# Note of PAM/PSK/QAM

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Multi-Dimensional Signal
---
- PAM: one-dimensional
- PSK: two-dimensional
- QAM: two-dimensional
- How to create three or higher dimensional signals?
- Subdivision of Time
Example: $N$ time slots can form $2N$ vector basis
elements (each slot has two $I/Q$ bearers.)

- Subdivision of Frequency
Example: $N$ frequency bands can form $2N$ vector basis
elements (each band has two $I/Q$ bearers.)

- Subdivision of both Time and Frequency Digital

PAM
===
## Signaling
$$s_{m}(t)=\Re\{A_{m}g(t)e^{2\pi f_{c}t}\}=A_{m}g(t)\cos(2\pi f_{c}t),\ t\in[0,T)
$$
where $A_{m}=(2m-1-M)d$ and $m=1,\ldots,M$

## Basis
$$\begin{aligned}
\phi_{1}(t)&=\frac{g(t)\cos(2\pi f_{c}t)}{\left \Vert g(t)\right \Vert\left \Vert\cos(2\pi f_{c}t)\right \Vert}\\
&=\frac{g(t)}{\left \Vert g(t)\right \Vert}\sqrt{2}\cos(2\pi f_{c}t)=\frac{g(t)}{\sqrt{\mathcal{E}_{g}}}\sqrt{2}\cos(2\pi f_{c}t)\\
\end{aligned}$$
$$s_{m}=\left[A_{m}\frac{\sqrt{\mathcal{E}_{g}}}{\sqrt{2}}\right]=\left[A_{m}\frac{\left \Vert g(t)\right \Vert}{\sqrt{2}}\right]
$$

## Energy

### Transmission energy of M-PAM Signals

$$\mathcal{E}_{m}=\int^{Ts}_{0}\left\lvert s_{m}(t) \right\rvert^{2}dt\approx \frac{A_{m}^{2}\left\Vert g(t)\right\Vert^{2}}{2}=\frac{1}{2}A_{m}^{2}\mathcal{E}_{g}
$$

### Average Signal Energy

$$\begin{aligned}\mathcal{E}_{avg}=E[\mathcal{E}_{m}]&\approx\frac{1}{M}\sum^{M}_{m=1}\frac{1}{2}A_{m}^{2}\mathcal{E}_{g}\\
&=\frac{\mathcal{E}_{g}}{2M}\sum^{M}_{m=1}A_{m}^{2}\\
&=\frac{2\mathcal{E}_{g}}{2M}\sum^{M/2}_{m=1}(2m-1)^{2}\\
&=\frac{\mathcal{E}_{g}}{M}\times \frac{M(M^{2}-1)}{6}\\
&=\frac{(M^{2}-1)}{6}\mathcal{E}_{g}\\
\end{aligned}
$$

### Average Bit Energy

$$\begin{aligned}\mathcal{E}_{bavg}=\frac{\mathcal{E}_{avg}}{\log_{2}M}=\frac{M^{2}-1}{6\log_{2}M}\mathcal{E}_{g}\ \ \ \left(\mathcal{E}_{g}=\frac{6\log_{2}M}{M^{2}-1}\mathcal{E}_{bavg}\right)
\end{aligned}
$$

## Distance

### Euclidean Distance

$$\begin{aligned}
d_{mn}&=\left\Vert s_{m}(t)-s_{n}(t) \right\Vert\\
&\approx\sqrt{\left\Vert s_{m}-s_{n} \right\Vert^{2}}\\
&=\sqrt{\left\Vert \sqrt{\frac{\mathcal{E}_{g}}{2}(A_{m}-A_{n})}\right\Vert^{2}}\\
&=\sqrt{\frac{\mathcal{E}_{g}}{2}}\left\lvert A_{m}-A_{n}\right\rvert
\end{aligned}
$$

