta \left ( \omega +2\pi \right ) \\ & =\frac{\pi }{j}e^{j\frac{\pi }{4}}\delta \left ( \omega -2\pi \right ) -\frac{\pi }{j}e^{-j\frac{\pi }{4}}\delta \left ( \omega +2\pi \right ) \end{align*}
Since this is periodic signal, then its Fourier transform is\[ X\left ( \omega \right ) =\sum _{k=-\infty }^{\infty }2\pi a_{k}\delta \left ( \omega -k\omega _{0}\right ) \] From \(1+\cos \left ( 6\pi t+\frac{\pi }{8}\right ) \) we see that \(\omega _{0}=6\pi \), hence\[ X\left ( \omega \right ) =\sum _{k=-\infty }^{\infty }2\pi a_{k}\delta \left ( \omega -6k\pi \right ) \] Writing \(1+\cos \left ( 6\pi t+\frac{\pi }{8}\right ) =1+\left ( \frac{1}{2}e^{j\left ( \omega _{0}t+\frac{\pi }{8}\right ) }+\frac{1}{2}e^{-j\left ( \omega _{0}t+\frac{\pi }{8}\right ) }\right ) =1+\left ( \frac{1}{2}e^{j\frac{\pi }{8}}e^{j\omega _{0}t}+\frac{1}{2}e^{-j\frac{\pi }{8}}e^{-j\omega _{0}t}\right ) \) shows that \(a_{1}=\frac{1}{2}e^{j\frac{\pi }{8}}\) and \(a_{-1}=\frac{1}{2}e^{-j\frac{\pi }{8}}\) and \(a_{0}=1\). Therefore the above becomes\begin{align*} X\left ( \omega \right ) & =2\pi a_{-1}\delta \left ( \omega +6\pi \right ) +2\pi a_{0}\delta \left ( \omega \right ) +2\pi a_{1}\delta \left ( \omega -6\pi \right ) \\ & =2\pi \left ( \frac{1}{2}e^{-j\frac{\pi }{8}}\right ) \delta \left ( \omega +6\pi \right ) +2\pi \delta \left ( \omega \right ) +2\pi \left ( \frac{1}{2}e^{j\frac{\pi }{8}}\right ) \delta \left ( \omega -6\pi \right ) \\ & =\pi e^{-j\frac{\pi }{8}}\delta \left ( \omega +6\pi \right ) +2\pi \delta \left ( \omega \right ) +\pi e^{j\frac{\pi }{8}}\delta \left ( \omega -6\pi \right ) \end{align*}
Use the Fourier transform synthesis equation (4.8) to determine the inverse Fourier transform of \(X\left ( j\omega \right ) =\left \vert X\left ( j\omega \right ) \right \vert e^{j\measuredangle X\left ( j\omega \right ) }\) where \(\left \vert X\left ( j\omega \right ) \right \vert =2\left ( u\left ( \omega +3\right ) -u\left ( \omega -3\right ) \right ) ,\measuredangle X\left ( j\omega \right ) =-\frac{3}{2}\omega +\pi \)
Solution
4.8 is \begin{equation} x\left ( t\right ) =\frac{1}{2\pi }\int _{-\infty }^{\infty }X\left ( \omega \right ) e^{j\omega t}d\omega \tag{4.8} \end{equation} Hence\begin{align*} x\left ( t\right ) & =\frac{1}{2\pi }\int _{-\infty }^{\infty }\left \vert X\left ( j\omega \right ) \right \vert e^{j\measuredangle X\left ( j\omega \right ) }e^{j\omega t}d\omega \\ & =\frac{1}{2\pi }\int _{-\infty }^{\infty }2\left ( u\left ( \omega +3\right ) -u\left ( \omega -3\right ) \right ) e^{j\left ( -\frac{3}{2}\omega +\pi \right ) }e^{j\omega t}d\omega \end{align*}
But \(u\left ( \omega +3\right ) -u\left ( \omega -3\right ) \) is one over \(\omega =-3\cdots 3\) and zero otherwise. The above simplifies to\begin{align*} x\left ( t\right ) & =\frac{1}{2\pi }\int _{-3}^{3}2e^{j\left ( -\frac{3}{2}\omega +\pi \right ) }e^{j\omega t}d\omega \\ & =\frac{1}{\pi }\int _{-3}^{3}e^{j\pi }e^{j-\frac{3}{2}\omega }e^{j\omega t}d\omega \\ & =\frac{e^{j\pi }}{\pi }\int _{-3}^{3}e^{j\left ( -\frac{3}{2}+t\right ) \omega }d\omega \end{align*}
