4.4 HW 4

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4.4.1 Problem 4.1(a), Chapter 4

Find Fourier transform of (a) $e^{-2(t-1)}u(t-1)$

Solution$$\begin{array}{rl}X\left(\omega \right)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\left(t\right)e^{-j\omega t}dt\\ & ={\int }_1^{\mathrm{\infty }}{e}^{-2(t-1)}{e}^{-i\omega t}dt\\ & ={\int }_1^{\mathrm{\infty }}{e}^{-2t}{e}^2{e}^{-i\omega t}dt\\ & =e^2{\int }_1^{\mathrm{\infty }}{e}^{-t(2+i\omega )}dt\\ & =\frac{e^2}{-(2+i\omega )}{\left[e^{-t(2+i\omega )}\right]}_1^{\mathrm{\infty }}\end{array}$$

Assuming $\mathrm{Im}\left(\omega \right)<2$ then$$\begin{array}{rl}X\left(\omega \right)& =\frac{e^2}{-(2+i\omega )}[0-e^{-2}{e}^{-i\omega }]\\ & =\frac{e^2}{(2+i\omega )}\left[e^{-2}{e}^{-i\omega }\right]\\ & =\frac{e^{-i\omega }}{(2+i\omega )}\end{array}$$

4.4.2 Problem 4.3, Chapter 4

Determine the Fourier transform of each of the following periodic signals (a) $\sin (2\pi t+\frac{\pi }{4})$ (b) $1+\cos (6\pi t+\frac{\pi }{8})$

Solution

4.4.2.1 Part a

Since this is periodic signal, then we can not use $X\left(\omega \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\left(t\right)e^{-j\omega t}dt$ which is for aperiodic signal. Instead we need to use 4.22 in the textbook which is $$X\left(\omega \right)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}2\pi a_k\delta (\omega -k{\omega }_0)$$ From $\sin (2\pi t+\frac{\pi }{4})$ we see that ${\omega }_0=2\pi$, hence$$X\left(\omega \right)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}2\pi a_k\delta (\omega -2k\pi )$$ Writing $\sin ({\omega }_{0}t+\frac{\pi }{4})=\frac{1}{2j}{e}^{j({\omega }_{0}t+\frac{\pi }{4})}-\frac{1}{2j}{e}^{-j({\omega }_{0}t+\frac{\pi }{4})}=\left(\frac{1}{2j}{e}^{j\frac{\pi }{4}}\right)e^{j{\omega }_{0}t}-\left(\frac{1}{2j}{e}^{-j\frac{\pi }{4}}\right)e^{-j{\omega }_{0}t}$ 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$$\begin{array}{rl}X\left(\omega \right)& =2\pi \left(\frac{1}{2j}{e}^{j\frac{\pi }{4}}\right)\delta (\omega -2\pi )+2\pi (-\frac{1}{2j}{e}^{-j\frac{\pi }{4}})\delta (\omega +2\pi )\\ & =\frac{\pi }{j}{e}^{j\frac{\pi }{4}}\delta (\omega -2\pi )-\frac{\pi }{j}{e}^{-j\frac{\pi }{4}}\delta (\omega +2\pi )\end{array}$$

4.4.2.2 Part b

Since this is periodic signal, then its Fourier transform is$$X\left(\omega \right)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}2\pi a_k\delta (\omega -k{\omega }_0)$$ From $1+\cos (6\pi t+\frac{\pi }{8})$ we see that ${\omega }_0=6\pi$, hence$$X\left(\omega \right)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}2\pi a_k\delta (\omega -6k\pi )$$ Writing $1+\cos (6\pi t+\frac{\pi }{8})=1+(\frac{1}{2}{e}^{j({\omega }_{0}t+\frac{\pi }{8})}+\frac{1}{2}{e}^{-j({\omega }_{0}t+\frac{\pi }{8})})=1+(\frac{1}{2}{e}^{j\frac{\pi }{8}}{e}^{j{\omega }_{0}t}+\frac{1}{2}{e}^{-j\frac{\pi }{8}}{e}^{-j{\omega }_{0}t})$ 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$$\begin{array}{rl}X\left(\omega \right)& =2\pi a_{-1}\delta (\omega +6\pi )+2\pi a_0\delta \left(\omega \right)+2\pi a_1\delta (\omega -6\pi )\\ & =2\pi \left(\frac{1}{2}{e}^{-j\frac{\pi }{8}}\right)\delta (\omega +6\pi )+2\pi \delta \left(\omega \right)+2\pi \left(\frac{1}{2}{e}^{j\frac{\pi }{8}}\right)\delta (\omega -6\pi )\\ & =\pi e^{-j\frac{\pi }{8}}\delta (\omega +6\pi )+2\pi \delta \left(\omega \right)+\pi e^{j\frac{\pi }{8}}\delta (\omega -6\pi )\end{array}$$

