5.2 cheat sheet

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$◼$ 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   $$\begin{array}{rl}x\left(t\right)& =\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k{e}^{jk{\omega }_{0}t}\\ a_k& =\frac{1}{T_0}{\int }_{T_0}x\left(t\right)e^{-jk{\omega }_{0}t}dt\end{array}$$

$◼$ Fourier series. Periodic signals, Discrete time

Let ${\mathrm{\Omega }}_0=\frac{2\pi }{N}$ be the fundamental frequency (rad/sample), and $N$ the fundamental period, then $$\begin{array}{rl}x\left[n\right]& =\sum _{k=0}^{N-1}{a}_k{e}^{jk{\mathrm{\Omega }}_{0}n}=\sum _{k=⟨N⟩}{a}_k{e}^{jk{\mathrm{\Omega }}_{0}n}\\ a_k& =\frac{1}{N}\sum _{n=0}^{N-1}x\left[n\right]e^{-jk{\mathrm{\Omega }}_{0}n}=\frac{1}{N}\sum _{n=⟨N⟩}x\left[n\right]e^{-jk{\mathrm{\Omega }}_{0}n}\end{array}$$

$◼$ Fourier transform. Non periodic signal, Continuous time.$$\begin{array}{rl}x\left(t\right)& =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}X\left(\omega \right)e^{i\omega t}d\omega \\ X\left(\omega \right)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\left(t\right)e^{-i\omega t}dt\end{array}$$

It is also possible to obtain a Fourier transform for periodic signal. For $x\left(t\right)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k{e}^{ik{\omega }_{0}t}$ its Fourier transform becomes (${\omega }_0=\frac{2\pi }{T_0}$)$$X\left(\omega \right)=2\pi \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k\delta (\omega -k{\omega }_0)$$ $◼$ Fourier transform. Non periodic signal, Discrete time.$$\begin{array}{rl}x\left[n\right]& =\frac{1}{2\pi }{\int }_{-\pi }^{\pi }X\left(\mathrm{\Omega }\right)e^{j\mathrm{\Omega }n}d\mathrm{\Omega }\\ X\left(\mathrm{\Omega }\right)& =\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left[n\right]e^{-j\mathrm{\Omega }n}\end{array}$$

It is also possible to obtain a Fourier transform for periodic discrete signal, where ${\mathrm{\Omega }}_0=\frac{2\pi }{N}$$$X\left(\mathrm{\Omega }\right)=2\pi \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k\delta (\mathrm{\Omega }-k{\mathrm{\Omega }}_0)$$ $◼$ 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 $$\begin{array}{rl}y\left(t\right)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}h\left(\tau \right)x(t-\tau )d\tau \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}h\left(\tau \right)e^{j\omega (t-\tau )}d\tau \\ & =e^{j\omega t}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}h\left(\tau \right)e^{-j\omega \tau }d\tau \\ & =e^{j\omega t}H\left(\omega \right)\end{array}$$

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.

$◼$ If input $x\left(t\right)=a\cos (5{\omega }_{0}t+\theta )$ and $H\left(\omega \right)$ is the Fourier transform of the system, then $$y\left(t\right)=a\left|H\left(5{\omega }_0\right)\right|\cos (5{\omega }_{0}t+\theta +\arg H\left(5{\omega }_0\right))$$ Same for discrete time.

$◼$ Modulation. $y\left(t\right)=x\left(t\right)h\left(t\right)$ in CTFT becomes $Y\left(\omega \right)=\frac{1}{2\pi }X\left(\omega \right)⊛H\left(\omega \right)$ where $X\left(\omega \right)⊛H\left(\omega \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}X\left(z\right)H(\omega -z)dz$. Notice the extra $\frac{1}{2\pi }$ factor.

$◼$ To find discrete period given a signal, write $x\left[n\right]=x[n+N]$ and then solve for $N$. See HW’s.

