\(\blacksquare \) Fourier series. Periodic signals, Continuous time
Let \(\omega _{0}=\frac{2\pi }{T_{0}}\) be the fundamental frequency (rad/sec), and \(T_{0}\) the fundamental period, then \begin{align*} x\left ( t\right ) & =\sum _{k=-\infty }^{\infty }a_{k}e^{jk\omega _{0}t}\\ a_{k} & =\frac{1}{T_{0}}\int _{T_{0}}x\left ( t\right ) e^{-jk\omega _{0}t}dt \end{align*}
\(\blacksquare \) Fourier series. Periodic signals, Discrete time
Let \(\Omega _{0}=\frac{2\pi }{N}\) be the fundamental frequency (rad/sample), and \(N\) the fundamental period, then \begin{align*} x\left [ n\right ] & =\sum _{k=0}^{N-1}a_{k}e^{jk\Omega _{0}n}=\sum _{k=\left \langle N\right \rangle }a_{k}e^{jk\Omega _{0}n}\\ a_{k} & =\frac{1}{N}\sum _{n=0}^{N-1}x\left [ n\right ] e^{-jk\Omega _{0}n}=\frac{1}{N}\sum _{n=\left \langle N\right \rangle }x\left [ n\right ] e^{-jk\Omega _{0}n} \end{align*}
\(\blacksquare \) Fourier transform. Non periodic signal, Continuous time.\begin{align*} x\left ( t\right ) & =\frac{1}{2\pi }\int _{-\infty }^{\infty }X\left ( \omega \right ) e^{i\omega t}d\omega \\ X\left ( \omega \right ) & =\int _{-\infty }^{\infty }x\left ( t\right ) e^{-i\omega t}dt \end{align*}
It is also possible to obtain a Fourier transform for periodic signal. For \(x\left ( t\right ) =\sum _{k=-\infty }^{\infty }a_{k}e^{ik\omega _{0}t}\) its Fourier transform becomes (\(\omega _{0}=\frac{2\pi }{T_{0}}\))\[ X\left ( \omega \right ) =2\pi \sum _{k=-\infty }^{\infty }a_{k}\delta \left ( \omega -k\omega _{0}\right ) \] \(\blacksquare \) Fourier transform. Non periodic signal, Discrete time.\begin{align*} x\left [ n\right ] & =\frac{1}{2\pi }\int _{-\pi }^{\pi }X\left ( \Omega \right ) e^{j\Omega n}d\Omega \\ X\left ( \Omega \right ) & =\sum _{n=-\infty }^{\infty }x\left [ n\right ] e^{-j\Omega n} \end{align*}
It is also possible to obtain a Fourier transform for periodic discrete signal, where \(\Omega _{0}=\frac{2\pi }{N}\)\[ X\left ( \Omega \right ) =2\pi \sum _{k=-\infty }^{\infty }a_{k}\delta \left ( \Omega -k\Omega _{0}\right ) \] \(\blacksquare \) When input to LTI system is \(x\left ( t\right ) =e^{j\omega t}\) and system has impulse response \(h\left ( t\right ) \) then output is \begin{align*} y\left ( t\right ) & =\int _{-\infty }^{\infty }h\left ( \tau \right ) x\left ( t-\tau \right ) d\tau \\ & =\int _{-\infty }^{\infty }h\left ( \tau \right ) e^{j\omega \left ( t-\tau \right ) }d\tau \\ & =e^{j\omega t}\int _{-\infty }^{\infty }h\left ( \tau \right ) e^{-j\omega \tau }d\tau \\ & =e^{j\omega t}H\left ( \omega \right ) \end{align*}
Where \(H\left ( \omega \right ) \) is the Fourier transform of \(h\left ( t\right ) \). In the above \(e^{j\omega t}\) is called eigenfucntions of the system and \(H\left ( \omega \right ) \) the eigenvalues.
\(\blacksquare \) If input \(x\left ( t\right ) =a\cos \left ( 5\omega _{0}t+\theta \right ) \) and \(H\left ( \omega \right ) \) is the Fourier transform of the system, then \[ y\left ( t\right ) =a\left \vert H\left ( 5\omega _{0}\right ) \right \vert \cos \left ( 5\omega _{0}t+\theta +\arg H\left ( 5\omega _{0}\right ) \right ) \] Same for discrete time.
