Determine the Fourier transform for \(-\pi \leq \omega <\pi \) in the case of each of the following periodic signals (a) \(\sin \left ( \frac{\pi }{3}n+\frac{\pi }{4}\right ) \) (b) \(2+\cos \left ( \frac{\pi }{6}n+\frac{\pi }{8}\right ) \)
solution
Since the signal is periodic, then the Fourier transform is given by \begin{equation} X\left ( \Omega \right ) =2\pi \sum _{k=-\infty }^{\infty }a_{k}\delta \left ( \Omega -k\Omega _{0}\right ) \tag{1} \end{equation} 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 \(\Omega _{0}=\frac{2\pi }{N}=\frac{\pi }{3}\). Now, using Euler relation\begin{align} \sin \left ( \frac{\pi }{3}n+\frac{\pi }{4}\right ) & =\frac{e^{\left ( \frac{\pi }{3}n+\frac{\pi }{4}\right ) j}-e^{-\left ( \frac{\pi }{3}n+\frac{\pi }{4}\right ) j}}{2j}\nonumber \\ & =\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 ) \tag{2} \end{align}
Comparing (2) to Fourier series expansion of periodic signal given by \begin{align*} x\left [ n\right ] & =\sum _{k=0}^{N-1}a_{k}e^{jk\Omega _{0}n}\\ & =\sum _{k=0}^{5}a_{k}e^{jk\Omega _{0}n}\\ & =\sum _{k=-2}^{3}a_{k}e^{jk\Omega _{0}n} \end{align*}
Since \(\Omega _{0}=\frac{\pi }{3}\) 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{align*} X\left ( \Omega \right ) & =2\pi \left ( a_{-1}\delta \left ( \Omega +\Omega _{0}\right ) +a_{1}\delta \left ( \Omega -\Omega _{0}\right ) \right ) \\ & =2\pi \left ( -\frac{1}{2j}e^{-j\frac{\pi }{4}}\delta \left ( \Omega +\frac{\pi }{3}\right ) +\frac{1}{2j}e^{j\frac{\pi }{4}}\delta \left ( \Omega -\frac{\pi }{3}\right ) \right ) \\ & =\frac{\pi }{j}\left ( -e^{-j\frac{\pi }{4}}\delta \left ( \Omega +\frac{\pi }{3}\right ) +e^{j\frac{\pi }{4}}\delta \left ( \Omega -\frac{\pi }{3}\right ) \right ) \end{align*}
Since the signal \(2+\cos \left ( \frac{\pi }{6}n+\frac{\pi }{8}\right ) \) is periodic, then the Fourier transform is given by \begin{equation} X\left ( \Omega \right ) =2\pi \sum _{k=-\infty }^{\infty }a_{k}\delta \left ( \Omega -k\Omega _{0}\right ) \tag{1} \end{equation} 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 \(\Omega _{0}=\frac{2\pi }{N}=\frac{\pi }{6}\). Now, using Euler relation \begin{align} 2+\cos \left ( \frac{\pi }{6}n+\frac{\pi }{8}\right ) & =2+\frac{e^{\left ( \frac{\pi }{6}n+\frac{\pi }{8}\right ) j}+e^{-\left ( \frac{\pi }{6}n+\frac{\pi }{8}\right ) j}}{2}\nonumber \\ & =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 ) \tag{2} \end{align}
Comparing (2) to Fourier series expansion of periodic signal given by \begin{align*} x\left [ n\right ] & =\sum _{k=0}^{N-1}a_{k}e^{jk\Omega _{0}n}\\ & =\sum _{k=0}^{11}a_{k}e^{jk\Omega _{0}n}\\ & =\sum _{k=-5}^{6}a_{k}e^{jk\Omega _{0}n} \end{align*}
Since \(\Omega _{0}=\frac{\pi }{6}\) 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{align*} X\left ( \Omega \right ) & =2\pi \left ( a_{0}\delta \left ( \Omega \right ) +a_{-1}\delta \left ( \Omega +\Omega _{0}\right ) +a_{1}\delta \left ( \Omega -\Omega _{0}\right ) \right ) \\ & =2\pi \left ( 2\delta \left ( \Omega \right ) +\frac{1}{2}e^{-j\frac{\pi }{8}}\delta \left ( \Omega +\frac{\pi }{6}\right ) +\frac{1}{2}e^{j\frac{\pi }{8}}\delta \left ( \Omega -\frac{\pi }{6}\right ) \right ) \\ & =4\pi \delta \left ( \Omega \right ) +\pi e^{-j\frac{\pi }{8}}\delta \left ( \Omega +\frac{\pi }{6}\right ) +\pi e^{j\frac{\pi }{8}}\delta \left ( \Omega -\frac{\pi }{6}\right ) \end{align*}
