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{align*} y\left [ n\right ] & =\delta \left [ n\right ] \circledast h\left [ n\right ] \\ & =\sum _{k=-\infty }^{\infty }\delta \left [ k\right ] h\left [ n-k\right ] \end{align*}
But \(\delta \left [ k\right ] =0\) for all \(k\) except when \(k=0\). Hence the above reduces to \begin{align*} y\left [ n\right ] & =h\left [ n\right ] \\ & =\left ( \frac{1}{2}\right ) ^{n}u\left [ n\right ] \end{align*}
Let \(x\left [ n\right ] =u\left [ n\right ] \), hence\begin{align*} y\left [ n\right ] & =u\left [ n\right ] \circledast h\left [ n\right ] \\ & =\sum _{k=-\infty }^{\infty }h\left [ k\right ] x\left [ n-k\right ] \end{align*}
Folding \(x\left [ -n\right ] \), 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 \(x\left [ -n\right ] \) is shifted to the right, then the convolution sum becomes \begin{align*} y\left [ n\right ] & =\sum _{k=0}^{n}h\left [ k\right ] \qquad n\geq 0\\ & =\sum _{k=0}^{n}\left ( \frac{1}{2}\right ) ^{k} \end{align*}
This is the partial sum, given by \(\frac{a^{1+n}-1}{a-1}\) where \(a=\frac{1}{2}<1\)\begin{align} \sum _{k=0}^{n}\left ( \frac{1}{2}\right ) ^{k} & =\frac{\left ( \frac{1}{2}\right ) ^{1+n}-1}{\frac{1}{2}-1}\nonumber \\ & =\frac{\left ( \frac{1}{2}\right ) ^{1+n}-1}{-\frac{1}{2}}\nonumber \\ & =2-2\left ( 2\right ) ^{-n-1}\nonumber \\ & =2-2^{-n}\tag{2} \end{align}
Therefore\[ y\left [ n\right ] =\left \{ \begin{array} [c]{ccc}2-2^{-n} & & n\geq 0\\ 0 & & n<0 \end{array} \right . \]
Part b Let \(x\left ( t\right ) =\delta \left ( t\right ) \), hence\begin{align*} y\left ( t\right ) & =u\left ( t\right ) \circledast h\left ( t\right ) \\ & =\int _{-\infty }^{\infty }x\left ( \tau \right ) h\left ( t-\tau \right ) d\tau \\ & =\int _{-\infty }^{\infty }\delta \left ( t\right ) h\left ( t-\tau \right ) d\tau \\ & =h\left ( t\right ) \\ & =e^{-0.5t} \end{align*}
Let \(x\left ( t\right ) =u\left ( t\right ) \), hence\begin{align*} y\left ( t\right ) & =u\left ( t\right ) \circledast h\left ( t\right ) \\ & =\int _{-\infty }^{\infty }x\left ( t-\tau \right ) h\left ( \tau \right ) d\tau \end{align*}
Folding \(u\left ( -t\right ) \), 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 \(u\left [ -n\right ] \) is shifted to the right, then the convolution becomes \begin{align*} y\left [ n\right ] & =\int _{0}^{t}h\left ( \tau \right ) d\tau \qquad t>0\\ & =\int _{0}^{t}e^{-0.5\tau }d\tau \\ & =\left ( \frac{e^{-0.5\tau }}{-0.5}\right ) _{0}^{t}\\ & =-2\left ( e^{-0.5t}-1\right ) \\ & =2-2e^{-0.5t}\\ & =2\left ( 1-e^{-0.5t}\right ) \end{align*}
Hence
\[ y\left ( t\right ) =\left \{ \begin{array} [c]{ccc}2\left ( 1-e^{-0.5t}\right ) & & t\geq 0\\ 0 & & t<0 \end{array} \right . \]
Given the frequency response of LTI system \(H\left ( \Omega \right ) \) for the following input signal, find the steady state expression of the output signal. (a) \(x\left [ n\right ] =2\cos \left ( \frac{\pi }{6}n+\frac{\pi }{5}\right ) \) (b) \(x\left [ n\right ] =5\sin \left ( \frac{\pi }{3}n+\frac{\pi }{8}\right ) \)
solution
