For the continuous-time periodic signal \(x\left ( t\right ) =2+\cos \left ( \frac{2\pi }{3}t\right ) +4\sin \left ( \frac{5\pi }{3}t\right ) \) determine the fundamental frequency \(\omega _{0}\) and the Fourier series coefficients \(a_{k}\) such that \(x\left ( t\right ) =\sum _{k=-\infty }^{\infty }a_{k}e^{jk\omega _{0}t}\)
Solution
The signal \(\cos \left ( \frac{2\pi }{3}t\right ) \) has period \(\frac{2\pi }{T_{1}}=\frac{2\pi }{3}\). Hence \(T_{1}=3\) and the signal \(\sin \left ( \frac{5\pi }{3}t\right ) \) has period \(\frac{2\pi }{T_{2}}=\frac{5\pi }{3}\) or \(T_{2}=\frac{6}{5}\). Therefore the LCM of \(3,\frac{6}{5}\) is \begin{align*} 3m & =\frac{6}{5}n\\ \frac{m}{n} & =\frac{2}{5} \end{align*}
Hence \(m=2\) and \(n=5\). Therefore \(T_{0}=6\). Therefore \begin{align*} \omega _{0} & =\frac{2\pi }{T_{0}}\\ & =\frac{2\pi }{6}\\ & =\frac{\pi }{3} \end{align*}
Hence \begin{equation} x\left ( t\right ) =\sum _{k=-\infty }^{\infty }a_{k}e^{jk\omega _{0}t}\tag{1} \end{equation} Where \begin{equation} a_{k}=\frac{1}{T_{0}}\int _{-\frac{T}{2}}^{\frac{T}{2}}x\left ( t\right ) e^{-jk\omega _{0}t}\tag{2} \end{equation} To find \(a_{k}\) for the given signal, instead of using the above integration formula, we could write the signal \(x\left ( t\right ) \) in exponential form using Euler relation and just read the \(a_{k}\) coefficients directly from the result. The signal \(x\left ( t\right ) \) can be written as\begin{align} x\left ( t\right ) & =2+\frac{e^{j\frac{2\pi }{3}t}+e^{-j\frac{2\pi }{3}t}}{2}+4\frac{e^{j\frac{5\pi }{3}t}-e^{-j\frac{5\pi }{3}t}}{2i}\nonumber \\ & =2+\frac{e^{j2\omega _{0}t}+e^{-j2\omega _{0}t}}{2}+4\frac{e^{j5\omega _{0}t}-e^{-j5\omega _{0}t}}{2i}\nonumber \\ & =2+\frac{1}{2}e^{j2\omega _{0}t}+\frac{1}{2}e^{-j2\omega _{0}t}+2ie^{j5\omega _{0}t}-2ie^{-j5\omega _{0}t}\tag{3} \end{align}
Comparing (3) to (1) shows that the coefficients are\begin{align*} a_{0} & =2\\ a_{2} & =\frac{1}{2}\\ a_{-2} & =\frac{1}{2}\\ a_{5} & =2j\\ a_{-5} & =-2j \end{align*}
Let \(x\left [ n\right ] \) be real and odd periodic signal with period \(N=7\) and Fourier coefficients \(a_{k}\). Given that \(a_{15}=j,a_{16}=2j,a_{17}=3j\), determine the values of \(a_{0},a_{-1},a_{-2},a_{-3}\).