4.4.3 Problem 4.5, Chapter 4

Use the Fourier transform synthesis equation (4.8) to determine the inverse Fourier transform of $X\left(j\omega \right)=\left|X\left(j\omega \right)\right|e^{j\measuredangle X\left(j\omega \right)}$ where $\left|X\left(j\omega \right)\right|=2(u(\omega +3)-u(\omega -3)),\measuredangle X\left(j\omega \right)=-\frac{3}{2}\omega +\pi$

Solution

4.8 is $$\begin{array}{}\text{(4.8)}& x\left(t\right)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}X\left(\omega \right)e^{j\omega t}d\omega \end{array}$$ Hence$$\begin{array}{rl}x\left(t\right)& =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left|X\left(j\omega \right)\right|e^{j\measuredangle X\left(j\omega \right)}{e}^{j\omega t}d\omega \\ & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}2(u(\omega +3)-u(\omega -3))e^{j(-\frac{3}{2}\omega +\pi )}{e}^{j\omega t}d\omega \end{array}$$

But $u(\omega +3)-u(\omega -3)$ is one over $\omega =-3\cdots 3$ and zero otherwise. The above simplifies to$$\begin{array}{rl}x\left(t\right)& =\frac{1}{2\pi }{\int }_{-3}^{3}2e^{j(-\frac{3}{2}\omega +\pi )}{e}^{j\omega t}d\omega \\ & =\frac{1}{\pi }{\int }_{-3}^3{e}^{j\pi }{e}^{j-\frac{3}{2}\omega }{e}^{j\omega t}d\omega \\ & =\frac{e^{j\pi }}{\pi }{\int }_{-3}^3{e}^{j(-\frac{3}{2}+t)\omega }d\omega \end{array}$$

But $e^{j\pi }=-1$ and $\int e^{j(-\frac{3}{2}+t)\omega }d\omega =\frac{e^{j(-\frac{3}{2}+t)\omega }}{j(-\frac{3}{2}+t)}$, hence the above becomes$$\begin{array}{rl}x\left(t\right)& =\frac{-1}{\pi }\frac{1}{j(-\frac{3}{2}+t)}{\left[e^{j(-\frac{3}{2}+t)\omega }\right]}_{-3}^3\\ & =\frac{-1}{\pi (-\frac{3}{2}+t)}\left(\frac{e^{j3(-\frac{3}{2}+t)}-e^{-j3(-\frac{3}{2}+t)}}{j}\right)\\ & =\frac{-1}{\pi (-\frac{3}{2}+t)}2(\sin \left(3(-\frac{3}{2}+t)\right))\\ & =\frac{-2}{\pi (t-\frac{3}{2})}\sin \left(3(t-\frac{3}{2})\right)\end{array}$$

4.4.4 Problem 4.11, Chapter 4

Given the relationships $y\left(t\right)=x\left(t\right)⊛h\left(t\right)$ and $g\left(t\right)=x\left(3t\right)⊛h\left(3t\right)$ 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 $y\left(t\right)⟺Y\left(\omega \right)$ then $y\left(at\right)⟺\frac{1}{a}Y\left(\frac{\omega }{a}\right)$. Therefore $x\left(3t\right)⟺\frac{1}{3}X\left(\frac{\omega }{3}\right)$ and $h\left(3t\right)⟺\frac{1}{3}H\left(\frac{\omega }{3}\right)$. Hence since $g\left(t\right)=x\left(3t\right)⊛h\left(3t\right)$ then $$\begin{array}{rl}G\left(\omega \right)& =\frac{1}{3}X\left(\frac{\omega }{3}\right)\frac{1}{3}H\left(\frac{\omega }{3}\right)\\ & =\frac{1}{3}\left(\frac{1}{3}X\left(\frac{\omega }{3}\right)H\left(\frac{\omega }{3}\right)\right)\end{array}$$

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 $\frac{1}{3}Y\left(\frac{\omega }{3}\right)⟺y\left(3t\right)$. Hence $A=\frac{1}{3},B=3$