### Minimun Euclidean Distance (d=1)

$$\begin{aligned}
d_{min}&=\sqrt{\frac{\mathcal{E}_{g}}{2}}\left\lvert A_{m}-A_{n}\right\rvert\\
&=\sqrt{\frac{\mathcal{E}_{g}}{2}}\times 2\\
&=\sqrt{2\mathcal{E}_{g}}&\left(\mathcal{E}_{g}=2d_{min}^{2}\right) \\
&=\sqrt{\frac{12\log_{2}M}{M^{2}-1}\mathcal{E}_{bavg}}&\left(\mathcal{E}_{bavg}=\frac{M^{2}-1}{12\log_{2}M}d_{min}^{2}\right)
\end{aligned}
$$

## Error Probability
Consider the M-ary PAM
$$\mathcal{S}=\left\{\pm\frac{1}{2} d_{min},\pm\frac{3}{2} d_{min},\cdots,\pm\frac{M-1}{2} d_{min}\right\}$$
There are two types of error events: assuming $n$ is zero-mean
Gaussian random variable with variance $\frac{N_{0}}{2}$

- Inner Points with error probability $P_{ei}$ :
$$\begin{aligned}P_{ei} &= Pr\left\{\left\lvert n \right\rvert > \frac{d_{min}}{2} \right\}\\
&=2Q\left(\frac{\frac{d_{min}}{2}-0}{\sqrt{\frac{N_{0}}{2}}}\right) &\left(\textrm{By $n$ is zero-mean and variance $\frac{N_{0}}{2}$}\right)\\
&=2Q\left(\frac{d_{min}}{\sqrt{2N_{0}}}\right)
\end{aligned}$$

- Outer Points with error probability $P_{eo}$ : only one end
causes errors:
$$\begin{aligned}P_{ei} &= Pr\left\{ n > \frac{d_{min}}{2} \right\}\\
&=Q\left(\frac{\frac{d_{min}}{2}-0}{\sqrt{\frac{N_{0}}{2}}}\right) &\left(\textrm{By $n$ is zero-mean and variance $\frac{N_{0}}{2}$}\right)\\
&=Q\left(\frac{d_{min}}{\sqrt{2N_{0}}}\right)
\end{aligned}
$$

### Symbol Error Probability

$$\begin{aligned}P_{e} &= \frac{1}{M}\sum^{M}_{m=1}Pr\left\{ error\ |\ m\ sent \right\}\\
&= \frac{1}{M}\left[ 2\left ( M-2 \right)Q\left(\frac{d_{min}}{\sqrt{2N_{0}}}\right)+ 2Q\left(\frac{d_{min}}{\sqrt{2N_{0}}}\right)\right]\\
&= \frac{2(M-1)}{M}Q\left(\frac{d_{min}}{\sqrt{2N_{0}}}\right)\\
&= \frac{2(M-1)}{M}Q\left(\sqrt{\frac{6\log_{2}M}{M^{2}-1}\frac{\mathcal{E}_{bavg}}{N_{0}}}\right)
\end{aligned}
$$

## Efficiency

$$\begin{aligned}P_{e} = \frac{2(M-1)}{M}Q\left(\sqrt{\frac{6\log_{2}M}{M^{2}-1}\frac{\mathcal{E}_{bavg}}{N_{0}}}\right)
\end{aligned}
$$

For large $k=\log_{2}M$
- To increase $k$ by 1 bit, we need to double $M$
- To keep the same $P_{e}$ we need $\mathcal{E}_{bavg}$ to roughly quadruple
- Increase $k$ by 1 bit $\implies$ increase $\mathcal{E}_{b}$ by roughly $6$ dB

PSK
===
## Signaling
$$\begin{aligned}s_{m}(t)&=\Re\{g(t)e^{2\pi f_{c}t}e^{\theta_{m}}\}=g(t)\cos(2\pi f_{c}t+\theta_{m})\\
&= g(t)\cos(\theta_{m})\cos(2\pi f_{c}t)-g(t)\sin(\theta_{m})\sin(2\pi f_{c}t),\ t\in[0,T)
\end{aligned}
$$
where $\theta_{m}=\frac{2\pi m}{M}$ and $m=0,\ldots,M-1$

