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## my digital communications cheat sheet

January 5, 2019   Compiled on May 23, 2020 at 3:35am

### 2 Some useful Fourier Transforms

 $$x\left ( t\right )$$ $$X\left ( f\right )$$ $$\sin \left ( 2\pi f_{c}t\right )$$ $$\frac{1}{2j}\left [ \delta \left ( f-f_{c}\right ) -\delta \left ( f+f_{c}\right ) \right ]$$ $$\cos \left ( 2\pi f_{c}t\right )$$ $$\frac{1}{2}\left [ \delta \left ( f-f_{c}\right ) +\delta \left ( f+f_{c}\right ) \right ]$$ $$\cos \left ( 2\pi f_{c}t+\theta \right )$$ $$\frac{1}{2}\left [ e^{j\theta }\delta \left ( f-f_{c}\right ) +e^{-j\theta }\delta \left ( f+f_{c}\right ) \right ]$$ $$\sin \left ( 2\pi f_{c}t+\theta \right )$$ $$\frac{1}{2}\left [ e^{j\theta }\delta \left ( f-f_{c}\right ) -e^{-j\theta }\delta \left ( f+f_{c}\right ) \right ]$$

### 3 Random process deﬁnitions

$$X\left ( t\right )$$ is Stationary: If all its statistics do not change with shift of origin

$$X\left ( t\right )$$ is Wide Sense Stationary: If the mean is constant, and $$R_{x}\left ( t,t+\tau \right ) =R_{x}\left ( \tau \right )$$

where autocorrelation $$R_{x}\left ( \tau \right )$$ is deﬁned as $$E\left [ x\left ( t\right ) x^{\ast }\left ( t+\tau \right ) \right ] .$$ Note, if $$X\left ( t\right )$$ is real, then $$R_{x}\left ( \tau \right )$$ is real and even

Note: $$R\left ( x\right )$$ must be WSS if it is ergodic$$.$$So ergodic process has constant mean.

### 4 How to determine Hilbert transform of a signal?

input $$x\left ( t\right )$$. Find $$\hat{x}\left ( t\right )$$ which is Hilbert transform of $$x\left ( t\right )$$ deﬁned as $$\hat{x}\left ( t\right ) =x\left ( t\right ) \otimes \frac{1}{\pi t}$$

An easy way is to ﬁrst ﬁnd $$\hat{G}\left ( f\right )$$ which is the Fourier transform of $$\hat{x}\left ( t\right )$$ and then inverse it to ﬁnd $$\hat{x}\left ( t\right )$$$\hat{G}\left ( f\right ) =-j\ sgn\left ( f\right ) \ G\left ( f\right )$ Where $$G\left ( f\right )$$ is Fourier transform of $$x\left ( t\right )$$

### 5 How to ﬁnd Power Spectrum (PSD) of a random signal $$x\left ( t\right )$$

input: random signal $$x\left ( t\right )$$

output: PSD of $$x\left ( t\right )$$

Algorithm:

1. Find autocorrelation $$R_{x}\left ( \tau \right )$$ of $$x\left ( t\right )$$
2. Find the Fourier Transform of $$R_{x}\left ( \tau \right )$$. The result is the PSD of $$x\left ( t\right )$$ called $$S_{x}\left ( f\right )$$

Another method (this below works if not random $$x\left ( t\right )$$) , why? can’t ﬁnd FT for random process?

1. Find Fourier Transform $$X\left ( f\right )$$ of $$x\left ( t\right )$$
2. Find the $$\left \vert X\left ( f\right ) \right \vert ^{2}=X\left ( f\right ) X^{\ast }\left ( f\right )$$

### 6 What is the relation between variance and power for a random signal $$x\left ( t\right )$$?

Variance is the sum of the total average normalized power and the DC power.$\sigma _{x}^{2}=\overset{\text{total Power}}{\overbrace{E\left [ x^{2}\left ( t\right ) \right ] }}+\overset{\text{DC power}}{\overbrace{E\left [ x\left ( t\right ) \right ] ^{2}}}$ For the a signal whose mean is zero,$\sigma _{x}^{2}=\overset{\text{total Power}}{\overbrace{E\left [ x^{2}\left ( t\right ) \right ] }}$ How to ﬁnd average, power, PEP, eﬀective value (or the RMS) of a periodic function?