\(\blacksquare \) Modulation. \(y\left ( t\right ) =x\left ( t\right ) h\left ( t\right ) \) in CTFT becomes \(Y\left ( \omega \right ) =\frac{1}{2\pi }X\left ( \omega \right ) \circledast H\left ( \omega \right ) \) where \(X\left ( \omega \right ) \circledast H\left ( \omega \right ) =\int _{-\infty }^{\infty }X\left ( z\right ) H\left ( \omega -z\right ) dz\). Notice the extra \(\frac{1}{2\pi }\) factor.
\(\blacksquare \) To find discrete period given a signal, write \(x\left [ n\right ] =x\left [ n+N\right ] \) and then solve for \(N\). See HW’s.
\(\blacksquare \) \(\sum _{n=0}^{\infty }a^{n}=\frac{1}{1-a}\) and \(\sum _{n=N}^{\infty }a^{n}=\frac{a^{N}}{1-a}\) and \(\sum _{n=0}^{N}a^{n}=\frac{a^{1+N}-1}{a-1}\), and \(\sum _{n=N_{1}}^{N_{2}}a^{n}=\frac{a^{N_{1}}-a^{N_{2}}+1}{1-a}\)
\(\blacksquare \) Fourier transform relations. \(y\left ( t\right ) \Longleftrightarrow Y\left ( \omega \right ) \) then \(y\left ( at\right ) \Longleftrightarrow \frac{1}{a}Y\left ( \frac{\omega }{a}\right ) \)
\(\blacksquare \) Euler relations.\(\ \cos x=\frac{e^{jx}+e^{-jx}}{2},\sin x=\frac{e^{jx}-e^{-jx}}{2j}\)
\(\blacksquare \) Circuit. Voltage cross resistor \(R\) is \(V\left ( t\right ) =Ri\left ( t\right ) \). Voltage cross inductor \(L\) is \(V\left ( t\right ) =L\frac{di}{dt}\) and current across capacitor \(C\) is \(i\left ( t\right ) =C\frac{dV}{dt}\)
\(\blacksquare \) Partial fractions.
| \(\frac{f\left ( x\right ) }{\left ( x-a\right ) \left ( x-b\right ) }\) | \(\frac{A}{x-a}+\frac{B}{x-b}\) |
| \(\frac{f\left ( x\right ) }{\left ( x-a\right ) ^{2}}\) | \(\frac{A}{x-a}+\frac{B}{\left ( x-a\right ) ^{2}}\) |
| \(\frac{f\left ( x\right ) }{\left ( x-a\right ) \left ( x^{2}+bx+c\right ) }\) | \(\frac{A}{x-a}+\frac{Bx+C}{x^{2}+bx+c}\) |
| \(\frac{f\left ( x\right ) }{\left ( x-a\right ) \left ( x+d\right ) ^{2}}\) | \(\frac{A}{x-a}+\frac{B}{x+d}+\frac{C}{\left ( x+d\right ) ^{2}}\) |
| \(\frac{f\left ( x\right ) }{\left ( x+d\right ) ^{2}}\) | \(\frac{A}{x+d}+\frac{B}{\left ( x+d\right ) ^{2}}\) |
| \(\frac{f\left ( x\right ) }{\left ( x-a\right ) \left ( x^{2}-b^{2}\right ) }\) | \(\frac{A}{x+d}+\frac{Bx+C}{x^{2}-b^{2}}\) |
| \(\frac{f\left ( x\right ) }{\left ( x^{2}-a\right ) \left ( x^{2}-b\right ) }\) | \(\frac{Ax+B}{x^{2}-a}+\frac{Cx+D}{x^{2}-b}\) |
| \(\frac{f\left ( x\right ) }{\left ( x^{2}-a\right ) ^{2}}\) | \(\frac{Ax+B}{x^{2}-a}+\frac{Cx+D}{\left ( x^{2}-a\right ) ^{2}}\) |
\(\blacksquare \) Parsevel’s. For non-periodic cont. time: \(\int _{-\infty }^{\infty }\left \vert x\left ( t\right ) \right \vert ^{2}dt=\frac{1}{2\pi }\int _{-\infty }^{\infty }\left \vert X\left ( \omega \right ) \right \vert ^{2}d\omega \). For periodic cont. time : \(\frac{1}{T}\int _{T}\left \vert x\left ( t\right ) \right \vert ^{2}dt=\sum _{k=-\infty }^{\infty }\left \vert a_{k}\right \vert ^{2}\). For discrete: \(\frac{1}{2\pi }\int _{-\infty }^{\infty }\left \vert X\left ( \Omega \right ) \right \vert ^{2}d\Omega =\sum _{n=-\infty }^{\infty }\left \vert x\left [ n\right ] \right \vert ^{2}\).