But \(e^{j\pi }=-1\) and \(\int e^{j\left ( -\frac{3}{2}+t\right ) \omega }d\omega =\frac{e^{j\left ( -\frac{3}{2}+t\right ) \omega }}{j\left ( -\frac{3}{2}+t\right ) }\), hence the above becomes\begin{align*} x\left ( t\right ) & =\frac{-1}{\pi }\frac{1}{j\left ( -\frac{3}{2}+t\right ) }\left [ e^{j\left ( -\frac{3}{2}+t\right ) \omega }\right ] _{-3}^{3}\\ & =\frac{-1}{\pi \left ( -\frac{3}{2}+t\right ) }\left ( \frac{e^{j3\left ( -\frac{3}{2}+t\right ) }-e^{-j3\left ( -\frac{3}{2}+t\right ) }}{j}\right ) \\ & =\frac{-1}{\pi \left ( -\frac{3}{2}+t\right ) }2\left ( \sin \left ( 3\left ( -\frac{3}{2}+t\right ) \right ) \right ) \\ & =\frac{-2}{\pi \left ( t-\frac{3}{2}\right ) }\sin \left ( 3\left ( t-\frac{3}{2}\right ) \right ) \end{align*}
Given the relationships \(y\left ( t\right ) =x\left ( t\right ) \circledast h\left ( t\right ) \) and \(g\left ( t\right ) =x\left ( 3t\right ) \circledast h\left ( 3t\right ) \) and given that \(x\left ( t\right ) \) has Fourier transform \(X\left ( j\omega \right ) \) and \(h\left ( t\right ) \) has Fourier transform \(H\left ( j\omega \right ) \), use Fourier transform properties to show that \(g(t)\) has the form \(g\left ( t\right ) =Ay\left ( Bt\right ) \). Determine the values of \(A\) and \(B\)
Solution
The main relation to use is that if \(y\left ( t\right ) \Longleftrightarrow Y\left ( \omega \right ) \) then \(y\left ( at\right ) \Longleftrightarrow \frac{1}{a}Y\left ( \frac{\omega }{a}\right ) \). Therefore \(x\left ( 3t\right ) \Longleftrightarrow \frac{1}{3}X\left ( \frac{\omega }{3}\right ) \) and \(h\left ( 3t\right ) \Longleftrightarrow \frac{1}{3}H\left ( \frac{\omega }{3}\right ) \). Hence since \(g\left ( t\right ) =x\left ( 3t\right ) \circledast h\left ( 3t\right ) \) then \begin{align*} G\left ( \omega \right ) & =\frac{1}{3}X\left ( \frac{\omega }{3}\right ) \frac{1}{3}H\left ( \frac{\omega }{3}\right ) \\ & =\frac{1}{3}\left ( \frac{1}{3}X\left ( \frac{\omega }{3}\right ) H\left ( \frac{\omega }{3}\right ) \right ) \end{align*}
But \(X\left ( \frac{\omega }{3}\right ) H\left ( \frac{\omega }{3}\right ) =Y\left ( \frac{\omega }{3}\right ) \). Therefore the above becomes
\[ G\left ( \omega \right ) =\frac{1}{3}\left ( \frac{1}{3}Y\left ( \frac{\omega }{3}\right ) \right ) \]
Inverse Fourier transform gives
\[ g\left ( t\right ) =\frac{1}{3}y\left ( 3t\right ) \]
Where in the above, we used \(\frac{1}{3}Y\left ( \frac{\omega }{3}\right ) \Longleftrightarrow y\left ( 3t\right ) \). Hence \(A=\frac{1}{3},B=3\)
Consider a causal LTI system with frequency response \(H\left ( j\omega \right ) =\frac{1}{j\omega +3}\). For a particular input \(x(t)\) this system is observed to produce the output \(y\left ( t\right ) =e^{-3t}u\left ( t\right ) -e^{-4t}u\left ( t\right ) \). Determine \(x\left ( t\right ) \).
Solution\[ y\left ( t\right ) =x\left ( t\right ) \circledast h\left ( t\right ) \] Taking Fourier transform gives\[ Y\left ( \omega \right ) =X\left ( \omega \right ) H\left ( \omega \right ) \] Hence\[ X\left ( \omega \right ) =\frac{Y\left ( \omega \right ) }{H\left ( \omega \right ) }\] But \(y\left ( t\right ) =e^{-3t}u\left ( t\right ) -e^{-4t}u\left ( t\right ) \), therefore, from table \(Y\left ( \omega \right ) =\frac{1}{3+j\omega }-\frac{1}{4+j\omega }=\frac{\left ( 4+j\omega \right ) -\left ( 3+j\omega \right ) }{\left ( 3+j\omega \right ) \left ( 4+j\omega \right ) }=\frac{1}{\left ( 3+j\omega \right ) \left ( 4+j\omega \right ) }\) and the above becomes\[ X\left ( \omega \right ) =\frac{\frac{1}{\left ( 3+j\omega \right ) \left ( 4+j\omega \right ) }}{H\left ( \omega \right ) }\] But we are given that \(H\left ( j\omega \right ) =\frac{1}{j\omega +3}\). The above simplifies to\begin{align*} X\left ( \omega \right ) & =\frac{\frac{1}{\left ( 3+j\omega \right ) \left ( 4+j\omega \right ) }}{\frac{1}{j\omega +3}}\\ & =\frac{1}{4+j\omega } \end{align*}