4.4.5 Problem 4.19, Chapter 4

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$$y\left(t\right)=x\left(t\right)⊛h\left(t\right)$$ 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 $Y\left(\omega \right)=\frac{1}{3+j\omega }-\frac{1}{4+j\omega }=\frac{(4+j\omega )-(3+j\omega )}{(3+j\omega )(4+j\omega )}=\frac{1}{(3+j\omega )(4+j\omega )}$ and the above becomes$$X\left(\omega \right)=\frac{\frac{1}{(3+j\omega )(4+j\omega )}}{H\left(\omega \right)}$$ But we are given that $H\left(j\omega \right)=\frac{1}{j\omega +3}$. The above simplifies to$$\begin{array}{rl}X\left(\omega \right)& =\frac{\frac{1}{(3+j\omega )(4+j\omega )}}{\frac{1}{j\omega +3}}\\ & =\frac{1}{4+j\omega }\end{array}$$

From tables$$x\left(t\right)=e^{-4t}u\left(t\right)$$

4.4.6 Problem 4.23, Chapter 4

Figure 4.57:Problem description

Solution

4.4.6.1 Part a

First we find the Fourier transform of $x_0\left(t\right)$. Since this is a periodic, then $x_0\left(t\right)⟺X_0\left(\omega \right)$ and $$\begin{array}{rl}{X}_0\left(\omega \right)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}_0\left(t\right)e^{-j\omega t}dt\\ & ={\int }_0^1{e}^{-t}{e}^{-j\omega t}dt\\ & ={\int }_0^1{e}^{-t(1+j\omega )}dt\\ & =\frac{-1}{1+j\omega }{\left[e^{-t(1+j\omega )}\right]}_0^1\\ & =\frac{-1}{1+j\omega }(e^{-(1+j\omega )}-1)\\ & =\frac{1-e^{-(1+j\omega )}}{1+j\omega }\end{array}$$

From table 4.1, property 4.3.5, Fourier transform of $x(-t)=X(-\omega )$. Hence $x_0(-t)⟺X_0(-\omega )$. Therefore, using the above result and taking its complex conjugate gives$$X_0(-\omega )=\frac{1-e^{-(1-j\omega )}}{1-j\omega }$$ Therefore the Fourier transform of $x_0\left(t\right)+x_0(-t)⟺X_1\left(\omega \right)=X_0\left(\omega \right)+X_0(-\omega )$ This is by linearity property. Hence$$\begin{array}{rl}{X}_1\left(\omega \right)& =X_0\left(\omega \right)+X_0(-\omega )\\ & =\frac{1-e^{-(1+j\omega )}}{1+j\omega }+\frac{1-e^{-(1-j\omega )}}{1-j\omega }\\ & =\frac{(1-j\omega )(1-e^{-(1+j\omega )})+(1+j\omega )(1-e^{-(1-j\omega )})}{(1+j\omega )(1-j\omega )}\\ & =\frac{1-e^{-(1+j\omega )}-j\omega (1-e^{-(1+j\omega )})+(1-e^{-(1-j\omega )})+j\omega (1-e^{-(1-j\omega )})}{1+{\omega }^2}\\ & =\frac{1-e^{-(1+j\omega )}-j\omega +j\omega e^{-(1+j\omega )}+(1-e^{-(1-j\omega )})+j\omega -j\omega e^{-(1-j\omega )}}{1+{\omega }^2}\\ & =\frac{1-e^{-(1+j\omega )}+j\omega e^{-(1+j\omega )}+1-e^{-(1-j\omega )}-j\omega e^{-(1-j\omega )}}{1+{\omega }^2}\\ & =\frac{1-e^{-1}{e}^{-j\omega }+j\omega e^{-1}{e}^{-j\omega }+1-e^{-1}{e}^{j\omega }-j\omega e^{-1}{e}^{j\omega }}{1+{\omega }^2}\\ & =\frac{1-e^{-1}(e^{j\omega }+e^{-j\omega })-j\omega e^{-1}(e^{j\omega }-e^{-j\omega })}{1+{\omega }^2}\\ & =\frac{1}{1+{\omega }^2}-\frac{e^{-1}}{1+{\omega }^2}(e^{j\omega }+e^{-j\omega })+\frac{\omega e^{-1}}{1+{\omega }^2}\frac{(e^{j\omega }-e^{-j\omega })}{j}\\ & =\frac{1}{1+{\omega }^2}-\frac{2}{e(1+{\omega }^2)}\cos \omega +\frac{2\omega }{e(1+{\omega }^2)}\sin \omega \end{array}$$

The following is a plot of the above

Figure 4.58:Plot of $X(\omega )$

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.