\(\blacksquare \) Properties Fourier series. If \(a_{k}=a_{-k}^{\ast }\) then \(x\left ( t\right ) \) is real. If \(a_{k}\) is even, then \(x\left ( t\right ) \) is even. For \(x\left ( t\right ) \) real and odd, then \(a_{k}\) are pure imaginary and odd. i.e. \(a_{k}=-a_{-k}\) , and \(a_{0}=0\).
\(\blacksquare \) More Fourier transform relations. Continuos time
| \(e^{-a\left \vert t\right \vert }\) | \(\frac{2a}{a^{2}+\omega ^{2}}\) |
| \(x\left ( t\right ) e^{-j\omega _{0}t}\) | \(X\left ( \omega +\omega _{0}\right ) \) |
| \(x\left ( t\right ) e^{j\omega _{0}t}\) | \(X\left ( \omega -\omega _{0}\right ) \) |
| \(\frac{\sin \left ( a\omega \right ) }{\omega }\) | Box from \(t=-a\cdots a\) |
Discrete time
| \(u\left [ n\right ] \) | \(\frac{1}{1-e^{-j\Omega }}\) |
| \(u\left [ n-1\right ] \) | \(e^{-j\Omega }U\left ( \Omega \right ) =e^{-j\Omega }\frac{1}{1-e^{-j\Omega }}\) |
| \(a^{n}u\left [ n\right ] \) | \(\frac{1}{1-ae^{-j\Omega }}\) |
| \(e^{j\Omega _{0}n}x\left [ n\right ] \) | \(X\left ( \Omega -\Omega _{0}\right ) \) |
From above we see that unit delay in discrete time means multiplying by \(e^{-j\Omega }\).
\(\blacksquare \) Difference equations. \(y\left [ n-1\right ] \Longleftrightarrow e^{-j\Omega }Y\left ( \Omega \right ) \). For example, given \(y\left [ n\right ] -ay\left [ n-1\right ] =x\left [ n\right ] \) then applying DFT gives \(Y\left ( \Omega \right ) -ae^{-j\Omega }Y\left ( \Omega \right ) =X\left ( \Omega \right ) \) or \(H\left ( \Omega \right ) =\frac{Y\left ( \Omega \right ) }{X\left ( \Omega \right ) }=\frac{1}{1-ae^{-j\Omega }}\). From tables, the inverse DFT of this is \(a^{n}u\left [ n\right ] \). Need to know partial fractions sometimes. For example given \(y\left [ n\right ] -\frac{3}{4}y\left [ n-1\right ] +\frac{1}{8}y\left [ n-2\right ] =2x\left [ n\right ] \) then \begin{align*} Y\left ( \Omega \right ) -\frac{3}{4}e^{-j\Omega }Y\left ( \Omega \right ) +\frac{1}{8}e^{-j2\Omega }Y\left ( \Omega \right ) & =2X\left ( \Omega \right ) \\ H\left ( \Omega \right ) & =\frac{Y\left ( \Omega \right ) }{X\left ( \Omega \right ) }\\ & =\frac{2}{\left ( 1-\frac{3}{4}e^{-j\Omega }+\frac{1}{8}e^{-j2\Omega }\right ) }\\ & =\frac{2}{\left ( 1-\frac{1}{2}e^{-j\Omega }\right ) \left ( 1-\frac{1}{4}e^{-j\Omega }\right ) } \end{align*}
And using partial fractions gives \(H\left ( \Omega \right ) =\frac{4}{1-\frac{1}{2}e^{-j\Omega }}-\frac{2}{1-\frac{1}{4}e^{-j\Omega }}\). Hence using above table gives \(h\left [ n\right ] =\left ( 4\left ( \frac{1}{2}\right ) ^{n}-2\left ( \frac{1}{4}\right ) ^{n}\right ) u\left [ n\right ] \)
\(\blacksquare \) \(\left \vert X\left ( \omega \right ) \right \vert ^{2}\) may be interpreted as the energy density spectrum of \(x\left ( t\right ) \). This means \(\frac{1}{2\pi }\left \vert X\left ( \omega \right ) \right \vert ^{2}d\omega \) is amount of energy in \(d\omega \) range of frequencies. i.e. between \(\omega \) and \(\omega +d\omega \). \(\left \vert X\left ( \omega \right ) \right \vert \) is called the gain of the system and \(\arg \left ( H\left ( \omega \right ) \right ) \) is called the phase shift of the system. When \(\arg \left ( H\left ( \omega \right ) \right ) \) is linear function in \(\omega \) then the effect in time domain is time shift. (delay).
