| \(x\left ( t\right ) \) | \(X\left ( f\right ) \) |
| \(\sin \left ( 2\pi f_{c}t\right ) \) | \(\frac {1}{2j}\left [ \delta \left ( f-f_{c}\right ) -\delta \left ( f+f_{c}\right ) \right ] \) |
| \(\cos \left ( 2\pi f_{c}t\right ) \) | \(\frac {1}{2}\left [ \delta \left ( f-f_{c}\right ) +\delta \left ( f+f_{c}\right ) \right ] \) |
| \(\cos \left ( 2\pi f_{c}t+\theta \right ) \) | \(\frac {1}{2}\left [ e^{j\theta }\delta \left ( f-f_{c}\right ) +e^{-j\theta }\delta \left ( f+f_{c}\right ) \right ] \) |
| \(\sin \left ( 2\pi f_{c}t+\theta \right ) \) | \(\frac {1}{2}\left [ e^{j\theta }\delta \left ( f-f_{c}\right ) -e^{-j\theta }\delta \left ( f+f_{c}\right ) \right ] \) |
\(X\left ( t\right ) \) 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.
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 \(x\left ( t\right ) \)
input: random signal \(x\left ( t\right ) \)
output: PSD of \(x\left ( t\right ) \)
Algorithm:
Another method (this below works if not random \(x\left ( t\right ) \)) , why? can’t find FT for random process?
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.
\[ PEP=\frac {1}{2}\tilde {x}_{\max }^{2}\left ( t\right ) \]
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 \(\cos \left ( t\right ) \))
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
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 \(\hat {x}\)
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
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
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
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
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\begin {equation} x\left ( t\right ) =\cos \left ( \omega _{c}t+\theta \left ( t\right ) \right ) \tag {1} \end {equation} 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\[ x_{PM}\left ( t\right ) =\cos \left ( \omega _{c}t+k_{p}m\left ( t\right ) \right ) \]
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*}
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 \(DSB-SS\), \(B_{T}=B.\)
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.
\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.
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*}
\[ s\left ( t\right ) =\frac {A_{c}}{2}\left [ m\left ( t\right ) \cos \omega _{c}t\mp m_{Q}\left ( t\right ) \sin \omega _{c}t\right ] \] \(m_{Q}\left ( t\right ) \) is the output of VSB filter when input is \(m\left ( t\right ) \)