$◼$ $\sum _{n=0}^{\mathrm{\infty }}{a}^n=\frac{1}{1-a}$ and $\sum _{n=N}^{\mathrm{\infty }}{a}^n=\frac{a^N}{1-a}$ and $\sum _{n=0}^N{a}^n=\frac{a^{1+N}-1}{a-1}$, and $\sum _{n=N_1}^{N_2}{a}^n=\frac{a^{N_1}-a^{N_2}+1}{1-a}$

$◼$ Fourier transform relations. $y\left(t\right)⟺Y\left(\omega \right)$ then $y\left(at\right)⟺\frac{1}{a}Y\left(\frac{\omega }{a}\right)$

$◼$ Euler relations.$\cos x=\frac{e^{jx}+e^{-jx}}{2},\sin x=\frac{e^{jx}-e^{-jx}}{2j}$

$◼$ 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 $i\left(t\right)=C\frac{dV}{dt}$

$◼$ Partial fractions.

$\frac{f\left(x\right)}{(x-a)(x-b)}$	$\frac{A}{x-a}+\frac{B}{x-b}$
$\frac{f\left(x\right)}{{(x-a)}^2}$	$\frac{A}{x-a}+\frac{B}{{(x-a)}^2}$
$\frac{f\left(x\right)}{(x-a)(x^2+bx+c)}$	$\frac{A}{x-a}+\frac{Bx+C}{x^2+bx+c}$
$\frac{f\left(x\right)}{(x-a){(x+d)}^2}$	$\frac{A}{x-a}+\frac{B}{x+d}+\frac{C}{{(x+d)}^2}$
$\frac{f\left(x\right)}{{(x+d)}^2}$	$\frac{A}{x+d}+\frac{B}{{(x+d)}^2}$
$\frac{f\left(x\right)}{(x-a)(x^2-b^2)}$	$\frac{A}{x+d}+\frac{Bx+C}{x^2-b^2}$
$\frac{f\left(x\right)}{(x^2-a)(x^2-b)}$	$\frac{Ax+B}{x^2-a}+\frac{Cx+D}{x^2-b}$
$\frac{f\left(x\right)}{{(x^2-a)}^2}$	$\frac{Ax+B}{x^2-a}+\frac{Cx+D}{{(x^2-a)}^2}$

$◼$ Parsevel’s. For non-periodic cont. time: ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|x\left(t\right)\right|}^{2}dt=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|X\left(\omega \right)\right|}^{2}d\omega$. For periodic cont. time : $\frac{1}{T}{\int }_T{\left|x\left(t\right)\right|}^{2}dt=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|a_k\right|}^2$. For discrete: $\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|X\left(\mathrm{\Omega }\right)\right|}^{2}d\mathrm{\Omega }=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|x\left[n\right]\right|}^2$.

$◼$ 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 $a_0=0$.

$◼$ More Fourier transform relations. Continuos time

$e^{-a\left|t\right|}$	$\frac{2a}{a^2+{\omega }^2}$
$x\left(t\right)e^{-j{\omega }_{0}t}$	$X(\omega +{\omega }_0)$
$x\left(t\right)e^{j{\omega }_{0}t}$	$X(\omega -{\omega }_0)$
$\frac{\sin \left(a\omega \right)}{\omega }$	Box from $t=-a\cdots a$

Discrete time

$u\left[n\right]$	$\frac{1}{1-e^{-j\mathrm{\Omega }}}$
$u[n-1]$	$e^{-j\mathrm{\Omega }}U\left(\mathrm{\Omega }\right)=e^{-j\mathrm{\Omega }}\frac{1}{1-e^{-j\mathrm{\Omega }}}$
$a^{n}u\left[n\right]$	$\frac{1}{1-a{e}^{-j\mathrm{\Omega }}}$
$e^{j{\mathrm{\Omega }}_{0}n}x\left[n\right]$	$X(\mathrm{\Omega }-{\mathrm{\Omega }}_0)$

From above we see that unit delay in discrete time means multiplying by $e^{-j\mathrm{\Omega }}$.