\(\blacksquare \) \(z\) transforms \(X\left ( z\right ) =\sum _{n=-\infty }^{\infty }x\left [ n\right ] z^{-n}\). If \(x\left [ n\right ] \rightarrow X\left ( z\right ) \) then \(x\left [ n-1\right ] \rightarrow z^{-1}X\left ( z\right ) \).
\(\blacksquare \) \(\frac{\sin \left ( ax\right ) }{ax}=\operatorname{sinc}\left ( \frac{ax}{\pi }\right ) \) and \(\frac{\sin \left ( x\right ) }{x}=\operatorname{sinc}\left ( \frac{x}{\pi }\right ) \). In class we use \(\frac{\sin \left ( \omega _{c}t\right ) }{\pi t}\). This has FT as rectangle from \(-\omega _{c}\) to \(\omega _{c}\) and amplitude \(1\).
\(\blacksquare \) in digital, sampling rate is in hz, but units is samples per second and not cycles per second as with analog.
\(\blacksquare \) \[ \Omega =\frac{\omega }{F_{s}}\] where \(F_{s}\) is sampling rate in samples per second, and \(\Omega \) is unnormalized digital frequency (radians per sample) and \(\omega \) is analog frequency (radians per second). This can also be written as \[ \Omega =\omega T_{s}\] where here \(T_{s}\) is seconds per sample (i.e. number of seconds to obtain one sample). Per sample is used to make the units come out OK.
\(\blacksquare \) Trig identities \begin{align*} \sin A\cos B & =\frac{1}{2}\left ( \sin \left ( A+B\right ) +\sin \left ( A-B\right ) \right ) \\ \cos A\cos B & =\frac{1}{2}\left ( \cos \left ( A+B\right ) +\cos \left ( A-B\right ) \right ) \\ \sin A\sin B & =\frac{1}{2}\left ( \cos \left ( A-B\right ) -\cos \left ( A+B\right ) \right ) \end{align*}
\(\blacksquare \) Group delay is given by \(-\frac{d}{d\omega }\left ( \arg \left ( H\left ( \omega \right ) \right ) \right ) \). For example, if \(H\left ( \omega \right ) =\frac{1}{2+j\omega }\) then \(\arg \left ( H\left ( \omega \right ) \right ) =-\arctan \left ( \frac{\omega }{2}\right ) \) which leads to group delay being \(\frac{2}{4+\omega ^{2}}\).
\(\blacksquare \) FT of \(\cos \left ( \omega _{c}t\right ) \,\) has delta at \(\pm \omega _{c}\) each of amplitude \(\pi \). And FT of \(\sin \left ( \omega _{c}t\right ) \,\) has delta at \(\omega _{c}\) of amplitude \(\frac{\pi }{j}\) and has delta at \(-\omega _{c}\) of amplitude \(\frac{-\pi }{j}\) and \(\frac{\sin \left ( \omega _{c}t\right ) }{\pi t}\) has FT as rectangle of amplitude \(1\) and width from \(-\omega _{c}\) to \(+\omega _{c}\).
