diff --git a/.gitignore b/.gitignore index 83384af..716b2ea 100644 --- a/.gitignore +++ b/.gitignore @@ -1 +1,2 @@ -copyrighted_images \ No newline at end of file +copyrighted_images +README-copyright.pdf \ No newline at end of file diff --git a/README.html b/README.html index 5752366..56f5237 100644 --- a/README.html +++ b/README.html @@ -140,6 +140,13 @@

Idiot's guide to ELEC4402 communication systems

+

This unit allows you to bring infinite physical notes (except books borrowed from the UWA library) to all tests and the final exam. You can't rely on what material they provide in the test/exam, it is very minimal to say the least. Hope this helps.

If you have issues or suggestions, raise them on GitHub. I accept pull requests for fixes or suggestions but the content must not be copyrighted under a non-GPL compatible license.

@@ -196,7 +203,7 @@ along with this program. If not, see http

Printable notes begins on next page (in PDF)

-

Fourier transform identities

+

Fourier transform identities and properties

@@ -250,10 +257,18 @@ along with this program. If not, see http + + + + + + + + @@ -306,6 +321,11 @@ along with this program. If not, see http + + + + + @@ -335,6 +355,16 @@ along with this program. If not, see http + + + + + + + + + +
12[δ(f−fc)+δ(f+fc)]\frac{1}{2}[\delta(f - f_c) + \delta(f + f_c)]
cos⁡(2πfct+θ)\cos(2\pi f_c t+\theta)12[δ(f−fc)exp⁡(jθ)+δ(f+fc)exp⁡(−jθ)]Use for coherent recv.\frac{1}{2}[\delta(f - f_c)\exp(j\theta) + \delta(f + f_c)\exp(-j\theta)]\quad\text{Use for coherent recv.}
sin⁡(2πfct)\sin(2\pi f_c t) 12j[δ(f−fc)−δ(f+fc)]\frac{1}{2j} [\delta(f - f_c) - \delta(f + f_c)]
sin⁡(2πfct+θ)\sin(2\pi f_c t+\theta)12j[δ(f−fc)exp⁡(jθ)−δ(f+fc)exp⁡(−jθ)]\frac{1}{2j} [\delta(f - f_c)\exp(j\theta) - \delta(f + f_c)\exp(-j\theta)]
sgn(t)\text{sgn}(t) 1jπf\frac{1}{j\pi f}
Differentiation wrt time
tg(t)tg(t)12πddfG(f)\frac{1}{2\pi}\frac{d}{df}G(f)\quadDifferentiation wrt frequency
g∗(t)g^*(t) G∗(−f)G^*(-f) Conjugate functionsaG(f)+bH(f)aG(f)+bH(f) Linearity a,ba,b constants
∫−∞∞x(t)y∗(t)dt\int_{-\infty}^\infty x(t)y^*(t)dt∫−∞∞X(f)Y∗(f)df\int_{-\infty}^\infty X(f)Y^*(f)dfParseval's theorem
Ex=∫−∞∞∣x(t)∣2dtE_x=\int_{-\infty}^\infty |x(t)|^2dtEx=∫−∞∞∣X(f)∣2dfE_x=\int_{-\infty}^\infty |X(f)|^2dfParseval's theorem
 @@ -539,15 +569,17 @@ h(t)&=h_I(t)\cos(2\pi f_c t)-h_Q(t)\sin(2\pi f_c t)\\

FM/PM

-

s(t)=Accos⁡[2πfct+kpm(t)]Phase modulated (PM)s(t)=Accos⁡[2πfct+2πkf∫0tm(τ)dτ]Frequency modulated (FM)s(t)=Accos⁡[2πfct+βsin⁡(2πfmt)]FM single toneβ=Δffm=kfAmModulation indexΔf=βfm=kfAmfm=max⁡t(kfm(t))−min⁡t(kfm(t))Maximum frequency deviationD=ΔfWmDeviation ratio, where Wm is bandwidth of m(t) (Use FT)\begin{align*} +

s(t)=Accos⁡[2πfct+kpm(t)]Phase modulated (PM)s(t)=Accos⁡(θi(t))=Accos⁡[2πfct+2πkf∫−∞tm(τ)dτ]Frequency modulated (FM)s(t)=Accos⁡[2πfct+βsin⁡(2πfmt)]FM single tonefi(t)=12πddtθi(t)Instantaneous frequency from instantaneous phaseΔf=βfm=kfAmfm=max⁡t(kfm(t))−min⁡t(kfm(t))Maximum frequency deviationΔf=max⁡t(fi(t))−min⁡t(fi(t))Maximum frequency deviationβ=Δffm=kfAmModulation indexD=ΔfWmDeviation ratio, where Wm is bandwidth of m(t) (Use FT)\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\left[2\pi f_c t + 2 \pi k_f \int_0^t m(\tau) d\tau\right]\quad\text{Frequency modulated (FM)}\\ + 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}\\ - \beta&=\frac{\Delta f}{f_m}=k_f A_m\quad\text{Modulation index}\\ + 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}\\ - D&=\frac{\Delta f}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)} + \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)