Use the Fourier transform synthesis equation (5.8) \begin{align} x\left [ n\right ] & =\frac{1}{2\pi }\int _{2\pi }X\left ( \Omega \right ) e^{j\Omega n}d\Omega \tag{5.8}\\ X\left ( \Omega \right ) & =\sum _{n=-\infty }^{\infty }x\left [ n\right ] e^{-j\Omega n} \tag{5.9} \end{align}
To determine the inverse Fourier transform of \(X\left ( \Omega \right ) =\left \vert X\left ( \Omega \right ) \right \vert e^{j\arg H\left ( \Omega \right ) }\), where \(\left \vert X\left ( \Omega \right ) \right \vert =\left \{ \begin{array} [c]{ccc}1 & & 0\leq \left \vert \Omega \right \vert <\frac{\pi }{4}\\ 0 & & \frac{\pi }{4}\leq \left \vert \Omega \right \vert <\pi \end{array} \right . \) and \(\arg H\left ( \Omega \right ) =\frac{-3\Omega }{2}\). Use your answer to determine the values of n for which \(x\left [ n\right ] =0\).
solution\begin{align*} x\left [ n\right ] & =\frac{1}{2\pi }\int _{2\pi }X\left ( \Omega \right ) e^{j\Omega n}d\Omega \\ & =\frac{1}{2\pi }\int _{2\pi }\left \vert X\left ( \Omega \right ) \right \vert e^{j\arg H\left ( \Omega \right ) }e^{j\Omega n}d\Omega \\ & =\frac{1}{2\pi }\int _{0}^{\frac{\pi }{4}}e^{j\arg H\left ( \Omega \right ) }e^{j\Omega n}d\Omega \\ & =\frac{1}{2\pi }\int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}e^{j\frac{-3\Omega }{2}}e^{j\Omega n}d\Omega \\ & =\frac{1}{2\pi }\int _{-\frac{\pi }{4}}^{\frac{\pi }{4}}e^{j\Omega \left ( \frac{-3}{2}+n\right ) }d\Omega \\ & =\frac{1}{2\pi }\frac{1}{j\left ( \frac{-3}{2}+n\right ) }\left [ e^{j\Omega \left ( \frac{-3}{2}+n\right ) }\right ] _{-\frac{\pi }{4}}^{\frac{\pi }{4}}\\ & =\frac{1}{2\pi }\frac{1}{j\left ( \frac{-3}{2}+n\right ) }\left [ e^{j\frac{\pi }{4}\left ( \frac{-3}{2}+n\right ) }-e^{-j\frac{\pi }{4}\left ( \frac{-3}{2}+n\right ) }\right ] \\ & =\frac{1}{\pi }\frac{1}{\left ( \frac{-3}{2}+n\right ) }\left [ \frac{e^{j\frac{\pi }{4}\left ( \frac{-3}{2}+n\right ) }-e^{-j\frac{\pi }{4}\left ( \frac{-3}{2}+n\right ) }}{2j}\right ] \\ & =\frac{1}{\pi }\frac{1}{\left ( \frac{-3}{2}+n\right ) }\sin \left ( \frac{\pi }{4}\left ( \frac{-3}{2}+n\right ) \right ) \\ & =\frac{1}{\pi }\frac{\sin \left ( \frac{\pi }{4}\left ( n-\frac{3}{2}\right ) \right ) }{n-\frac{3}{2}} \end{align*}
Now the above is zero when \(\sin \left ( \frac{\pi }{4}\left ( n-\frac{3}{2}\right ) \right ) =0\) or \(\frac{\pi }{4}\left ( n-\frac{3}{2}\right ) =m\pi \) 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 \(\sin \left ( \frac{\pi }{4}\left ( n-\frac{3}{2}\right ) \right ) =0\). The other option is to look at denominator of \(\frac{\sin \left ( \frac{\pi }{4}\left ( n-\frac{3}{2}\right ) \right ) }{n-\frac{3}{2}}\) and ask where is that \(\infty \). This happens when \(n\rightarrow \pm \infty \) and only then \(x\left [ n\right ] =0\).