## Basis
$$\begin{aligned}
\phi_{1}(t)&=\frac{g(t)\cos(2\pi f_{c}t)}{\left \Vert g(t)\right \Vert\left \Vert\cos(2\pi f_{c}t)\right \Vert}=\frac{g(t)}{\left \Vert g(t)\right \Vert}\sqrt{2}\cos(2\pi f_{c}t)
\end{aligned}
$$
$$\begin{aligned}
\phi_{2}(t)&=-\frac{g(t)\sin(2\pi f_{c}t)}{\left \Vert g(t)\right \Vert\left \Vert\sin(2\pi f_{c}t)\right \Vert}=-\frac{g(t)}{\left \Vert g(t)\right \Vert}\sqrt{2}\sin(2\pi f_{c}t)
\end{aligned}
$$
$$s_{m}=\left[\frac{\left \Vert g(t)\right \Vert}{\sqrt{2}}\cos(\theta_{m}),\frac{\left \Vert g(t)\right \Vert}{\sqrt{2}}\sin(\theta_{m}) \right]
= \frac{\left \Vert g(t)\right \Vert}{\sqrt{2}}\left[\cos(\theta_{m}),\sin(\theta_{m}) \right]
$$

## Energy

### Transmission energy of M-PAM Signals

$$\mathcal{E}_{m}=\int^{Ts}_{0}\left\lvert s_{m}(t) \right\rvert^{2}dt\approx \frac{\left\Vert g(t)\right\Vert^{2}}{2}\left[\cos^{2}(\theta_{m})+\sin^{2}(\theta_{m})\right]=\frac{\mathcal{E}_{g}}{2}
$$
Advantage: equal energy for every channel symbol.

### Bit Energy

$$\begin{aligned}\mathcal{E}_{b}=\frac{\mathcal{E}_{m}}{\log_{2}M}=\frac{\mathcal{E}_{g}}{2\log_{2}M}\ \ \ \left(\mathcal{E}_{g}=2\mathcal{E}_{b}\log_{2}M\right)
\end{aligned}
$$

## Distance

### Euclidean Distance

$$\begin{aligned}
d_{mn}&=\left\Vert s_{m}(t)-s_{n}(t) \right\Vert\\
&\approx\sqrt{\left\Vert s_{m}-s_{n} \right\Vert^{2}}\\
&=\frac{\left\Vert g(t) \right\Vert}{\sqrt{2}}\sqrt{\left\lvert \cos(\theta_{m})-\cos(\theta_{n})\right\rvert^{2}+\left\lvert \sin(\theta_{m})-\sin(\theta_{n}) \right\rvert^{2}}\\
&= \left\Vert g(t) \right\Vert\sqrt{1-\cos(\theta_{m}-\theta_{n})}\\
&= \sqrt{\mathcal{E}_{g}\left[1-\cos\left(\frac{2\pi}{M}(m-n)\right)\right]}
\end{aligned}
$$

### Minimun Euclidean Distance (d=1)

$$\begin{aligned}
d_{min}&=\sqrt{\mathcal{E}_{g}\left[1-\cos\left(\frac{2\pi}{M}\right)\right]}\\
&=\sqrt{2\mathcal{E}_{g}\sin^{2}\frac{\pi}{M}} &\left(\textrm{By }\sin\theta=\frac{1-\cos2\theta}{2}\right)\\
&=2\sqrt{\left(\log_{2}M\times\sin^{2}\frac{\pi}{M}\right)\mathcal{E}_{b}}&\left(\textrm{By } \mathcal{E}_{g}=2\mathcal{E}_{b}\log_{2}M\right)\\
&\approx 2\sqrt{\frac{\pi^{2}\log_{2}M}{M^{2}}\mathcal{E}_{b}}&\left(\textrm{By large } M,\textrm{ we have }\sin\frac{\pi}{M}\approx\frac{\pi}{M}\right)
\end{aligned}
$$

## Error Probability
Signal constellation for $M$-ary PSK is
$$
\mathcal{S}=\left\{s_{m}=\sqrt{E}\left(\cos\left(\frac{2\pi m}{M}\right),\sin\left(\frac{2\pi m}{M}\right) \right): m=0,1,\cdots,M-1\right\}
$$