Part a \[ x\left [ n\right ] =2\cos \left ( \frac{\pi }{6}n+\frac{\pi }{5}\right ) \] To find the fundamental period, \(\cos \left ( \frac{\pi }{6}n+\frac{\pi }{5}\right ) =\cos \left ( \frac{\pi }{6}\left ( n+N\right ) +\frac{\pi }{5}\right ) =\cos \left ( \left ( \frac{\pi }{6}n+\frac{\pi }{5}\right ) +\frac{\pi }{6}N\right ) \). Hence need \(\frac{\pi }{6}N=m2\pi \) or \(\frac{m}{N}=\frac{1}{12}\). Hence. \(N=12\). Therefore \[ \Omega _{0}=\frac{2\pi }{12}\] And the input is \(x\left [ n\right ] =2\cos \left ( \Omega _{0}n+\frac{\pi }{5}\right ) \). Hence the output is\[ y\left [ n\right ] =2\left \vert H\left ( \Omega _{0}\right ) \right \vert \cos \left ( \Omega _{0}n+\frac{\pi }{5}+\arg H\left ( \Omega _{0}\right ) \right ) \]
Part b \[ x\left [ n\right ] =5\sin \left ( \frac{\pi }{3}n+\frac{\pi }{8}\right ) \] To find the fundamental period, \(\sin \left ( \frac{\pi }{3}n+\frac{\pi }{8}\right ) =\sin \left ( \frac{\pi }{3}\left ( n+N\right ) +\frac{\pi }{8}\right ) =\sin \left ( \frac{\pi }{3}n+\frac{\pi }{8}+\frac{\pi }{3}N\right ) \). Hence need \(\frac{\pi }{3}N=m2\pi \) or \(\frac{m}{N}=\frac{1}{6}\). Hence. \(N=6\). Therefore\[ \Omega _{0}=\frac{2\pi }{6}\] And the input is \(x\left [ n\right ] =5\sin \left ( \Omega _{0}n+\frac{\pi }{8}\right ) \). Hence the output is\[ y\left [ n\right ] =5\left \vert H\left ( \Omega _{0}\right ) \right \vert \sin \left ( \Omega _{0}n+\frac{\pi }{8}+\arg H\left ( \Omega _{0}\right ) \right ) \]
Compute Fourier series coeff. for the following signals. (a) \(x\left [ n\right ] =2\sin \left ( \frac{\pi }{3}n+\frac{\pi }{2}\right ) +3\cos \left ( \frac{\pi }{6}n+\frac{\pi }{5}\right ) \). (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{align} x\left [ n\right ] & =\sum _{k=-\infty }^{\infty }a_{k}e^{jk\left ( \frac{2\pi }{N}\right ) n}\tag{1}\\ a_{k} & =\frac{1}{N}\sum _{n=0}^{N-1}x\left [ n\right ] e^{-jk\left ( \frac{2\pi }{N}\right ) n}\tag{2} \end{align}
In this problem
\[ x\left [ n\right ] =2\sin \left ( \frac{\pi }{3}n+\frac{\pi }{2}\right ) +3\cos \left ( \frac{\pi }{6}n+\frac{\pi }{5}\right ) \] To find the common fundamental period. \(\sin \left ( \frac{\pi }{3}n+\frac{\pi }{2}\right ) =\sin \left ( \frac{\pi }{3}\left ( n+N\right ) +\frac{\pi }{2}\right ) =\sin \left ( \left ( \frac{\pi }{3}n+\frac{\pi }{2}\right ) +\frac{\pi }{3}N\right ) \). Hence \(\frac{\pi }{3}N=m2\pi \) or \(\frac{m}{N}=\frac{1}{6}\). hence \(N=6\) for first signal. For second signal \(\cos \left ( \frac{\pi }{6}n+\frac{\pi }{5}\right ) \) 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 \[ \Omega _{0}=\frac{2\pi }{12}\] Hence (2) becomes\begin{align*} 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\Omega _{0}n} \end{align*}
