3.1 practice exams

3.1.1 Midterm 1, oct 2001

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3.1.2 My solution to Midterm 1, oct 2001

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3.1.2.1 Problem 1

Obtain impulse and step response for LTI described by (a) $h\left[n\right]={\left(\frac{1}{2}\right)}^{n}u\left[n\right]$ (b) $h\left(t\right)=e^{-\frac{1}{2}t}u\left(t\right)$

solution

Part a Let $x\left[n\right]=\delta \left[n\right]$, hence $$\begin{array}{rl}y\left[n\right]& =\delta \left[n\right]⊛h\left[n\right]\\ & =\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\delta \left[k\right]h[n-k]\end{array}$$

But $\delta \left[k\right]=0$ for all $k$ except when $k=0$. Hence the above reduces to $$\begin{array}{rl}y\left[n\right]& =h\left[n\right]\\ & ={\left(\frac{1}{2}\right)}^{n}u\left[n\right]\end{array}$$

Let $x\left[n\right]=u\left[n\right]$, hence$$\begin{array}{rl}y\left[n\right]& =u\left[n\right]⊛h\left[n\right]\\ & =\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}h\left[k\right]x[n-k]\end{array}$$

Folding $x[-n]$, we see that for $n<0$ then there no overlap with $h\left[n\right]$. Hence $y\left[n\right]=0$ for $n<0$. As $x[-n]$ is shifted to the right, then the convolution sum becomes $$\begin{array}{rl}y\left[n\right]& =\sum _{k=0}^{n}h\left[k\right]n\ge 0\\ & =\sum _{k=0}^n{\left(\frac{1}{2}\right)}^k\end{array}$$

This is the partial sum, given by $\frac{a^{1+n}-1}{a-1}$ where $a=\frac{1}{2}<1$$$\begin{array}{rl}\sum _{k=0}^n{\left(\frac{1}{2}\right)}^k& =\frac{{\left(\frac{1}{2}\right)}^{1+n}-1}{\frac{1}{2}-1}\\ & =\frac{{\left(\frac{1}{2}\right)}^{1+n}-1}{-\frac{1}{2}}\\ & =2-2{\left(2\right)}^{-n-1}\\ \text{(2)}& & =2-2^{-n}\end{array}$$

Therefore$$y\left[n\right]=\{\begin{array}{ccc}2-2^{-n}& & n\ge 0\\ 0& & n<0\end{array}$$

Part b Let $x\left(t\right)=\delta \left(t\right)$, hence$$\begin{array}{rl}y\left(t\right)& =u\left(t\right)⊛h\left(t\right)\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\left(\tau \right)h(t-\tau )d\tau \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\delta \left(t\right)h(t-\tau )d\tau \\ & =h\left(t\right)\\ & =e^{-0.5t}\end{array}$$

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

Folding $u(-t)$, we see that for $t<0$ then there no overlap with $h\left(\tau \right)=e^{-0.5\tau }u\left(\tau \right)$. Hence $y\left(t\right)=0$ for $t<0$. As $u[-n]$ is shifted to the right, then the convolution becomes $$\begin{array}{rl}y\left[n\right]& ={\int }_0^{t}h\left(\tau \right)d\tau t>0\\ & ={\int }_0^t{e}^{-0.5\tau }d\tau \\ & ={\left(\frac{e^{-0.5\tau }}{-0.5}\right)}_0^t\\ & =-2(e^{-0.5t}-1)\\ & =2-2e^{-0.5t}\\ & =2(1-e^{-0.5t})\end{array}$$

Hence

$$y\left(t\right)=\{\begin{array}{ccc}2(1-e^{-0.5t})& & t\ge 0\\ 0& & t<0\end{array}$$

3.1.2.2 Problem 2

Given the frequency response of LTI system $H\left(\mathrm{\Omega }\right)$ for the following input signal, find the steady state expression of the output signal. (a) $x\left[n\right]=2\cos (\frac{\pi }{6}n+\frac{\pi }{5})$ (b) $x\left[n\right]=5\sin (\frac{\pi }{3}n+\frac{\pi }{8})$