Let $$x\left ( t\right )$$ be a periodic function, of period $$T$$, then $\text{average\ of}\ x\left ( t\right ) =\left \langle x\left ( t\right ) \right \rangle =\frac{1}{T}\int _{0}^{T}x\left ( t\right ) dt$ The average power is$p_{av}=\left \langle x^{2}\left ( t\right ) \right \rangle =\frac{1}{T}\int _{0}^{T}\left \vert x\left ( t\right ) \right \vert ^{2}dt$ Eﬀective value, or the RMS value is$x_{rms}\left ( t\right ) =\sqrt{\left \langle x^{2}\left ( t\right ) \right \rangle }=\sqrt{p_{av}}=\sqrt{\frac{1}{T}\int _{0}^{T}x^{2}\left ( t\right ) dt}$ For example, for $$x\left ( t\right ) =A\cos \left ( x\right ) ,\left \langle x\left ( t\right ) \right \rangle =0,P_{av}=\frac{A^{2}}{2},x_{rms}\left ( t\right ) =0.707A$$

To ﬁnd PEP (which is the peak envelope power), ﬁnd the complex envelope $$\tilde{x}\left ( t\right )$$, then ﬁnd the average power of it. i.e.

$PEP=\frac{1}{2}\tilde{x}_{\max }^{2}\left ( t\right )$

### 7 How to ﬁnd the SNR for sampling quantization?

Suppose we have a message $$m\left ( t\right )$$ that is sampled. Assume we have $$n$$ bits to use for encoding the sample levels. Hence there are $$2^{n}$$ levels of quantizations. We want to ﬁnd the ration of the signal to the noise power. Noise here is generated due to quantization (i.e. due to the rounding oﬀ values of $$m\left ( t\right )$$ during sampling).

This is the algorithm:

Input: $$n$$, the number of bits for encoding, $$m_{p}$$ absolute maximum value of the message $$m\left ( t\right )$$, the pdf $$f_{X}\left ( t\right )$$ of the message $$m\left ( t\right )$$ is $$m\left ( t\right )$$ is random message or $$m\left ( t\right )$$ function if it is deterministic (such as $$\cos \left ( t\right )$$)

1. Find the quantization step size $$S=\frac{2m_{p}}{2^{2}}$$
2. Find $$P_{av}$$ of the error is $$\frac{1}{12}S^{2}$$ where $$S$$ is the step size found in (1), hence $$P_{av}=\frac{1}{12}S^{2}=\frac{1}{12}\left ( \frac{2m_{p}}{2^{2}}\right ) ^{2}=\frac{m_{p}^{2}}{3\times 2^{2n}}$$
3. If $$m\left ( t\right )$$ is deterministic ﬁnd $$p_{av}=\left \langle m^{2}\left ( t\right ) \right \rangle =\frac{1}{T}\int _{0}^{T}\left \vert m\left ( t\right ) \right \vert ^{2}dt$$
4. If $$m\left ( t\right )$$ is random, ﬁnd $$p_{av}=E\left ( m\left ( t\right ) \right ) ={\int }m^{2}\left ( t\right ) f_{X}\left ( t\right ) dt$$, this is called the second moment of the pdf
5. $$SNR=\frac{E\left ( m\left ( t\right ) \right ) }{\frac{m_{p}^{2}}{3\times 2^{2n}}}$$

Hence ﬁnd $$SNR$$ for noise quantisation comes down to ﬁnding the power in the message $$m\left ( t\right )$$.

Examples: For sinusoidal message $$m\left ( t\right )$$, $$SNR_{db}=6n+1.761$$. For random $$m\left ( t\right )$$ with PDF which is uniform distributed $$SNR_{db}=6n$$, for random $$m\left ( t\right )$$ which is AWGN. Do this later

### 8 How to determine coding of a number from quantization?

Given an analog value say $$x$$ and given a maximum absolute possible value to be $$m_{p}$$, and given the number of bits available for coding to be $$N$$, the following are the algorithm to generate the quantized version of $$x$$, called $$\hat{x}$$

#### 8.1 sign magnitude

Input: $$x,m_{p},N$$

output: $$\hat{x}$$

Let $$\Delta =\frac{m_{p}}{2^{N-1}}$$ called the step size

Let $$q=round\left [ \frac{abs\left ( x\right ) }{\Delta }\right ]$$ which is the quantization level

If $$q\geq 2^{N-1}-1$$ then $$q=2^{N-1}$$ end if

if $$x<0$$ then $$code=q+2^{N-1}$$ else $$code=q$$ endif

return $$code$$ in base 2

#### 8.2 ones complement

Input: $$x,m_{p},N$$

output: $$\hat{x}$$

Let $$\Delta =\frac{m_{p}}{2^{N-1}}$$ called the step size

Let $$q=round\left [ \frac{abs\left ( x\right ) }{\Delta }\right ]$$ which is the quantization level