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 \(\Omega _{0}\) gives\begin{align*} x\left [ n\right ] & =2\sin \left ( 2\Omega _{0}n+\frac{\pi }{2}\right ) +3\cos \left ( \Omega _{0}n+\frac{\pi }{5}\right ) \\ & =2\left ( \frac{e^{j\left ( 2\Omega _{0}n+\frac{\pi }{2}\right ) }-e^{-j\left ( 2\Omega _{0}n+\frac{\pi }{2}\right ) }}{2j}\right ) +3\left ( \frac{e^{j\left ( \Omega _{0}n+\frac{\pi }{5}\right ) }+e^{-j\left ( \Omega _{0}n+\frac{\pi }{5}\right ) }}{2}\right ) \\ & =\frac{2}{2j}\left ( e^{j\frac{\pi }{2}}e^{j2\Omega _{0}n}-e^{-j\frac{\pi }{2}}e^{-j2\Omega _{0}n}\right ) +\frac{3}{2}\left ( e^{j\frac{\pi }{5}}e^{j\Omega _{0}n}+e^{-j\frac{\pi }{5}}e^{-j\Omega _{0}n}\right ) \\ & =\frac{1}{j}e^{j\frac{\pi }{2}}e^{j2\Omega _{0}n}-\frac{1}{j}e^{-j\frac{\pi }{2}}e^{-j2\Omega _{0}n}+\frac{3}{2}e^{j\frac{\pi }{5}}e^{j\Omega _{0}n}+\frac{3}{2}e^{-j\frac{\pi }{5}}e^{-j\Omega _{0}n} \end{align*}
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{align*} 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{align*}
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{align*} 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}\left ( -j\right ) =1 \end{align*}
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{align*} x\left ( t\right ) & =\sum _{k=-\infty }^{\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{align*}
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{align*} 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{align*}
Comparing the above to\[ x\left ( t\right ) =\sum _{k=-\infty }^{\infty }a_{k}e^{jk\omega _{0}t}\] Shows that for \(k=2,a_{k}=1\) and for \(k=3,a_{k}=1\) and \(a_{k}=0\) for all other \(k\).
Given the magnitude and phase profile of this filter, find impulse response.
solution
We are given \(H\left ( \Omega \right ) \) and need to find \(h\left [ n\right ] \). i.e. the inverse Fourier transform \begin{align*} h\left [ n\right ] & =\frac{1}{2\pi }\int _{-\pi }^{\pi }H\left ( \Omega \right ) e^{j\Omega n}d\Omega \\ & =\frac{1}{2\pi }\int _{-\pi }^{\pi }\left \vert H\left ( \Omega \right ) \right \vert e^{j\arg H\left ( \Omega \right ) }e^{j\Omega n}d\Omega \end{align*}
But \(\left \vert H\left ( \Omega \right ) \right \vert =1\) and \(\arg H\left ( \Omega \right ) =-\Omega \) as given. The above reduces to\begin{align*} h\left [ n\right ] & =\frac{1}{2\pi }\int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}e^{-j\Omega }e^{j\Omega n}d\Omega \\ & =\frac{1}{2\pi }\int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}e^{-j\Omega \left ( 1-n\right ) }d\Omega \\ & =\left ( \frac{1}{2\pi }\right ) \frac{1}{-j\left ( 1-n\right ) }\left [ e^{-j\Omega \left ( 1-n\right ) }\right ] _{-\frac{\pi }{4}}^{\frac{\pi }{4}}\\ & =\left ( \frac{1}{2\pi }\right ) \frac{1}{-j\left ( 1-n\right ) }\left ( e^{-j\frac{\pi }{4}\left ( 1-n\right ) }-e^{j\frac{\pi }{4}\left ( 1-n\right ) }\right ) \\ & =\frac{1}{\pi }\frac{1}{\left ( 1-n\right ) }\left ( \frac{e^{j\frac{\pi }{4}\left ( 1-n\right ) }-e^{-j\frac{\pi }{4}\left ( 1-n\right ) }}{2j}\right ) \\ & =\frac{1}{\pi \left ( 1-n\right ) }\sin \left ( \frac{\pi }{4}\left ( 1-n\right ) \right ) \\ & =\frac{-1}{\pi \left ( 1-n\right ) }\sin \left ( \frac{\pi }{4}\left ( n-1\right ) \right ) \\ & =\frac{1}{\pi \left ( n-1\right ) }\sin \left ( \frac{\pi }{4}\left ( n-1\right ) \right ) \end{align*}
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.