Solution
For discrete signal\begin{align*} x\left [ n\right ] & =\sum _{k=0}^{N-1}a_{k}e^{jk\omega _{0}n}\\ & =\sum _{k=0}^{N-1}a_{k}e^{jk\frac{2\pi }{N}n} \end{align*}
Where\begin{align*} a_{k} & =\frac{1}{N}\sum _{n=0}^{N-1}x\left [ n\right ] e^{-jk\omega _{0}n}\\ & =\frac{1}{N}\sum _{n=0}^{N-1}x\left [ n\right ] e^{-jk\frac{2\pi }{N}n} \end{align*}
Since the signal \(x\left [ n\right ] \) is real, then we know that \(a_{k}=a_{-k}^{\ast }\). And since \(x\left [ n\right ] \) is odd then we know that \(a_{k}\) is purely imaginary and odd. The Fourier coefficients repeat every \(N\) samples which is \(7\). Hence \(a_{15}=a_{9}=a_{1}\) and \(a_{16}=a_{9}=a_{2}\) and \(a_{17}=a_{10}=a_{3}\). And since \(a_{k}\) is odd then\begin{align*} a_{0} & =0\\ a_{1} & =-a_{-1}\\ a_{2} & =-a_{-2}\\ a_{3} & =-a_{-3} \end{align*}
But we know from above that \(a_{1}=a_{15}=j\) and \(a_{2}=a_{16}=2j\) and \(a_{3}=a_{17}=3j\) then the above gives\begin{align*} a_{0} & =0\\ a_{-1} & =-j\\ a_{-2} & =-2j\\ a_{-3} & =-3j \end{align*}
Solution
The output of discrete LTI system when the input is \(x\left [ n\right ] =a_{n}e^{jn\omega }\) is given by \(y\left [ n\right ] =a_{n}H\left ( e^{j\omega }\right ) e^{jn\omega }\) where \(H\left ( e^{j\omega }\right ) \) is given to us in the problem statement. Hence, to find \(y\left [ n\right ] \) we need to express each input in its Fourier series representation in order to determine the \(a_{n}\).
Here \(x_{1}\left [ n\right ] =\left ( -1\right ) ^{n}=\left ( e^{j\pi }\right ) ^{n}=e^{jn\pi }\). To find the period\(N\), let \(x_{1}\left [ n\right ] =x_{1}\left [ n+N\right ] \) or \begin{align*} e^{jn\pi } & =e^{j\left ( n+N\right ) \pi }\\ & =e^{jn\pi }e^{jN\pi } \end{align*}
Hence \(N=2\). Therefore \(\omega _{0}=\frac{2\pi }{N}=\frac{2\pi }{2}=\pi \) and \(x_{1}\left [ n\right ] =\sum _{k=0}^{N-1}a_{k}e^{jk\omega _{0}n}=a_{0}+a_{1}e^{j\pi n}\). Comparing this to \(e^{jn\pi }\) shows that \begin{align*} a_{0} & =0\\ a_{1} & =1 \end{align*}
Now that we found the Fourier coefficients for \(x_{1}\left [ n\right ] \) then the output is\begin{align*} y_{1}\left [ n\right ] & =\sum _{k=0}^{N-1}a_{n}H\left ( jk\omega _{0}\right ) e^{jnk\omega _{0}}\\ & =a_{0}H\left ( 0\right ) e^{0}+a_{1}H\left ( j\pi \right ) e^{jn\pi } \end{align*}
But \(a_{0}=1,a_{1}=1\) and the above becomes\[ y_{1}\left [ n\right ] =H\left ( j\pi \right ) e^{jn\pi }\] From the graph of \(H\left ( jk\omega _{0}\right ) \) given, we see that at \(\omega =\pi ,H\left ( j\pi \right ) =0\). Therefore\[ y_{1}\left [ n\right ] =0 \]