4.4.6.2 Part b

We found $X_0\left(\omega \right)=\frac{1-e^{-(1+j\omega )}}{1+j\omega }$ above. Hence $-x_0(-t)⟺-X_0(-\omega )=-\frac{1-e^{-(1-j\omega )}}{1-j\omega }=\frac{1-e^{-(1-j\omega )}}{j\omega -1}$. Therefore$$\begin{array}{rl}{X}_2\left(\omega \right)& =X_0\left(\omega \right)-X_0(-\omega )\\ & =\frac{1-e^{-(1+j\omega )}}{1+j\omega }+\frac{1-e^{-(1-j\omega )}}{j\omega -1}\\ & =\frac{(j\omega -1)(1-e^{-(1+j\omega )})+(1+j\omega )(1-e^{-(1-j\omega )})}{(1+j\omega )(j\omega -1)}\\ & =\frac{j\omega (1-e^{-(1+j\omega )})-(1-e^{-(1+j\omega )})+(1-e^{-(1-j\omega )})+j\omega (1-e^{-(1-j\omega )})}{j\omega -1-{\omega }^2-j\omega }\\ & =\frac{j\omega -j\omega e^{-(1+j\omega )}-1+e^{-(1+j\omega )}+1-e^{-(1-j\omega )}+j\omega -j\omega e^{-(1-j\omega )}}{-(1+{\omega }^2)}\\ & =\frac{2j\omega -j\omega e^{-(1+j\omega )}+e^{-(1+j\omega )}-e^{-(1-j\omega )}-j\omega e^{-(1-j\omega )}}{-(1+{\omega }^2)}\\ & =\frac{2j\omega -j\omega e^{-1}{e}^{-j\omega }+e^{-1}{e}^{-j\omega }-e^{-1}{e}^{j\omega }-j\omega e^{-1}{e}^{j\omega }}{j\omega -{\omega }^2-2}\\ & =\frac{2j\omega -j\omega e^{-1}(e^{j\omega }+e^{-j\omega })-e^{-1}(e^{j\omega }-e^{-j\omega })}{-(1+{\omega }^2)}\\ & =\frac{2j\omega }{-(1+{\omega }^2)}-j\omega e^{-1}\frac{(e^{j\omega }+e^{-j\omega })}{-(1+{\omega }^2)}-e^{-1}\frac{e^{j\omega }-e^{-j\omega }}{-(1+{\omega }^2)}\\ & =\frac{2j\omega }{-(1+{\omega }^2)}-2j\omega e^{-1}\frac{\cos \omega }{-(1+{\omega }^2)}-2j{e}^{-1}\frac{\sin \left(\omega \right)}{-(1+{\omega }^2)}\end{array}$$

Hence$$\begin{array}{rl}{X}_2\left(\omega \right)& =j(\frac{-2\omega }{(1+{\omega }^2)}+2\omega \frac{\cos \omega }{e(1+{\omega }^2)}+2\frac{\sin \left(\omega \right)}{e(1+{\omega }^2)})\\ & =\frac{j}{e(1+{\omega }^2)}(-2e\omega +2\omega \cos \omega +2\sin \left(\omega \right))\end{array}$$

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

Figure 4.59:Plot of imaginary part of $X(\omega )$

4.4.7 Problem 4.26, Chapter 4

Figure 4.60:Problem description

Solution

4.4.7.1 Part a

Part i $$y\left(t\right)=x\left(t\right)⊛h\left(t\right)$$ Taking Fourier transform the above, convolution becomes multiplication$$Y\left(\omega \right)=X\left(\omega \right)H\left(\omega \right)$$ Given that $x\left(t\right)=t{e}^{-2t}u\left(t\right)$, then from tables $X\left(\omega \right)=\frac{1}{{(2+j\omega )}^2}$ 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$$Y\left(\omega \right)=\frac{1}{{(2+j\omega )}^2(4+j\omega )}$$ Doing partial fractions. Let $s=j\omega$ then $$\begin{array}{rl}\frac{1}{{(2+s)}^2(4+s)}& =\frac{A}{{(2+s)}^2}+\frac{B}{2+s}+\frac{C}{4+s}\\ & =\frac{A(4+s)+B\left((2+s)(4+s)\right)+C{(2+s)}^2}{{(2+s)}^2(4+s)}\end{array}$$

Expanding numerator$$\begin{array}{rl}1& =4A+8B+4C+B{s}^2+C{s}^2+As+6Bs+4Cs\\ 1& =(4A+8B+4C)+s(A+6B+4C)+s^2(B+C)\end{array}$$

