Modulation Formats

Introduction

Bit Rate Rb bps
Symbol Rate Rs sym/sec
# bits per symbol K
# different symbols 2^K=M
Bit Duration Tb = 1/Rb
Symbol duration Ts = K*Tb
Energy of mth signal Em
Probability of transmiting mth signal Pm
Average Energy Eavg= $\sum_{1}^{M} E_m P_m$
Average bit energy Eb,avg= Eavg/K
Average Power Pavg=Eb,avg/Tb

Memoryless modulation scheme

Consider a data stream denoted by b0 ,b1,….,b2. The baseband signal transimitted is mathematically expressed as,

(1)
\begin{eqnarray} S_l(t) &=& \sum b_n g(t-nT_s) \\ S(t) &=& \Re \{S_l(t) e^ {j2\pi f_c t} \} \end{eqnarray}

Modulation Schemes with Memory

At time l input is $I_l$ and the modulator is in state $S_l$. Suppose there are $2^Q$ states, $Q=(L-1)K$.

(2)
\begin{align} S_{m_l}(t) = f (S_{l-1}, I_l), \hspace{3mm} S_l= g(S_{l-1},I_l) \end{align}

Pulse Amplitude Modulation( M-ary-PAM)


Baseband Transmission

In this modulation shceme, the m signals(base-band) that are mapped to blocks of bits are given by,

(3)
\begin{eqnarray} s_m(t) = A_m P(t) \hspace{5mm} 1 \leq m \leq M \\ \end{eqnarray}
(4)
\begin{align} A_m = 2m-1-M \hspace{5mm} m=1,2,...,M \end{align}

P(t) is chosen to have the orrect spectral charecteristics. let Ep be the energy of the pulse.

4.JPG

The following table summerizes the various properties of this modulation shceme. We assume that the apriori probability of mth signal is Pm = 1/M.

E_m $A_m ^2 E_p$
Eavg $\left(\frac {M^2-1} {3} \right )E_p$
Eb-avg $\left(\frac {M^2-1} {3 \log_2 M} \right ) E_p$
Basis , $\phi(t)$ $\frac{P(t)}{\sqrt{E_p}}$
$d_{min}^2$ $4E_p$
Power efficiency, $\eta$ $\frac{d_{min}^2}{4E_{b,avg}}$
$\eta$ $\frac{3log_2 M}{ m^2 -1}$

Passband Transmission

The symbols are mathematically represented as,

(5)
\begin{eqnarray} S_m(t) &=& \Re \left[ A_m g(t) e^{(j2\pi f_c t)}\left] \\ &=& A_m g(t) \cos(2\pi f_c t) \end{eqnarray}

In the above expression, Am=2m-1-M and g(t) is the real baseband pulse.Inorder to calculate the properties of this modualtion shceme in passband we substitute Ep= Eg/2 in the baseband case.

5.jpg
E_m $\frac{A_m ^2}{2} E_g$
Eavg $\left(\frac {M^2-1} {6} \right )E_g$
Eb-avg $\left(\frac {M^2-1} {6 \log_2 M} \right ) E_g$
Basis , $\phi(t)$ $\sqrt\frac{2}{E_g} g(t) \cos (2\pi f_c t)$
$d_{min}^2$ $2E_g$
Power efficiency, $\eta$ $\frac{d_{min}^2}{4E_{b,avg}}$
$\eta$ $\frac{3log_2 M}{ m^2 -1}$

We note that the power-efficiency is same in both the cases.


Bit to Symbol mapping

Typically we use gray coding. This improves the performance of the modualtion shceme.

6.jpg

USB,SSB Modulation

In a PAM system we have,

(6)
\begin{eqnarray} S_m(t) &=& \Re \{ A_m g(t) e^{j2\pi f_c t} \}\\ S_l(t) &=& A_m g(t) \end{eqnarray}

If g(t) is real, the frequency content of Sl(t) is symmetric about origin.In the following figure we depict spectrum of the Lowpass equivalent signals, passband signal,The upper sideband of Sl(t). We observe that one side of the spectrum is enough to recover the entire spectrum.Therefore we can use DSB or SSB modulation to reduce the bandwidth of the transmitted signal.

