4.5 HW 5

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4.5.1 Problem 5.3, Chapter 5

Determine the Fourier transform for $-\pi \le \omega <\pi$ in the case of each of the following periodic signals (a) $\sin (\frac{\pi }{3}n+\frac{\pi }{4})$ (b) $2+\cos (\frac{\pi }{6}n+\frac{\pi }{8})$

solution

4.5.1.1 Part a

Since the signal is periodic, then the Fourier transform is given by $$\begin{array}{}\text{(1)}& X\left(\mathrm{\Omega }\right)=2\pi \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k\delta (\mathrm{\Omega }-k{\mathrm{\Omega }}_0)\end{array}$$ Where $a_k$ are the Fourier series coefficients of $x\left[n\right]$. To determine $a_k$ we can expression $x\left[n\right]$ using Euler relation. To find the period, $\frac{\pi }{3}N=m2\pi$. Hence $\frac{m}{N}=\frac{1}{6}$. Hence $$N=6$$ Therefore ${\mathrm{\Omega }}_0=\frac{2\pi }{N}=\frac{\pi }{3}$. Now, using Euler relation$$\begin{array}{rl}\sin (\frac{\pi }{3}n+\frac{\pi }{4})& =\frac{e^{(\frac{\pi }{3}n+\frac{\pi }{4})j}-e^{-(\frac{\pi }{3}n+\frac{\pi }{4})j}}{2j}\\ \text{(2)}& & =\frac{1}{2j}{e}^{j\frac{\pi }{4}}\left(e^{j\frac{\pi }{3}n}\right)-\frac{1}{2j}{e}^{-j\frac{\pi }{4}}\left(e^{-j\frac{\pi }{3}n}\right)\end{array}$$

Comparing (2) to Fourier series expansion of periodic signal given by $$\begin{array}{rl}x\left[n\right]& =\sum _{k=0}^{N-1}{a}_k{e}^{jk{\mathrm{\Omega }}_{0}n}\\ & =\sum _{k=0}^5{a}_k{e}^{jk{\mathrm{\Omega }}_{0}n}\\ & =\sum _{k=-2}^3{a}_k{e}^{jk{\mathrm{\Omega }}_{0}n}\end{array}$$

Since ${\mathrm{\Omega }}_0=\frac{\pi }{3}$ then the above becomes$$x\left[n\right]=\sum _{k=-2}^3{a}_k{e}^{jk\frac{\pi }{3}n}$$ Comparing the above with (2) shows that $a_1=\frac{1}{2j}{e}^{j\frac{\pi }{4}}$ and $a_{-1}=-\frac{1}{2j}{e}^{-j\frac{\pi }{4}}$ and all other $a_k=0$ for $k=-2,0,2,3$. Hence (1) becomes$$\begin{array}{rl}X\left(\mathrm{\Omega }\right)& =2\pi (a_{-1}\delta (\mathrm{\Omega }+{\mathrm{\Omega }}_0)+a_1\delta (\mathrm{\Omega }-{\mathrm{\Omega }}_0))\\ & =2\pi (-\frac{1}{2j}{e}^{-j\frac{\pi }{4}}\delta (\mathrm{\Omega }+\frac{\pi }{3})+\frac{1}{2j}{e}^{j\frac{\pi }{4}}\delta (\mathrm{\Omega }-\frac{\pi }{3}))\\ & =\frac{\pi }{j}(-e^{-j\frac{\pi }{4}}\delta (\mathrm{\Omega }+\frac{\pi }{3})+e^{j\frac{\pi }{4}}\delta (\mathrm{\Omega }-\frac{\pi }{3}))\end{array}$$