\begin{align*} X\left ( \Omega \right ) & =\sum _{n=-\infty }^{\infty }x\left [ n\right ] e^{-j\Omega n}\\ & =\sum _{n=-\infty }^{\infty }\left ( \frac{1}{2}\right ) ^{n}\cos \left ( \frac{\pi n}{2}\right ) u[n]e^{-j\Omega n}\\ & =\sum _{n=0}^{\infty }\left ( \frac{1}{2}\right ) ^{n}\cos \left ( \frac{\pi n}{2}\right ) e^{-j\Omega n} \end{align*}
But \(\cos \left ( \frac{\pi n}{2}\right ) =\frac{1}{2}\left ( e^{j\frac{\pi n}{2}}+e^{-j\frac{\pi n}{2}}\right ) \) and the above becomes
\begin{align*} X\left ( \Omega \right ) & =\sum _{n=0}^{\infty }\left ( \frac{1}{2}\right ) ^{n}\frac{1}{2}\left ( e^{j\frac{\pi n}{2}}+e^{-j\frac{\pi n}{2}}\right ) e^{-j\Omega n}\\ & =\frac{1}{2}\left ( \sum _{n=0}^{\infty }\left ( \frac{1}{2}\right ) ^{n}e^{j\frac{\pi n}{2}}e^{-j\Omega n}+\sum _{n=0}^{\infty }\left ( \frac{1}{2}\right ) ^{n}e^{-j\frac{\pi n}{2}}e^{-j\Omega n}\right ) \\ & =\frac{1}{2}\left ( \sum _{n=0}^{\infty }\left ( \frac{1}{2}e^{j\left ( \frac{\pi }{2}-\Omega \right ) }\right ) ^{n}+\sum _{n=0}^{\infty }\left ( \frac{1}{2}e^{j\left ( -\frac{\pi }{2}-\Omega \right ) }\right ) ^{n}\right ) \end{align*}
Since \(\frac{1}{2}e^{j\left ( \frac{\pi }{2}-\Omega \right ) }<1\) then we can use \(\sum _{n=0}^{\infty }a^{n}=\frac{1}{1-a}\) for both terms and the above becomes
\begin{align*} X\left ( \Omega \right ) & =\frac{1}{2}\left ( \frac{1}{1-\frac{1}{2}e^{j\left ( \frac{\pi }{2}-\Omega \right ) }}+\frac{1}{1-\frac{1}{2}e^{j\left ( -\frac{\pi }{2}-\Omega \right ) }}\right ) \\ & =\frac{1}{2}\left ( \frac{1}{1-\frac{1}{2}e^{j\frac{\pi }{2}}e^{-j\Omega }}+\frac{1}{1-\frac{1}{2}e^{-j\frac{\pi }{2}}e^{-j\Omega }}\right ) \end{align*}
But \(e^{j\frac{\pi }{2}}=j\) and \(e^{-j\frac{\pi }{2}}=-j\) and the above becomes
\begin{align*} X\left ( \Omega \right ) & =\frac{1}{2}\left ( \frac{1}{1-\frac{1}{2}je^{-j\Omega }}+\frac{1}{1+\frac{1}{2}je^{-j\Omega }}\right ) \\ & =\frac{1}{2}\left ( \frac{1+\frac{1}{2}je^{-j\Omega }+1-\frac{1}{2}je^{-j\Omega }}{\left ( 1-\frac{1}{2}je^{-j\Omega }\right ) \left ( 1+\frac{1}{2}je^{-j\Omega }\right ) }\right ) \\ & =\frac{1}{2}\left ( \frac{2}{1+\frac{1}{2}je^{-j\Omega }-\frac{1}{2}je^{-j\Omega }-\frac{1}{4}j^{2}e^{-2j\Omega }}\right ) \\ & =\frac{1}{1+\frac{1}{4}e^{-2j\Omega }} \end{align*}
\(\blacksquare \) Z transforms
| \(u\left [ n\right ] \) | \(Z\) |
| \(a^{n}u\left [ n\right ] \) | \(\frac{1}{1-az^{-1}}\) |
| \(a^{n-1}u\left [ n-1\right ] \) | \(z^{-1}\frac{1}{1-az^{-1}}\) |
| \(a^{n-2}u\left [ n-2\right ] \) | \(z^{-2}\frac{1}{1-az^{-1}}\) |
If the ROC outside the out most pole, then right-handed signal. (Causal). If the ROC isinside the inner most pole, then left-handed signal (non causal).