7.jpg

Notice that,

(7)
\begin{align} F\left[j \hat{g}(t) \right] = \mbox{Sgn}(f) G(f) \end{align}

Therefore ,

(8)
\begin{align} F\left[ g(t) \underline{+} j * \hat{g}(t) \right] = G(f) \underline{+}\mbox{ Sgn}(f)G(f) \end{align}

Quadrature Amplitude Modulation

Passband tranmission

Mathematically,

(9)
\begin{eqnarray} S_m(t) &=& A_{mc} g(t) \cos (2 \pi f_c t) - A_{ms} g(t) \sin(2 \pi f_c t) \\ &=& \Re \left[ \left(A_{mc} \underline{+} j A_{ms} \right) g(t) e^{j2 \pi f_c t} \right] \end{eqnarray}

For any real baseband pulse shape g(t),

(10)
\begin{align} < g(t)\cos(2\pi f_c t), g(t)\sin(2\pi f_c t)> = 0 \end{align}

Therefore the basis for this signal space are,

(11)
\begin{eqnarray} \phi_1(t) &=& \sqrt{\frac{2}{E_g}} g(t)\cos(2\pi f_c t) \\ \phi_2(t) &=& \sqrt{\frac{2}{E_g}} g(t)\sin(2\pi f_c t) \end{eqnarray}

The scalars for repesenting the mth signal are $\left( A _{mc} \sqrt{\frac{1}{2} E_g}, A_{ms} \sqrt{\frac{1}{2} E_g}\right)$.

8.jpg

Phase modualted signals

Pass-band

Mathematically,

(12)
\begin{eqnarray} S_m(t) &=& \Re \left[ g(t) e^{j 2 \pi \left( \frac{m-1}{M} \right) } e^{j 2 \pi f_c t} \right] \\ &=& g(t)\cos \left( (2\pi f_c t) + \frac{2\pi (m-1)}{M} \right) \\ &=& g(t)\cos \left( 2\pi f_c t \right) \cos \left( \frac{2\pi (m-1)}{M} \right) - g(t)\sin \left( 2\pi f_c t \right) \sin\left( \frac{2\pi (m-1)}{M} \right) \end{eqnarray}

We observe that the basis for such a signal space,

(13)
\begin{eqnarray} \phi_1(t) &=& \sqrt{\frac{2}{E_g}} g(t)\cos(2\pi f_c t) \\ \phi_2(t) &=& \sqrt{\frac{2}{E_g}} g(t)\sin(2\pi f_c t) \end{eqnarray}
9.jpg

and the scalars for representing each signal are $\sqrt{\frac{1}{2} E_g}\cos\left(\frac{2\pi (m-1)}{M} \right ), \sqrt{\frac{1}{2} E_g}\sin\left(\frac{2\pi (m-1)}{M} \right)$.

E_m $\frac {E_g} {2}$
Eavg $\frac {E_g} {2}$
Eb-avg $\frac {E_g} { 2\log_2 M}$
$d_{min}^2$ $2E_g \sin^2 \left( \frac{2 \pi}{M}\right)$
$\eta$ $log_2 M \sin^2 \left( \frac{2 \pi}{M}\right)$

Base-band

(14)
\begin{eqnarray} S_{lm}(t) &= & \left [g(t) e^{j 2 \pi \left(\frac{m-1}{M}\right)} \right]\\ &=& \frac{1}{\sqrt{E_g}}g(t) \sqrt{E_g} e^{j 2 \pi \left(\frac{m-1}{M}\right)} \end{eqnarray}

Energy = Eg. Therefore the pass band energy = Eg/2.


Frequency Shift keying

In this modualtion scheme we represent Siganls as,

(15)
\begin{align} S_m(t) = \sqrt{\frac{2E}{T}} \cos \left( 2\pi f_c t + 2\pi m\Delta f t \right) \end{align}

assuming g(t) as arectangular pulse from 0 to T.

(16)
\begin{align} S_{ml}(t)= \sqrt{\frac{2E}{T}} e^{2\pi m\Delta f t} \end{align}
(17)
\begin{align} \rho_{km}= \frac{ \frac{2E}{t}\int_0^t e^{2\pi(k-m)\Delta f t}}{\frac{2E}{t}} dt \end{align}
(18)
\begin{align} \Re \{\rho_{km}\}= \frac{\sin(2\pi (k-m)\Delta f T)}{2\pi (k-m)\Delta f} \end{align}
(19)
\begin{align} \Re \{\rho_{km}\}=0 \mbox{ if $ \Delta f =\frac {n}{2T(k-m)}$, $ n\in Z $ } \end{align}

Thus the minimum separation required is 1/2T.

10.JPG

Orthogonality for arbitrary initial phase.