solution

Part a $$x\left[n\right]=2\cos (\frac{\pi }{6}n+\frac{\pi }{5})$$ To find the fundamental period, $\cos (\frac{\pi }{6}n+\frac{\pi }{5})=\cos (\frac{\pi }{6}(n+N)+\frac{\pi }{5})=\cos ((\frac{\pi }{6}n+\frac{\pi }{5})+\frac{\pi }{6}N)$. Hence need $\frac{\pi }{6}N=m2\pi$ or $\frac{m}{N}=\frac{1}{12}$. Hence. $N=12$. Therefore $${\mathrm{\Omega }}_0=\frac{2\pi }{12}$$ And the input is $x\left[n\right]=2\cos ({\mathrm{\Omega }}_{0}n+\frac{\pi }{5})$. Hence the output is$$y\left[n\right]=2\left|H\left({\mathrm{\Omega }}_0\right)\right|\cos ({\mathrm{\Omega }}_{0}n+\frac{\pi }{5}+\arg H\left({\mathrm{\Omega }}_0\right))$$

Part b $$x\left[n\right]=5\sin (\frac{\pi }{3}n+\frac{\pi }{8})$$ To find the fundamental period, $\sin (\frac{\pi }{3}n+\frac{\pi }{8})=\sin (\frac{\pi }{3}(n+N)+\frac{\pi }{8})=\sin (\frac{\pi }{3}n+\frac{\pi }{8}+\frac{\pi }{3}N)$. Hence need $\frac{\pi }{3}N=m2\pi$ or $\frac{m}{N}=\frac{1}{6}$. Hence. $N=6$. Therefore$${\mathrm{\Omega }}_0=\frac{2\pi }{6}$$ And the input is $x\left[n\right]=5\sin ({\mathrm{\Omega }}_{0}n+\frac{\pi }{8})$. Hence the output is$$y\left[n\right]=5\left|H\left({\mathrm{\Omega }}_0\right)\right|\sin ({\mathrm{\Omega }}_{0}n+\frac{\pi }{8}+\arg H\left({\mathrm{\Omega }}_0\right))$$

3.1.2.3 Problem 3

Compute Fourier series coeff. for the following signals. (a) $x\left[n\right]=2\sin (\frac{\pi }{3}n+\frac{\pi }{2})+3\cos (\frac{\pi }{6}n+\frac{\pi }{5})$. (b) $x\left(t\right)=e^{j2\pi t}+e^{j3\pi t}$

solution

Part a For discrete periodic signal, the Fourier series coeff. $a_k$ is given by$$\begin{array}{}\text{(1)}& x\left[n\right]& =\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k{e}^{jk\left(\frac{2\pi }{N}\right)n}\text{(2)}& a_k& =\frac{1}{N}\sum _{n=0}^{N-1}x\left[n\right]e^{-jk\left(\frac{2\pi }{N}\right)n}\end{array}$$

In this problem

$$x\left[n\right]=2\sin (\frac{\pi }{3}n+\frac{\pi }{2})+3\cos (\frac{\pi }{6}n+\frac{\pi }{5})$$ To find the common fundamental period. $\sin (\frac{\pi }{3}n+\frac{\pi }{2})=\sin (\frac{\pi }{3}(n+N)+\frac{\pi }{2})=\sin ((\frac{\pi }{3}n+\frac{\pi }{2})+\frac{\pi }{3}N)$. Hence $\frac{\pi }{3}N=m2\pi$ or $\frac{m}{N}=\frac{1}{6}$. hence $N=6$ for first signal. For second signal $\cos (\frac{\pi }{6}n+\frac{\pi }{5})$ we obtain $\frac{\pi }{6}N=m2\pi$ or $\frac{m}{N}=\frac{1}{12}$ or $N=12$. hence the least common multiplier between $6$ and $12$ is $N=12$. Therefore $${\mathrm{\Omega }}_0=\frac{2\pi }{12}$$ Hence (2) becomes$$\begin{array}{rl}{a}_k& =\frac{1}{12}\sum _{n=0}^{11}x\left[n\right]e^{-jk\left(\frac{2\pi }{12}\right)n}\\ & =\frac{1}{12}\sum _{n=0}^{11}x\left[n\right]e^{-jk{\mathrm{\Omega }}_{0}n}\end{array}$$