s(t)=Accos⁡[2πfct+βsin⁡(2πfmt)]⇔s(t)=Ac∑n=−∞∞Jn(β)cos⁡[2π(fc+nfm)t]\begin{align*} @@ -577,21 +609,6 @@ B &= \begin{cases} \end{align*}

-

Δf\Delta f of arbitrary modulating signal

-

Find instantaneous frequency fFMf_\text{FM}.

-

MM: Number of pairs of significant sidebands

-

s(t)=Accos⁡(θFM(t))fFM(t)=12πdθFM(t)dtAm=max⁡t∣m(t)∣Δf=max⁡t(fFM(t))−fcWm=max(frequencies in θFM(t)...)Example: sinc(At+t)+2cos⁡(2πt)=sin⁡(2π((At+t)/2))π(At+t)+2cos⁡(2πt)→Wm=max⁡(A+12,1)D=ΔfWmBT=2(D+1)Wm\begin{align*} -s(t)&=A_c\cos(\theta_\text{FM}(t))\\ -f_\text{FM}(t) &= \frac{1}{2\pi}\frac{d\theta_\text{FM}(t)}{dt}\\ -A_m &= \max_t|m(t)|\\ -\Delta f &= \max_t(f_\text{FM}(t)) - f_c\\ -W_m &= \text{max}(\text{frequencies in $\theta_\text{FM}(t)$...}) \\ -\text{Example: }&\text{sinc}(At+t)+2\cos(2\pi t)=\frac{\sin(2\pi((At+t)/2))}{\pi(At+t)}+2\cos(2\pi t)\to W_m=\max\left(\frac{A+1}{2},1\right)\\ -D &= \frac{\Delta f}{W_m}\\ -B_T &= 2(D+1)W_m -\end{align*} -

-

Complex envelope

s(t)=Accos⁡(2πfct+βsin⁡(2πfmt))⇔s~(t)=Acexp⁡(jβsin⁡(2πfmt))s(t)=Re[s~(t)exp⁡(j2πfct)]s~(t)=Ac∑n=−∞∞Jn(β)exp⁡(j2πfmt)\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))\\ @@ -616,22 +633,39 @@ B_T &= 2(D+1)W_m

Power, energy and autocorrelation

-

GWGN(f)=N02Gx(f)=∣H(f)∣2Gw(f) (PSD)Gx(f)=G(f)Gw(f) (PSD)Gx(f)=lim⁡T→∞∣XT(f)∣2T (PSD)Gx(f)=F[Rx(τ)] (WSS)P=σ2=∫RGx(f)dfP=σ2=lim⁡t→∞1T∫−T/2T/2∣x(t)∣2dtP[Acos⁡(2πft+ϕ)]=A22Power of sinusoid E=∫−∞∞∣x(t)∣2dt=∣X(f)∣2Rx(τ)=F(Gx(f))PSD to Autocorrelation\begin{align*} +

GWGN(f)=N02Gx(f)=∣H(f)∣2Gw(f) (PSD)Gx(f)=G(f)Gw(f) (PSD)Gx(f)=lim⁡T→∞∣XT(f)∣2T (PSD)Gx(f)=F[Rx(τ)] (WSS)Px=σx2=∫RGx(f)dfFor zero meanPx=σx2=lim⁡t→∞1T∫−T/2T/2∣x(t)∣2dtFor zero meanP[Acos⁡(2πft+ϕ)]=A22Power of sinusoid Ex=∫−∞∞∣x(t)∣2dt=∫−∞∞∣X(f)∣2dfParseval’s theoremRx(τ)=F(Gx(f))PSD to 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&=\sigma^2=\int_\mathbb{R}G_x(f)df\\ - P&=\sigma^2=\lim_{t\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt\\ + 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&=\int_{-\infty}^{\infty}|x(t)|^2dt=|X(f)|^2\\ + 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.