Here \(x_{2}\left [ n\right ] =1+\sin \left ( \frac{3\pi }{8}n+\frac{\pi }{4}\right ) \). The first step is to find the period \(N\)\begin{align*} x_{2}\left [ n\right ] & =x_{2}\left [ n+N\right ] \\ 1+\sin \left ( \frac{3\pi }{8}n+\frac{\pi }{4}\right ) & =1+\sin \left ( \frac{3\pi }{8}\left ( n+N\right ) +\frac{\pi }{4}\right ) \\ & =1+\sin \left ( \frac{3\pi }{8}n+\frac{3\pi }{8}N+\frac{\pi }{4}\right ) \\ & =1+\sin \left ( \left ( \frac{3\pi }{8}n+\frac{\pi }{4}\right ) +\frac{3\pi }{8}N\right ) \end{align*}
Hence \(\frac{3\pi }{8}N=2\pi m\) or \(\frac{N}{m}=\frac{16}{3}\). Since these are relatively prime, then \(N=16\) is the fundamental period. Therefore\[ x_{2}\left [ n\right ] =\sum _{k=0}^{N-1}a_{k}e^{jk\omega _{0}n}\] where \(\omega _{0}=\frac{2\pi }{N}=\frac{2\pi }{16}=\frac{\pi }{8}\). The above becomes\begin{equation} x_{2}\left [ n\right ] =\sum _{k=0}^{15}a_{k}e^{jk\frac{\pi }{8}n}\tag{1} \end{equation} But \begin{align} 1+\sin \left ( \frac{3\pi }{8}n+\frac{\pi }{4}\right ) & =1+\frac{e^{j\left ( \frac{3\pi }{8}n+\frac{\pi }{4}\right ) }-e^{-j\left ( \frac{3\pi }{8}n+\frac{\pi }{4}\right ) }}{2j}\nonumber \\ & =1+\frac{1}{2j}e^{j\frac{3\pi }{8}n}e^{j\frac{\pi }{4}}-\frac{1}{2j}e^{-j\frac{3\pi }{8}n}e^{-j\frac{\pi }{4}}\tag{2} \end{align}
Comparing (1) and (2) shows that \(a_{0}=1,a_{3}=\frac{1}{2j}e^{j\frac{\pi }{4}},a_{-3}=-\frac{1}{2j}e^{-j\frac{\pi }{4}}\). But \(a_{-3}=a_{-3+16}=a_{13}\) due to periodicity (and since we want to keep the index from \(0\) to \(15\). Therefore\begin{align*} a_{0} & =1\\ a_{3} & =\frac{1}{2j}e^{j\frac{\pi }{4}}\\ a_{13} & =-\frac{1}{2j}e^{-j\frac{\pi }{4}} \end{align*}
And all other \(a_{k}=0\). Now that we found the Fourier coefficient, then the response \(y_{2}\left [ n\right ] \) is found from\begin{align*} y_{2}\left [ n\right ] & =\sum _{k=0}^{N-1}a_{n}H\left ( jk\omega _{0}\right ) e^{jkn\omega _{0}}\\ & =a_{0}H\left ( 0\right ) +a_{3}H\left ( j3\frac{\pi }{8}\right ) e^{j3\frac{\pi }{8}n}+a_{13}H\left ( j13\frac{\pi }{8}\right ) e^{j13\frac{\pi }{8}n}\\ & =H\left ( 0\right ) +\left ( \frac{1}{2j}e^{j\frac{\pi }{4}}\right ) H\left ( j\frac{3\pi }{8}\right ) e^{j\frac{3\pi }{8}n}+\left ( -\frac{1}{2j}e^{-j\frac{\pi }{4}}\right ) H\left ( j\frac{13\pi }{8}\right ) e^{j\frac{13\pi }{8}n} \end{align*}
From the graph of \(H\left ( jk\omega _{0}\right ) \) given, we see that at \(\omega =0,H\left ( 0\right ) =0\) and at \(\omega =\frac{3\pi }{8},H\left ( j\frac{3\pi }{8}\right ) =1\) and that at \(\omega =\frac{13\pi }{8},H\left ( j\frac{13\pi }{8}\right ) =1\). Hence the above becomes\[ y_{2}\left [ n\right ] =\left ( \frac{1}{2j}e^{j\frac{\pi }{4}}\right ) e^{j\frac{3\pi }{8}n}+\left ( -\frac{1}{2j}e^{-j\frac{\pi }{4}}\right ) e^{j\frac{13\pi }{8}n}\] But \(e^{j\frac{13\pi }{8}n}=e^{j\frac{-3\pi }{8}n}\) since period is \(N=16\). Therefore the above simplifies to\begin{align*} y_{2}\left [ n\right ] & =\left ( \frac{1}{2j}e^{j\frac{\pi }{4}}\right ) e^{j\frac{3\pi }{8}n}+\left ( -\frac{1}{2j}e^{-j\frac{\pi }{4}}\right ) e^{j\frac{-3\pi }{8}n}\\ & =\frac{e^{j\left ( \frac{\pi }{4}+\frac{3\pi }{8}n\right ) }-e^{-j\left ( \frac{\pi }{4}+\frac{3\pi }{8}n\right ) }}{2j}\\ & =\sin \left ( \frac{3\pi }{8}n+\frac{\pi }{4}\right ) \end{align*}