Comparing coefficients$$\begin{array}{rl}1& =4A+8B+4C\\ 0& =A+6B+4C\\ 0& =B+C\end{array}$$

Or$$\left(\begin{array}{ccc}4& 8& 4\\ 1& 6& 4\\ 0& 1& 1\end{array}\right)\left(\begin{array}{c}A\\ B\\ C\end{array}\right)=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$$ Gaussian elimination. Multiplying second row by 4 and subtracting result from first row gives$$\left(\begin{array}{ccc}4& 8& 4\\ 0& 16& 12\\ 0& 1& 1\end{array}\right)\left(\begin{array}{c}A\\ B\\ C\end{array}\right)=\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)$$ Multiplying third row by 16 and subtracting result from second row gives$$\left(\begin{array}{ccc}4& 8& 4\\ 0& 16& 12\\ 0& 0& 4\end{array}\right)\left(\begin{array}{c}A\\ B\\ C\end{array}\right)=\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)$$ 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 $4A+8(-\frac{1}{4})+4\left(\frac{1}{4}\right)=1$ or $4A-2+1=1$. Hence $A=\frac{1}{2}$. Therefore partial fractions gives$$\frac{1}{{(2+s)}^2(4+s)}=\frac{\left(\frac{1}{2}\right)}{{(2+s)}^2}+\frac{(-\frac{1}{4})}{2+s}+\frac{\left(\frac{1}{4}\right)}{4+s}$$ Replacing $s$ back with $j\omega$$$Y\left(\omega \right)=\frac{1}{2}\frac{1}{{(2+j\omega )}^2}-\frac{1}{4}\frac{1}{2+j\omega }+\frac{1}{4}\frac{1}{4+j\omega }$$ Applying inverse Fourier transform, using table gives$$y\left(t\right)=\frac{1}{2}t{e}^{-2t}u\left(t\right)-\frac{1}{4}{e}^{-2t}u\left(t\right)+\frac{1}{4}{e}^{-4t}u\left(t\right)$$

Part ii $$y\left(t\right)=x\left(t\right)⊛h\left(t\right)$$ Taking Fourier transform the above, convolution becomes multiplication$$Y\left(\omega \right)=X\left(\omega \right)H\left(\omega \right)$$ Given that $x\left(t\right)=t{e}^{-2t}u\left(t\right)$, then from tables $X\left(\omega \right)=\frac{1}{{(2+j\omega )}^2}$ and given that $h\left(t\right)=t{e}^{-4t}u\left(t\right)$ then from table 4.2 $H\left(\omega \right)=\frac{1}{{(4+j\omega )}^2}$. Hence the above becomes$$Y\left(\omega \right)=\frac{1}{{(2+j\omega )}^2{(4+j\omega )}^2}$$ Doing partial fractions. Let $s=j\omega$ then $$\begin{array}{rl}\frac{1}{{(2+s)}^2{(4+s)}^2}& =\frac{A}{{(2+s)}^2}+\frac{B}{2+s}+\frac{C}{{(4+s)}^2}+\frac{D}{(4+s)}\\ & =\frac{A{(4+s)}^2+B\left((2+s){(4+s)}^2\right)+C{(2+s)}^2+D\left({(2+s)}^2(4+s)\right)}{{(2+s)}^2{(4+s)}^2}\end{array}$$

Expanding numerator$$\begin{array}{rl}1& =16A+32B+4C+16D+20sD+A{s}^2+10B{s}^2+B{s}^3+C{s}^2+8s^{2}D+s^{3}D+8As+32Bs+4Cs\\ 1& =(16A+32B+4C+16D)+s(20D+8A+32B+4C)+s^2(A+10B+C+8D)+s^3(B+D)\end{array}$$

Comparing coefficients$$\begin{array}{rl}1& =16A+32B+4C+16D\\ 0& =8A+32B+4C+20D\\ 0& =A+10B+C+8D\\ 0& =B+D\end{array}$$