From tables\[ x\left ( t\right ) =e^{-4t}u\left ( t\right ) \]
Solution
First we find the Fourier transform of \(x_{0}\left ( t\right ) \). Since this is a periodic, then \(x_{0}\left ( t\right ) \Longleftrightarrow X_{0}\left ( \omega \right ) \) and \begin{align*} X_{0}\left ( \omega \right ) & =\int _{-\infty }^{\infty }x_{0}\left ( t\right ) e^{-j\omega t}dt\\ & =\int _{0}^{1}e^{-t}e^{-j\omega t}dt\\ & =\int _{0}^{1}e^{-t\left ( 1+j\omega \right ) }dt\\ & =\frac{-1}{1+j\omega }\left [ e^{-t\left ( 1+j\omega \right ) }\right ] _{0}^{1}\\ & =\frac{-1}{1+j\omega }\left ( e^{-\left ( 1+j\omega \right ) }-1\right ) \\ & =\frac{1-e^{-\left ( 1+j\omega \right ) }}{1+j\omega } \end{align*}
From table 4.1, property 4.3.5, Fourier transform of \(x\left ( -t\right ) =X\left ( -\omega \right ) \). Hence \(x_{0}\left ( -t\right ) \Longleftrightarrow X_{0}\left ( -\omega \right ) \). Therefore, using the above result and taking its complex conjugate gives\[ X_{0}\left ( -\omega \right ) =\frac{1-e^{-\left ( 1-j\omega \right ) }}{1-j\omega }\] Therefore the Fourier transform of \(x_{0}\left ( t\right ) +x_{0}\left ( -t\right ) \Longleftrightarrow X_{1}\left ( \omega \right ) =X_{0}\left ( \omega \right ) +X_{0}\left ( -\omega \right ) \) This is by linearity property. Hence\begin{align*} X_{1}\left ( \omega \right ) & =X_{0}\left ( \omega \right ) +X_{0}\left ( -\omega \right ) \\ & =\frac{1-e^{-\left ( 1+j\omega \right ) }}{1+j\omega }+\frac{1-e^{-\left ( 1-j\omega \right ) }}{1-j\omega }\\ & =\frac{\left ( 1-j\omega \right ) \left ( 1-e^{-\left ( 1+j\omega \right ) }\right ) +\left ( 1+j\omega \right ) \left ( 1-e^{-\left ( 1-j\omega \right ) }\right ) }{\left ( 1+j\omega \right ) \left ( 1-j\omega \right ) }\\ & =\frac{1-e^{-\left ( 1+j\omega \right ) }-j\omega \left ( 1-e^{-\left ( 1+j\omega \right ) }\right ) +\left ( 1-e^{-\left ( 1-j\omega \right ) }\right ) +j\omega \left ( 1-e^{-\left ( 1-j\omega \right ) }\right ) }{1+\omega ^{2}}\\ & =\frac{1-e^{-\left ( 1+j\omega \right ) }-j\omega +j\omega e^{-\left ( 1+j\omega \right ) }+\left ( 1-e^{-\left ( 1-j\omega \right ) }\right ) +j\omega -j\omega e^{-\left ( 1-j\omega \right ) }}{1+\omega ^{2}}\\ & =\frac{1-e^{-\left ( 1+j\omega \right ) }+j\omega e^{-\left ( 1+j\omega \right ) }+1-e^{-\left ( 1-j\omega \right ) }-j\omega e^{-\left ( 1-j\omega \right ) }}{1+\omega ^{2}}\\ & =\frac{1-e^{-1}e^{-j\omega }+j\omega e^{-1}e^{-j\omega }+1-e^{-1}e^{j\omega }-j\omega e^{-1}e^{j\omega }}{1+\omega ^{2}}\\ & =\frac{1-e^{-1}\left ( e^{j\omega }+e^{-j\omega }\right ) -j\omega e^{-1}\left ( e^{j\omega }-e^{-j\omega }\right ) }{1+\omega ^{2}}\\ & =\frac{1}{1+\omega ^{2}}-\frac{e^{-1}}{1+\omega ^{2}}\left ( e^{j\omega }+e^{-j\omega }\right ) +\frac{\omega e^{-1}}{1+\omega ^{2}}\frac{\left ( e^{j\omega }-e^{-j\omega }\right ) }{j}\\ & =\frac{1}{1+\omega ^{2}}-\frac{2}{e\left ( 1+\omega ^{2}\right ) }\cos \omega +\frac{2\omega }{e\left ( 1+\omega ^{2}\right ) }\sin \omega \end{align*}
The following is a plot of the above
We see that \(X_{1}\left ( \omega \right ) \) is even and real. This agrees with table 4.1, property 4.3.3 which says that for real \(x\left ( t\right ) \) which is even, then its Fourier transform is real and even.