But instead of using the above formula, an easier way is to rewrite $x\left[n\right]$  using Euler relation and use (1) to read off $a_k$ directly from the result. Writing $x\left[n\right]$ in terms of the fundamental frequency ${\mathrm{\Omega }}_0$ gives$$\begin{array}{rl}x\left[n\right]& =2\sin (2{\mathrm{\Omega }}_{0}n+\frac{\pi }{2})+3\cos ({\mathrm{\Omega }}_{0}n+\frac{\pi }{5})\\ & =2\left(\frac{e^{j(2{\mathrm{\Omega }}_{0}n+\frac{\pi }{2})}-e^{-j(2{\mathrm{\Omega }}_{0}n+\frac{\pi }{2})}}{2j}\right)+3\left(\frac{e^{j({\mathrm{\Omega }}_{0}n+\frac{\pi }{5})}+e^{-j({\mathrm{\Omega }}_{0}n+\frac{\pi }{5})}}{2}\right)\\ & =\frac{2}{2j}(e^{j\frac{\pi }{2}}{e}^{j2{\mathrm{\Omega }}_{0}n}-e^{-j\frac{\pi }{2}}{e}^{-j2{\mathrm{\Omega }}_{0}n})+\frac{3}{2}(e^{j\frac{\pi }{5}}{e}^{j{\mathrm{\Omega }}_{0}n}+e^{-j\frac{\pi }{5}}{e}^{-j{\mathrm{\Omega }}_{0}n})\\ & =\frac{1}{j}{e}^{j\frac{\pi }{2}}{e}^{j2{\mathrm{\Omega }}_{0}n}-\frac{1}{j}{e}^{-j\frac{\pi }{2}}{e}^{-j2{\mathrm{\Omega }}_{0}n}+\frac{3}{2}{e}^{j\frac{\pi }{5}}{e}^{j{\mathrm{\Omega }}_{0}n}+\frac{3}{2}{e}^{-j\frac{\pi }{5}}{e}^{-j{\mathrm{\Omega }}_{0}n}\end{array}$$

Now we can read the Fourier coefficients by comparing the above to Eq(1).

This gives for $k=2,a_2=\frac{1}{j}{e}^{j\frac{\pi }{2}}$ and for $k=-2,a_{-2}=-\frac{1}{j}{e}^{-j\frac{\pi }{2}}$ and for $k=1,a_1=\frac{3}{2}{e}^{j\frac{\pi }{5}}$ and for $k=-1,a_{-1}=\frac{3}{2}{e}^{-j\frac{\pi }{5}}$$$\begin{array}{rl}{a}_1& =\frac{3}{2}{e}^{j\frac{\pi }{5}}\\ a_{-1}& =\frac{3}{2}{e}^{-j\frac{\pi }{5}}\\ a_2& =\frac{1}{j}{e}^{j\frac{\pi }{2}}\\ a_{-2}& =-\frac{1}{j}{e}^{-j\frac{\pi }{2}}\end{array}$$

But $e^{j\frac{\pi }{2}}=j\sin \frac{\pi }{2}=j$ and $e^{-j\frac{\pi }{2}}=-j\sin \frac{\pi }{2}=-j$. Hence the above becomes$$\begin{array}{rl}{a}_1& =\frac{3}{2}{e}^{j\frac{\pi }{5}}\\ a_{-1}& =\frac{3}{2}{e}^{-j\frac{\pi }{5}}\\ a_2& =\frac{1}{j}j=1\\ a_{-2}& =-\frac{1}{j}(-j)=1\end{array}$$

And $a_k=0$ for all other $k$.

Part b For continuos time periodic signal $x\left(t\right)$, the Fourier series coeff. $a_k$ is given by$$\begin{array}{rl}x\left(t\right)& =\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k{e}^{jk{\omega }_{0}t}\\ a_k& =\frac{1}{T}{\int }_{T}x\left(t\right)e^{-jk{\omega }_{0}t}dt\end{array}$$

In this problem$$x\left(t\right)=e^{j2\pi t}+e^{j3\pi t}$$ The period of $e^{j2\pi t}$ is $1$ and the period of $e^{j3\pi t}$ is $\frac{2}{3}$. Hence least common multiplier is $T_0=2$ seconds. ${\omega }_0=\frac{2\pi }{2}=\pi$ rad/sec. Both of the above terms can now be written $$\begin{array}{rl}x\left(t\right)& =e^{j\frac{4\pi }{T_0}t}+e^{j\frac{6\pi }{T_0}t}\\ & =e^{j2{\omega }_{0}t}+e^{j3{\omega }_{0}t}\end{array}$$

Comparing the above to$$x\left(t\right)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k{e}^{jk{\omega }_{0}t}$$ Shows that  for $k=2,a_k=1$ and for $k=3,a_k=1$ and $a_k=0$ for all other $k$.