- By symmetry we can assume $s_{0}=(\sqrt{\mathcal{E}},0)$ ($\mathcal{E}=\mathcal{E}_{g}$) was transmitted.
- The received signal vector $r$ is
$$
r= \left(\sqrt{\mathcal{E}}+n_{1},n_{2}\right)
$$
- $\mathcal{E}_{b}=\frac{\mathcal{E}}{\log_{2}M}$

### Symbol Error Probability

- When $M=2$, binary PSK is antipodal:
$$
P_{e}=Q\left(\sqrt{\frac{2\mathcal{E}_{b}}{N_{0}}}\right)
$$
- When $M=4$, it is QPSK:
$$
P_{e}=1-\left[1-Q\left(\sqrt{\frac{2\mathcal{E}_{b}}{N_{0}}}\right)\right]^{2}
$$
- For large $M$ we can approximate $P_{e}$ as
$$
P_{e}\approx 2Q\left(\sqrt{\frac{2\pi^{2}\log_{2}M}{M^{2}}\frac{\mathcal{E}_{b}}{N_{0}}}\right)
$$

## Efficiency
For large $M$ we can approximate $P_{e}$ as
$$
P_{e}\approx 2Q\left(\sqrt{\frac{2\pi^{2}\log_{2}M}{M^{2}}\frac{\mathcal{E}_{b}}{N_{0}}}\right)
$$

- To increase $k$ by 1 bit, we need to double $M$
- To keep the same $P_{e}$ we need $\mathcal{E}_{b}$ to roughly quadruple
- Increase $k$ by 1 bit $\implies$ increase $\mathcal{E}_{b}$ by roughly $6$ dB

QAM
===
## Signaling

$$
s_{m}(t)=A_{m,i}g(t)\cos(2\pi f_{c}t)-A_{m,q}g(t)\sin(2\pi f_{c}t)
$$
where $A_{m,i},A_{m,q}\in\left\{\left(2a-1-\sqrt{M}\right):a=1,2,\ldots,\sqrt{M}\right\}$

## Basis
$$\begin{aligned}
\phi_{1}(t)&=\frac{g(t)\cos(2\pi f_{c}t)}{\left \Vert g(t)\right \Vert\left \Vert\cos(2\pi f_{c}t)\right \Vert}=\frac{g(t)}{\left \Vert g(t)\right \Vert}\sqrt{2}\cos(2\pi f_{c}t)
\end{aligned}
$$
$$\begin{aligned}
\phi_{2}(t)&=-\frac{g(t)\sin(2\pi f_{c}t)}{\left \Vert g(t)\right \Vert\left \Vert\sin(2\pi f_{c}t)\right \Vert}=-\frac{g(t)}{\left \Vert g(t)\right \Vert}\sqrt{2}\sin(2\pi f_{c}t)
\end{aligned}
$$
$$s_{m}=\left[\frac{\left \Vert g(t)\right \Vert}{\sqrt{2}}A_{m,i},\frac{\left \Vert g(t)\right \Vert}{\sqrt{2}}A_{m,q} \right]
= \frac{\left \Vert g(t)\right \Vert}{\sqrt{2}}\left[A_{m,i},A_{m,q} \right]
$$

## Energy

### Transmission energy of M-PAM Signals

$$\mathcal{E}_{m}=\int^{Ts}_{0}\left\lvert s_{m}(t) \right\rvert^{2}dt\approx \frac{\left\Vert g(t)\right\Vert^{2}}{2}\left(A_{m,i}^{2},A_{m,q}^{2}\right)
$$

### Average symbol energy
$$
\begin{aligned}
\mathcal{E}_{avg}&=\frac{1}{M}\frac{\mathcal{E}_{g}}{2}\sum^{\sqrt{M}}_{m=1}\sum^{\sqrt{M}}_{n=1}\left(A_{m,i}^{2}+A_{n,q}^{2}\right)\\
&=\frac{\mathcal{E}_{g}}{2M}\times \frac{2M(M-1)}{3}\\
&=\frac{M-1}{3}\mathcal{E}_{g}
\end{aligned}
$$