We found \(X_{0}\left ( \omega \right ) =\frac{1-e^{-\left ( 1+j\omega \right ) }}{1+j\omega }\) above. Hence \(-x_{0}\left ( -t\right ) \Longleftrightarrow -X_{0}\left ( -\omega \right ) =-\frac{1-e^{-\left ( 1-j\omega \right ) }}{1-j\omega }=\frac{1-e^{-\left ( 1-j\omega \right ) }}{j\omega -1}\). Therefore\begin{align*} X_{2}\left ( \omega \right ) & =X_{0}\left ( \omega \right ) -X_{0}\left ( -\omega \right ) \\ & =\frac{1-e^{-\left ( 1+j\omega \right ) }}{1+j\omega }+\frac{1-e^{-\left ( 1-j\omega \right ) }}{j\omega -1}\\ & =\frac{\left ( j\omega -1\right ) \left ( 1-e^{-\left ( 1+j\omega \right ) }\right ) +\left ( 1+j\omega \right ) \left ( 1-e^{-\left ( 1-j\omega \right ) }\right ) }{\left ( 1+j\omega \right ) \left ( j\omega -1\right ) }\\ & =\frac{j\omega \left ( 1-e^{-\left ( 1+j\omega \right ) }\right ) -\left ( 1-e^{-\left ( 1+j\omega \right ) }\right ) +\left ( 1-e^{-\left ( 1-j\omega \right ) }\right ) +j\omega \left ( 1-e^{-\left ( 1-j\omega \right ) }\right ) }{j\omega -1-\omega ^{2}-j\omega }\\ & =\frac{j\omega -j\omega e^{-\left ( 1+j\omega \right ) }-1+e^{-\left ( 1+j\omega \right ) }+1-e^{-\left ( 1-j\omega \right ) }+j\omega -j\omega e^{-\left ( 1-j\omega \right ) }}{-\left ( 1+\omega ^{2}\right ) }\\ & =\frac{2j\omega -j\omega e^{-\left ( 1+j\omega \right ) }+e^{-\left ( 1+j\omega \right ) }-e^{-\left ( 1-j\omega \right ) }-j\omega e^{-\left ( 1-j\omega \right ) }}{-\left ( 1+\omega ^{2}\right ) }\\ & =\frac{2j\omega -j\omega e^{-1}e^{-j\omega }+e^{-1}e^{-j\omega }-e^{-1}e^{j\omega }-j\omega e^{-1}e^{j\omega }}{j\omega -\omega ^{2}-2}\\ & =\frac{2j\omega -j\omega e^{-1}\left ( e^{j\omega }+e^{-j\omega }\right ) -e^{-1}\left ( e^{j\omega }-e^{-j\omega }\right ) }{-\left ( 1+\omega ^{2}\right ) }\\ & =\frac{2j\omega }{-\left ( 1+\omega ^{2}\right ) }-j\omega e^{-1}\frac{\left ( e^{j\omega }+e^{-j\omega }\right ) }{-\left ( 1+\omega ^{2}\right ) }-e^{-1}\frac{e^{j\omega }-e^{-j\omega }}{-\left ( 1+\omega ^{2}\right ) }\\ & =\frac{2j\omega }{-\left ( 1+\omega ^{2}\right ) }-2j\omega e^{-1}\frac{\cos \omega }{-\left ( 1+\omega ^{2}\right ) }-2je^{-1}\frac{\sin \left ( \omega \right ) }{-\left ( 1+\omega ^{2}\right ) } \end{align*}
Hence\begin{align*} X_{2}\left ( \omega \right ) & =j\left ( \frac{-2\omega }{\left ( 1+\omega ^{2}\right ) }+2\omega \frac{\cos \omega }{e\left ( 1+\omega ^{2}\right ) }+2\frac{\sin \left ( \omega \right ) }{e\left ( 1+\omega ^{2}\right ) }\right ) \\ & =\frac{j}{e\left ( 1+\omega ^{2}\right ) }\left ( -2e\omega +2\omega \cos \omega +2\sin \left ( \omega \right ) \right ) \end{align*}
We see that \(X_{2}\left ( \omega \right ) \) is pure imaginary. This agrees with table 4.1, property 4.3.3 which says that for real \(x\left ( t\right ) \) which is odd, then its Fourier transform is pure imaginary and odd.
The following is a plot of the above which shows that the imaginary part of \(X_{2}\left ( \omega \right ) \) is odd
Solution
Part i \[ y\left ( t\right ) =x\left ( t\right ) \circledast h\left ( t\right ) \] Taking Fourier transform the above, convolution becomes multiplication\[ Y\left ( \omega \right ) =X\left ( \omega \right ) H\left ( \omega \right ) \] Given that \(x\left ( t\right ) =te^{-2t}u\left ( t\right ) \), then from tables \(X\left ( \omega \right ) =\frac{1}{\left ( 2+j\omega \right ) ^{2}}\) and given that \(h\left ( t\right ) =e^{-4t}u\left ( t\right ) \) then from table 4.2 \(H\left ( \omega \right ) =\frac{1}{4+j\omega }\). Hence the above becomes\[ Y\left ( \omega \right ) =\frac{1}{\left ( 2+j\omega \right ) ^{2}\left ( 4+j\omega \right ) }\] Doing partial fractions. Let \(s=j\omega \) then \begin{align*} \frac{1}{\left ( 2+s\right ) ^{2}\left ( 4+s\right ) } & =\frac{A}{\left ( 2+s\right ) ^{2}}+\frac{B}{2+s}+\frac{C}{4+s}\\ & =\frac{A\left ( 4+s\right ) +B\left ( \left ( 2+s\right ) \left ( 4+s\right ) \right ) +C\left ( 2+s\right ) ^{2}}{\left ( 2+s\right ) ^{2}\left ( 4+s\right ) } \end{align*}