If $$q\geq 2^{N-1}-1$$ then $$q=2^{N-1}-1$$ end if

If $$x>0$$ then $$code=q$$ else $$code=\left ( 2^{N}-1\right ) -q$$ endif

return $$code$$ in base 2

#### 8.3 oﬀset binary

Input: $$x,m_{p},N$$

output: $$\hat{x}$$

Let $$\Delta =\frac{m_{p}}{2^{N-1}}$$ called the step size

Let $$q=round\left [ \frac{abs\left ( x\right ) }{\Delta }\right ]$$ which is the quantization level

If $$x\geq -\frac{\Delta }{2}$$ then

if $$q\geq 2^{N-1}-1$$ then

$$\ \ \ \ \ q=2^{N-1}-1$$

end if

$$code=2^{N-1}+q$$

else

if $$q\geq 2^{N-1}-1$$ then

$$\ \ \ \ \ \ q=2^{N-1}$$

end if

$$code=2^{N-1}-q$$

end if

return $$code$$ in base 2

#### 8.4 2’s complement

Input: $$x,m_{p},N$$

output: $$\hat{x}$$

Let $$\Delta =\frac{m_{p}}{2^{N-1}}$$ called the step size

Let $$q=round\left [ \frac{abs\left ( x\right ) }{\Delta }\right ]$$ which is the quantization level

If $$x\geq -\frac{\Delta }{2}$$ then

if $$q\geq 2^{N-1}-1$$ then

$$\ \ \ \ \ q=2^{N-1}-1$$

end if

$$code=q$$

else

if $$q\geq 2^{N-1}-1$$ then

$$\ \ \ \ \ \ q=2^{N-1}$$

end if

$$code=2^{N}-q$$

end if

return $$code$$ in base 2

### 9 How to derive the Phase and Frequency modulation signals?

For any bandpass signal, we can write it as$x\left ( t\right ) =\operatorname{Re}\left ( \tilde{x}\left ( t\right ) e^{j\omega _{c}t}\right )$ Where $$\tilde{x}\left ( t\right )$$ is the complex envelope of $$x\left ( t\right )$$. For PM and FM, the baseband modulated signal, $$\tilde{x}\left ( t\right )$$ has the form $$A_{c}e^{j\theta \left ( t\right ) }$$ Hence the above becomes\begin{align*} x\left ( t\right ) & =\operatorname{Re}\left ( A_{c}e^{j\theta \left ( t\right ) }e^{j\omega _{c}t}\right ) \\ & =A_{c}\left ( \cos \omega _{c}t\cos \theta \left ( t\right ) -\sin \omega _{c}t\sin \theta \left ( t\right ) \right ) \end{align*}

But $$\cos \left ( A+B\right ) =\cos A\cos B-\sin A\sin B$$, hence the above becomes$$x\left ( t\right ) =\cos \left ( \omega _{c}t+\theta \left ( t\right ) \right ) \tag{1}$$ The above is the general form for PM and FM. Now, for PM, $$\theta \left ( t\right ) =k_{p}m\left ( t\right )$$ and for FM, $$\theta \left ( t\right ) =k_{f}\int _{0}^{t}m\left ( t_{1}\right ) dt_{1}$$. Hence, substituting in (1) we obtain$x_{FM}\left ( t\right ) =\cos \left ( \omega _{c}t+k_{f}\int _{0}^{t}m\left ( t_{1}\right ) dt_{1}\right )$ and$x_{PM}\left ( t\right ) =\cos \left ( \omega _{c}t+k_{p}m\left ( t\right ) \right )$

### 10 How to obtain the phase deviation and the frequency deviation for angle modulated signal?

From the general form for angle modulated signal (see above note)$x\left ( t\right ) =\cos \left ( \omega _{c}t+\theta \left ( t\right ) \right )$ The phase deviation is $$\theta \left ( t\right )$$. And the maximum phase deviation is simply the maximum of $$\theta \left ( t\right )$$

Now, to ﬁnd the frequency deviation, we need a little bit more work. Start with$f_{i}=f_{c}+\Delta f$ Where $$f_{i}$$ is the instantaneous frequency in Hz. But \begin{align*} f_{i} & =\frac{1}{2\pi }\frac{d}{dt}\left ( \omega _{c}t+\theta \left ( t\right ) \right ) \\ & =f_{c}+\overset{\Delta f}{\overbrace{\frac{1}{2\pi }\frac{d}{dt}\theta \left ( t\right ) }} \end{align*}