3.1.2.4 Problem 4

Given the magnitude and phase profile of this filter, find impulse response.

solution

We are given $H\left(\mathrm{\Omega }\right)$ and need to find $h\left[n\right]$. i.e. the inverse Fourier transform $$\begin{array}{rl}h\left[n\right]& =\frac{1}{2\pi }{\int }_{-\pi }^{\pi }H\left(\mathrm{\Omega }\right)e^{j\mathrm{\Omega }n}d\mathrm{\Omega }\\ & =\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left|H\left(\mathrm{\Omega }\right)\right|e^{j\arg H\left(\mathrm{\Omega }\right)}{e}^{j\mathrm{\Omega }n}d\mathrm{\Omega }\end{array}$$

But $\left|H\left(\mathrm{\Omega }\right)\right|=1$ and $\arg H\left(\mathrm{\Omega }\right)=-\mathrm{\Omega }$ as given. The above reduces to$$\begin{array}{rl}h\left[n\right]& =\frac{1}{2\pi }{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{e}^{-j\mathrm{\Omega }}{e}^{j\mathrm{\Omega }n}d\mathrm{\Omega }\\ & =\frac{1}{2\pi }{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{e}^{-j\mathrm{\Omega }(1-n)}d\mathrm{\Omega }\\ & =\left(\frac{1}{2\pi }\right)\frac{1}{-j(1-n)}{\left[e^{-j\mathrm{\Omega }(1-n)}\right]}_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\\ & =\left(\frac{1}{2\pi }\right)\frac{1}{-j(1-n)}(e^{-j\frac{\pi }{4}(1-n)}-e^{j\frac{\pi }{4}(1-n)})\\ & =\frac{1}{\pi }\frac{1}{(1-n)}\left(\frac{e^{j\frac{\pi }{4}(1-n)}-e^{-j\frac{\pi }{4}(1-n)}}{2j}\right)\\ & =\frac{1}{\pi (1-n)}\sin \left(\frac{\pi }{4}(1-n)\right)\\ & =\frac{-1}{\pi (1-n)}\sin \left(\frac{\pi }{4}(n-1)\right)\\ & =\frac{1}{\pi (n-1)}\sin \left(\frac{\pi }{4}(n-1)\right)\end{array}$$

3.1.3 Midterm 1, oct 2018

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3.1.4 My solution to Midterm 1, oct 2018

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3.1.4.1 Problem 1

Given the discrete time system with input $x\left[n\right]$ and impulse response $h\left[n\right]$ obtain the output sequence $y\left[n\right]$ by applying discrete convolution.

$x\left[n\right]=[1,0,-1,1],h\left[n\right]=[1,1,-1,-1]$. Both sequences are positive and start with $n=0$ position.

Solution$$y\left[n\right]=x\left[n\right]⊛h\left[n\right]$$ Folding $x[-n]$. When $n=0,y\left[0\right]=1$. When $n=1$, then $y\left[n\right]=\left(0\right)\left(1\right)+\left(1\right)\left(1\right)=1$. When $n=2,y\left[n\right]=(-1)\left(1\right)+\left(0\right)\left(1\right)+\left(1\right)(-1)=-2$. When $n=3,y\left[n\right]=\left(1\right)\left(1\right)+(-1)\left(1\right)+\left(0\right)(-1)+\left(1\right)(-1)=1-1-1=-1$. When $n=4,y\left[n\right]=1+1=2$. When $n=5,y\left[n\right]=-1+1=0$, when $n=6,y\left[6\right]=-1$. When $n>6,y\left[n\right]=0$.

Hence $$y\left[n\right]=[1,1,-2,-1,2,0,-1]$$

3.1.4.2 Problem 2

The impulse response of LTI system is given by $h\left(t\right)=u\left(t\right)-2u(t-1)+u(t-2)$ where $u\left(t\right)$ is unit step signal. Determine the output of this $y\left(t\right)$ where its input $x\left(t\right)$ is given as $x\left(t\right)=e^{-t}u(t+1)$.

Solution

By folding $x\left(t\right)$. See key solution. Used same method.