4.5.1.2 Part b

Since the signal $2+\cos (\frac{\pi }{6}n+\frac{\pi }{8})$ is periodic, then the Fourier transform is given by $$\begin{array}{}\text{(1)}& X\left(\mathrm{\Omega }\right)=2\pi \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k\delta (\mathrm{\Omega }-k{\mathrm{\Omega }}_0)\end{array}$$ Where $a_k$ are the Fourier series coefficients of $x\left[n\right]$. To determine $a_k$ we can expression $x\left[n\right]$ using Euler relation. To find the period, $\frac{\pi }{6}N=m2\pi$. Hence $\frac{m}{N}=\frac{1}{12}$. Hence $$N=12$$ Therefore ${\mathrm{\Omega }}_0=\frac{2\pi }{N}=\frac{\pi }{6}$. Now, using Euler relation $$\begin{array}{rl}2+\cos (\frac{\pi }{6}n+\frac{\pi }{8})& =2+\frac{e^{(\frac{\pi }{6}n+\frac{\pi }{8})j}+e^{-(\frac{\pi }{6}n+\frac{\pi }{8})j}}{2}\\ \text{(2)}& & =2+\frac{1}{2}{e}^{j\frac{\pi }{8}}\left(e^{j\frac{\pi }{6}n}\right)+\frac{1}{2}{e}^{-j\frac{\pi }{8}}\left(e^{-j\frac{\pi }{6}n}\right)\end{array}$$

Comparing (2) to Fourier series expansion of periodic signal given by $$\begin{array}{rl}x\left[n\right]& =\sum _{k=0}^{N-1}{a}_k{e}^{jk{\mathrm{\Omega }}_{0}n}\\ & =\sum _{k=0}^{11}{a}_k{e}^{jk{\mathrm{\Omega }}_{0}n}\\ & =\sum _{k=-5}^6{a}_k{e}^{jk{\mathrm{\Omega }}_{0}n}\end{array}$$

Since ${\mathrm{\Omega }}_0=\frac{\pi }{6}$ then the above becomes$$x\left[n\right]=\sum _{k=-5}^6{a}_k{e}^{jk\frac{\pi }{6}n}$$ Comparing the above with (2) shows that $a_0=2,a_1=\frac{1}{2}{e}^{j\frac{\pi }{8}}$ and $a_{-1}=\frac{1}{2}{e}^{-j\frac{\pi }{8}}$ and all other $a_k=0$. Hence (1) becomes$$\begin{array}{rl}X\left(\mathrm{\Omega }\right)& =2\pi (a_0\delta \left(\mathrm{\Omega }\right)+a_{-1}\delta (\mathrm{\Omega }+{\mathrm{\Omega }}_0)+a_1\delta (\mathrm{\Omega }-{\mathrm{\Omega }}_0))\\ & =2\pi (2\delta \left(\mathrm{\Omega }\right)+\frac{1}{2}{e}^{-j\frac{\pi }{8}}\delta (\mathrm{\Omega }+\frac{\pi }{6})+\frac{1}{2}{e}^{j\frac{\pi }{8}}\delta (\mathrm{\Omega }-\frac{\pi }{6}))\\ & =4\pi \delta \left(\mathrm{\Omega }\right)+\pi e^{-j\frac{\pi }{8}}\delta (\mathrm{\Omega }+\frac{\pi }{6})+\pi e^{j\frac{\pi }{8}}\delta (\mathrm{\Omega }-\frac{\pi }{6})\end{array}$$

4.5.2 Problem 5.5, Chapter 5

Use the Fourier transform synthesis equation (5.8) $$\begin{array}{}\text{(5.8)}& x\left[n\right]& =\frac{1}{2\pi }{\int }_{2\pi }X\left(\mathrm{\Omega }\right)e^{j\mathrm{\Omega }n}d\mathrm{\Omega }\text{(5.9)}& X\left(\mathrm{\Omega }\right)& =\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left[n\right]e^{-j\mathrm{\Omega }n}\end{array}$$

To determine the inverse Fourier transform of $X\left(\mathrm{\Omega }\right)=\left|X\left(\mathrm{\Omega }\right)\right|e^{j\arg H\left(\mathrm{\Omega }\right)}$, where $\left|X\left(\mathrm{\Omega }\right)\right|=\{\begin{array}{ccc}1& & 0\le \left|\mathrm{\Omega }\right|<\frac{\pi }{4}\\ 0& & \frac{\pi }{4}\le \left|\mathrm{\Omega }\right|<\pi \end{array}$ and $\arg H\left(\mathrm{\Omega }\right)=\frac{-3\mathrm{\Omega }}{2}$. Use your answer to determine the values of n for which $x\left[n\right]=0$.