\(x\left [ n\right ] =\left [ 1,0,-1,1\right ] ,h\left [ n\right ] =\left [ 1,1,-1,-1\right ] \). Both sequences are positive and start with \(n=0\) position.
Solution\[ y\left [ n\right ] =x\left [ n\right ] \circledast h\left [ n\right ] \] Folding \(x\left [ -n\right ] \). 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 \(n=2,y\left [ n\right ] =\left ( -1\right ) \left ( 1\right ) +\left ( 0\right ) \left ( 1\right ) +\left ( 1\right ) \left ( -1\right ) =-2\). When \(n=3,y\left [ n\right ] =\left ( 1\right ) \left ( 1\right ) +\left ( -1\right ) \left ( 1\right ) +\left ( 0\right ) \left ( -1\right ) +\left ( 1\right ) \left ( -1\right ) =1-1-1=-1\). 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 \[ y\left [ n\right ] =\left [ 1,1,-2,-1,2,0,-1\right ] \]
The impulse response of LTI system is given by \(h\left ( t\right ) =u\left ( t\right ) -2u\left ( t-1\right ) +u\left ( t-2\right ) \) 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 \(x\left ( t\right ) =e^{-t}u\left ( t+1\right ) \).
Solution
By folding \(x\left ( t\right ) \). See key solution. Used same method.
A discrete periodic sequence is given as \(x\left [ n\right ] =2\sin \left ( \frac{3\pi }{8}n+\frac{\pi }{2}\right ) +\cos \left ( \frac{\pi }{4}n+\frac{\pi }{3}\right ) \). (a) Find fundamental frequency of this signal. (b) Fourier series coefficients for \(x\left [ n\right ] \). (c) if \(x\left [ n\right ] =\cos \left ( \frac{\pi }{4}n+\frac{\pi }{3}\right ) \) is an input to system with frequency response \(H\left ( \Omega \right ) =\frac{1-e^{-j\Omega }}{2+e^{-2j\Omega }}\), obtain expression for \(y\left [ n\right ] \)
Solution
Part a For \(\sin \left ( \frac{3\pi }{8}n+\frac{\pi }{2}\right ) \), we need \(\frac{3}{8}\pi N=m2\pi \) or \(\frac{m}{N}=\frac{3}{16}\). Since relatively prime, hence \(N=16\). For \(\cos \left ( \frac{\pi }{4}n+\frac{\pi }{3}\right ) \) 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 \(\Omega _{0}=\frac{2\pi }{N}=\frac{\pi }{8}\).
Part b Since input is periodic, then\begin{equation} x\left [ n\right ] =\sum _{k=-\infty }^{\infty }a_{k}e^{jk\Omega _{0}n} \tag{1} \end{equation} 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 \(\Omega _{0}\) as\[ x\left [ n\right ] =2\sin \left ( 3\Omega _{0}n+\frac{\pi }{2}\right ) +\cos \left ( 2\Omega _{0}n+\frac{\pi }{3}\right ) \] Hence\begin{align*} x\left [ n\right ] & =2\left ( \frac{e^{j\left ( 3\Omega _{0}n+\frac{\pi }{2}\right ) }-e^{-j\left ( 3\Omega _{0}n+\frac{\pi }{2}\right ) }}{2j}\right ) +\frac{e^{j\left ( 2\Omega _{0}n+\frac{\pi }{3}\right ) }+e^{-j\left ( 2\Omega _{0}n+\frac{\pi }{3}\right ) }}{2}\\ & =\frac{1}{j}e^{j\left ( 3\Omega _{0}n+\frac{\pi }{2}\right ) }-\frac{1}{j}e^{-j\left ( 3\Omega _{0}n+\frac{\pi }{2}\right ) }+\frac{1}{2}e^{j\left ( 2\Omega _{0}n+\frac{\pi }{3}\right ) }+\frac{1}{2}e^{-j\left ( 2\Omega _{0}n+\frac{\pi }{3}\right ) }\\ & =\frac{1}{j}e^{j\frac{\pi }{2}}e^{j3\Omega _{0}n}-\frac{1}{j}e^{-j\frac{\pi }{2}}e^{-j3\Omega _{0}n}+\frac{1}{2}e^{j\frac{\pi }{3}}e^{j2\Omega _{0}n}+\frac{1}{2}e^{-j\frac{\pi }{3}}e^{-j2\Omega _{0}n} \end{align*}
