Idiot's guide to ELEC4402 communication systems

Fourier transform identities and properties

$$\begin{align*} u(t) &= \begin{cases} 1, & t > 0 \\ \frac{1}{2}, & t = 0 \\ 0, & t < 0 \end{cases}&\text{Unit Step Function}\\ \text{sgn}(t) &= \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}&\text{Signum Function}\\ \text{sinc}(2Wt) &= \frac{\sin(2\pi W t)}{2\pi W t}&\text{sinc Function}\\ \text{rect}(t) = \Pi(t) &= \begin{cases} 1, & -0.5 < t < 0.5 \\ 0, & \lvert t \rvert > 0.5 \end{cases}&\text{Rectangular/Gate Function}\\ g(t)*h(t)=(g*h)(t)&=\int_\infty^\infty g(\tau)h(t-\tau)d\tau&\text{Convolution}\\ \end{align*}$$

Fourier transform of continuous time periodic signal

Required for some questions on sampling:

Transform a continuous time-periodic signal $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:

$$X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}$$

Calculate $C_n$ coefficient as follows from $x_p(t)$:

$$\begin{align*} C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\ &=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$} \end{align*}$$

$\text{rect}$ function

Bessel function

$$\begin{align*} \sum_{n\in\mathbb{Z}}{J_n}^2(\beta)&=1\\ J_n(\beta)&=(-1)^nJ_{-n}(\beta) \end{align*}$$

White noise

$$\begin{align*} R_W(\tau)&=\frac{N_0}{2}\delta(\tau)=\frac{kT}{2}\delta(\tau)=\sigma^2\delta(\tau)\\ G_w(f)&=\frac{N_0}{2}\\ N_0&=kT\\ G_y(f)&=|H(f)|^2G_w(f)\\ G_y(f)&=G(f)G_w(f)\\ \end{align*}$$

WSS

$$\begin{align*} \mu_X(t) &= \mu_X\text{ Constant}\\ R_{XX}(t_1,t_2)&=R_X(t_1-t_2)=R_X(\tau)\\ E[X(t_1)X(t_2)]&=E[X(t)X(t+\tau)] \end{align*}$$

Ergodicity

$$\begin{align*} \braket{X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t)dt\\ \braket{X(t+\tau)X(t)}_T&=\frac{1}{2T}\int_{-T}^{T}x(t+\tau)x(t)dt\\ E[\braket{X(t)}_T]&=\frac{1}{2T}\int_{-T}^{T}x(t)dt=\frac{1}{2T}\int_{-T}^{T}m_Xdt=m_X\\ \end{align*}$$

Note: A WSS random process needs to be both ergodic in mean and autocorrelation to be considered an ergodic process

Other identities

$$\begin{align*} f*(g*h) &=(f*g)*h\quad\text{Convolution associative}\\ a(f*g) &= (af)*g \quad\text{Convolution associative}\\ \sum_{x=-\infty}^\infty(f(x a)\delta(\omega-x b))&=f\left(\frac{\omega a}{b}\right) \end{align*}$$

Other trig

$$\begin{align*} \cos2\theta=2 \cos^2 \theta-1&\Leftrightarrow\frac{\cos2\theta+1}{2}=\cos^2\theta\\ e^{-j\alpha}-e^{j\alpha}&=-2j \sin(\alpha)\\ e^{-j\alpha}+e^{j\alpha}&=2 \cos(\alpha)\\ \cos(-A)&=\cos(A)\\ \sin(-A)&=-\sin(A)\\ \sin(A+\pi/2)&=\cos(A)\\ \sin(A-\pi/2)&=-\cos(A)\\ \cos(A-\pi/2)&=\sin(A)\\ \cos(A+\pi/2)&=-\sin(A)\\ \int_{x\in\mathbb{R}}\text{sinc}(A x) &= \frac{1}{|A|}\\ \end{align*}$$