Solution
Input voltage is \(x\left ( t\right ) \). Hence drop in voltage around circuit is \[ x\left ( t\right ) =Ri\left ( t\right ) +L\frac{di}{dt}+y\left ( t\right ) \] Now we need to relate the current \(i\left ( t\right ) \) to \(y\left ( t\right ) \). Since current across the capacitor is given by \(i\left ( t\right ) =C\frac{dy}{dt}\) then replacing \(i\left ( t\right ) \) in the above by \(C\frac{dy}{dt}\) gives the diffeential equation\[ x\left ( t\right ) =RC\frac{dy}{dt}+LC\frac{d^{2}y}{dt^{2}}+y\left ( t\right ) \] Or\[ LCy^{\prime \prime }\left ( t\right ) +RCy^{\prime }\left ( t\right ) +y\left ( t\right ) =x\left ( t\right ) \] But \(L=1,R=1,C=1\) therefore\[ y^{\prime \prime }\left ( t\right ) +y^{\prime }\left ( t\right ) +y\left ( t\right ) =x\left ( t\right ) \]
Let the input \(x\left ( t\right ) =e^{j\omega t}\). Therefore \(y\left ( t\right ) =H\left ( \omega \right ) e^{j\omega t}\) where \(H\left ( \omega \right ) \) is the frequency response (Book writes this as \(H\left ( e^{j\omega }\right ) \) but \(H\left ( \omega \right ) \) is simpler notation).
Hence \begin{align*} y^{\prime }\left ( t\right ) & =H\left ( \omega \right ) j\omega e^{j\omega t}\\ y^{\prime \prime }\left ( t\right ) & =H\left ( \omega \right ) \left ( j\omega \right ) ^{2}e^{j\omega t}\\ & =-H\left ( \omega \right ) \omega ^{2}e^{j\omega t} \end{align*}
Substituting the above into the ODE gives\[ -H\left ( \omega \right ) \omega ^{2}e^{j\omega t}+H\left ( \omega \right ) j\omega e^{j\omega t}+H\left ( \omega \right ) e^{j\omega t}=e^{j\omega t}\] Dividing by \(e^{j\omega t}\neq 0\) results in\[ -H\left ( \omega \right ) \omega ^{2}+H\left ( \omega \right ) j\omega +H\left ( \omega \right ) =1 \] Solving for \(H\left ( \omega \right ) \) gives
\begin{align} H\left ( \omega \right ) \left ( -\omega ^{2}+j\omega +1\right ) & =1\tag{1}\\ H\left ( \omega \right ) & =\frac{1}{-\omega ^{2}+j\omega +1}\nonumber \end{align}
Since we now know \(H\left ( \omega \right ) \) then the output \(y\left ( t\right ) \) when the input is \(x\left ( t\right ) =\sin \left ( t\right ) \) is given by\begin{equation} y\left ( t\right ) =\sum _{k=-\infty }^{\infty }a_{k}H\left ( k\omega _{0}\right ) e^{jk\omega _{0}t}\tag{2} \end{equation} Where \(a_{k}\) are the Fourier coefficients of \(\sin \left ( t\right ) \) and \(\omega _{0}\) is the fundamental frequency of \(x\left ( t\right ) \). Since \(\sin \left ( t\right ) =\sin \left ( \frac{2\pi }{T}t\right ) \) then \(\frac{2\pi }{T}=1\) and \(T=2\pi \). Hence \(\omega _{0}=1\). And since \(\sin \left ( t\right ) =\frac{1}{2j}\left ( e^{jt}-e^{-jt}\right ) \) then \(a_{1}=\frac{1}{2j},a_{-1}=-\frac{1}{2j}\). Eq. (2) becomes\begin{align} y\left ( t\right ) & =a_{-1}H\left ( -\omega _{0}\right ) e^{-j\omega _{0}t}+a_{1}H\left ( \omega _{0}\right ) e^{j\omega _{0}t}\nonumber \\ & =-\frac{1}{2j}H\left ( -1\right ) e^{-jt}+\frac{1}{2j}H\left ( 1\right ) e^{jt}\tag{3} \end{align}
Now we need to find \(H\left ( -1\right ) ,H\left ( 1\right ) \). From (1) \begin{align*} H\left ( -1\right ) & =\frac{1}{-\left ( -1\right ) ^{2}-j\left ( -1\right ) +1}\\ & =\frac{1}{-1+j+1}\\ & =\frac{1}{j} \end{align*}
And\begin{align*} H\left ( +1\right ) & =\frac{1}{-\left ( +1\right ) ^{2}-j\left ( +1\right ) +1}\\ & =\frac{1}{-1-j+1}\\ & =\frac{1}{j} \end{align*}
Therefore (3) becomes\begin{align*} y\left ( t\right ) & =-\frac{1}{2j}\frac{1}{j}e^{-jt}+\frac{1}{2j}\frac{1}{j}e^{jt}\\ & =-\frac{1}{2j^{2}}e^{-jt}+\frac{1}{2j^{2}}e^{jt}\\ & =\frac{1}{2}e^{-jt}-\frac{1}{2}e^{jt}\\ & =-\left ( \frac{1}{2}e^{jt}-\frac{1}{2}e^{-jt}\right ) \end{align*}
Hence \[ y\left ( t\right ) =-\cos \left ( t\right ) \]
Solution
First signal
The signal in P3.28(a) has period \(N=7\). Therefore \(x\left [ n\right ] =\sum _{k=0}^{N-1}a_{k}e^{jn\left ( k\omega _{0}\right ) }\). We need to determine \(a_{k}.\) Since \(\omega _{0}=\frac{2\pi }{N}=\frac{2\pi }{7}\), then\begin{align*} a_{k} & =\frac{1}{N}\sum _{n=0}^{N-1}x\left [ n\right ] e^{-jk\omega _{0}n}\\ & =\frac{1}{N}\sum _{n=0}^{N-1}x\left [ n\right ] e^{-jk\frac{2\pi }{N}n}\\ & =\frac{1}{7}\sum _{n=0}^{6}x\left [ n\right ] e^{-jk\frac{2\pi }{7}n} \end{align*}
We first notice that \(x\left [ n\right ] =0\) for \(n=5,6\) and \(x\left [ n\right ] =1\) otherwise. Hence the above sum simplifies to\[ a_{k}=\frac{1}{7}\sum _{n=0}^{4}e^{-jk\frac{2\pi }{7}n}\] Using the relation \(\sum _{n=0}^{M-1}a^{n}=\left \{ \begin{array} [c]{ccc}M & & a=1\\ \frac{1-a^{M}}{1-a} & & a\neq 1 \end{array} \right . \) to simplify the above where now \(M=5\) gives\begin{align*} a_{k} & =\frac{1}{7}\frac{1-\left ( e^{-jk\frac{2\pi }{7}}\right ) ^{5}}{1-e^{-jk\frac{2\pi }{7}}}\qquad k=0,1,\cdots 6\\ & =\frac{1}{7}\frac{1-e^{-jk\frac{10\pi }{7}}}{1-e^{-jk\frac{2\pi }{7}}} \end{align*}
This is plot of \(\left \vert a_{k}\right \vert \)
This is plot of the phase of \(a_{k}\)
second signal
The signal in P3.28(b) has period \(N=6\). Therefore \(x\left [ n\right ] =\sum _{k=0}^{N-1}a_{k}e^{jn\left ( k\omega _{0}\right ) }\). We need to determine \(a_{k}\), where \(\omega _{0}=\frac{2\pi }{N}=\frac{2\pi }{6}\). Hence\begin{align*} a_{k} & =\frac{1}{N}\sum _{n=0}^{N-1}x\left [ n\right ] e^{-jk\omega _{0}n}\\ & =\frac{1}{6}\sum _{n=0}^{5}x\left [ n\right ] e^{-jk\frac{2\pi }{6}n} \end{align*}