$◼$ Difference equations. $y[n-1]⟺e^{-j\mathrm{\Omega }}Y\left(\mathrm{\Omega }\right)$. For example, given $y\left[n\right]-ay[n-1]=x\left[n\right]$ then applying DFT gives $Y\left(\mathrm{\Omega }\right)-a{e}^{-j\mathrm{\Omega }}Y\left(\mathrm{\Omega }\right)=X\left(\mathrm{\Omega }\right)$ or $H\left(\mathrm{\Omega }\right)=\frac{Y\left(\mathrm{\Omega }\right)}{X\left(\mathrm{\Omega }\right)}=\frac{1}{1-a{e}^{-j\mathrm{\Omega }}}$. From tables, the inverse DFT of this is $a^{n}u\left[n\right]$. Need to know partial fractions sometimes. For example given $y\left[n\right]-\frac{3}{4}y[n-1]+\frac{1}{8}y[n-2]=2x\left[n\right]$ then $$\begin{array}{rl}Y\left(\mathrm{\Omega }\right)-\frac{3}{4}{e}^{-j\mathrm{\Omega }}Y\left(\mathrm{\Omega }\right)+\frac{1}{8}{e}^{-j2\mathrm{\Omega }}Y\left(\mathrm{\Omega }\right)& =2X\left(\mathrm{\Omega }\right)\\ H\left(\mathrm{\Omega }\right)& =\frac{Y\left(\mathrm{\Omega }\right)}{X\left(\mathrm{\Omega }\right)}\\ & =\frac{2}{(1-\frac{3}{4}{e}^{-j\mathrm{\Omega }}+\frac{1}{8}{e}^{-j2\mathrm{\Omega }})}\\ & =\frac{2}{(1-\frac{1}{2}{e}^{-j\mathrm{\Omega }})(1-\frac{1}{4}{e}^{-j\mathrm{\Omega }})}\end{array}$$

And using partial fractions gives $H\left(\mathrm{\Omega }\right)=\frac{4}{1-\frac{1}{2}{e}^{-j\mathrm{\Omega }}}-\frac{2}{1-\frac{1}{4}{e}^{-j\mathrm{\Omega }}}$. Hence using above table gives $h\left[n\right]=(4{\left(\frac{1}{2}\right)}^n-2{\left(\frac{1}{4}\right)}^n)u\left[n\right]$

$◼$ ${\left|X\left(\omega \right)\right|}^2$ may be interpreted as the energy density spectrum of $x\left(t\right)$. This means $\frac{1}{2\pi }{\left|X\left(\omega \right)\right|}^{2}d\omega$ is amount of energy in $d\omega$ range of frequencies. i.e. between $\omega$ and $\omega +d\omega$. $\left|X\left(\omega \right)\right|$ 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).

$◼$ $z$ transforms $X\left(z\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left[n\right]z^{-n}$. If $x\left[n\right]\to X\left(z\right)$ then $x[n-1]\to z^{-1}X\left(z\right)$.

$◼$ $\frac{\sin \left(ax\right)}{ax}=\mathrm{sinc}\left(\frac{ax}{\pi }\right)$ and $\frac{\sin \left(x\right)}{x}=\mathrm{sinc}\left(\frac{x}{\pi }\right)$. 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 $1$.

$◼$ in digital, sampling rate is in hz, but units is samples per second and not cycles per second as with analog.

$◼$ $$\mathrm{\Omega }=\frac{\omega }{F_s}$$ where $F_s$ is sampling rate in samples per second, and $\mathrm{\Omega }$ is unnormalized digital frequency (radians per sample) and $\omega$ is analog frequency (radians per second). This can also be written as $$\mathrm{\Omega }=\omega T_s$$ 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.

$◼$ Trig identities $$\begin{array}{rl}\sin A\cos B& =\frac{1}{2}(\sin (A+B)+\sin (A-B))\\ \cos A\cos B& =\frac{1}{2}(\cos (A+B)+\cos (A-B))\\ \sin A\sin B& =\frac{1}{2}(\cos (A-B)-\cos (A+B))\end{array}$$

$◼$ Group delay is given by $-\frac{d}{d\omega }(\arg \left(H\left(\omega \right)\right))$. 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 $\frac{2}{4+{\omega }^2}$.