$$\begin{align*} \cos(A+B) &= \cos (A) \cos (B)-\sin (A) \sin (B) \\ \sin(A+B) &= \sin (A) \cos (B)+\cos (A) \sin (B) \\ \cos(A)\cos(B) &= \frac{1}{2} (\cos (A-B)+\cos (A+B)) \\ \cos(A)\sin(B) &= \frac{1}{2} (\sin (A+B)-\sin (A-B)) \\ \sin(A)\sin(B) &= \frac{1}{2} (\cos (A-B)-\cos (A+B)) \\ \end{align*}$$

$$\begin{align*} \cos(A)+\cos(B) &= 2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\ \cos(A)-\cos(B) &= -2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \sin \left(\frac{A}{2}+\frac{B}{2}\right) \\ \sin(A)+\sin(B) &= 2 \sin \left(\frac{A}{2}+\frac{B}{2}\right) \cos \left(\frac{A}{2}-\frac{B}{2}\right) \\ \sin(A)-\sin(B) &= 2 \sin \left(\frac{A}{2}-\frac{B}{2}\right) \cos \left(\frac{A}{2}+\frac{B}{2}\right) \\ \cos(A)+\sin(B)&= -2 \sin \left(\frac{A}{2}-\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}+\frac{B}{2}+\frac{\pi }{4}\right) \\ \cos(A)-\sin(B)&= -2 \sin \left(\frac{A}{2}+\frac{B}{2}-\frac{\pi }{4}\right) \sin \left(\frac{A}{2}-\frac{B}{2}+\frac{\pi }{4}\right) \\ \end{align*}$$

IQ/Complex envelope

Def. $\tilde{g}(t)=g_I(t)+jg_Q(t)$ as the complex envelope. Best to convert to $e^{j\theta}$ form.

Convert complex envelope representation to time-domain representation of signal

$$\begin{align*} g(t)&=g_I(t)\cos(2\pi f_c t)-g_Q(t)\sin(2\pi f_c t)\\ &=\text{Re}[\tilde{g}(t)\exp{(j2\pi f_c t)}]\\ &=A(t)\cos(2\pi f_c t+\phi(t))\\ A(t)&=|g(t)|=\sqrt{g_I^2(t)+g_Q^2(t)}\quad\text{Amplitude}\\ \phi(t)&\quad\text{Phase}\\ g_I(t)&=A(t)\cos(\phi(t))\quad\text{In-phase component}\\ g_Q(t)&=A(t)\sin(\phi(t))\quad\text{Quadrature-phase component}\\ \end{align*}$$

For transfer function

$$\begin{align*} h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\ &=2\text{Re}[\tilde{h}(t)\exp{(j2\pi f_c t)}]\\ \Rightarrow\tilde{h}(t)&=h_I(t)/2+jh_Q(t)/2=A(t)/2\exp{(j\phi(t))} \end{align*}$$

AM

CAM

$$\begin{align*} m_a &= \frac{\min_t|k_a m(t)|}{A_c} \quad\text{$k_a$ is the amplitude sensitivity ($\text{volt}^{-1}$), $m_a$ is the modulation index.}\\ m_a &= \frac{A_\text{max}-A_\text{min}}{A_\text{max}+A_\text{min}}\quad\text{ (Symmetrical $m(t)$)}\\ m_a&=k_a A_m \quad\text{ (Symmetrical $m(t)$)}\\ x(t)&=A_c\cos(2\pi f_c t)\left[1+k_a m(t)\right]=A_c\cos(2\pi f_c t)\left[1+m_a m(t)/A_c\right], \\ &\text{where $m(t)=A_m\hat m(t)$ and $\hat m(t)$ is the normalized modulating signal}\\ P_c &=\frac{ {A_c}^2}{2}\quad\text{Carrier power}\\ P_x &=\frac{1}{4}{m_a}^2{A_c}^2\\ \eta&=\frac{\text{Signal Power}}{\text{Total Power}}=\frac{P_x}{P_x+P_c}\\ B_T&=2f_m=2B \end{align*}$$