Expanding numerator\begin{align*} 1 & =4A+8B+4C+Bs^{2}+Cs^{2}+As+6Bs+4Cs\\ 1 & =\left ( 4A+8B+4C\right ) +s\left ( A+6B+4C\right ) +s^{2}\left ( B+C\right ) \end{align*}
Comparing coefficients\begin{align*} 1 & =4A+8B+4C\\ 0 & =A+6B+4C\\ 0 & =B+C \end{align*}
Or\[\begin{pmatrix} 4 & 8 & 4\\ 1 & 6 & 4\\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} A\\ B\\ C \end{pmatrix} =\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} \] Gaussian elimination. Multiplying second row by 4 and subtracting result from first row gives\[\begin{pmatrix} 4 & 8 & 4\\ 0 & 16 & 12\\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} A\\ B\\ C \end{pmatrix} =\begin{pmatrix} 1\\ -1\\ 0 \end{pmatrix} \] Multiplying third row by 16 and subtracting result from second row gives\[\begin{pmatrix} 4 & 8 & 4\\ 0 & 16 & 12\\ 0 & 0 & 4 \end{pmatrix}\begin{pmatrix} A\\ B\\ C \end{pmatrix} =\begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix} \] Backsubstitution. Last row gives \(C=\frac{1}{4}\). Second row gives \(16B+12C=-1\) or \(16B=-1-12\left ( \frac{1}{4}\right ) \), hence \(16B=-4\), Hence \(B=-\frac{1}{4}\). First row gives \(4A+8B+4C=1\) or \(4A+8\left ( -\frac{1}{4}\right ) +4\left ( \frac{1}{4}\right ) =1\) or \(4A-2+1=1\). Hence \(A=\frac{1}{2}\). Therefore partial fractions gives\[ \frac{1}{\left ( 2+s\right ) ^{2}\left ( 4+s\right ) }=\frac{\left ( \frac{1}{2}\right ) }{\left ( 2+s\right ) ^{2}}+\frac{\left ( -\frac{1}{4}\right ) }{2+s}+\frac{\left ( \frac{1}{4}\right ) }{4+s}\] Replacing \(s\) back with \(j\omega \)\[ Y\left ( \omega \right ) =\frac{1}{2}\frac{1}{\left ( 2+j\omega \right ) ^{2}}-\frac{1}{4}\frac{1}{2+j\omega }+\frac{1}{4}\frac{1}{4+j\omega }\] Applying inverse Fourier transform, using table gives\[ y\left ( t\right ) =\frac{1}{2}te^{-2t}u\left ( t\right ) -\frac{1}{4}e^{-2t}u\left ( t\right ) +\frac{1}{4}e^{-4t}u\left ( t\right ) \]
Part ii \[ y\left ( t\right ) =x\left ( t\right ) \circledast h\left ( t\right ) \] Taking Fourier transform the above, convolution becomes multiplication\[ Y\left ( \omega \right ) =X\left ( \omega \right ) H\left ( \omega \right ) \] Given that \(x\left ( t\right ) =te^{-2t}u\left ( t\right ) \), then from tables \(X\left ( \omega \right ) =\frac{1}{\left ( 2+j\omega \right ) ^{2}}\) and given that \(h\left ( t\right ) =te^{-4t}u\left ( t\right ) \) then from table 4.2 \(H\left ( \omega \right ) =\frac{1}{\left ( 4+j\omega \right ) ^{2}}\). Hence the above becomes\[ Y\left ( \omega \right ) =\frac{1}{\left ( 2+j\omega \right ) ^{2}\left ( 4+j\omega \right ) ^{2}}\] Doing partial fractions. Let \(s=j\omega \) then \begin{align*} \frac{1}{\left ( 2+s\right ) ^{2}\left ( 4+s\right ) ^{2}} & =\frac{A}{\left ( 2+s\right ) ^{2}}+\frac{B}{2+s}+\frac{C}{\left ( 4+s\right ) ^{2}}+\frac{D}{\left ( 4+s\right ) }\\ & =\frac{A\left ( 4+s\right ) ^{2}+B\left ( \left ( 2+s\right ) \left ( 4+s\right ) ^{2}\right ) +C\left ( 2+s\right ) ^{2}+D\left ( \left ( 2+s\right ) ^{2}\left ( 4+s\right ) \right ) }{\left ( 2+s\right ) ^{2}\left ( 4+s\right ) ^{2}} \end{align*}
Expanding numerator\begin{align*} 1 & =16A+32B+4C+16D+20sD+As^{2}+10Bs^{2}+\allowbreak Bs^{3}+Cs^{2}+8s^{2}D+s^{3}D+8As+32Bs+4Cs\\ 1 & =\left ( 16A+32B+4C+16D\right ) +s\left ( 20D+8A+32B+4C\right ) +s^{2}\left ( A+10B+C+8D\right ) +s^{3}\left ( B+D\right ) \end{align*}
Comparing coefficients\begin{align*} 1 & =16A+32B+4C+16D\\ 0 & =8A+32B+4C+20D\\ 0 & =A+10B+C+8D\\ 0 & =B+D \end{align*}