### 11 How to quickly determine SNR$$_{i}$$ from $$SNR_{c}$$?

First ﬁnd $$SNR_{c}$$, for to ﬁnd $$SNR_{i}$$ use the following

$$SNR_{i}=SNR_{c}\frac{B}{B_{T}}$$, where $$B_{T}$$ is the transmission bandwidth, and $$B$$ is the baseband bandwidth. For $$AM$$, $$B_{T}=2B$$. For $$DSB-SC$$, $$B_{T}=2B$$. For $$DSB-SS$$, $$B_{T}=B.$$

### 12 How to determine ﬁgure of merit for DSB-SC using coherent detector?

Figure of merit, $$\gamma$$ is deﬁned as $$\frac{SNR_{o}}{SNR_{c}}$$ where $$SNR_{o}$$ is the signal-to-noise ratio on output from modulator, and $$SNR_{c}$$ is signal-to-noise ratio for the channel, assuming channel has AWGN added. The following diagram shows the calculations. I used a coherent demodulator.

Question: Verify the above.

### 14 How to determine ﬁgure of merit for AM using envelope detector?

\begin{align*} s_{1}\left ( t\right ) & =A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) \cos \omega _{c}t\\ s_{2}\left ( t\right ) & =A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) \cos \omega _{c}t+w\left ( t\right ) \end{align*}

And\begin{align*} SNR_{c} & =\frac{\left \langle \left ( A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) \cos \omega _{c}t\right ) ^{2}\right \rangle }{BN_{0}}\\ & =\frac{\frac{A_{c}^{2}}{2}\left \langle \left ( 1+k_{a}m\left ( t\right ) \right ) ^{2}\right \rangle }{BN_{0}}\\ & =\frac{\frac{A_{c}^{2}}{2}\left \langle 1+k_{a}^{2}m^{2}\left ( t\right ) +2k_{a}m\left ( t\right ) \right \rangle }{BN_{0}} \end{align*}

Now assuming $$\left \langle m\left ( t\right ) \right \rangle =0$$, the above simpliﬁes to$SNR_{c}=\frac{\frac{A_{c}^{2}}{2}\left ( 1+k_{a}^{2}P_{m}\right ) }{BN_{0}}$ Hence\begin{align*} SNR_{i} & =SNR_{c}\frac{B}{B_{T}}\\ & =\frac{\frac{A_{c}^{2}}{2}\left ( 1+k_{a}^{2}P_{m}\right ) }{BN_{0}}\frac{B}{2B}\\ & =\frac{\frac{A_{c}^{2}}{2}\left ( 1+k_{a}^{2}P_{m}\right ) }{2BN_{0}} \end{align*}

Now ﬁnd $$s_{3}\left ( t\right )$$\begin{align*} s_{3}\left ( t\right ) & =A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) \cos \omega _{c}t+\overset{\text{narrow band noise}}{\overbrace{n\left ( t\right ) }}\\ & =A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) \cos \omega _{c}t+\left [ n_{I}\left ( t\right ) \cos \omega _{c}t-n_{Q}\left ( t\right ) \sin \omega _{c}t\right ] \\ & =\overset{\text{in phase}}{\overbrace{\left [ A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) +n_{I}\left ( t\right ) \right ] }}\cos \omega _{c}t-\overset{\text{quadrature}}{\overbrace{n_{Q}\left ( t\right ) }}\sin \omega _{c}t \end{align*}

Now, to ﬁnd $$s_{4}\left ( t\right )$$, which is the envelope of $$s_{3}\left ( t\right ) .$$\begin{align*} s_{4}\left ( t\right ) & =\text{envelope}\left ( s_{3}\left ( t\right ) \right ) \\ & =\sqrt{\left ( s_{3}\right ) _{I}^{2}+\left ( s_{3}\right ) _{Q}^{2}}\\ & =\sqrt{\left ( A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) +n_{I}\left ( t\right ) \right ) ^{2}+n_{Q}^{2}\left ( t\right ) } \end{align*}