$B_T$: Signal bandwidth $B$: Bandwidth of modulating wave

Overmodulation (resulting in phase reversals at crossing points): $m_a>1$

DSB-SC

$$\begin{align*} x_\text{DSB}(t) &= A_c \cos{(2\pi f_c t)} m(t)\\ B_T&=2f_m=2B \end{align*}$$

FM/PM

$$\begin{align*} s(t) &= A_c\cos\left[2\pi f_c t + k_p m(t)\right]\quad\text{Phase modulated (PM)}\\ s(t) &= A_c\cos(\theta_i(t))=A_c\cos\left[2\pi f_c t + 2 \pi k_f \int_{-\infty}^t m(\tau) d\tau\right]\quad\text{Frequency modulated (FM)}\\ s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right]\quad\text{FM single tone}\\ f_i(t) &= \frac{1}{2\pi}\frac{d}{dt}\theta_i(t)\quad\text{Instantaneous frequency from instantaneous phase}\\ \Delta f&=\beta f_m=k_f A_m f_m = \max_t(k_f m(t))- \min_t(k_f m(t))\quad\text{Maximum frequency deviation}\\ \Delta f&=\max_t(f_i(t))- \min_t(f_i(t))\quad\text{Maximum frequency deviation}\\ \beta&=\frac{\Delta f}{f_m}=k_f A_m\quad\text{Modulation index}\\ D&=\frac{\Delta f}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)}\\ \end{align*}$$

Bessel form and magnitude spectrum (single tone)

$$\begin{align*} s(t) &= A_c\cos\left[2\pi f_c t + \beta \sin(2\pi f_m t)\right] \Leftrightarrow s(t)= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c+nf_m)t] \end{align*}$$

FM signal power

$$\begin{align*} P_\text{av}&=\frac{ {A_c}^2}{2}\\ P_\text{band\_index}&=\frac{ {A_c}^2{J_\text{band\_index}}^2(\beta)}{2}\\ \text{band\_index}&=0\implies f_c+0f_m\\ \text{band\_index}&=1\implies f_c+1f_m,\dots\\ \end{align*}$$

Carson's rule to find $B$ (98% power bandwidth rule)

$$\begin{align*} B &= 2Mf_m = 2(\beta + 1)f_m\\ &= 2(\Delta f+f_m)\\ &= 2(k_f A_m+f_m)\\ &= 2(D+1)W_m\\ B &= \begin{cases} 2(\Delta f+f_m) & \text{FM, sinusoidal message}\\ 2(\Delta\phi + 1)f_m & \text{PM, sinusoidal message} \end{cases}\\ \end{align*}$$

Complex envelope

$$\begin{align*} s(t)&=A_c\cos(2\pi f_c t+\beta\sin(2\pi f_m t)) \Leftrightarrow \tilde{s}(t) = A_c\exp(j\beta\sin(2\pi f_m t))\\ s(t)&=\text{Re}[\tilde{s}(t)\exp{(j2\pi f_c t)}]\\ \tilde{s}(t) &= A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\exp(j2\pi f_m t) \end{align*}$$

Band

Power, energy and autocorrelation

$$\begin{align*} G_\text{WGN}(f)&=\frac{N_0}{2}\\ G_x(f)&=|H(f)|^2G_w(f)\text{ (PSD)}\\ G_x(f)&=G(f)G_w(f)\text{ (PSD)}\\ G_x(f)&=\lim_{T\to\infty}\frac{|X_T(f)|^2}{T}\text{ (PSD)}\\ G_x(f)&=\mathfrak{F}[R_x(\tau)]\text{ (WSS)}\\ P_x&={\sigma_x}^2=\int_\mathbb{R}G_x(f)df\quad\text{For zero mean}\\ P_x&={\sigma_x}^2=\lim_{t\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt\quad\text{For zero mean}\\ P[A\cos(2\pi f t+\phi)]&=\frac{A^2}{2}\quad\text{Power of sinusoid }\\ E_x&=\int_{-\infty}^{\infty}|x(t)|^2dt=\int_{-\infty}^{\infty}|X(f)|^2df\quad\text{Parseval's theorem}\\ R_x(\tau) &= \mathfrak{F}(G_x(f))\quad\text{PSD to Autocorrelation} \end{align*}$$