The above orthogonality is not true for an arbitrary Phase, i.e, if

(20)
\begin{align} S_m(t) = e^ {j 2\pi m \Delta f t + \theta_m} \end{align}
(21)
\begin{eqnarray} <S_m(t),S_k(t)> &=& \int_0^T e^{j2\pi(m-k)\Delta f t} e^{j(\theta_m - \theta_k)} dt\\ &=& e^{j(\theta_m-\theta_k)} \int_0^T e^{j2\pi(m-k)\Delta f t} dt \end{eqnarray}

In this case we want the entire integral to be zero rather than just the real part. Therefore by setting the integral to zero we derive the codition that $\Delta f$ should satisfy, which is,

(22)
\begin{align} \Delta f = \frac{1}{(m-k)T} \end{align}

To gurantee orthogonality for all m,k we must have,

(23)
\begin{align} \Delta f = \frac{n}{T} \end{align}

Thus the minimum separation required is 1/T.


Orthogonal Signals

Multidimensional signals

There are other ways to produce orthogonal signals.

11.JPG

Further within each interval we ca perform QAM Modulation or M-ary FSK to increase the number of signals being transmitted. The frequecy separation for the FSK being 1/2T.

12.jpg

Pulse Position Modulation

13.JPG

It must be noted that the vector spcae approach has the same signal space structure for FSK and PPM modualtion schemes.


Hadamard Codes

(24)
\begin{align} H_1 = \left(1 \right), H_2 = \[ \left( \begin{array}{ccc} 1 & 1 \\ 1 & -1 \end{array} \right), H_4 = \[ \left( \begin{array}{ccc} H_2 & H_2 \\ H_2 & -H_2 \end{array} \right), H_{2n} = \[ \left( \begin{array}{ccc} H_n & H_n \\ H_n & -H_n \end{array} \right) \] \end{align}
(25)
\begin{align} S_m = \sum_{i=1}^n H_{m,i} g(t-iT_c)\cos(2\pi f_c t) \sqrt{\frac{2E}{T_c}} \end{align}

Simplex Signal sets

Suppose we have a set of M orthogonal waveforms {s1(t),s2(t),……..,sM(t)} with vector representation {s1,s2,s3,…….,sM}.

(26)
\begin{align} E = \|s_m\|^2, \end{align}
(27)
\begin{eqnarray} <\underline{s}_i,\underline{s}_j> &=& E \hspace{5mm} i=j \\ &=& 0 \hspace{5mm} otherwise \end{eqnarray}

Let,

(28)
\begin{align} \underline{\overline{s}}= \frac{1}{M} \sum \underline{s_i} \end{align}

We form a new set of signals where $s_m\prime = s_m - \underline{\overline{s}}$

(29)
\begin{align} \|s_m \prime \|^2 = \|s_m - \underline{\overline{s}} \|^2 = E\left(1-\frac{1}{M}\right) \end{align}
(30)
\begin{eqnarray} <\underline{s}_i,\underline{s}_j> = \frac{-1}{M-1} \end{eqnarray}

Orthogonality is not necessarily important from a performance(coherent detection) point of veiw.


Biorthogonal signals

We add negeative of the Orthogonal signals.

16.jpg

Modualtion format from binary codes

Example:

(31)
\begin{align} C = \[ \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array} \right)\] \end{align}
(32)
\begin{align} S_m(t) = \sum_{i=1}^n \left(2c_{mi}-1\right)g(t-iT_c) \cos(2\pi f_c t) \end{align}
15.JPG
(33)
\begin{align} g(t)= \sqrt{\frac{2E_c}{Tc}} \hspace{4mm} 0<t\leq T_c \end{align}
(34)
\begin{align} \phi_i(t) = \sqrt{\frac{2}{Tc}} \cos( 2\pi f_c t) \hspace{ 5mm} (i-1)T_c< t\leq iT_c \end{align}
(35)
\begin{align} E_s = 3E_c , \hspace{4mm} E_b = \frac{E_s}{2} = \frac{3E_c}{2} \\ \end{align}
(36)
\begin{equation} d_{min}^2= 8E_c \end{equation}
(37)
\begin{align} \eta = \frac{8}{6}= \frac{3}{4} \end{align}

Compare this with some other 4-ary one dimensional modulation frmat e.g 4 PAM (Passband).

(38)
\begin{align} \eta = \frac{3log_2 M}{ m^2 -1} = \frac{6}{15} \end{align}
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License