solution$$\begin{array}{rl}x\left[n\right]& =\frac{1}{2\pi }{\int }_{2\pi }X\left(\mathrm{\Omega }\right)e^{j\mathrm{\Omega }n}d\mathrm{\Omega }\\ & =\frac{1}{2\pi }{\int }_{2\pi }\left|X\left(\mathrm{\Omega }\right)\right|e^{j\arg H\left(\mathrm{\Omega }\right)}{e}^{j\mathrm{\Omega }n}d\mathrm{\Omega }\\ & =\frac{1}{2\pi }{\int }_0^{\frac{\pi }{4}}{e}^{j\arg H\left(\mathrm{\Omega }\right)}{e}^{j\mathrm{\Omega }n}d\mathrm{\Omega }\\ & =\frac{1}{2\pi }{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{e}^{j\frac{-3\mathrm{\Omega }}{2}}{e}^{j\mathrm{\Omega }n}d\mathrm{\Omega }\\ & =\frac{1}{2\pi }{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{e}^{j\mathrm{\Omega }(\frac{-3}{2}+n)}d\mathrm{\Omega }\\ & =\frac{1}{2\pi }\frac{1}{j(\frac{-3}{2}+n)}{\left[e^{j\mathrm{\Omega }(\frac{-3}{2}+n)}\right]}_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\\ & =\frac{1}{2\pi }\frac{1}{j(\frac{-3}{2}+n)}[e^{j\frac{\pi }{4}(\frac{-3}{2}+n)}-e^{-j\frac{\pi }{4}(\frac{-3}{2}+n)}]\\ & =\frac{1}{\pi }\frac{1}{(\frac{-3}{2}+n)}\left[\frac{e^{j\frac{\pi }{4}(\frac{-3}{2}+n)}-e^{-j\frac{\pi }{4}(\frac{-3}{2}+n)}}{2j}\right]\\ & =\frac{1}{\pi }\frac{1}{(\frac{-3}{2}+n)}\sin \left(\frac{\pi }{4}(\frac{-3}{2}+n)\right)\\ & =\frac{1}{\pi }\frac{\sin \left(\frac{\pi }{4}(n-\frac{3}{2})\right)}{n-\frac{3}{2}}\end{array}$$

Now the above is zero when $\sin \left(\frac{\pi }{4}(n-\frac{3}{2})\right)=0$ or $\frac{\pi }{4}(n-\frac{3}{2})=m\pi$ for integer $m$. Hence $n-\frac{3}{2}=4m$. Or $n=4m+\frac{3}{2}$. Since $m$ is integer, and since $n$ must be an integer as well, then there is no finite $n$ where $\sin \left(\frac{\pi }{4}(n-\frac{3}{2})\right)=0$.  The other option is to look at denominator of $\frac{\sin \left(\frac{\pi }{4}(n-\frac{3}{2})\right)}{n-\frac{3}{2}}$ and ask where is that $\mathrm{\infty }$. This happens when $n\to \pm \mathrm{\infty }$ and only then $x\left[n\right]=0$.

4.5.3 Problem 5.9, Chapter 5

The following four facts are given about a real signal $x\left[n\right]$ with Fourier transform $X\left(\mathrm{\Omega }\right)$

1. $x\left[n\right]=0$ for $n>0$
2. $x\left[0\right]>0$
3. $\mathrm{Im}\left(X\left(\mathrm{\Omega }\right)\right)=\sin \mathrm{\Omega }-\sin \left(2\mathrm{\Omega }\right)$
4. $\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{\left|X\left(\mathrm{\Omega }\right)\right|}^{2}d\mathrm{\Omega }=3$

Determine $x\left[n\right]$

solution

From tables we know that the odd part of $x\left[n\right]$ has Fourier transform which is $j\mathrm{Im}\left(X\left(\mathrm{\Omega }\right)\right)$. Hence using (3) above, this means that odd part of $x\left[n\right]$ has Fourier transform of $j(\sin \mathrm{\Omega }-\sin \left(2\mathrm{\Omega }\right))$ or $j(\frac{e^{j\mathrm{\Omega }}-e^{-j\mathrm{\Omega }}}{2j}-\frac{e^{j2\mathrm{\Omega }}-e^{-j2\mathrm{\Omega }}}{2j})$ or $\frac{1}{2}(e^{j\mathrm{\Omega }}-e^{-j\mathrm{\Omega }}-e^{j2\mathrm{\Omega }}+e^{-j2\mathrm{\Omega }})$. From tables, we know find the inverse Fourier transform of this. Hence odd part of $x\left[n\right]$ is $\frac{1}{2}(\delta [n+1]-\delta [n-1]-\delta [n+2]+\delta [n-2])$. So now we know what the odd part of $x\left[n\right]$ is.