The following four facts are given about a real signal \(x\left [ n\right ] \) with Fourier transform \(X\left ( \Omega \right ) \)
Determine \(x\left [ n\right ] \)
solution
From tables we know that the odd part of \(x\left [ n\right ] \) has Fourier transform which is \(j\operatorname{Im}\left ( X\left ( \Omega \right ) \right ) \). Hence using (3) above, this means that odd part of \(x\left [ n\right ] \) has Fourier transform of \(j\left ( \sin \Omega -\sin \left ( 2\Omega \right ) \right ) \) or \(j\left ( \frac{e^{j\Omega }-e^{-j\Omega }}{2j}-\frac{e^{j2\Omega }-e^{-j2\Omega }}{2j}\right ) \) or \(\frac{1}{2}\left ( e^{j\Omega }-e^{-j\Omega }-e^{j2\Omega }+e^{-j2\Omega }\right ) \). From tables, we know find the inverse Fourier transform of this. Hence odd part of \(x\left [ n\right ] \) is \(\frac{1}{2}\left ( \delta \left [ n+1\right ] -\delta \left [ n-1\right ] -\delta \left [ n+2\right ] +\delta \left [ n-2\right ] \right ) \). 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 \(\frac{1}{2}\left ( \delta \left [ n+1\right ] -\delta \left [ n+2\right ] \right ) \).
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 \(x\left [ n\right ] =2\left ( \frac{1}{2}\left ( \delta \left [ n+1\right ] -\delta \left [ n+2\right ] \right ) \right ) \) for \(n<0\). Hence\[ x\left [ n\right ] =\delta \left [ n+1\right ] -\delta \left [ n+2\right ] \qquad n<0 \] Finally, using (4) above, \[ \frac{1}{2\pi }\int _{-\pi }^{\pi }\left \vert X\left ( \Omega \right ) \right \vert ^{2}d\Omega =3=\sum _{n=-\infty }^{\infty }\left \vert x\left [ n\right ] \right \vert ^{2}=\sum _{n=-\infty }^{0}\left \vert x\left [ n\right ] \right \vert ^{2}\] Hence\begin{align*} 3 & =\left \vert \delta \left [ -1\right ] \right \vert ^{2}+\left \vert \delta \left [ -2\right ] \right \vert ^{2}+\left \vert x\left [ 0\right ] \right \vert ^{2}\\ & =1+1+x\left [ n\right ] ^{2}\\ x\left [ n\right ] ^{2} & =3-2\\ & =1 \end{align*}
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\[ x\left [ n\right ] =\delta \left [ n+1\right ] -\delta \left [ n+2\right ] +\delta \left [ n\right ] \qquad n\leq 0 \]
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 \[ H\left ( \Omega \right ) =\frac{-12+5e^{-j\Omega }}{12-7e^{-j\Omega }+e^{-j2\Omega }}\] 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 \(H\left ( \Omega \right ) =H_{1}\left ( \Omega \right ) +H_{2}\left ( \Omega \right ) \). Hence \begin{equation} H_{2}\left ( \Omega \right ) =H\left ( \Omega \right ) -H_{1}\left ( \Omega \right ) \tag{1} \end{equation} But \begin{align*} H_{1}\left ( \Omega \right ) & =\sum _{n=-\infty }^{\infty }h_{1}\left [ n\right ] e^{-j\Omega n}\\ & =\sum _{n=0}^{\infty }\left ( \frac{1}{3}\right ) ^{n}e^{-j\Omega n}\\ & =\sum _{n=0}^{\infty }\left ( \frac{1}{3}e^{-j\Omega }\right ) ^{n}\\ & =\sum _{n=0}^{\infty }a^{n}=\frac{1}{1-a}=\frac{1}{1-\frac{1}{3}e^{-j\Omega }}\\ & =\frac{3}{3-e^{-j\Omega }} \end{align*}