### Average bit energy

$$\begin{aligned}\mathcal{E}_{bavg}=\frac{\mathcal{E}_{avg}}{\log_{2}M}=\frac{M-1}{3\log_{2}M}\mathcal{E}_{g}\ \ \ \left(\mathcal{E}_{g}=\frac{3\log_{2}M}{M-1}\mathcal{E}_{b}\right)
\end{aligned}
$$

## Distance

### Euclidean Distance

$$\begin{aligned}
d_{mn}&=\left\Vert s_{m}(t)-s_{n}(t) \right\Vert\\
&\approx\sqrt{\left\Vert s_{m}-s_{n} \right\Vert^{2}}\\
&=\frac{\left\Vert g(t) \right\Vert}{\sqrt{2}}\sqrt{\left\lvert A_{m,i}-A_{n,i}\right\rvert^{2}+\left\lvert A_{m,q}-A_{n,q} \right\rvert^{2}}\\
\end{aligned}
$$

### Minimun Euclidean Distance (d=1)

$$\begin{aligned}
d_{min}&=\min\left\Vert s_{m}(t)-s_{n}(t) \right\Vert\\
&\approx \frac{\left\Vert g(t) \right\Vert}{\sqrt{2}}\sqrt{\left\lvert 2\right\rvert^{2}+\left\lvert 2 \right\rvert^{2}}\\
&=\sqrt{2\mathcal{E}_{g}} &\left(\mathcal{E}_{g}=2d_{min}^{2}\right) \\
&=\sqrt{\frac{6\log_{2}M}{M-1}\mathcal{E}_{bavg}} &\left(\mathcal{E}_{bavg}=\frac{M-1}{6\log_{2}M}d_{min}^{2}\right)
\end{aligned}
$$
## Error Probability
- $M$ is usually a square number
- $M$-ary QAM is composed of two independent $\sqrt{M}$-ary PAM
$$
\begin{aligned}
&\mathcal{S}_{PAM}=\left\{\pm\frac{1}{2}d_{min},\pm\frac{3}{2}d_{min},\cdots,\pm\frac{\sqrt{M}-1}{2}d_{min}\right\}\\
&\mathcal{S}_{QAM}=\left\{\left(x,y\right):x,y\in\mathcal{S}_{PAM}\right\}
\end{aligned}
$$
$$
\mathcal{S}_{M-QAM}=\mathcal{S}_{\sqrt{M}-PAM}\times \mathcal{S}_{\sqrt{M}-PAM}
$$

### Symbol Error Probability

$$
P_{e,\sqrt{M}-{PAM}}=2\left(1-\frac{1}{\sqrt{M}}\right)Q\left(\frac{d_{min}}{\sqrt{2N_{0}}}\right)
$$

$$
\begin{aligned}
P_{e,M-{QAM}}&=1-\left(1-P_{e,\sqrt{M}-PAM}\right)^{2}\\
&= 2P_{e,\sqrt{M}-PAM}\left(1-\frac{1}{2}P_{e,\sqrt{M}-PAM}\right)\\
&= 4\left(1-\frac{1}{N}\right)Q\left(\frac{d_{min}}{\sqrt{2N_{0}}}\right)\times\left(1-\left(1-\frac{1}{\sqrt{M}}\right)Q\left(\frac{d_{min}}{\sqrt{2N_{0}}}\right)\right)\\
&\leq 4Q\left(\frac{d_{min}}{\sqrt{2N_{0}}}\right)=4Q\left(\sqrt{\frac{3\log_{2}M}{M-1}\frac{\mathcal{E}_{bavg}}{N_{0}}}\right)
\end{aligned}
$$

## Efficiency

$$
\begin{aligned}
P_{e,M-{QAM}}\approx 4Q\left(\sqrt{\frac{3\log_{2}M}{M-1}\frac{\mathcal{E}_{bavg}}{N_{0}}}\right)
\end{aligned}
$$

- To increase $k=\log_{2}M$ by 1 bit, we need to double $M$
- To keep the same $P_{e}$ we need to double $\mathcal{E}_{bavg}$
- Increase $k$ by 1 bit $\implies$ increase $\mathcal{E}_{b}$ by roughly $3$ dB
- QAM is more power efficient than PAM and PSK.

###### tags: `Digital Communication` `Book`