+

y(t)=m(t)cos⁡(2πfct+θ)=m(t)cos⁡2(2πfct+θ) After IF mixing.=m(t)12(1+cos⁡(4fct+2θ))=12m(t)+12cos⁡(4fct+2θ)  ⟹  Su(f)=(12)2Su(f)After LPF.\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:

+

GWGN(f)=N02Gx(f)=∣H(f)∣2Gw(f)Note the square in ∣H(f)∣2Px=σx2=∫RGx(f)dfOften perform graphical integration\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*} +

+

CNRin=PinPnoiseCNRin,FM=A22WN0SNRFM=3A2kf2P2N0W3SNR(dB)=10log⁡10(SNR)Decibels from ratio\begin{align*} \text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\ \text{CNR}_\text{in,FM} &= \frac{A^2}{2WN_0}\\ @@ -677,8 +711,8 @@ B_T &= 2(D+1)W_m

By Bob K - Own work, CC0, https://commons.wikimedia.org/w/index.php?curid=94674142

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

Cannot add directly due to copyright!

-

sampling -sampling

+sampling +

sampling

Quantizer

Δ=xMax−xMin2kfor k-bit quantizer (V/lsb)\begin{align*} \Delta &= \frac{x_\text{Max}-x_\text{Min}}{2^k} \quad\text{for $k$-bit quantizer (V/lsb)}\\ @@ -697,8 +731,8 @@ B_T &= 2(D+1)W_m

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

Cannot add directly due to copyright!

-

quantizer -quantizer

+quantizer +

quantizer

Line codes

binary_codes

Rb→Bit rateD→Symbol rate | Rd | 1/TbA→maV(f)→Pulse shapeVrectangle(f)=Tsinc(fT×DutyCycle)GMunipolarNRZ(f)=(M2−1)A2D12∣V(f)∣2+(M−1)24(DA)2∑l=−∞∞∣V(lD)∣2δ(f−lD)GMpolarNRZ(f)=(M2−1)A2D3∣V(f)∣2GunipolarNRZ(f)=A24Rb(sinc2(fRb)+Rbδ(f)),NB0=RbGpolarNRZ(f)=A2Rbsinc2(fRb)GunipolarNRZ(f)=A24Rb(sinc2(fRb)+Rbδ(f))GunipolarRZ(f)=A216(∑l=−∞∞δ(f−lTb)∣sinc(duty×l)∣2+Tb∣sinc(duty×fTb)∣2),NB0=2Rb\begin{align*} @@ -1046,7 +1080,8 @@ tBitstream[{1, 1, -1, -1, -1, -1, 1, 1, -1, -1}, 1, "Q(t)"]

Matched filter

1. Filter function

-

Find transfer function h(t)h(t) of matched filter and apply to an input:

+

Find transfer function h(t)h(t) of matched filter and apply to an input:
+Note that x(T−t)x(T-t) is equivalent to horizontally flipping x(t)x(t) around x=T/2x=T/2.

h(t)=s1(T−t)−s2(T−t)h(t)=s∗(T−t)((.)* is the conjugate)son(t)=h(t)∗sn(t)=∫∞∞h(τ)sn(t−τ)dτFilter outputno(t)=h(t)∗n(t)Noise at filter output\begin{align*} h(t)&=s_1(T-t)-s_2(T-t)\\ h(t)&=s^*(T-t) \qquad\text{((.)* is the conjugate)}\\ @@ -1055,7 +1090,7 @@ tBitstream[{1, 1, -1, -1, -1, -1, 1, 1, -1, -1}, 1, "Q(t)"] \end{align*}

-

2. Bit error rate

+

2. Bit error rate of matched filter

Bit error rate (BER) from matched filter outputs and filter output noise

Q(x)=12−12erf(x2)⇔erf(x2)=1−2Q(x)Eb=d2=∫−∞∞∣s1(t)−s2(t)∣2dtEnergy per bit/DistanceT=1/RbRb: BitrateEb=PavT=Pav/RbEnergy per bitPav=Eb/T=EbRbAverage powerP(W)=10P(dB)10PRX(W)=PTX(W)⋅10Ploss(dB)10Ploss is expressed with negative sign e.g. "-130 dB"BERMatchedFilter=Q(d22N0)=Q(Eb2N0)BERunipolarNRZ|BASK=Q(d2N0)=Q(EbN0)BERpolarNRZ|BPSK=Q(2d2N0)=Q(2EbN0)\begin{align*} Q(x)&=\frac{1}{2}-\frac{1}{2}\text{erf}\left(\frac{x}{\sqrt{2}}\right)\Leftrightarrow\text{erf}\left(\frac{x}{\sqrt{2}}\right)=1-2Q(x)\\ @@ -1823,8 +1858,24 @@ h400000v40h-400000z"/>​

ISI, channel model

+

Raised cosine (RC) pulse

+

Raised cosine pulse

+

0≤α≤10\leq\alpha\leq1 +

+ +

⚠ NOTE might not be safe to assume T′=TT'=T, if you can solve the question without TT then use that method.