3.1.4.3 Problem 3

A discrete periodic sequence is given as $x\left[n\right]=2\sin (\frac{3\pi }{8}n+\frac{\pi }{2})+\cos (\frac{\pi }{4}n+\frac{\pi }{3})$. (a) Find fundamental frequency of this signal. (b) Fourier series coefficients for $x\left[n\right]$. (c) if $x\left[n\right]=\cos (\frac{\pi }{4}n+\frac{\pi }{3})$ is an input to system with frequency response $H\left(\mathrm{\Omega }\right)=\frac{1-e^{-j\mathrm{\Omega }}}{2+e^{-2j\mathrm{\Omega }}}$, obtain expression for $y\left[n\right]$

Solution

Part a For $\sin (\frac{3\pi }{8}n+\frac{\pi }{2})$, we need $\frac{3}{8}\pi N=m2\pi$ or $\frac{m}{N}=\frac{3}{16}$. Since relatively prime, hence $N=16$. For $\cos (\frac{\pi }{4}n+\frac{\pi }{3})$ we need $\frac{\pi }{4}N=m2\pi$ or $\frac{m}{N}=\frac{1}{8}$. Hence $N=8$. The least common multiplier is $16$. Hence fundamental period is $N=16$. Therefore ${\mathrm{\Omega }}_0=\frac{2\pi }{N}=\frac{\pi }{8}$.

Part b Since input is periodic, then$$\begin{array}{}\text{(1)}& x\left[n\right]=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k{e}^{jk{\mathrm{\Omega }}_{0}n}\end{array}$$ By writing the input, using Euler relation, we can compare the input to the above and read off $a_k$. First we rewrite the input using common ${\mathrm{\Omega }}_0$ as$$x\left[n\right]=2\sin (3{\mathrm{\Omega }}_{0}n+\frac{\pi }{2})+\cos (2{\mathrm{\Omega }}_{0}n+\frac{\pi }{3})$$ Hence$$\begin{array}{rl}x\left[n\right]& =2\left(\frac{e^{j(3{\mathrm{\Omega }}_{0}n+\frac{\pi }{2})}-e^{-j(3{\mathrm{\Omega }}_{0}n+\frac{\pi }{2})}}{2j}\right)+\frac{e^{j(2{\mathrm{\Omega }}_{0}n+\frac{\pi }{3})}+e^{-j(2{\mathrm{\Omega }}_{0}n+\frac{\pi }{3})}}{2}\\ & =\frac{1}{j}{e}^{j(3{\mathrm{\Omega }}_{0}n+\frac{\pi }{2})}-\frac{1}{j}{e}^{-j(3{\mathrm{\Omega }}_{0}n+\frac{\pi }{2})}+\frac{1}{2}{e}^{j(2{\mathrm{\Omega }}_{0}n+\frac{\pi }{3})}+\frac{1}{2}{e}^{-j(2{\mathrm{\Omega }}_{0}n+\frac{\pi }{3})}\\ & =\frac{1}{j}{e}^{j\frac{\pi }{2}}{e}^{j3{\mathrm{\Omega }}_{0}n}-\frac{1}{j}{e}^{-j\frac{\pi }{2}}{e}^{-j3{\mathrm{\Omega }}_{0}n}+\frac{1}{2}{e}^{j\frac{\pi }{3}}{e}^{j2{\mathrm{\Omega }}_{0}n}+\frac{1}{2}{e}^{-j\frac{\pi }{3}}{e}^{-j2{\mathrm{\Omega }}_{0}n}\end{array}$$

But $e^{j\frac{\pi }{2}}=j\sin \frac{\pi }{2}=j$ and $e^{-j\frac{\pi }{2}}=-j\sin \frac{\pi }{2}=-j$ and $e^{j\frac{\pi }{3}}=\cos \left(\frac{\pi }{3}\right)+j\sin \left(\frac{\pi }{3}\right)=\frac{1}{2}\sqrt{3}j+\frac{1}{2}$ and $e^{-j\frac{\pi }{3}}=\cos \left(\frac{\pi }{3}\right)-j\sin \left(\frac{\pi }{3}\right)=\frac{1}{2}-\frac{1}{2}\sqrt{3}j$. Hence the above simplifies to$$\begin{array}{}\text{(2)}& x\left[n\right]=e^{j3{\mathrm{\Omega }}_{0}n}+e^{-j3{\mathrm{\Omega }}_{0}n}+\frac{1}{4}(1+\sqrt{3}j)e^{j2{\mathrm{\Omega }}_{0}n}-\frac{1}{4}(1-\sqrt{3}j)e^{-j2{\mathrm{\Omega }}_{0}n}\end{array}$$ Comparing (2) to (1) shows that$$\begin{array}{rl}{a}_3& =1\\ a_{-3}& =1\\ a_2& =\frac{1}{2}{e}^{j\frac{\pi }{3}}=\frac{1}{4}(1+\sqrt{3}j)\\ a_{-2}& =\frac{1}{2}{e}^{-j\frac{\pi }{3}}=-\frac{1}{4}(1-\sqrt{3}j)\end{array}$$