Or$$\left(\begin{array}{cccc}16& 32& 4& 16\\ 8& 32& 4& 20\\ 1& 10& 1& 8\\ 0& 1& 0& 1\end{array}\right)\left(\begin{array}{c}A\\ B\\ C\\ D\end{array}\right)=\left(\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right)$$ Gaussian elimination. replacing row 2 by result of subtracting $1/2$ times row 1 from row 2$$\left(\begin{array}{cccc}16& 32& 4& 16\\ 0& 16& 2& 12\\ 1& 10& 1& 8\\ 0& 1& 0& 1\end{array}\right)\left(\begin{array}{c}A\\ B\\ C\\ D\end{array}\right)=\left(\begin{array}{c}1\\ -\frac{1}{2}\\ 0\\ 0\end{array}\right)$$ Replacing row 3 by result of subtracting $\frac{1}{16}$ times row 1 from row 3$$\left(\begin{array}{cccc}16& 32& 4& 16\\ 0& 16& 2& 12\\ 0& 8& \frac{3}{4}& 7\\ 0& 1& 0& 1\end{array}\right)\left(\begin{array}{c}A\\ B\\ C\\ D\end{array}\right)=\left(\begin{array}{c}1\\ -\frac{1}{2}\\ -\frac{1}{16}\\ 0\end{array}\right)$$ Replacing row 3 by result of subtracting $\frac{1}{2}$ times row 2 from row 3$$\left(\begin{array}{cccc}16& 32& 4& 16\\ 0& 16& 2& 12\\ 0& 0& \frac{-1}{4}& 1\\ 0& 1& 0& 1\end{array}\right)\left(\begin{array}{c}A\\ B\\ C\\ D\end{array}\right)=\left(\begin{array}{c}1\\ -\frac{1}{2}\\ \frac{3}{16}\\ 0\end{array}\right)$$ Replacing row 4 by result of subtracting $\frac{1}{16}$ times row 2 from row 4$$\left(\begin{array}{cccc}16& 32& 4& 16\\ 0& 16& 2& 12\\ 0& 0& \frac{-1}{4}& 1\\ 0& 0& \frac{-1}{8}& \frac{1}{4}\end{array}\right)\left(\begin{array}{c}A\\ B\\ C\\ D\end{array}\right)=\left(\begin{array}{c}1\\ -\frac{1}{2}\\ \frac{3}{16}\\ \frac{1}{32}\end{array}\right)$$ Replacing row 4 by result of subtracting $\frac{1}{2}$ times row 3 from row 4$$\left(\begin{array}{cccc}16& 32& 4& 16\\ 0& 16& 2& 12\\ 0& 0& \frac{-1}{4}& 1\\ 0& 0& 0& \frac{-1}{4}\end{array}\right)\left(\begin{array}{c}A\\ B\\ C\\ D\end{array}\right)=\left(\begin{array}{c}1\\ -\frac{1}{2}\\ \frac{3}{16}\\ \frac{-1}{16}\end{array}\right)$$ Backsubstitution phase:$$\begin{array}{rl}\frac{-1}{4}D& =\frac{-1}{16}\\ D& =\frac{1}{4}\end{array}$$

Third row gives $\frac{-1}{4}C+D=\frac{3}{16}$ or $\frac{-1}{4}C+\frac{1}{4}=\frac{3}{16}\to C=\frac{1}{4}$. Second row gives $16B+2C+12D=-\frac{1}{2}$ or $16B+2\left(\frac{1}{4}\right)+12\left(\frac{1}{4}\right)=-\frac{1}{2}\to B=-\frac{1}{4}$. First row gives $16A+32B+4C+16D=1$ or $16A+32(-\frac{1}{4})+4\left(\frac{1}{4}\right)+16\left(\frac{1}{4}\right)=1,\to A=\frac{1}{4}$.

Therefore partial fractions gives$$\begin{array}{rl}\frac{1}{{(2+s)}^2{(4+s)}^2}& =\frac{A}{{(2+s)}^2}+\frac{B}{2+s}+\frac{C}{{(4+s)}^2}+\frac{D}{(4+s)}\\ & =\frac{1}{4}\frac{1}{{(2+s)}^2}-\frac{1}{4}\frac{1}{2+s}+\frac{1}{4}\frac{1}{{(4+s)}^2}+\frac{1}{4}\frac{1}{(4+s)}\end{array}$$

Replace $s$ back with $j\omega$$$Y\left(\omega \right)=\frac{1}{4}\frac{1}{{(2+j\omega )}^2}-\frac{1}{4}\frac{1}{2+j\omega }+\frac{1}{4}\frac{1}{{(4+j\omega )}^2}+\frac{1}{4}\frac{1}{(4+j\omega )}$$ Applying inverse Fourier transform, using table gives$$y\left(t\right)=\frac{1}{4}t{e}^{-2t}u\left(t\right)-\frac{1}{4}{e}^{-2t}u\left(t\right)+\frac{1}{4}t{e}^{-4t}u\left(t\right)+\frac{1}{4}{e}^{-4t}u\left(t\right)$$