Nyquist criterion for zero ISI

-

TODO:

+ +

D>2WUse W from table below depending on modulation scheme.BNyquist=W1+αα=Excess BWBNyquist=Babs−BNyquistBNyquist\begin{align*} + D &> 2W\quad\text{Use $W$ from table below depending on modulation scheme.}\\ + B_\text{Nyquist} &= \frac{W}{1+\alpha}\\ + \alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\ +\end{align*} +

+

Nomenclature

D→Symbol Rate, Max. Signalling RateT→Symbol DurationM→Symbol set sizeW→Bandwidth\begin{align*} D&\rightarrow\text{Symbol Rate, Max. Signalling Rate}\\ @@ -1834,12 +1885,7 @@ h400000v40h-400000z"/>​

-

Raised cosine (RC) pulse

-

Raised cosine pulse

-

0≤α≤10\leq\alpha\leq1 -

- -

⚠ NOTE might not be safe to assume T′=TT'=T, if you can solve the question without TT then use that method.

+

Bandwidth WW and bit error rate of modulation schemes

To solve this type of question:

  1. Use the formula for DD below
    2. @@ -1848,12 +1894,20 @@ h400000v40h-400000z"/>​ -Linear modulation (MM-PSK, BPSK, MM-QAM) -NRZ unipolar encoding, BASK +Linear modulation +Half +BPSK, QPSK, MM-PSK, MM-QAM, ASK, FSK +MM-PAM, PAM + + +RZ unipolar, Manchester +NRZ Unipolar, NRZ Polar, Bipolar RZ + + W=Babs-absW=B_\text{\color{green}abs-abs} W=BabsW=B_\text{\color{green}abs} @@ -1867,28 +1921,13 @@ h400000v40h-400000z"/>​ -

    Rb bit/s=(D symbol/s)×(k bit/symbol)M symbol/set=2kEb=PT=Pav/RbEnergy per bit\begin{align*} +

    Rb bit/s=(D symbol/s)×(k bit/symbol)M symbol/set=2kT s/symbol=1/(D symbol/s)Eb=PT=Pav/RbEnergy per bit\begin{align*} R_b\text{ bit/s}&=(D\text{ symbol/s})\times(k\text{ bit/symbol})\\ M\text{ symbol/set}&=2^k\\ + T\text{ s/symbol}&=1/(D\text{ symbol/s})\\ E_b&=PT=P_\text{av}/R_b\quad\text{Energy per bit}\\ \end{align*} -

    - -

    Nyquist stuff

    -

    TODO: Condition for 0 ISI

    -

    Pr(kT)={1k=00k≠0P_r(kT)=\begin{cases} - 1 & k=0\\ - 0 & k\neq0 -\end{cases} -

    - -

    Other

    -

    Excess BW=Babs−BNyquist=1+α2T−12T=α2TFOR NRZ (Use correct Babs)α=Excess BWBNyquist=Babs−BNyquistBNyquistT=1/D\begin{align*} - \text{Excess BW}&=B_\text{abs}-B_\text{Nyquist}=\frac{1+\alpha}{2T}-\frac{1}{2T}=\frac{\alpha}{2T}\quad\text{FOR NRZ (Use correct $B_\text{abs}$)}\\ - \alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\ - T&=1/D -\end{align*} -

    +

    Table of bandpass signalling and BER

    @@ -2198,14 +2237,14 @@ M347 1759 V0 H263 V1759 v1800 v1759 h84z"/>

    Channel capacity of weakly symmetric channel

    -

    C→Channel capacity (bits/channels used)N→Output alphabet sizep→Probability vector, any row of the transition matrixC=log⁡2(N)−H(p)Capacity for weakly symmetric and symmetric channelsR<C for error-free transmission\begin{align*} +

    C→Channel capacity (bits/channels used)N→Output alphabet sizep→Probability vector, any row of the transition matrixC=log⁡2(N)−H(p)Capacity for weakly symmetric and symmetric channelsRb<C for error-free transmission\begin{align*} C &\to\text{Channel capacity (bits/channels used)}\\ N &\to\text{Output alphabet size}\\ \mathbf{p} &\to\text{Probability vector, any row of the transition matrix}\\ C &= \log_2(N)-H(\mathbf{p})\quad\text{Capacity for weakly symmetric and symmetric channels}\\ - R &< C \text{ for error-free transmission} + R_b &< C \text{ for error-free transmission} \end{align*} -

    +

    Channel capacity of an AWGN channel

    yi=xi+nini∼N(0,N0/2)y_i=x_i+n_i\quad n_i\thicksim N(0,N_0/2) @@ -2214,17 +2253,29 @@ M347 1759 V0 H263 V1759 v1800 v1759 h84z"/>C=12log⁡2(1+PavN0/2)C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right)

    -

    Channel capacity of a bandwidth AWGN channel

    -

    Note: Define XOR (⊕\oplus) as exclusive OR, or modulo-2 addition.