Now, assuming $$A_{c}\gg \left \vert n_{I}\left ( t\right ) \right \vert$$ and $$A_{c}\gg \left \vert n_{Q}\left ( t\right ) \right \vert$$, then the above simpliﬁes to$s_{4}\left ( t\right ) =A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) +n_{I}\left ( t\right )$ now apply the DC blocker, we obtain$s_{5}\left ( t\right ) =A_{c}k_{a}m\left ( t\right ) +n_{I}\left ( t\right )$$SNR_{o}=\frac{\left \langle \left ( A_{c}k_{a}m\left ( t\right ) \right ) ^{2}\right \rangle }{E\left [ n_{I}^{2}\left ( t\right ) \right ] }=\frac{A_{c}^{2}k_{a}^{2}P_{m}}{2BN_{0}}$$\gamma =\frac{SNR_{o}}{SNR_{c}}=\frac{\frac{A_{c}^{2}k_{a}^{2}P_{m}}{2BN_{0}}}{\frac{\frac{A_{c}^{2}}{2}\left ( 1+k_{a}^{2}P_{m}\right ) }{BN_{0}}}=\frac{k_{a}^{2}P_{m}}{1+k_{a}^{2}P_{m}}$ We notice, that for Large $$SNR_{i}$$, this detector gives the same result as coherent detector.

For small $$SNR_{i}$$, it is better to use the coherent detector than the envelope detector.

### 15 How to determine ﬁgure of merit for SSB using coherent detector?

The diﬀerence here is that SSB signal has transmission bandwidth $$B_{T}=B$$ and not $$2B$$ as in all the previous signals. Assume we are working with upper sideband. Analysis is the same for lower sideband.$s_{1}\left ( t\right ) =k\left [ m\left ( t\right ) \cos \omega _{c}t-\hat{m}\left ( t\right ) \sin \omega _{c}t\right ]$ Where $$k$$ is a constant. Usually $$\frac{A_{c}}{2}$$ but we will leave it as $$k$$ for now. $$\hat{m}\left ( t\right )$$ is the Hilbert transform of $$m\left ( t\right )$$$s_{2}\left ( t\right ) =k\left [ m\left ( t\right ) \cos \omega _{c}t-\hat{m}\left ( t\right ) \sin \omega _{c}t\right ] +w\left ( t\right )$\begin{align*} SNR_{c} & =\frac{\left \langle \left ( k\left [ m\left ( t\right ) \cos \omega _{c}t-\hat{m}\left ( t\right ) \sin \omega _{c}t\right ] \right ) ^{2}\right \rangle }{BN_{0}}\\ & =\frac{k^{2}\left \langle m^{2}\left ( t\right ) \cos ^{2}\omega _{c}t\right \rangle +k^{2}\left \langle \hat{m}^{2}\left ( t\right ) \sin ^{2}\omega _{c}t\right \rangle -2k^{2}\left \langle m\left ( t\right ) \hat{m}\left ( t\right ) \cos \left ( \omega _{c}t\right ) \sin \left ( \omega _{c}t\right ) \right \rangle }{BN_{0}}\\ & =\frac{k^{2}\left \langle m^{2}\left ( t\right ) \right \rangle \left \langle \cos ^{2}\omega _{c}t\right \rangle +k^{2}\left \langle m^{2}\left ( t\right ) \right \rangle \left \langle \sin ^{2}\omega _{c}t\right \rangle -2k^{2}\left \langle m\left ( t\right ) \hat{m}\left ( t\right ) \cos \left ( \omega _{c}t\right ) \sin \left ( \omega _{c}t\right ) \right \rangle }{BN_{0}} \end{align*}

Assume $$\left \langle m\left ( t\right ) \right \rangle =0$$, we obtain\begin{align*} SNR_{c} & =\frac{k^{2}\frac{P_{m}}{2}+k^{2}\frac{P_{m}}{2}}{BN_{0}}\\ & =\frac{k^{2}P_{m}}{BN_{0}} \end{align*}

$s_{3}\left ( t\right ) =k\left [ m\left ( t\right ) \cos \omega _{c}t-\hat{m}\left ( t\right ) \sin \omega _{c}t\right ] +n\left ( t\right )$ Hence\begin{align*} SNR_{i} & =SNR_{c}\frac{B}{B_{T}}\\ & =\frac{k^{2}P_{m}}{BN_{0}}\frac{B}{B}\\ & =\frac{k^{2}P_{m}}{BN_{0}} \end{align*}