We first notice that \(x\left [ n\right ] =0\) for \(n=4,5\) and \(x\left [ n\right ] =1\) otherwise, Hence the above sum simplifies to\[ a_{k}=\frac{1}{6}\sum _{n=0}^{3}e^{-jk\frac{2\pi }{6}n}\] Using the relation \(\sum _{n=0}^{M-1}a^{n}=\left \{ \begin{array} [c]{ccc}M & & a=1\\ \frac{1-a^{M}}{1-a} & & a\neq 1 \end{array} \right . \) to simplify the above, where now \(M=4\) gives\begin{align*} a_{k} & =\frac{1}{6}\frac{1-\left ( e^{-jk\frac{2\pi }{6}}\right ) ^{4}}{1-e^{-jk\frac{2\pi }{6}}}\qquad k=0,1,\cdots 5\\ & =\frac{1}{6}\frac{1-e^{-jk\frac{8\pi }{6}}}{1-e^{-jk\frac{2\pi }{6}}} \end{align*}
This is plot of \(\left \vert a_{k}\right \vert \)
This is plot of the phase of \(a_{k}\)
Third signal
The signal in P3.28(c) also has period \(N=6\). Therefore \(x\left [ n\right ] =\sum _{k=0}^{N-1}a_{k}e^{jn\left ( k\omega _{0}\right ) }\). We need to determine \(a_{k}\). Given that \(\omega _{0}=\frac{2\pi }{N}\) then\begin{align*} a_{k} & =\frac{1}{N}\sum _{n=0}^{N-1}x\left [ n\right ] e^{-jk\omega _{0}n}\\ & =\frac{1}{6}\sum _{n=0}^{5}x\left [ n\right ] e^{-jk\frac{2\pi }{6}n} \end{align*}
Where \(x\left [ 0\right ] =1,x\left [ 1\right ] =2,x\left [ 2\right ] =-1,x\left [ 3\right ] =0,x\left [ 4\right ] =-1,x\left [ 5\right ] =2\). Hence the above sum becomes\begin{align*} a_{k} & =\frac{1}{6}\left ( 1+2e^{-jk\frac{2\pi }{6}}-e^{-jk\frac{2\pi }{6}2}+0-e^{-jk\frac{2\pi }{6}4}+2e^{-jk\frac{2\pi }{6}5}\right ) \\ & =\frac{1}{6}\left ( 1+2e^{-jk\frac{2\pi }{6}}-e^{-jk\frac{4\pi }{6}}-e^{-jk\frac{8\pi }{6}}+2e^{-jk\frac{10\pi }{6}}\right ) \end{align*}
This is plot of \(\left \vert a_{k}\right \vert \)
This is plot of the phase of \(a_{k}\)
\[ x\left [ n\right ] =\sin \left ( 2\pi \frac{n}{3}\right ) \cos \left ( \pi \frac{n}{2}\right ) \] The first step is to find \(N\), the fundamental period. Since \(\sin \left ( A\right ) \cos \left ( B\right ) =\frac{1}{2}\left ( \sin \left ( A+B\right ) +\sin \left ( A-B\right ) \right ) \) then\begin{align*} x\left [ n\right ] & =\frac{1}{2}\left ( \sin \left ( 2\pi \frac{n}{3}+\pi \frac{n}{2}\right ) +\sin \left ( 2\pi \frac{n}{3}-\pi \frac{n}{2}\right ) \right ) \\ & =\frac{1}{2}\left ( \sin \left ( \frac{7}{6}\pi n\right ) +\sin \left ( \frac{1}{6}\pi n\right ) \right ) \end{align*}
To find the period of \(\sin \left ( \frac{7}{6}\pi n\right ) =\sin \left ( \frac{7}{6}\pi \left ( n+N\right ) \right ) \) or \(\sin \left ( \frac{7}{6}\pi n\right ) =\sin \left ( \frac{7}{6}\pi n+\frac{7}{6}\pi N\right ) \). Hence \(\frac{7}{6}\pi N=2\pi m\) which gives \(\frac{m}{N}=\frac{7}{12}\). Hence \(N=12\).