    -

    Ps→Bandwidth limited average poweryi=bandpassW(xi)+nini∼N(0,N0/2)C=Wlog⁡2(1+PsN0W)C=Wlog⁡2(1+SNR)SNR=Ps/(N0W)\begin{align*} +

    Channel capacity of a bandwidth limited AWGN channel

    +

    Ps→Bandwidth limited average poweryi=bandpassW(xi)+nini∼N(0,N0/2)C=Wlog⁡2(1+PsN0W)C=Wlog⁡2(1+SNR)SNR=Ps/(N0W)\begin{align*} P_s&\to\text{Bandwidth limited average power}\\ y_i&=\text{bandpass}_W(x_i)+n_i\quad n_i\thicksim N(0,N_0/2)\\ C&=W\log_2\left(1+\frac{P_s}{N_0 W}\right)\\ - C&=W\log_2(1+\text{SNR})\quad\text{SNR}=P_s/(N_0 W) + C&=W\log_2(1+\text{SNR})\\ + \text{SNR}&=P_s/(N_0 W) \end{align*} -

    +

    + +

    Shannon limit

    +

    Rb<C  ⟹  Rb<Wlog⁡2(1+PsN0W)For bandwidth limited AWGN channelEbN0>2η−1ηSNR per bit required for error-free transmissionη=RbWSpectral efficiency (bit/(s-Hz))η≫1Bandwidth limitedη≪1Power limited\begin{align*} + R_b &< C\\ + \implies R_b &< W\log_2\left(1+\frac{P_s}{N_0 W}\right)\quad\text{For bandwidth limited AWGN channel}\\ + \frac{E_b}{N_0} &> \frac{2^\eta-1}{\eta}\quad\text{SNR per bit required for error-free transmission}\\ + \eta &= \frac{R_b}{W}\quad\text{Spectral efficiency (bit/(s-Hz))}\\ + \eta &\gg 1\quad\text{Bandwidth limited}\\ + \eta &\ll 1\quad\text{Power limited} +\end{align*} +

    Channel code

    +

    Note: Define XOR (⊕\oplus) as exclusive OR, or modulo-2 addition.

    @@ -2259,8 +2310,8 @@ M347 1759 V0 H263 V1759 v1800 v1759 h84z"/>
  2. Zero vector must be present at least once
  3. The XOR of any codeword pair in the code must result in a codeword that is already present in the code table.
    4. +
  4. dmin=wmind_\text{min}=w_\text{min} (Implied by (1) and (2).)
    5. -

    For a linear block code, dmin=wmind_\text{min}=w_\text{min}

    Code generation

    Each generator vector is a binary string of size nn. There are kk generator vectors in G\mathbf{G}.

    gi=[gi,0…gi,n−2gi,n−1]g0=[1010]Example for n=4G=[g0g1⋮gk−1]=[g0,0…g0,n−2g0,n−1g1,0…g1,n−2g1,n−1⋮⋱⋮⋮gk−1,0…gk−1,n−2gk−1,n−1]\begin{align*} diff --git a/README.md b/README.md index fdb3ff6..1d030ce 100644 --- a/README.md +++ b/README.md @@ -1,5 +1,13 @@ # Idiot's guide to ELEC4402 communication systems + + This unit allows you to bring infinite physical notes (except books borrowed from the UWA library) to all tests and the final exam. You can't rely on what material they provide in the test/exam, it is very _minimal_ to say the least. Hope this helps. @@ -70,7 +78,7 @@ along with this program. If not, see .