Therefore from (1)\[ H_{2}\left ( \Omega \right ) =\frac{-12+5e^{-j\Omega }}{12-7e^{-j\Omega }+e^{-j2\Omega }}-\frac{3}{3-e^{-j\Omega }}\] Let \(e^{-j\Omega }=x\) to simplify notation. The above becomes\begin{align*} H_{2}\left ( \Omega \right ) & =\frac{-12+5x}{12-7x+x^{2}}-\frac{3}{3-x}\\ & =\frac{-12+5x}{\left ( x-3\right ) \left ( x-4\right ) }+\frac{3}{\left ( x-3\right ) }\\ & =\frac{-12+5x+3\left ( x-4\right ) }{\left ( x-3\right ) \left ( x-4\right ) }\\ & =\frac{-12+5x+3x-12}{\left ( x-3\right ) \left ( x-4\right ) }\\ & =\frac{8x-24}{\left ( x-3\right ) \left ( x-4\right ) }\\ & =\frac{8\left ( x-3\right ) }{\left ( x-3\right ) \left ( x-4\right ) }\\ & =\frac{8}{\left ( x-4\right ) }\\ & =-2\left ( \frac{1}{1-\frac{1}{4}x}\right ) \end{align*}
Hence\[ H_{2}\left ( \Omega \right ) =-2\left ( \frac{1}{1-\frac{1}{4}e^{-j\Omega }}\right ) \] from tables, \(a^{n}u\left [ n\right ] \Longleftrightarrow \frac{1}{1-ae^{-j\Omega }}\) for \(\left \vert a\right \vert <1\). Comparing this to the above gives \[ h_{2}\left [ n\right ] =-2\left ( \frac{1}{4}\right ) ^{n}u\left [ n\right ] \]
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
\[ y\left [ n\right ] -\frac{1}{6}y\left [ n-1\right ] -\frac{1}{6}y\left [ n-2\right ] =x\left [ n\right ] \]
(a) Determine the frequency response \(H\left ( \Omega \right ) \) for the system \(S\). (b) Determine the impulse response \(h[n]\) for the system \(S\).
solution
Taking DFT of the difference equation gives\begin{align*} Y\left ( \Omega \right ) -\frac{1}{6}e^{-j\Omega }Y\left ( \Omega \right ) -\frac{1}{6}e^{-j2\Omega }Y\left ( \Omega \right ) & =X\left ( \Omega \right ) \\ Y\left ( \Omega \right ) \left ( 1-\frac{1}{6}e^{-j\Omega }-\frac{1}{6}e^{-j2\Omega }\right ) & =X\left ( \Omega \right ) \\ \frac{Y\left ( \Omega \right ) }{X\left ( \Omega \right ) } & =\frac{1}{1-\frac{1}{6}e^{-j\Omega }-\frac{1}{6}e^{-j2\Omega }} \end{align*}
Let \(e^{-j\Omega }=x\) to simplify the notation, then \[ \frac{Y\left ( \Omega \right ) }{X\left ( \Omega \right ) }=\frac{1}{1-\frac{1}{6}x-\frac{1}{6}x^{2}}=\frac{6}{6-x-x^{2}}=\frac{-6}{x^{2}+x-6}=\frac{-6}{\left ( x-2\right ) \left ( x+3\right ) }\] Hence\begin{align*} H\left ( \Omega \right ) & =\frac{Y\left ( \Omega \right ) }{X\left ( \Omega \right ) }\\ & =\frac{-6}{\left ( e^{-j\Omega }-2\right ) \left ( e^{-j\Omega }+3\right ) } \end{align*}