Or\[\begin{pmatrix} 16 & 32 & 4 & 16\\ 8 & 32 & 4 & 20\\ 1 & 10 & 1 & 8\\ 0 & 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} A\\ B\\ C\\ D \end{pmatrix} =\begin{pmatrix} 1\\ 0\\ 0\\ 0 \end{pmatrix} \] Gaussian elimination. replacing row 2 by result of subtracting \(1/2\) times row 1 from row 2\[\begin{pmatrix} 16 & 32 & 4 & 16\\ 0 & 16 & 2 & 12\\ 1 & 10 & 1 & 8\\ 0 & 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} A\\ B\\ C\\ D \end{pmatrix} =\begin{pmatrix} 1\\ -\frac{1}{2}\\ 0\\ 0 \end{pmatrix} \] Replacing row 3 by result of subtracting \(\frac{1}{16}\) times row 1 from row 3\[\begin{pmatrix} 16 & 32 & 4 & 16\\ 0 & 16 & 2 & 12\\ 0 & 8 & \frac{3}{4} & 7\\ 0 & 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} A\\ B\\ C\\ D \end{pmatrix} =\begin{pmatrix} 1\\ -\frac{1}{2}\\ -\frac{1}{16}\\ 0 \end{pmatrix} \] Replacing row 3 by result of subtracting \(\frac{1}{2}\) times row 2 from row 3\[\begin{pmatrix} 16 & 32 & 4 & 16\\ 0 & 16 & 2 & 12\\ 0 & 0 & \frac{-1}{4} & 1\\ 0 & 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} A\\ B\\ C\\ D \end{pmatrix} =\begin{pmatrix} 1\\ -\frac{1}{2}\\ \frac{3}{16}\\ 0 \end{pmatrix} \] Replacing row 4 by result of subtracting \(\frac{1}{16}\) times row 2 from row 4\[\begin{pmatrix} 16 & 32 & 4 & 16\\ 0 & 16 & 2 & 12\\ 0 & 0 & \frac{-1}{4} & 1\\ 0 & 0 & \frac{-1}{8} & \frac{1}{4}\end{pmatrix}\begin{pmatrix} A\\ B\\ C\\ D \end{pmatrix} =\begin{pmatrix} 1\\ -\frac{1}{2}\\ \frac{3}{16}\\ \frac{1}{32}\end{pmatrix} \] Replacing row 4 by result of subtracting \(\frac{1}{2}\) times row 3 from row 4\[\begin{pmatrix} 16 & 32 & 4 & 16\\ 0 & 16 & 2 & 12\\ 0 & 0 & \frac{-1}{4} & 1\\ 0 & 0 & 0 & \frac{-1}{4}\end{pmatrix}\begin{pmatrix} A\\ B\\ C\\ D \end{pmatrix} =\begin{pmatrix} 1\\ -\frac{1}{2}\\ \frac{3}{16}\\ \frac{-1}{16}\end{pmatrix} \] Backsubstitution phase:\begin{align*} \frac{-1}{4}D & =\frac{-1}{16}\\ D & =\frac{1}{4} \end{align*}
Third row gives \(\frac{-1}{4}C+D=\frac{3}{16}\) or \(\frac{-1}{4}C+\frac{1}{4}=\frac{3}{16}\rightarrow C=\frac{1}{4}\). Second row gives \(16B+2C+12D=-\frac{1}{2}\) or \(16B+2\left ( \frac{1}{4}\right ) +12\left ( \frac{1}{4}\right ) =-\frac{1}{2}\rightarrow B=-\frac{1}{4}\). First row gives \(16A+32B+4C+16D=1\) or \(16A+32\left ( -\frac{1}{4}\right ) +4\left ( \frac{1}{4}\right ) +16\left ( \frac{1}{4}\right ) =1,\rightarrow A=\frac{1}{4}\).
Therefore partial fractions gives\begin{align*} \frac{1}{\left ( 2+s\right ) ^{2}\left ( 4+s\right ) ^{2}} & =\frac{A}{\left ( 2+s\right ) ^{2}}+\frac{B}{2+s}+\frac{C}{\left ( 4+s\right ) ^{2}}+\frac{D}{\left ( 4+s\right ) }\\ & =\frac{1}{4}\frac{1}{\left ( 2+s\right ) ^{2}}-\frac{1}{4}\frac{1}{2+s}+\frac{1}{4}\frac{1}{\left ( 4+s\right ) ^{2}}+\frac{1}{4}\frac{1}{\left ( 4+s\right ) } \end{align*}
Replace \(s\) back with \(j\omega \)\[ Y\left ( \omega \right ) =\frac{1}{4}\frac{1}{\left ( 2+j\omega \right ) ^{2}}-\frac{1}{4}\frac{1}{2+j\omega }+\frac{1}{4}\frac{1}{\left ( 4+j\omega \right ) ^{2}}+\frac{1}{4}\frac{1}{\left ( 4+j\omega \right ) }\] Applying inverse Fourier transform, using table gives\[ y\left ( t\right ) =\frac{1}{4}te^{-2t}u\left ( t\right ) -\frac{1}{4}e^{-2t}u\left ( t\right ) +\frac{1}{4}te^{-4t}u\left ( t\right ) +\frac{1}{4}e^{-4t}u\left ( t\right ) \]
Part iii \[ y\left ( t\right ) =x\left ( t\right ) \circledast h\left ( t\right ) \] Taking Fourier transform the above, convolution becomes multiplication\[ Y\left ( \omega \right ) =X\left ( \omega \right ) H\left ( \omega \right ) \] Given that \(x\left ( t\right ) =e^{-t}u\left ( t\right ) \), then from tables \(X\left ( \omega \right ) =\frac{1}{\left ( 1+j\omega \right ) }\) and given that \(h\left ( t\right ) =e^{t}u\left ( -t\right ) \) then \(H\left ( \omega \right ) =\frac{1}{1-j\omega }\). Hence the above becomes\[ Y\left ( \omega \right ) =\frac{1}{\left ( 1+j\omega \right ) }\frac{1}{\left ( 1-j\omega \right ) }\] Doing partial fractions. Let \(s=j\omega \) then \[ \frac{1}{\left ( 1+s\right ) }\frac{1}{\left ( 1-s\right ) }=\frac{A}{\left ( 1+s\right ) }+\frac{B}{1-s}\] Hence \(A=\left ( \frac{1}{\left ( 1-s\right ) }\right ) _{s=-1}=\frac{1}{2}\) and \(B=\left ( \frac{1}{\left ( 1+s\right ) }\right ) _{s=1}=\frac{1}{2}\). Hence \[ Y\left ( \omega \right ) =\frac{1}{2}\frac{1}{\left ( 1+j\omega \right ) }+\frac{1}{2}\frac{1}{1-j\omega }\] Therefore\[ y\left ( t\right ) =\frac{1}{2}e^{-t}u\left ( t\right ) +\frac{1}{2}e^{t}u\left ( -t\right ) \] The above means for \(t<0,y\left ( t\right ) =\frac{1}{2}e^{t}\) and for \(t>0,y\left ( t\right ) =\frac{1}{2}e^{-t}\).