\begin{align*} s_{4}\left ( t\right ) & =\left [ k\left [ m\left ( t\right ) \cos \omega _{c}t-\hat{m}\left ( t\right ) \sin \omega _{c}t\right ] +n\left ( t\right ) \right ] A_{c}^{^{\prime }}\cos \omega _{c}t\\ & =A_{c}^{^{\prime }}km\left ( t\right ) \cos ^{2}\omega _{c}t-A_{c}^{^{\prime }}k\hat{m}\left ( t\right ) \sin \omega _{c}t\cos \omega _{c}t+A_{c}^{^{\prime }}\left [ n_{I}\left ( t\right ) \cos \omega _{c}t-n_{Q}\left ( t\right ) \sin \omega _{c}t\right ] \cos \omega _{c}t\\ & =A_{c}^{^{\prime }}km\left ( t\right ) \left ( \frac{1}{2}+\frac{1}{2}\cos 2\omega _{c}t\right ) -A_{c}^{^{\prime }}k\hat{m}\left ( t\right ) \frac{1}{2}\left ( \sin \left ( 0\right ) +\sin \left ( 2\omega _{c}t\right ) \right ) \\ & +A_{c}^{^{\prime }}\left [ n_{I}\left ( t\right ) \cos ^{2}\omega _{c}t-n_{Q}\left ( t\right ) \sin \omega _{c}t\cos \omega _{c}t\right ] \\ & =\frac{1}{2}A_{c}^{^{\prime }}km\left ( t\right ) +\frac{1}{2}A_{c}^{^{\prime }}km\left ( t\right ) \cos 2\omega _{c}t-A_{c}^{^{\prime }}k\hat{m}\left ( t\right ) \frac{1}{2}\sin \left ( 2\omega _{c}t\right ) \\ & +A_{c}^{^{\prime }}\left [ n_{I}\left ( t\right ) \left ( \frac{1}{2}+\frac{1}{2}\cos 2\omega _{c}t\right ) -n_{Q}\left ( t\right ) \frac{1}{2}\left ( \sin \left ( 0\right ) +\sin \left ( 2\omega _{c}t\right ) \right ) \right ] \\ & =\frac{1}{2}A_{c}^{^{\prime }}km\left ( t\right ) +\frac{1}{2}A_{c}^{^{\prime }}km\left ( t\right ) \cos 2\omega _{c}t-\frac{1}{2}A_{c}^{^{\prime }}k\hat{m}\left ( t\right ) \sin \left ( 2\omega _{c}t\right ) \\ & +\frac{A_{c}^{^{\prime }}}{2}n_{I}\left ( t\right ) +\frac{A_{c}^{^{\prime }}}{2}n_{I}\left ( t\right ) \cos 2\omega _{c}t-\frac{A_{c}^{^{\prime }}}{2}n_{Q}\left ( t\right ) \sin 2\omega _{c}t \end{align*}

After low pass ﬁlter, we obtain$s_{5}\left ( t\right ) =\frac{1}{2}A_{c}^{^{\prime }}km\left ( t\right ) +\frac{A_{c}^{^{\prime }}}{2}n_{I}\left ( t\right )$ Hence, \begin{align*} SNR_{o} & =\frac{\left \langle \left ( \frac{1}{2}A_{c}^{^{\prime }}km\left ( t\right ) \right ) ^{2}\right \rangle }{E\left ( \left [ \frac{A_{c}^{^{\prime }}}{2}n_{I}\left ( t\right ) \right ] ^{2}\right ) }\\ & =\frac{\frac{1}{4}\left ( A_{c}^{^{\prime }}\right ) ^{2}k^{2}P_{m}}{\frac{1}{4}\left ( A_{c}^{^{\prime }}\right ) ^{2}N_{0}B}\\ & =\frac{k^{2}P_{m}}{N_{0}B} \end{align*}

Hence\begin{align*} \frac{SNR_{o}}{SNR_{i}} & =\frac{\frac{k^{2}P_{m}}{N_{0}B}}{\frac{k^{2}P_{m}}{BN_{0}}}\\ & =1 \end{align*}

Hence\begin{align*} \gamma & =\frac{SNR_{o}}{SNR_{c}}\\ & =\frac{\frac{k^{2}P_{m}}{N_{0}B}}{\frac{k^{2}P_{m}}{BN_{0}}}\\ & =1 \end{align*}

### 16 How to determine ﬁgure of merit for VSB using coherent detector?

$s\left ( t\right ) =\frac{A_{c}}{2}\left [ m\left ( t\right ) \cos \omega _{c}t\mp m_{Q}\left ( t\right ) \sin \omega _{c}t\right ]$ $$m_{Q}\left ( t\right )$$ is the output of VSB ﬁlter when input is $$m\left ( t\right )$$