Applying partial fractions\[ H\left ( \Omega \right ) =\frac{-6}{\left ( e^{-j\Omega }-2\right ) \left ( e^{-j\Omega }+3\right ) }=\frac{A}{\left ( x-2\right ) }+\frac{B}{\left ( x+3\right ) }\] Hence \(A=-\frac{6}{5},B=\frac{6}{5}\). Therefore\begin{align*} H\left ( \Omega \right ) & =-\frac{6}{5}\frac{1}{e^{-j\Omega }-2}+\frac{6}{5}\frac{1}{e^{-j\Omega }+3}\\ & =-\frac{3}{5}\frac{1}{\frac{1}{2}e^{-j\Omega }-1}+\frac{2}{5}\frac{1}{\frac{1}{3}e^{-j\Omega }+1}\\ & =\frac{3}{5}\frac{1}{1-\frac{1}{2}e^{-j\Omega }}+\frac{2}{5}\frac{1}{1+\frac{1}{3}e^{-j\Omega }} \end{align*}
Taking the inverse DFT using tables gives\begin{align*} h\left [ n\right ] & =\frac{3}{5}\left ( \frac{1}{2}\right ) ^{n}u\left [ n\right ] +\frac{2}{5}\left ( -\frac{1}{3}\right ) ^{n}u\left [ n\right ] \\ & =\left ( \frac{3}{5}\left ( \frac{1}{2}\right ) ^{n}+\frac{2}{5}\left ( -\frac{1}{3}\right ) ^{n}\right ) u\left [ n\right ] \end{align*}
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\[ h\left ( n\right ) =\frac{W}{\pi }\operatorname{sinc}\left ( \frac{Wn}{\pi }\right ) =\frac{\sin \left ( Wn\right ) }{\pi n}\] (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
Given \(h\left ( n\right ) =\frac{W}{\pi }\operatorname{sinc}\left ( \frac{Wn}{\pi }\right ) =\frac{\sin \left ( Wn\right ) }{\pi n}\). We will show that \(H\left ( \Omega \right ) \) is the rectangle function by reverse. Assuming that \(H\left ( \Omega \right ) =\begin{array} [c]{ccc}1 & & \left \vert \Omega \right \vert <2W\\ 0 & & \text{otherwise}\end{array} \) therefore\begin{align*} x\left [ n\right ] & =\frac{1}{2\pi }\int _{-W}^{W}X\left ( \Omega \right ) e^{j\Omega n}d\omega \\ & =\frac{1}{2\pi }\int _{-W}^{W}e^{j\Omega n}d\omega \\ & =\frac{1}{2\pi }\frac{e^{jWn}-e^{-jWn}}{jn}\\ & =\frac{1}{\pi n}\sin \left ( Wn\right ) \end{align*}
Which is the \(h\left [ n\right ] \) given. Therefore, the above shows that \(\frac{\sin \left ( Wn\right ) }{\pi n}\) has DFT of \(H\left ( \Omega \right ) \) as the rectangle function. Here is sketch
\[ x\left [ n\right ] =\sin \left ( \frac{\pi n}{8}\right ) -2\cos \left ( \frac{\pi n}{4}\right ) \] (i) \(h\left [ n\right ] =\frac{\sin \left ( \frac{\pi n}{6}\right ) }{\pi n}\). Hence \(y\left [ n\right ] =x\left [ n\right ] \circledast h\left [ n\right ] \). Or \(Y\left ( \Omega \right ) =X\left ( \Omega \right ) H\left ( \Omega \right ) \), 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 \[ y\left [ n\right ] =\sin \left ( \frac{n\pi }{8}\right ) \]
Here is plot of \(y\left [ n\right ] \) for \(n=-8\cdots 8\)
(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\[ y\left [ n\right ] =2\sin \left ( \frac{n\pi }{8}\right ) -2\cos \left ( \frac{n\pi }{4}\right ) \]
Here is plot of \(y\left [ n\right ] \) for \(n=-8\cdots 8\)