Part iii $$y\left(t\right)=x\left(t\right)⊛h\left(t\right)$$ 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 $X\left(\omega \right)=\frac{1}{(1+j\omega )}$ and given that $h\left(t\right)=e^{t}u(-t)$ then $H\left(\omega \right)=\frac{1}{1-j\omega }$. Hence the above becomes$$Y\left(\omega \right)=\frac{1}{(1+j\omega )}\frac{1}{(1-j\omega )}$$ Doing partial fractions. Let $s=j\omega$ then $$\frac{1}{(1+s)}\frac{1}{(1-s)}=\frac{A}{(1+s)}+\frac{B}{1-s}$$ Hence $A={\left(\frac{1}{(1-s)}\right)}_{s=-1}=\frac{1}{2}$ and $B={\left(\frac{1}{(1+s)}\right)}_{s=1}=\frac{1}{2}$. Hence $$Y\left(\omega \right)=\frac{1}{2}\frac{1}{(1+j\omega )}+\frac{1}{2}\frac{1}{1-j\omega }$$ Therefore$$y\left(t\right)=\frac{1}{2}{e}^{-t}u\left(t\right)+\frac{1}{2}{e}^{t}u(-t)$$ The above means for $t<0,y\left(t\right)=\frac{1}{2}{e}^t$ and for $t>0,y\left(t\right)=\frac{1}{2}{e}^{-t}$.

4.4.7.2 Part b

$x\left(t\right)=e^{-(t-2)}u(t-2)$, $h\left(t\right)=u(t+1)-u(t-3)$. Let us first find $X\left(\omega \right)$ and $H\left(\omega \right)$$$\begin{array}{rl}X\left(\omega \right)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-(t-2)}u(t-2)e^{-j\omega t}dt\\ & ={\int }_2^{\mathrm{\infty }}{e}^{-(t-2)}{e}^{-j\omega t}dt\\ & ={\int }_2^{\mathrm{\infty }}{e}^{-t}{e}^2{e}^{-j\omega t}dt\\ & =e^2{\int }_2^{\mathrm{\infty }}{e}^{-t(1+j\omega )}dt\\ & =-\frac{e^2}{1+j\omega }{\left[e^{-t(1+j\omega )}\right]}_2^{\mathrm{\infty }}\\ & =-\frac{e^2}{1+j\omega }(0-e^{-2(1+j\omega )})\\ & =\frac{e^2{e}^{-2(1+j\omega )}}{1+j\omega }\\ & =\frac{e^{-2-2j\omega +2}}{1+j\omega }\\ & =\frac{e^{-2j\omega }}{1+j\omega }\end{array}$$

And$$\begin{array}{rl}H\left(\omega \right)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}u(t+1)-u(t-3)e^{-j\omega t}dt\\ & ={\int }_{-1}^3{e}^{-j\omega t}dt\\ & =\frac{1}{j\omega }{\left(e^{-j\omega t}\right)}_{-1}^3\\ & =\frac{1}{j\omega }(e^{-3j\omega }-e^{j\omega t})\\ & =\frac{1}{j\omega }{e}^{-j\omega }(e^{-j2\omega }-e^{j2\omega t})\\ & =-\frac{1}{j\omega }{e}^{-j\omega }(e^{j2\omega t}-e^{-j2\omega })\\ & =-\frac{1}{\omega }{e}^{-j\omega }\left(\frac{e^{j2\omega t}-e^{-j2\omega }}{j}\right)\\ & =-\frac{2}{\omega }{e}^{-j\omega }\sin \left(2\omega \right)\end{array}$$

Hence $$\begin{array}{rl}Y\left(\omega \right)& =H\left(\omega \right)X\left(\omega \right)\\ & =\frac{e^{-2j\omega }}{1+j\omega }(-\frac{2}{\omega }{e}^{-j\omega }\sin \left(2\omega \right))\\ & =\frac{-2}{\omega (1+j\omega )}{e}^{-2j\omega }{e}^{-j\omega }\sin \left(2\omega \right)\\ \text{(1)}& & =\frac{-2}{\omega (1+j\omega )}{e}^{-3j\omega }\sin \left(2\omega \right)\end{array}$$

Now, $$\begin{array}{rl}y\left(t\right)& =x\left(t\right)⊛h\left(t\right)\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\left(\tau \right)h(t-\tau )d\tau \end{array}$$