$◼$ 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 $+{\omega }_c$.

$$\begin{array}{rl}X\left(\mathrm{\Omega }\right)& =\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x\left[n\right]e^{-j\mathrm{\Omega }n}\\ & =\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^n\cos \left(\frac{\pi n}{2}\right)u[n]e^{-j\mathrm{\Omega }n}\\ & =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^n\cos \left(\frac{\pi n}{2}\right)e^{-j\mathrm{\Omega }n}\end{array}$$

But $\cos \left(\frac{\pi n}{2}\right)=\frac{1}{2}(e^{j\frac{\pi n}{2}}+e^{-j\frac{\pi n}{2}})$ and the above becomes

$$\begin{array}{rl}X\left(\mathrm{\Omega }\right)& =\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^n\frac{1}{2}(e^{j\frac{\pi n}{2}}+e^{-j\frac{\pi n}{2}})e^{-j\mathrm{\Omega }n}\\ & =\frac{1}{2}(\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^n{e}^{j\frac{\pi n}{2}}{e}^{-j\mathrm{\Omega }n}+\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^n{e}^{-j\frac{\pi n}{2}}{e}^{-j\mathrm{\Omega }n})\\ & =\frac{1}{2}(\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{2}{e}^{j(\frac{\pi }{2}-\mathrm{\Omega })}\right)}^n+\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{2}{e}^{j(-\frac{\pi }{2}-\mathrm{\Omega })}\right)}^n)\end{array}$$

Since $\frac{1}{2}{e}^{j(\frac{\pi }{2}-\mathrm{\Omega })}<1$ then we can use $\sum _{n=0}^{\mathrm{\infty }}{a}^n=\frac{1}{1-a}$ for both terms and the above becomes

$$\begin{array}{rl}X\left(\mathrm{\Omega }\right)& =\frac{1}{2}(\frac{1}{1-\frac{1}{2}{e}^{j(\frac{\pi }{2}-\mathrm{\Omega })}}+\frac{1}{1-\frac{1}{2}{e}^{j(-\frac{\pi }{2}-\mathrm{\Omega })}})\\ & =\frac{1}{2}(\frac{1}{1-\frac{1}{2}{e}^{j\frac{\pi }{2}}{e}^{-j\mathrm{\Omega }}}+\frac{1}{1-\frac{1}{2}{e}^{-j\frac{\pi }{2}}{e}^{-j\mathrm{\Omega }}})\end{array}$$

But $e^{j\frac{\pi }{2}}=j$ and $e^{-j\frac{\pi }{2}}=-j$ and the above becomes

$$\begin{array}{rl}X\left(\mathrm{\Omega }\right)& =\frac{1}{2}(\frac{1}{1-\frac{1}{2}j{e}^{-j\mathrm{\Omega }}}+\frac{1}{1+\frac{1}{2}j{e}^{-j\mathrm{\Omega }}})\\ & =\frac{1}{2}\left(\frac{1+\frac{1}{2}j{e}^{-j\mathrm{\Omega }}+1-\frac{1}{2}j{e}^{-j\mathrm{\Omega }}}{(1-\frac{1}{2}j{e}^{-j\mathrm{\Omega }})(1+\frac{1}{2}j{e}^{-j\mathrm{\Omega }})}\right)\\ & =\frac{1}{2}\left(\frac{2}{1+\frac{1}{2}j{e}^{-j\mathrm{\Omega }}-\frac{1}{2}j{e}^{-j\mathrm{\Omega }}-\frac{1}{4}{j}^2{e}^{-2j\mathrm{\Omega }}}\right)\\ & =\frac{1}{1+\frac{1}{4}{e}^{-2j\mathrm{\Omega }}}\end{array}$$

$◼$ Z transforms

$u\left[n\right]$	$Z$
$a^{n}u\left[n\right]$	$\frac{1}{1-a{z}^{-1}}$
$a^{n-1}u[n-1]$	$z^{-1}\frac{1}{1-a{z}^{-1}}$
$a^{n-2}u[n-2]$	$z^{-2}\frac{1}{1-a{z}^{-1}}$

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).