Part c The output is$$\begin{array}{}\text{(1)}& y\left[n\right]={\left|H\left(\mathrm{\Omega }\right)\right|}_{\mathrm{\Omega }=\frac{\pi }{4}}\cos (\frac{\pi }{4}n+\frac{\pi }{3}+\arg H{\left(\mathrm{\Omega }\right)}_{\mathrm{\Omega }=\frac{\pi }{4}})\end{array}$$ But$$\begin{array}{rl}\left|H\left(\mathrm{\Omega }\right)\right|& =\left|\frac{1-e^{-j\mathrm{\Omega }}}{2+e^{-2j\mathrm{\Omega }}}\right|\\ & =\frac{|1-e^{-j\mathrm{\Omega }}|}{|2+e^{-2j\mathrm{\Omega }}|}\\ & =\frac{\sqrt{{(1-\cos \mathrm{\Omega })}^2+{\sin }^2\mathrm{\Omega }}}{\sqrt{{(2+\cos 2\mathrm{\Omega })}^2+{\sin }^2\left(2\mathrm{\Omega }\right)}}\end{array}$$

When $\mathrm{\Omega }=\frac{\pi }{4}$ the above becomes$$\begin{array}{rl}\left|H\left(\frac{\pi }{4}\right)\right|& =\frac{\sqrt{{(1-\cos \left(\frac{\pi }{4}\right))}^2+{\sin }^2\left(\frac{\pi }{4}\right)}}{\sqrt{{(2+\cos \left(\frac{\pi }{2}\right))}^2+{\sin }^2\left(\frac{\pi }{2}\right)}}\\ & =\frac{\sqrt{{(1-\cos \left(\frac{\pi }{4}\right))}^2+{\sin }^2\left(\frac{\pi }{4}\right)}}{\sqrt{4+1}}\\ & =\frac{\sqrt{\frac{3}{2}-\sqrt{2}+\frac{1}{2}}}{\sqrt{5}}\\ & =\frac{\sqrt{2-\sqrt{2}}}{\sqrt{5}}\\ & =0.34228\end{array}$$

And $$\begin{array}{rl}\arg H\left(\mathrm{\Omega }\right)& =\arg \frac{1-e^{-j\mathrm{\Omega }}}{2+e^{-2j\mathrm{\Omega }}}\\ & =\arg \frac{(1-\cos \mathrm{\Omega })+j\sin \mathrm{\Omega }}{(2+\cos \left(2\mathrm{\Omega }\right))-j\sin \left(2\mathrm{\Omega }\right)}\\ & =\arctan \left(\frac{\sin \mathrm{\Omega }}{1-\cos \mathrm{\Omega }}\right)-\arctan \left(\frac{-\sin \left(2\mathrm{\Omega }\right)}{2+\cos \left(2\mathrm{\Omega }\right)}\right)\end{array}$$

When $\mathrm{\Omega }=\frac{\pi }{4}$ the above becomes$$\begin{array}{rl}\arg H\left(\frac{\pi }{4}\right)& =\arctan \left(\frac{\sin \left(\frac{\pi }{4}\right)}{1-\cos \left(\frac{\pi }{4}\right)}\right)-\arctan \left(\frac{-\sin \left(\frac{\pi }{2}\right)}{2+\cos \left(\frac{\pi }{2}\right)}\right)\\ & =\arctan \left(\frac{\sin \left(\frac{\pi }{4}\right)}{1-\cos \left(\frac{\pi }{4}\right)}\right)-\arctan \left(\frac{-1}{2}\right)\\ & =\arctan \left(\frac{\sin \left(\frac{\pi }{4}\right)}{1-\cos \left(\frac{\pi }{4}\right)}\right)+\arctan \left(\frac{1}{2}\right)\\ & =\arctan \left(2.4142\right)+0.46365\\ & =1.1781+0.46365\\ & =1.6418\text{ rad}\end{array}$$

Hence (1) becomes$$y\left[n\right]=0.3423\cos (\frac{\pi }{4}n+\frac{\pi }{3}+1.6418)$$

3.1.5 Final exam practice exam 1

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3.1.6 Final exam practice exam 2

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