Noise performance

Coherent detection system.

$$\begin{align*} y(t) &= m(t)\cos(2\pi f_c t+\theta) = m(t)\cos^2(2\pi f_c t+\theta)\text{ After IF mixing.}\\ &= m(t)\frac{1}{2}(1+\cos(4 f_c t+2\theta))\\ &= \frac{1}{2}m(t)+\frac{1}{2}\cos(4 f_c t+2\theta)\\ \implies S_u(f) &= \left(\frac{1}{2}\right)^2S_u(f)\quad\text{After LPF.} \end{align*}$$

Use formualas from previous section, Power, energy and autocorrelation.
Use these formulas in particular:

$$\begin{align*} G_\text{WGN}(f)&=\frac{N_0}{2}\\ G_x(f)&=|H(f)|^2G_w(f)&\text{Note the square in $|H(f)|^2$}\\ P_x&={\sigma_x}^2=\int_\mathbb{R}G_x(f)df&\text{Often perform graphical integration}\\ \end{align*}$$

$$\begin{align*} \text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\ \text{CNR}_\text{in,FM} &= \frac{A^2}{2WN_0}\\ \text{SNR}_\text{FM} &= \frac{3A^2k_f^2P}{2N_0W^3}\\ \text{SNR(dB)} &= 10\log_{10}(\text{SNR}) \quad\text{Decibels from ratio} \end{align*}$$

Sampling

$$\begin{align*} t&=nT_s\\ T_s&=\frac{1}{f_s}\\ x_s(t)&=x(t)\delta_s(t)=x(t)\sum_{n\in\mathbb{Z}}\delta(t-nT_s)=\sum_{n\in\mathbb{Z}}x(nT_s)\delta(t-nT_s)\\ X_s(f)&=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-\frac{n}{T_s}\right)=X(f)*\sum_{n\in\mathbb{Z}}\delta\left(f-n f_s\right)\\ B&>\frac{1}{2}f_s, 2B>f_s\rightarrow\text{Aliasing}\\ \end{align*}$$

Procedure to reconstruct sampled signal

Analog signal $x'(t)$ which can be reconstructed from a sampled signal $x_s(t)$: Put $x_s(t)$ through LPF with maximum frequency of $f_s/2$ and minimum frequency of $-f_s/2$. Anything outside of the BPF will be attenuated, therefore $n$ which results in frequencies outside the BPF will evaluate to $0$ and can be ignored.

Example: $f_s=5000\implies \text{LPF}\in[-2500,2500]$

Then iterate for $n=0,1,-1,2,-2,\dots$ until the first iteration where the result is 0 since all terms are eliminated by the LPF.

TODO: Add example

Then add all terms and transform $\bar X_s(f)$ back to time domain to get $x_s(t)$

Fourier transform of continuous time periodic signal (1)

Required for some questions on sampling:

Transform a continuous time-periodic signal $x_p(t)=\sum_{n=-\infty}^\infty x(t-nT_s)$ with period $T_s$:

$$X_p(f)=\sum_{n=-\infty}^\infty C_n\delta(f-nf_s)\quad f_s=\frac{1}{T_s}$$

Calculate $C_n$ coefficient as follows from $x_p(t)$:

$$\begin{align*} C_n&=\frac{1}{T_s} \int_{T_s} x_p(t)\exp(-j2\pi f_s t)dt\\ &=\frac{1}{T_s} X(nf_s)\quad\color{red}\text{(TODO: Check)}\quad\color{white}\text{$x(t-nT_s)$ is contained in the interval $T_s$} \end{align*}$$

Nyquist criterion for zero-ISI

Do not transmit more than $2B$ samples per second over a channel of $B$ bandwidth.