But since $x\left[n\right]=0$ for $n>0$ then the odd part of $x\left[n\right]$ reduces to $\frac{1}{2}(\delta [n+1]-\delta [n+2])$.

But we also know that any function can be expressed as the sum of its odd part and its even part. But since $x\left[n\right]=0$ for $n>0$ then this means $x\left[n\right]=2\left(\frac{1}{2}(\delta [n+1]-\delta [n+2])\right)$ for $n<0$. Hence$$x\left[n\right]=\delta [n+1]-\delta [n+2]n<0$$ Finally, using (4) above, $$\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{\left|X\left(\mathrm{\Omega }\right)\right|}^{2}d\mathrm{\Omega }=3=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|x\left[n\right]\right|}^2=\sum _{n=-\mathrm{\infty }}^0{\left|x\left[n\right]\right|}^2$$ Hence$$\begin{array}{rl}3& ={\left|\delta [-1]\right|}^2+{\left|\delta [-2]\right|}^2+{\left|x\left[0\right]\right|}^2\\ & =1+1+x{\left[n\right]}^2\\ x{\left[n\right]}^2& =3-2\\ & =1\end{array}$$

Therefore $x\left[n\right]=1$ or $x\left[n\right]=-1$. But from (2)  $x\left[0\right]>0$. Hence $x\left[0\right]$.  Therefore$$x\left[n\right]=\delta [n+1]-\delta [n+2]+\delta \left[n\right]n\le 0$$

4.5.4 Problem 5.13, Chapter 5

An LTI system with impulse response $h_1\left[n\right]={\left(\frac{1}{3}\right)}^{n}u\left[n\right]$ is connected in parallel with another causal LTI system with impulse response $h_2\left[n\right]$.  The resulting parallel

interconnection has the frequency response $$H\left(\mathrm{\Omega }\right)=\frac{-12+5e^{-j\mathrm{\Omega }}}{12-7e^{-j\mathrm{\Omega }}+e^{-j2\mathrm{\Omega }}}$$ Determine $h_2\left[n\right]$.

solution

Since the connection is parallel, then $h\left[n\right]=h_1\left[n\right]+h_2\left[n\right]$. Or $H\left(\mathrm{\Omega }\right)=H_1\left(\mathrm{\Omega }\right)+H_2\left(\mathrm{\Omega }\right)$. Hence $$\begin{array}{}\text{(1)}& H_2\left(\mathrm{\Omega }\right)=H\left(\mathrm{\Omega }\right)-H_1\left(\mathrm{\Omega }\right)\end{array}$$ But $$\begin{array}{rl}{H}_1\left(\mathrm{\Omega }\right)& =\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{h}_1\left[n\right]e^{-j\mathrm{\Omega }n}\\ & =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{3}\right)}^n{e}^{-j\mathrm{\Omega }n}\\ & =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{3}{e}^{-j\mathrm{\Omega }}\right)}^n\\ & =\sum _{n=0}^{\mathrm{\infty }}{a}^n=\frac{1}{1-a}=\frac{1}{1-\frac{1}{3}{e}^{-j\mathrm{\Omega }}}\\ & =\frac{3}{3-e^{-j\mathrm{\Omega }}}\end{array}$$

Therefore from (1)$$H_2\left(\mathrm{\Omega }\right)=\frac{-12+5e^{-j\mathrm{\Omega }}}{12-7e^{-j\mathrm{\Omega }}+e^{-j2\mathrm{\Omega }}}-\frac{3}{3-e^{-j\mathrm{\Omega }}}$$ Let $e^{-j\mathrm{\Omega }}=x$ to simplify notation. The above becomes$$\begin{array}{rl}{H}_2\left(\mathrm{\Omega }\right)& =\frac{-12+5x}{12-7x+x^2}-\frac{3}{3-x}\\ & =\frac{-12+5x}{(x-3)(x-4)}+\frac{3}{(x-3)}\\ & =\frac{-12+5x+3(x-4)}{(x-3)(x-4)}\\ & =\frac{-12+5x+3x-12}{(x-3)(x-4)}\\ & =\frac{8x-24}{(x-3)(x-4)}\\ & =\frac{8(x-3)}{(x-3)(x-4)}\\ & =\frac{8}{(x-4)}\\ & =-2\left(\frac{1}{1-\frac{1}{4}x}\right)\end{array}$$