The period of \(\sin \left ( \frac{1}{6}\pi n\right ) =\sin \left ( \frac{1}{6}\pi \left ( n+N\right ) \right ) \) or \(\sin \left ( \frac{1}{6}\pi n\right ) =\sin \left ( \frac{1}{6}\pi n+\frac{1}{6}\pi N\right ) \). Hence \(\frac{1}{6}\pi N=2\pi m\) or \(\frac{m}{N}=\frac{1}{12}\). Hence common period is \(N=12\). Now that we know \(N\) then \[ a_{k}=\frac{1}{N}\sum _{n=0}^{N-1}x\left [ n\right ] e^{-jk\omega _{0}n}\] Where \(\omega _{0}=\frac{2\pi }{12}\). The above becomes\begin{align*} a_{k} & =\frac{1}{12}\sum _{n=0}^{11}\sin \left ( 2\pi \frac{n}{3}\right ) \cos \left ( \pi \frac{n}{2}\right ) e^{-jk\frac{2\pi }{12}n}\\ 12a_{k} & =0+\sin \left ( 2\pi \frac{1}{3}\right ) \cos \left ( \pi \frac{1}{2}\right ) e^{-jk\frac{2\pi }{12}}+\sin \left ( 2\pi \frac{2}{3}\right ) \cos \left ( \pi \frac{2}{2}\right ) e^{-jk\frac{2\pi }{12}2}+\sin \left ( 2\pi \frac{3}{3}\right ) \cos \left ( \pi \frac{3}{2}\right ) e^{-jk\frac{2\pi }{12}3}\\ & +\sin \left ( 2\pi \frac{4}{3}\right ) \cos \left ( \pi \frac{4}{2}\right ) e^{-jk\frac{2\pi }{12}4}+\sin \left ( 2\pi \frac{5}{3}\right ) \cos \left ( \pi \frac{5}{2}\right ) e^{-jk\frac{2\pi }{12}5}+\sin \left ( 2\pi \frac{6}{3}\right ) \cos \left ( \pi \frac{6}{2}\right ) e^{-jk\frac{2\pi }{12}6}\\ & +\sin \left ( 2\pi \frac{7}{3}\right ) \cos \left ( \pi \frac{7}{2}\right ) e^{-jk\frac{2\pi }{12}7}+\sin \left ( 2\pi \frac{8}{3}\right ) \cos \left ( \pi \frac{8}{2}\right ) e^{-jk\frac{2\pi }{12}8}+\sin \left ( 2\pi \frac{9}{3}\right ) \cos \left ( \pi \frac{9}{2}\right ) e^{-jk\frac{2\pi }{12}9}\\ & +\sin \left ( 2\pi \frac{10}{3}\right ) \cos \left ( \pi \frac{10}{2}\right ) e^{-jk\frac{2\pi }{12}10}+\sin \left ( 2\pi \frac{11}{3}\right ) \cos \left ( \pi \frac{11}{2}\right ) e^{-jk\frac{2\pi }{12}11} \end{align*}
Which simplifies to (many terms go to zero)\[ 12a_{k}=\frac{1}{2}\sqrt{3}e^{-jk\frac{4\pi }{12}}+\frac{1}{2}\sqrt{3}e^{-jk\frac{8\pi }{12}}-\frac{1}{2}\sqrt{3}e^{-jk\frac{2\pi }{12}8}-\frac{1}{2}\sqrt{3}e^{-jk\frac{2\pi }{12}10}\] Hence \[ a_{k}=\frac{\sqrt{3}}{24}\left ( e^{-jk\frac{4\pi }{12}}+e^{-jk\frac{8\pi }{12}}-e^{-jk\frac{16\pi }{12}}-e^{-jk\frac{20\pi }{12}}\right ) \] Evaluating these for \(k=0\cdots N-1\) gives