    -## Fourier transform identities +## Fourier transform identities and properties | Time domain $x(t)$ | Frequency domain $X(f)$ | | --------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------- | @@ -85,25 +93,30 @@ along with this program. If not, see . | $\delta(t - t_0)$ | $\exp(-j2\pi f t_0)$ | | $\exp(j2\pi f_c t)$ | $\delta(f - f_c)$ | | $\cos(2\pi f_c t)$ | $\frac{1}{2}[\delta(f - f_c) + \delta(f + f_c)]$ | +| $\cos(2\pi f_c t+\theta)$ | $\frac{1}{2}[\delta(f - f_c)\exp(j\theta) + \delta(f + f_c)\exp(-j\theta)]\quad\text{Use for coherent recv.}$ | | $\sin(2\pi f_c t)$ | $\frac{1}{2j} [\delta(f - f_c) - \delta(f + f_c)]$ | +| $\sin(2\pi f_c t+\theta)$ | $\frac{1}{2j} [\delta(f - f_c)\exp(j\theta) - \delta(f + f_c)\exp(-j\theta)]$ | | $\text{sgn}(t)$ | $\frac{1}{j\pi f}$ | | $\frac{1}{\pi t}$ | $-j \text{sgn}(f)$ | | $u(t)$ | $\frac{1}{2} \delta(f) + \frac{1}{j2\pi f}$ | | $\sum_{n=-\infty}^{\infty} \delta(t - nT_0)$ | $\frac{1}{T_0} \sum_{n=-\infty}^{\infty} \delta\left(f - \frac{n}{T_0}\right)=f_0 \sum_{n=-\infty}^{\infty} \delta\left(f - n f_0\right)$ | -| Time domain $x(t)$ | Frequency domain $X(f)$ | Property | -| ------------------------------- | ------------------------------------------------ | ------------------------- | -| $g(t-a)$ | $\exp(-j2\pi fa)G(f)$ | Time shifting | -| $\exp(-j2\pi f_c t)g(t)$ | $G(f-f_c)$ | Frequency shifting | -| $g(bt)$ | $\frac{G(f/b)}{\|b\|}$ | Time scaling | -| $g(bt-a)$ | $\frac{1}{\|b\|}\exp(-j2\pi a(f/b))\cdot G(f/b)$ | Time scaling and shifting | -| $\frac{d}{dt}g(t)$ | $j2\pi fG(f)\quad$ | Differentiation wrt time | -| $g^*(t)$ | $G^*(-f)$ | Conjugate functions | -| $G(t)$ | $g(-f)$ | Duality | -| $\int_{-\infty}^t g(\tau)d\tau$ | $\frac{1}{j2\pi f}G(f)+\frac{G(0)}{2}\delta(f)$ | Integration wrt time | -| $g(t)h(t)$ | $G(f)*H(f)$ | Time multiplication | -| $g(t)*h(t)$ | $G(f)H(f)$ | Time convolution | -| $ag(t)+bh(t)$ | $aG(f)+bH(f)$ | Linearity $a,b$ constants | +| Time domain $x(t)$ | Frequency domain $X(f)$ | Property | +| ---------------------------------------- | ------------------------------------------------ | ----------------------------- | +| $g(t-a)$ | $\exp(-j2\pi fa)G(f)$ | Time shifting | +| $\exp(-j2\pi f_c t)g(t)$ | $G(f-f_c)$ | Frequency shifting | +| $g(bt)$ | $\frac{G(f/b)}{\|b\|}$ | Time scaling | +| $g(bt-a)$ | $\frac{1}{\|b\|}\exp(-j2\pi a(f/b))\cdot G(f/b)$ | Time scaling and shifting | +| $\frac{d}{dt}g(t)$ | $j2\pi fG(f)\quad$ | Differentiation wrt time | +| $tg(t)$ | $\frac{1}{2\pi}\frac{d}{df}G(f)\quad$ | Differentiation wrt frequency | +| $g^*(t)$ | $G^*(-f)$ | Conjugate functions | +| $G(t)$ | $g(-f)$ | Duality | +| $\int_{-\infty}^t g(\tau)d\tau$ | $\frac{1}{j2\pi f}G(f)+\frac{G(0)}{2}\delta(f)$ | Integration wrt time | +| $g(t)h(t)$ | $G(f)*H(f)$ | Time multiplication | +| $g(t)*h(t)$ | $G(f)H(f)$ | Time convolution | +| $ag(t)+bh(t)$ | $aG(f)+bH(f)$ | Linearity $a,b$ constants | +| $\int_{-\infty}^\infty x(t)y^*(t)dt$ | $\int_{-\infty}^\infty X(f)Y^*(f)df$ | Parseval's theorem | +| $E_x=\int_{-\infty}^\infty \|x(t)\|^2dt$ | $E_x=\int_{-\infty}^\infty \|X(f)\|^2df$ | Parseval's theorem | | Description | Property | | ----------------------------------- | ----------------- | @@ -338,11 +351,13 @@ Overmodulation (resulting in phase reversals at crossing points): $m_a>1$ ```math \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\left[2\pi f_c t + 2 \pi k_f \int_0^t m(\tau) d\tau\right]\quad\text{Frequency modulated (FM)}\\ + 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}\\ - \beta&=\frac{\Delta f}{f_m}=k_f A_m\quad\text{Modulation index}\\ + 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}\\ - D&=\frac{\Delta f}{W_m}\quad\text{Deviation ratio, where $W_m$ is bandwidth of $m(t)$ (Use FT)} + \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*} ``` @@ -388,27 +403,6 @@ B &= \begin{cases} -#### $\Delta f$ of arbitrary modulating signal - -Find instantaneous frequency $f_\text{FM}$. - -$M$: Number of **pairs** of significant sidebands - -```math -\begin{align*} -s(t)&=A_c\cos(\theta_\text{FM}(t))\\ -f_\text{FM}(t) &= \frac{1}{2\pi}\frac{d\theta_\text{FM}(t)}{dt}\\ -A_m &= \max_t|m(t)|\\ -\Delta f &= \max_t(f_\text{FM}(t)) - f_c\\ -W_m &= \text{max}(\text{frequencies in $\theta_\text{FM}(t)$...