Hence$$H_2\left(\mathrm{\Omega }\right)=-2\left(\frac{1}{1-\frac{1}{4}{e}^{-j\mathrm{\Omega }}}\right)$$ from tables, $a^{n}u\left[n\right]⟺\frac{1}{1-a{e}^{-j\mathrm{\Omega }}}$ for $\left|a\right|<1$. Comparing this to the above gives $$h_2\left[n\right]=-2{\left(\frac{1}{4}\right)}^{n}u\left[n\right]$$

4.5.5 Problem 5.19, Chapter 5

Consider a causal and stable LTI system $S$ whose input $x\left[n\right]$ and output $y\left[n\right]$ are related through the second-order difference equation

$$y\left[n\right]-\frac{1}{6}y[n-1]-\frac{1}{6}y[n-2]=x\left[n\right]$$

(a) Determine the frequency response $H\left(\mathrm{\Omega }\right)$ for the system $S$. (b) Determine the impulse response $h[n]$ for the system $S$.

solution

4.5.5.1 part a

Taking DFT of the difference equation gives$$\begin{array}{rl}Y\left(\mathrm{\Omega }\right)-\frac{1}{6}{e}^{-j\mathrm{\Omega }}Y\left(\mathrm{\Omega }\right)-\frac{1}{6}{e}^{-j2\mathrm{\Omega }}Y\left(\mathrm{\Omega }\right)& =X\left(\mathrm{\Omega }\right)\\ Y\left(\mathrm{\Omega }\right)(1-\frac{1}{6}{e}^{-j\mathrm{\Omega }}-\frac{1}{6}{e}^{-j2\mathrm{\Omega }})& =X\left(\mathrm{\Omega }\right)\\ \frac{Y\left(\mathrm{\Omega }\right)}{X\left(\mathrm{\Omega }\right)}& =\frac{1}{1-\frac{1}{6}{e}^{-j\mathrm{\Omega }}-\frac{1}{6}{e}^{-j2\mathrm{\Omega }}}\end{array}$$

Let $e^{-j\mathrm{\Omega }}=x$ to simplify the notation, then $$\frac{Y\left(\mathrm{\Omega }\right)}{X\left(\mathrm{\Omega }\right)}=\frac{1}{1-\frac{1}{6}x-\frac{1}{6}{x}^2}=\frac{6}{6-x-x^2}=\frac{-6}{x^2+x-6}=\frac{-6}{(x-2)(x+3)}$$ Hence$$\begin{array}{rl}H\left(\mathrm{\Omega }\right)& =\frac{Y\left(\mathrm{\Omega }\right)}{X\left(\mathrm{\Omega }\right)}\\ & =\frac{-6}{(e^{-j\mathrm{\Omega }}-2)(e^{-j\mathrm{\Omega }}+3)}\end{array}$$

4.5.5.2 part b

Applying partial fractions$$H\left(\mathrm{\Omega }\right)=\frac{-6}{(e^{-j\mathrm{\Omega }}-2)(e^{-j\mathrm{\Omega }}+3)}=\frac{A}{(x-2)}+\frac{B}{(x+3)}$$ Hence $A=-\frac{6}{5},B=\frac{6}{5}$. Therefore$$\begin{array}{rl}H\left(\mathrm{\Omega }\right)& =-\frac{6}{5}\frac{1}{e^{-j\mathrm{\Omega }}-2}+\frac{6}{5}\frac{1}{e^{-j\mathrm{\Omega }}+3}\\ & =-\frac{3}{5}\frac{1}{\frac{1}{2}{e}^{-j\mathrm{\Omega }}-1}+\frac{2}{5}\frac{1}{\frac{1}{3}{e}^{-j\mathrm{\Omega }}+1}\\ & =\frac{3}{5}\frac{1}{1-\frac{1}{2}{e}^{-j\mathrm{\Omega }}}+\frac{2}{5}\frac{1}{1+\frac{1}{3}{e}^{-j\mathrm{\Omega }}}\end{array}$$

Taking the inverse DFT using tables gives$$\begin{array}{rl}h\left[n\right]& =\frac{3}{5}{\left(\frac{1}{2}\right)}^{n}u\left[n\right]+\frac{2}{5}{(-\frac{1}{3})}^{n}u\left[n\right]\\ & =(\frac{3}{5}{\left(\frac{1}{2}\right)}^n+\frac{2}{5}{(-\frac{1}{3})}^n)u\left[n\right]\end{array}$$