Insert here figure 8.3 from M F Mesiya - Contemporary Communication Systems (Add image to images/sampling.png)

Cannot add directly due to copyright!

Quantizer

$$\begin{align*} \Delta &= \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\\ \end{align*}$$

Quantization noise

$$\begin{align*} e &:= y-x\quad\text{Quantization error}\\ \mu_E &= E[E] = 0\quad\text{Zero mean}\\ {\sigma_E}^2&=E[E^2]-0^2=\int_{-\Delta/2}^{\Delta/2}e^2\times\left(\frac{1}{\Delta}\right) de\quad\text{Where $E\thicksim 1/\Delta$ uniform over $(-\Delta/2,\Delta/2)$}\\ \text{SQNR}&=\frac{\text{Signal power}}{\text{Quantization noise}}\\ \text{SQNR(dB)}&=10\log_{10}(\text{SQNR}) \end{align*}$$

Insert here figure 8.17 from M F Mesiya - Contemporary Communication Systems (Add image to images/quantizer.png)

Cannot add directly due to copyright!

Line codes

$$\begin{align*} R_b&\rightarrow\text{Bit rate}\\ D&\rightarrow\text{Symbol rate | }R_d\text{ | }1/T_b\\ A&\rightarrow m_a\\ V(f)&\rightarrow\text{Pulse shape}\\ V_\text{rectangle}(f)&=T\text{sinc}(fT\times\text{DutyCycle})\\ G_\text{MunipolarNRZ}(f)&=\frac{(M^2-1)A^2D}{12}|V(f)|^2+\frac{(M-1)^2}{4}(DA)^2\sum_{l=-\infty}^{\infty}|V(lD)|^2\delta(f-lD)\\ G_\text{MpolarNRZ}(f)&=\frac{(M^2-1)A^2D}{3}|V(f)|^2\\ G_\text{unipolarNRZ}(f)&=\frac{A^2}{4R_b}\left(\text{sinc}^2\left(\frac{f}{R_b}\right)+R_b\delta(f)\right), \text{NB}_0=R_b\\ G_\text{polarNRZ}(f)&=\frac{A^2}{R_b}\text{sinc}^2\left(\frac{f}{R_b}\right)\\ G_\text{unipolarNRZ}(f)&=\frac{A^2}{4R_b}\left(\text{sinc}^2\left(\frac{f}{R_b}\right)+R_b\delta(f)\right)\\ G_\text{unipolarRZ}(f)&=\frac{A^2}{16} \left(\sum _{l=-\infty }^{\infty } \delta \left(f-\frac{l}{T_b}\right) \left| \text{sinc}(\text{duty} \times l) \right| {}^2+T_b \left| \text{sinc}\left(\text{duty} \times f T_b\right) \right| {}^2\right), \text{NB}_0=2R_b \end{align*}$$

Modulation and basis functions

BASK

Basis functions

$$\begin{align*} \varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\ \end{align*}$$

Symbol mapping

$$b_n:\{1,0\}\to a_n:\{1,0\}$$

2 possible waveforms

$$\begin{align*} s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{2E_b}\varphi_1(t)\\ s_1(t)&=0\\ &\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + 0)=\frac{ {A_c}^2}{4}T_b$} \end{align*}$$

Distance is $d=\sqrt{2E_b}$

BPSK

Basis functions

$$\begin{align*} \varphi_1(t) &= \sqrt{\frac{2}{T_b}}\cos(2\pi f_c t)\quad0\leq t\leq T_b\\ \end{align*}$$