Folding $h\left(t\right)$. For $t<1$, then $y\left(t\right)=0$. For $2<1+t<6$ or $1<t<5$$$\begin{array}{rl}y\left(t\right)& ={\int }_2^{1+t}x\left(\tau \right)d\tau \\ & ={\int }_2^{1+t}{e}^{-(\tau -2)}d\tau \\ & =-{\left(e^{-(\tau -2)}\right)}_2^{1+t}\\ & =-(e^{-(1+t-2)}-e^{-(2-2)})\\ & =-(e^{-(-1+t)}-1)\\ & =1-e^{1-t}\end{array}$$

For $6<1+t$ or $t>5$$$\begin{array}{rl}y\left(t\right)& ={\int }_{t-3}^{t+1}x\left(\tau \right)d\tau \\ & ={\int }_{t-3}^{t+1}{e}^{-(\tau -2)}d\tau \\ & =-{\left(e^{-(\tau -2)}\right)}_{t-3}^{1+t}\\ & =-(e^{-(1+t-2)}-e^{-(t-3-2)})\\ & =-(e^{-(-1+t)}-e^{-(t-5)})\\ & =-(e^{1-t}-e^{5-t})\\ & =e^{5-t}-e^{1-t}\end{array}$$

Hence$$y\left(t\right)=\{\begin{array}{ccc}0& & t<1\\ 1-e^{1-t}& & 1<t<5\\ e^{5-t}-e^{1-t}& & t>5\end{array}$$ The above can be written as $y\left(t\right)$$$\begin{array}{rl}y\left(t\right)& =(1-e^{1-t})(u(t-1)-u(t-5))+(e^{5-t}-e^{1-t})u(t-5)\\ & =u(t-1)-u(t-5)-e^{1-t}(u(t-1)-u(t-5))+(e^{5-t}-e^{1-t})u(t-5)\\ & =u(t-1)-u(t-5)-e^{1-t}u(t-1)+e^{1-t}u(t-5)+e^{5-t}u(t-5)-e^{1-t}u(t-5)\\ \text{(2)}& & =u(t-1)-u(t-5)-e^{1-t}u(t-1)+e^{5-t}u(t-5)\end{array}$$

Taking the Fourier transform of the above from tables $$\begin{array}{rl}u(t-1)& ⟺\frac{1}{j\omega }{e}^{-j\omega }+\pi \delta \left(\omega \right)\\ u(t-5)& ⟺\frac{1}{j\omega }{e}^{-5j\omega }+\pi \delta \left(\omega \right)\\ e^{1-t}u(t-1)& ⟺\frac{e^{-j\omega }}{1+j\omega }\\ e^{5-t}u(t-5)& ⟺\frac{e^{-5j\omega }}{1+j\omega }\end{array}$$

Hence $Y\left(\omega \right)$ is$$\begin{array}{rl}Y\left(\omega \right)& =\frac{1}{j\omega }{e}^{-j\omega }+\pi \delta \left(\omega \right)-(\frac{1}{j\omega }{e}^{-5j\omega }+\pi \delta \left(\omega \right))-\frac{e^{-j\omega }}{1+j\omega }+\frac{e^{-5j\omega }}{1+j\omega }\\ & =\frac{1}{j\omega }{e}^{-j\omega }-\frac{1}{j\omega }{e}^{-5j\omega }-\frac{e^{-j\omega }}{1+j\omega }+\frac{e^{-5j\omega }}{1+j\omega }\\ & =\frac{e^{-3j\omega }}{\omega }(\frac{1}{j}{e}^{2j\omega }-\frac{1}{j}{e}^{-2j\omega })-\frac{e^{-j\omega }}{1+j\omega }+\frac{e^{-5j\omega }}{1+j\omega }\\ & =\frac{e^{-3j\omega }}{\omega }2(\sin 2\omega )-\frac{e^{-3j\omega }}{1+j\omega }(e^{2j\omega }-e^{-2j\omega })\\ & =\frac{e^{-3j\omega }}{\omega }2(\sin 2\omega )-2\frac{e^{-3j\omega }}{j+\omega }\sin 2\omega \\ & =2e^{-3j\omega }\sin 2\omega (\frac{1}{\omega }-\frac{1}{j+\omega })\\ & =2e^{-3j\omega }\sin 2\omega \left(\frac{j+\omega -\omega }{\omega (\omega +j)}\right)\\ & =2e^{-3j\omega }\sin 2\omega \frac{j}{\omega (\omega +j)}\\ & =-2e^{-3j\omega }\sin 2\omega \frac{1}{\omega (1+j\omega )}\end{array}$$

Comparing the above to (1) shows they are the same.

Hence this shows that Fourier transform of $x\left(t\right)⊛h\left(t\right)$ gives same answer as $H\left(\omega \right)X\left(\omega \right)$

4.4.8 key solution

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