4.5.6 Problem 5.30, Chapter 5

In Chapter 4, we indicated that the continuous-time LTI system with impulse response$$h\left(t\right)=\frac{W}{\pi }\sin c\left(\frac{Wt}{\pi }\right)=\frac{\sin \left(Wt\right)}{\pi t}$$ plays a very important role in LTI system analysis. The same is true of the discrete time LTI system with impulse response$$h\left(n\right)=\frac{W}{\pi }\mathrm{sinc}\left(\frac{Wn}{\pi }\right)=\frac{\sin \left(Wn\right)}{\pi n}$$ (a) Determine and sketch the frequency response for the system with impulse response $h\left[n\right]$. (b) Consider the signal $x\left[n\right]=\sin \left(\frac{\pi n}{8}\right)-2\cos \left(\frac{\pi n}{4}\right)$. Suppose that this signal is the input to LTI systems with the following impulse responses. Determine the output in each case (i) $h\left[n\right]=\frac{\sin \left(\frac{\pi n}{6}\right)}{\pi n}$. (ii) $h\left[n\right]=\frac{\sin \left(\frac{\pi n}{6}\right)}{\pi n}+\frac{\sin \left(\frac{\pi n}{2}\right)}{\pi n}$

solution

4.5.6.1 Part a

Given $h\left(n\right)=\frac{W}{\pi }\mathrm{sinc}\left(\frac{Wn}{\pi }\right)=\frac{\sin \left(Wn\right)}{\pi n}$. We will show that $H\left(\mathrm{\Omega }\right)$ is the rectangle function by reverse. Assuming that $H\left(\mathrm{\Omega }\right)=\begin{array}{ccc}1& & \left|\mathrm{\Omega }\right|<2W\\ 0& & \text{otherwise}\end{array}$ therefore$$\begin{array}{rl}x\left[n\right]& =\frac{1}{2\pi }{\int }_{-W}^{W}X\left(\mathrm{\Omega }\right)e^{j\mathrm{\Omega }n}d\omega \\ & =\frac{1}{2\pi }{\int }_{-W}^W{e}^{j\mathrm{\Omega }n}d\omega \\ & =\frac{1}{2\pi }\frac{e^{jWn}-e^{-jWn}}{jn}\\ & =\frac{1}{\pi n}\sin \left(Wn\right)\end{array}$$

Which is the $h\left[n\right]$ given. Therefore, the above shows that $\frac{\sin \left(Wn\right)}{\pi n}$ has DFT of $H\left(\mathrm{\Omega }\right)$ as the rectangle function. Here is sketch

Figure 4.61:Plot of $H(\mathrm{\Omega })$

4.5.6.2 Part b

$$x\left[n\right]=\sin \left(\frac{\pi n}{8}\right)-2\cos \left(\frac{\pi n}{4}\right)$$ (i) $h\left[n\right]=\frac{\sin \left(\frac{\pi n}{6}\right)}{\pi n}$. Hence $y\left[n\right]=x\left[n\right]⊛h\left[n\right]$. Or $Y\left(\mathrm{\Omega }\right)=X\left(\mathrm{\Omega }\right)H\left(\mathrm{\Omega }\right)$, and then we find $y\left[n\right]$ by taking the inverse discrete Fourier transform. Here is the result and the code used. The result is  $$y\left[n\right]=\sin \left(\frac{n\pi }{8}\right)$$

Figure 4.62:Code used to generate $y[n]$

Here is plot of $y\left[n\right]$ for $n=-8\cdots 8$

Figure 4.63:Plot of above $y[n]$

(ii)  $h\left[n\right]=\frac{\sin \left(\frac{\pi n}{6}\right)}{\pi n}+\frac{\sin \left(\frac{\pi n}{2}\right)}{\pi n}$. Here is the result and the code used. The result is$$y\left[n\right]=2\sin \left(\frac{n\pi }{8}\right)-2\cos \left(\frac{n\pi }{4}\right)$$

Figure 4.64:Code used to generate $y[n]$

Here is plot of $y\left[n\right]$ for $n=-8\cdots 8$

Figure 4.65:Plot of above $y[n]$

4.5.7 key solution

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