}) \\ -\text{Example: }&\text{sinc}(At+t)+2\cos(2\pi t)=\frac{\sin(2\pi((At+t)/2))}{\pi(At+t)}+2\cos(2\pi t)\to W_m=\max\left(\frac{A+1}{2},1\right)\\ -D &= \frac{\Delta f}{W_m}\\ -B_T &= 2(D+1)W_m -\end{align*} -``` - - - ### Complex envelope ```math @@ -436,20 +430,44 @@ B_T &= 2(D+1)W_m 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&=\sigma^2=\int_\mathbb{R}G_x(f)df\\ - P&=\sigma^2=\lim_{t\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt\\ + 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&=\int_{-\infty}^{\infty}|x(t)|^2dt=|X(f)|^2\\ + 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. + +```math +\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](#power-energy-and-autocorrelation).\ +Use these formulas in particular: + +```math +\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*} +``` + + + ```math \begin{align*} \text{CNR}_\text{in} &= \frac{P_\text{in}}{P_\text{noise}}\\ @@ -526,7 +544,8 @@ Do not transmit more than $2B$ samples per second over a channel of $B$ bandwidt Cannot add directly due to copyright! -![sampling](copyrighted_images/sampling.png) +sampling + ![sampling](images/sampling.png) ## Quantizer @@ -557,7 +576,8 @@ Cannot add directly due to copyright! Cannot add directly due to copyright! -![quantizer](copyrighted_images/quantizer.png) +quantizer + ![quantizer](images/quantizer.png) ## Line codes @@ -748,7 +768,8 @@ Remember that $T=2T_b$ ### 1. Filter function -Find transfer function $h(t)$ of matched filter and apply to an input: +Find transfer function $h(t)$ of matched filter and apply to an input:\ +Note that $x(T-t)$ is equivalent to horizontally flipping $x(t)$ around $x=T/2$. ```math \begin{align*} @@ -761,7 +782,7 @@ Find transfer function $h(t)$ of matched filter and apply to an input: -### 2. Bit error rate +### 2. Bit error rate of matched filter Bit error rate (BER) from matched filter outputs and filter output noise @@ -886,9 +907,34 @@ Adapted from table 6.1 M F Mesiya - Contemporary Communication Systems ## ISI, channel model +### Raised cosine (RC) pulse + +![Raised cosine pulse](images/RC.drawio.svg) + +```math +0\leq\alpha\leq1 +``` + + + +⚠ NOTE might not be safe to assume $T'=T$, if you can solve the question without $T$ then use that method. + ### Nyquist criterion for zero ISI -TODO: + + +```math +\begin{align*} + D &> 2W\quad\text{Use $W$ from table below depending on modulation scheme.}\\ + B_\text{Nyquist} &= \frac{W}{1+\alpha}\\ + \alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\ +\end{align*} +``` + + ### Nomenclature @@ -903,25 +949,17 @@ TODO: -### Raised cosine (RC) pulse - -![Raised cosine pulse](images/RC.drawio.svg) - -```math -0\leq\alpha\leq1 -``` - - - -⚠ NOTE might not be safe to assume $T'=T$, if you can solve the question without $T$ then use that method. +### Bandwidth $W$ and bit error rate of modulation schemes To solve this type of question: 1. Use the formula for $D$ below 2. Consult the BER table below to get the BER which relates the noise of the channel $N_0$ to $E_b$ and to $R_b$. -| Linear modulation ($M$-PSK, BPSK, $M$-QAM) | NRZ unipolar encoding, BASK | +| Linear modulation | Half | | --------------------------------------------------- | -------------------------------------------------- | +| BPSK, QPSK, $M$-PSK, $M$-QAM, ASK, FSK | $M$-PAM, PAM | +| RZ unipolar, Manchester | NRZ Unipolar, NRZ Polar, Bipolar RZ | | $W=B_\text{\color{green}abs-abs}$ | $W=B_\text{\color{green}abs}$ | | $W=B_\text{abs-abs}=\frac{1+\alpha}{T}=(1+\alpha)D$ | $W=B_\text{abs}=\frac{1+\alpha}{2T}=(1+\alpha)D/2$ | | $D=\frac{W\text{ symbol/s}}{1+\alpha}$ | $D=\frac{2W\text{ symbol/s}}{1+\alpha}$ | @@ -930,37 +968,13 @@ To solve this type of question: \begin{align*} R_b\text{ bit/s}&=(D\text{ symbol/s})\times(k\text{ bit/symbol})\\ M\text{ symbol/set}&=2^k\\ + T\text{ s/symbol}&=1/(D\text{ symbol/s})\\ E_b&=PT=P_\text{av}/R_b\quad\text{Energy per bit}\\ \end{align*} ``` -### Nyquist stuff - -#### TODO: Condition for 0 ISI - -```math -P_r(kT)=\begin{cases} - 1 & k=0\\ - 0 & k\neq0 -\end{cases} -``` - - - -#### Other - -```math -\begin{align*} - \text{Excess BW}&=B_\text{abs}-B_\text{Nyquist}=\frac{1+\alpha}{2T}-\frac{1}{2T}=\frac{\alpha}{2T}\quad\text{FOR NRZ (Use correct $B_\text{abs}$)}\\ - \alpha&=\frac{\text{Excess BW}}{B_\text{Nyquist}}=\frac{B_\text{abs}-B_\text{Nyquist}}{B_\text{Nyquist}}\\ - T&=1/D -\end{align*} -``` - - - ### Table of bandpass signalling and BER | **Binary Bandpass Signaling** | **$B_\text{null-null}$ (Hz)** | **$B_\text{abs-abs}\color{red}=2B_\text{abs}$ (Hz)** | **BER with Coherent Detection** | **BER with Noncoherent Detection** | @@ -1113,7 +1127,7 @@ Output has probabiltiy distribution $p_Y(b_j)=P(y=b_j)$ N &\to\text{Output alphabet size}\\ \mathbf{p} &\to\text{Probability vector, any row of the transition matrix}\\ C &= \log_2(N)-H(\mathbf{p})\quad\text{Capacity for weakly symmetric and symmetric channels}\\ - R &< C \text{ for error-free transmission} + R_b &< C \text{ for error-free transmission} \end{align*} ``` @@ -1133,16 +1147,30 @@ C=\frac{1}{2}\log_2\left(1+\frac{P_\text{av}}{N_0/2}\right) -#### Channel capacity of a bandwidth AWGN channel - -Note: Define XOR ($\oplus$) as exclusive OR, or modulo-2 addition. +#### Channel capacity of a bandwidth limited AWGN channel ```math \begin{align*} P_s&\to\text{Bandwidth limited average power}\\ y_i&=\text{bandpass}_W(x_i)+n_i\quad n_i\thicksim N(0,N_0/2)\\ C&=W\log_2\left(1+\frac{P_s}{N_0 W}\right)\\ - C&=W\log_2(1+\text{SNR})\quad\text{SNR}=P_s/(N_0 W) + C&=W\log_2(1+\text{SNR})\\ + \text{SNR}&=P_s/(N_0 W) +\end{align*} +``` + + + +#### Shannon limit + +```math +\begin{align*} + R_b &< C\\ + \implies R_b &< W\log_2\left(1+\frac{P_s}{N_0 W}\right)\quad\text{For bandwidth limited AWGN channel}\\ + \frac{E_b}{N_0} &> \frac{2^\eta-1}{\eta}\quad\text{SNR per bit required for error-free transmission}\\ + \eta &= \frac{R_b}{W}\quad\text{Spectral efficiency (bit/(s-Hz))}\\ + \eta &\gg 1\quad\text{Bandwidth limited}\\ + \eta &\ll 1\quad\text{Power limited} \end{align*} ``` @@ -1150,6 +1178,8 @@ Note: Define XOR ($\oplus$) as exclusive OR, or modulo-2 addition. ## Channel code +Note: Define XOR ($\oplus$) as exclusive OR, or modulo-2 addition. + | | | | | ---------------- | --------------------------------- | ---------------------------------------------------------------------------------------------------------------------------- | | Hamming weight | $w_H(x)$ | Number of `'1'` in codeword $x$ | @@ -1168,8 +1198,7 @@ A linear block code must be a subspace and satisfy both: 1. Zero vector must be present at least once 2. The XOR of any codeword pair in the code must result in a codeword that is already present in the code table. - -For a linear block code, $d_\text{min}=w_\text{min}$ +3. $d_\text{min}=w_\text{min}$ (Implied by (1) and (2).) ### Code generation diff --git a/README.pdf b/README.pdf index b4d6b8c..8e02372 100644 Binary files a/README.pdf and b/README.pdf differ