Symbol mapping

$$b_n:\{1,0\}\to a_n:\{1,\color{green}-1\color{white}\}$$

2 possible waveforms

$$\begin{align*} s_1(t)&=A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=\sqrt{E_b}\varphi_1(t)\\ s_1(t)&=-A_c\sqrt{\frac{T_b}{2}}\varphi_1(t)=-\sqrt{E_b}\varphi_2(t)\\ &\text{Since $E_b=E_\text{average}=\frac{1}{2}(\frac{ {A_c}^2}{2}\times T_b + \frac{ {A_c}^2}{2}\times T_b)=\frac{ {A_c}^2}{2}T_b$} \end{align*}$$

Distance is $d=2\sqrt{E_b}$

QPSK ($M=4$ PSK)

Basis functions

$$\begin{align*} T &= 2 T_b\quad\text{Time per symbol for two bits $T_b$}\\ \varphi_1(t) &= \sqrt{\frac{2}{T}}\cos(2\pi f_c t)\quad0\leq t\leq T\\ \varphi_2(t) &= \sqrt{\frac{2}{T}}\sin(2\pi f_c t)\quad0\leq t\leq T\\ \end{align*}$$

4 possible waveforms

$$\begin{align*} s_1(t)&=\sqrt{E_s/2}\left[\varphi_1(t)+\varphi_2(t)\right]\\ s_2(t)&=\sqrt{E_s/2}\left[\varphi_1(t)-\varphi_2(t)\right]\\ s_3(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)+\varphi_2(t)\right]\\ s_4(t)&=\sqrt{E_s/2}\left[-\varphi_1(t)-\varphi_2(t)\right]\\ \end{align*}$$

Note on energy per symbol: Since $|s_i(t)|=A_c$, have to normalize distance as follows:

$$\begin{align*} s_i(t)&=A_c\sqrt{T/2}/\sqrt{2}\times\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ &=\sqrt{T{A_c}^2/4}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ &=\sqrt{E_s/2}\left[\alpha_{1i}\varphi_1(t)+\alpha_{2i}\varphi_2(t)\right]\\ \end{align*}$$

Signal

$$\begin{align*} \text{Symbol mapping: }& \left\{1,0\right\}\to\left\{1,-1\right\}\\ I(t) &= b_{2n}\varphi_1(t)\quad\text{Even bits}\\ Q(t) &= b_{2n+1}\varphi_2(t)\quad\text{Odd bits}\\ x(t) &= A_c[I(t)\cos(2\pi f_c t)-Q(t)\sin(2\pi f_c t)] \end{align*}$$

Example of waveform

tBitstream[bitstream_, Tb_, title_] :=
  Module[{timeSteps, gridLines, plot},
   timeSteps =
    Flatten[Table[{(n - 1)     Tb, bitstream[[n]]}, {n, 1,
        Length[bitstream]}] /. {t_, v_} :> { {t, v}, {t + Tb, v}}, 1];
   gridLines = {Join[
      Table[{n  Tb, Dashed}, {n, 1, 2  Length[bitstream], 2}],
      Table[{n  Tb, Thin}, {n, 0, 2  Length[bitstream], 2}]], None};
   plot =
    Labeled[ListLinePlot[timeSteps, InterpolationOrder -> 0,
      PlotRange -> Full, GridLines -> gridLines, PlotStyle -> Thick,
      Ticks -> {Table[{n     Tb,
          Row[{n, "\!\(\*SubscriptBox[\(T\), \(b\)]\)"}]}, {n, 0,
          Length[bitstream]}], {-1, 0, 1}},
      LabelStyle -> Directive[Bold, 12],
      PlotRangePadding -> {Scaled[.05]}, AspectRatio -> 0.1,
      ImageSize -> Large], {Style[title, "Text", 16]}, {Right}]];
tBitstream[{0, 1, 0, 0, 1, 0, 1, 1, 1, 0}, 1, "Bitstream Step Plot"]
tBitstream[{-1, -1, -1, -1, 1, 1, 1, 1, 1, 1}, 1, "I(t)"]
tBitstream[{1, 1, -1, -1, -1, -1, 1, 1, -1, -1}, 1, "Q(t)"]

Remember that $T=2T_b$