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{equation} x\left [ n\right ] =e^{j3\Omega _{0}n}+e^{-j3\Omega _{0}n}+\frac{1}{4}\left ( 1+\sqrt{3}j\right ) e^{j2\Omega _{0}n}-\frac{1}{4}\left ( 1-\sqrt{3}j\right ) e^{-j2\Omega _{0}n} \tag{2} \end{equation} Comparing (2) to (1) shows that\begin{align*} a_{3} & =1\\ a_{-3} & =1\\ a_{2} & =\frac{1}{2}e^{j\frac{\pi }{3}}=\frac{1}{4}\left ( 1+\sqrt{3}j\right ) \\ a_{-2} & =\frac{1}{2}e^{-j\frac{\pi }{3}}=-\frac{1}{4}\left ( 1-\sqrt{3}j\right ) \end{align*}
Part c The output is\begin{equation} y\left [ n\right ] =\left \vert H\left ( \Omega \right ) \right \vert _{\Omega =\frac{\pi }{4}}\cos \left ( \frac{\pi }{4}n+\frac{\pi }{3}+\arg H\left ( \Omega \right ) _{\Omega =\frac{\pi }{4}}\right ) \tag{1} \end{equation} But\begin{align*} \left \vert H\left ( \Omega \right ) \right \vert & =\left \vert \frac{1-e^{-j\Omega }}{2+e^{-2j\Omega }}\right \vert \\ & =\frac{\left \vert 1-e^{-j\Omega }\right \vert }{\left \vert 2+e^{-2j\Omega }\right \vert }\\ & =\frac{\sqrt{\left ( 1-\cos \Omega \right ) ^{2}+\sin ^{2}\Omega }}{\sqrt{\left ( 2+\cos 2\Omega \right ) ^{2}+\sin ^{2}\left ( 2\Omega \right ) }} \end{align*}
When \(\Omega =\frac{\pi }{4}\) the above becomes\begin{align*} \left \vert H\left ( \frac{\pi }{4}\right ) \right \vert & =\frac{\sqrt{\left ( 1-\cos \left ( \frac{\pi }{4}\right ) \right ) ^{2}+\sin ^{2}\left ( \frac{\pi }{4}\right ) }}{\sqrt{\left ( 2+\cos \left ( \frac{\pi }{2}\right ) \right ) ^{2}+\sin ^{2}\left ( \frac{\pi }{2}\right ) }}\\ & =\frac{\sqrt{\left ( 1-\cos \left ( \frac{\pi }{4}\right ) \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}}\\ & =0.342\,28 \end{align*}
And \begin{align*} \arg H\left ( \Omega \right ) & =\arg \frac{1-e^{-j\Omega }}{2+e^{-2j\Omega }}\\ & =\arg \frac{\left ( 1-\cos \Omega \right ) +j\sin \Omega }{\left ( 2+\cos \left ( 2\Omega \right ) \right ) -j\sin \left ( 2\Omega \right ) }\\ & =\arctan \left ( \frac{\sin \Omega }{1-\cos \Omega }\right ) -\arctan \left ( \frac{-\sin \left ( 2\Omega \right ) }{2+\cos \left ( 2\Omega \right ) }\right ) \end{align*}
When \(\Omega =\frac{\pi }{4}\) the above becomes\begin{align*} \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{align*}
Hence (1) becomes\[ y\left [ n\right ] =0.3423\cos \left ( \frac{\pi }{4}n+\frac{\pi }{3}+1.6418\right ) \]