\(x\left ( t\right ) =e^{-\left ( t-2\right ) }u\left ( t-2\right ) \), \(h\left ( t\right ) =u\left ( t+1\right ) -u\left ( t-3\right ) \). Let us first find \(X\left ( \omega \right ) \) and \(H\left ( \omega \right ) \)\begin{align*} X\left ( \omega \right ) & =\int _{-\infty }^{\infty }e^{-\left ( t-2\right ) }u\left ( t-2\right ) e^{-j\omega t}dt\\ & =\int _{2}^{\infty }e^{-\left ( t-2\right ) }e^{-j\omega t}dt\\ & =\int _{2}^{\infty }e^{-t}e^{2}e^{-j\omega t}dt\\ & =e^{2}\int _{2}^{\infty }e^{-t\left ( 1+j\omega \right ) }dt\\ & =-\frac{e^{2}}{1+j\omega }\left [ e^{-t\left ( 1+j\omega \right ) }\right ] _{2}^{\infty }\\ & =-\frac{e^{2}}{1+j\omega }\left ( 0-e^{-2\left ( 1+j\omega \right ) }\right ) \\ & =\frac{e^{2}e^{-2\left ( 1+j\omega \right ) }}{1+j\omega }\\ & =\frac{e^{-2-2j\omega +2}}{1+j\omega }\\ & =\frac{e^{-2j\omega }}{1+j\omega } \end{align*}
And\begin{align*} H\left ( \omega \right ) & =\int _{-\infty }^{\infty }u\left ( t+1\right ) -u\left ( t-3\right ) e^{-j\omega t}dt\\ & =\int _{-1}^{3}e^{-j\omega t}dt\\ & =\frac{1}{j\omega }\left ( e^{-j\omega t}\right ) _{-1}^{3}\\ & =\frac{1}{j\omega }\left ( e^{-3j\omega }-e^{j\omega t}\right ) \\ & =\frac{1}{j\omega }e^{-j\omega }\left ( e^{-j2\omega }-e^{j2\omega t}\right ) \\ & =-\frac{1}{j\omega }e^{-j\omega }\left ( e^{j2\omega t}-e^{-j2\omega }\right ) \\ & =-\frac{1}{\omega }e^{-j\omega }\left ( \frac{e^{j2\omega t}-e^{-j2\omega }}{j}\right ) \\ & =-\frac{2}{\omega }e^{-j\omega }\sin \left ( 2\omega \right ) \end{align*}
Hence \begin{align} Y\left ( \omega \right ) & =H\left ( \omega \right ) X\left ( \omega \right ) \nonumber \\ & =\frac{e^{-2j\omega }}{1+j\omega }\left ( -\frac{2}{\omega }e^{-j\omega }\sin \left ( 2\omega \right ) \right ) \nonumber \\ & =\frac{-2}{\omega \left ( 1+j\omega \right ) }e^{-2j\omega }e^{-j\omega }\sin \left ( 2\omega \right ) \nonumber \\ & =\frac{-2}{\omega \left ( 1+j\omega \right ) }e^{-3j\omega }\sin \left ( 2\omega \right ) \tag{1} \end{align}
Now, \begin{align*} \,y\left ( t\right ) & =x\left ( t\right ) \circledast h\left ( t\right ) \\ & =\int _{-\infty }^{\infty }x\left ( \tau \right ) h\left ( t-\tau \right ) d\tau \end{align*}
Folding \(h\left ( t\right ) \). For \(t<1\), then \(y\left ( t\right ) =0\). For \(2<1+t<6\) or \(1<t<5\)\begin{align*} \,y\left ( t\right ) & =\int _{2}^{1+t}x\left ( \tau \right ) d\tau \\ & =\int _{2}^{1+t}e^{-\left ( \tau -2\right ) }d\tau \\ & =-\left ( e^{-\left ( \tau -2\right ) }\right ) _{2}^{1+t}\\ & =-\left ( e^{-\left ( 1+t-2\right ) }-e^{-\left ( 2-2\right ) }\right ) \\ & =-\left ( e^{-\left ( -1+t\right ) }-1\right ) \\ & =1-e^{1-t} \end{align*}
For \(6<1+t\) or \(t>5\)\begin{align*} \,y\left ( t\right ) & =\int _{t-3}^{t+1}x\left ( \tau \right ) d\tau \\ & =\int _{t-3}^{t+1}e^{-\left ( \tau -2\right ) }d\tau \\ & =-\left ( e^{-\left ( \tau -2\right ) }\right ) _{t-3}^{1+t}\\ & =-\left ( e^{-\left ( 1+t-2\right ) }-e^{-\left ( t-3-2\right ) }\right ) \\ & =-\left ( e^{-\left ( -1+t\right ) }-e^{-\left ( t-5\right ) }\right ) \\ & =-\left ( e^{1-t}-e^{5-t}\right ) \\ & =e^{5-t}-e^{1-t} \end{align*}