| \(k\) | \(a_{k}\) |
| \(0\) | \(0\) |
| \(1\) | \(\frac{-j}{4}\) |
| \(2\) | \(0\) |
| \(3\) | \(0\) |
| \(4\) | \(0\) |
| \(5\) | \(\frac{j}{4}\) |
| \(6\) | \(0\) |
| \(7\) | \(\frac{-j}{4}\) |
| \(8\) | \(0\) |
| \(9\) | \(0\) |
| \(10\) | \(0\) |
| \(11\) | \(\frac{j}{4}\) |
Hence the \(\left \vert a_{k}\right \vert \) and phase are
| \(k\) | \(a_{k}\) | \(\left \vert a_{k}\right \vert \) | phase (degree) |
| \(0\) | \(0\) | \(0\) | \(0\) |
| \(1\) | \(\frac{-j}{4}\) | \(\frac{1}{4}\) | \(-90\) |
| \(2\) | \(0\) | \(0\) | \(0\) |
| \(3\) | \(0\) | \(0\) | \(0\) |
| \(4\) | \(0\) | \(0\) | \(0\) |
| \(5\) | \(\frac{j}{4}\) | \(\frac{1}{4}\) | \(90\) |
| \(6\) | \(0\) | \(0\) | \(0\) |
| \(7\) | \(\frac{-j}{4}\) | \(\frac{1}{4}\) | \(-90\) |
| \(8\) | \(0\) | \(0\) | \(0\) |
| \(9\) | \(0\) | \(0\) | \(0\) |
| \(10\) | \(0\) | \(0\) | \(0\) |
| \(11\) | \(\frac{j}{4}\) | \(\frac{1}{4}\) | \(90\) |
Consider the signal \(x\left ( t\right ) =\cos \left ( 2\pi t\right ) \) since \(x\left ( t\right ) \) is periodic with a fundamental period of \(1\), it is also periodic with a period of \(N\), where \(N\) is any positive integer. What are the Fourier series coefficients of \(x(t)\) if we regard it as a periodic signal with period \(3\)?
Solution
The Fourier series coefficients for \(\cos \left ( 2\pi t\right ) \) are found from Euler relation. Since \(\omega _{0}=2\pi \) rad/sec, then \[ \cos \left ( \omega _{0}t\right ) =\frac{1}{2}e^{j\omega _{0}t}+\frac{1}{2}e^{-j\omega _{0}t}\] Comparing the above to \[ x\left ( t\right ) =\sum _{k=-\infty }^{\infty }a_{k}e^{jk\omega _{0}t}\] Show that \(a_{1}=\frac{1}{2}\) and \(a_{-1}=\frac{1}{2}\) and all other \(a_{k}=0\).
Similarly, if the period happened to be \(3\), then \(\omega _{0}=\frac{2\pi }{3}\) and now \(x(t)\) can be written as \(\cos \left ( 2\pi t\right ) =\cos \left ( 3\omega _{0}t\right ) \). Therefore doing the same as above gives\[ \cos \left ( 3\omega _{0}t\right ) =\frac{1}{2}e^{j3\omega _{0}t}+\frac{1}{2}e^{-j3\omega _{0}t}\] Comparing the above to \(x\left ( t\right ) =\sum _{k=-\infty }^{\infty }a_{k}e^{jk\omega _{0}t}\) shows that \(a_{3}=\frac{1}{2}\) and \(a_{-3}=\frac{1}{2}\) and all other \(a_{k}=0\).