Hence\[ y\left ( t\right ) =\left \{ \begin{array} [c]{ccc}0 & & t<1\\ 1-e^{1-t} & & 1<t<5\\ e^{5-t}-e^{1-t} & & t>5 \end{array} \right . \] The above can be written as \(y\left ( t\right ) \)\begin{align} y\left ( t\right ) & =\left ( 1-e^{1-t}\right ) \left ( u\left ( t-1\right ) -u\left ( t-5\right ) \right ) +\left ( e^{5-t}-e^{1-t}\right ) u\left ( t-5\right ) \nonumber \\ & =u\left ( t-1\right ) -u\left ( t-5\right ) -e^{1-t}\left ( u\left ( t-1\right ) -u\left ( t-5\right ) \right ) +\left ( e^{5-t}-e^{1-t}\right ) u\left ( t-5\right ) \nonumber \\ & =u\left ( t-1\right ) -u\left ( t-5\right ) -e^{1-t}u\left ( t-1\right ) +e^{1-t}u\left ( t-5\right ) +e^{5-t}u\left ( t-5\right ) -e^{1-t}u\left ( t-5\right ) \nonumber \\ & =u\left ( t-1\right ) -u\left ( t-5\right ) -e^{1-t}u\left ( t-1\right ) +e^{5-t}u\left ( t-5\right ) \tag{2} \end{align}
Taking the Fourier transform of the above from tables \begin{align*} u\left ( t-1\right ) & \Longleftrightarrow \frac{1}{j\omega }e^{-j\omega }+\pi \delta \left ( \omega \right ) \\ u\left ( t-5\right ) & \Longleftrightarrow \frac{1}{j\omega }e^{-5j\omega }+\pi \delta \left ( \omega \right ) \\ e^{1-t}u\left ( t-1\right ) & \Longleftrightarrow \frac{e^{-j\omega }}{1+j\omega }\\ e^{5-t}u\left ( t-5\right ) & \Longleftrightarrow \frac{e^{-5j\omega }}{1+j\omega } \end{align*}
Hence \(Y\left ( \omega \right ) \) is\begin{align*} Y\left ( \omega \right ) & =\frac{1}{j\omega }e^{-j\omega }+\pi \delta \left ( \omega \right ) -\left ( \frac{1}{j\omega }e^{-5j\omega }+\pi \delta \left ( \omega \right ) \right ) -\frac{e^{-j\omega }}{1+j\omega }+\frac{e^{-5j\omega }}{1+j\omega }\\ & =\frac{1}{j\omega }e^{-j\omega }-\frac{1}{j\omega }e^{-5j\omega }-\frac{e^{-j\omega }}{1+j\omega }+\frac{e^{-5j\omega }}{1+j\omega }\\ & =\frac{e^{-3j\omega }}{\omega }\left ( \frac{1}{j}e^{2j\omega }-\frac{1}{j}e^{-2j\omega }\right ) -\frac{e^{-j\omega }}{1+j\omega }+\frac{e^{-5j\omega }}{1+j\omega }\\ & =\frac{e^{-3j\omega }}{\omega }2\left ( \sin 2\omega \right ) -\frac{e^{-3j\omega }}{1+j\omega }\left ( e^{2j\omega }-e^{-2j\omega }\right ) \\ & =\frac{e^{-3j\omega }}{\omega }2\left ( \sin 2\omega \right ) -2\frac{e^{-3j\omega }}{j+\omega }\sin 2\omega \\ & =2e^{-3j\omega }\sin 2\omega \left ( \frac{1}{\omega }-\frac{1}{j+\omega }\right ) \\ & =2e^{-3j\omega }\sin 2\omega \left ( \frac{j+\omega -\omega }{\omega \left ( \omega +j\right ) }\right ) \\ & =2e^{-3j\omega }\sin 2\omega \frac{j}{\omega \left ( \omega +j\right ) }\\ & =-2e^{-3j\omega }\sin 2\omega \frac{1}{\omega \left ( 1+j\omega \right ) } \end{align*}
Comparing the above to (1) shows they are the same.
Hence this shows that Fourier transform of \(x\left ( t\right ) \circledast h\left ( t\right ) \) gives same answer as \(H\left ( \omega \right ) X\left ( \omega \right ) \)