Week 14

We have shown that convolution in the time domain is multiplication in the frequency domain:

$$(f*g)(t) \ftp F(\omega)G(\omega)$$

Consider the Modulation Property of the Continuous-Time Fourier Transform:

\begin{nicebox}
\begin{align*}
f(t)g(t) \ftp \frac{1}{2\pi} (F*G)(\omega) 
\end{align*}
\end{nicebox}

\begin{proof}
\begin{align*}
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}(F*G)(\omega)e^{i\omega t}d\omega \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}F(W)G(\omega-W) dW e^{i\omega t}d\omega \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)\int_{-\infty}^{\infty}G(\omega-W) e^{i\omega t} d\omega dW \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)\int_{-\infty}^{\infty}G(\Omega) e^{i(\Omega+W) t} d\Omega dW \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)\lr{\int_{-\infty}^{\infty}G(\Omega) e^{i\Omega t} d\Omega }e^{iWt}dW \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)e^{iWt}dW \lr{\int_{-\infty}^{\infty}G(\Omega) e^{i\Omega t} d\Omega}\\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\lr{2\pi f(t)}\lr{2\pi g(t)} \\
\iftrans{(F*G)(\omega)} &= 2\pi f(t)g(t) \\
\iftrans{\frac{1}{2\pi}(F*G)(\omega)} &= f(t)g(t) \\
\end{align*}
\end{proof}


Which shows that multiplication in the time domain is the product of a scalar and the convolution in the frequency domain.

Modulation Property of CTFT

 Consider an AM modulator where $y(t) = x(t)g(t)$, where $g(t) = e^{i\omega_0 t}$, where $\omega_0$ is large and positive:

% \img{images/modulators/am.ps}

\begin{center}
\begin{pspicture}(-3,-3)(3,3)

% in from x
\rput(-3.2,0){ $x(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(-2.7,0)(-0.25,0)
% out to y
\rput(3.2,0){$y(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(0.25,0)(2.7,0)
% up arrow
\rput(0,-2){ $g(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(0,-1.65)(0,-0.25)

% multiplier
\pscircle(0,0){0.25}
\psline(-0.175,0.175)(0.175,-0.175)
\psline(0.175,0.175)(-0.175,-0.175)

% box
\pspolygon(-1.5,-2.65)(-1.5,1)(1.5,1)(1.5,-2.65)

\end{pspicture}
\end{center}

Lets say the spectrum of the input signal is given by the following figure:


% \img{images/triangles/triangle.ps}

\begin{center}
\begin{pspicture}(-3.2,-3.2)(3,3)

% axis
\psline{->}(-3,0)(3,0)
\psline(-2,0)(0,1)
\psline(0,1)(2,0)


\rput(-2,-0.25){$-A$}
\rput(0,-0.25){0}
\rput(2,-0.25){$A$}

\rput(3.25,-0.25){$\omega$}
\rput(-2,1){$X(\omega)$}

\rput(1.12,1){$(1)$}


\end{pspicture}
\end{center}


We can determine that $Y(\omega) = \frac{1}{2\pi}(X*G)(\omega)$. Since $g(t) = e^{i\omega_0 t}$, it represents a single frequency in the frequency domain. In other words, the spectrum of $g(t)$, namely, $G(\omega)$, is a single dirac delta shifted by $\omega_0$ with an amplitude of $2\pi$ (see previous sections for derivation).

% \img{images/dirac/2pi.ps}


\begin{center}
\begin{pspicture}(-1,-1)(3,3)

\psline{->}(-2,0)(2,0)
\rput(2,-0.25){$\omega$}
\rput(2,1){$G(\omega)$}

\psline{->}(0,0)(0,1)
\rput(0.3,1){$(2\pi)$}
\rput(0,-0.25){$\omega_0$}


\end{pspicture}
\end{center} 

In other words, 

$$ g(t) = e^{i\omega_0 t} \ftp 2\pi \delta(\omega-\omega_0) $$

When doing amplitude modulation with a single complex exponential of frequency $\omega_0$, we find that the frequency response $Y(\omega) = X(\omega-\omega_0)$. This is because $(X*G)(\omega)=2\pi X(\omega - \omega_0)$, and the $2\pi$ cancels out. Here is how it works out:

\begin{align*}
Y(\omega) &= \frac{1}{2\pi}(X*G)(\omega) \\
Y(\omega) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\Omega)G(\omega-\Omega)d\Omega \\
Y(\omega) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\Omega)2\pi\delta(\omega-\Omega-\omega_0)d\Omega \\
Y(\omega) &= \int_{-\infty}^{\infty}X(\Omega) \delta(\omega-\omega_0-\Omega)d\Omega \\
Y(\omega) &= X(\omega-\omega_0) 
\end{align*}

Thus we have a spectrum for $Y(\omega)$ that looks like

% \img{images/triangles/yomega.ps}

\begin{center}
\begin{pspicture}(-3.2,-3.2)(3,3)

% axis
\psline{->}(-3,0)(3,0)
\psline(-2,0)(0,1)
\psline(0,1)(2,0)


\rput(-2,-0.25){$\omega_0-A$}
\rput(0,-0.25){$\omega_0$}
\rput(2,-0.25){$\omega_0+A$}

\rput(3.25,-0.25){$\omega$}
\rput(-2,1){$Y(\omega)=X(\omega-\omega_0)$}

\rput(1.12,1){$(1)$}


\end{pspicture}
\end{center}

Complex Exponential Carrier

Generally an antenna must be approximately a fourth of the size of a cycle of the wave it transmits. The speed of light can be given in terms of a frequency $f$ in Hertz multiplied by the corresponding wavelength $\lambda$:
$$c=f\lambda = 2.9979 \times 10^8 m/s $$

Lets say the spectrum of ones voice fell along a range from $-3.3kHz$ to $3.3kHz$, then we can approximate the length of the antenna needed by first determining the wavelength:

\begin{align*}
\lambda &= c/f \\
\lambda &\approx 98 km\\
\end{align*}

This tells us we would need about $25km$ of antenna to transmit this persons voice. This is why we use amplitude modulation---to transmit a signal at higher frequencies.


Lets consider an amplitude modulator with a sinusoidal carrier given by 

% \img{images/modulators/amcos.ps}

\begin{center}
\begin{pspicture}(-3,-3)(3,2)

% in from x
\rput(-3.2,0){$x(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(-2.7,0)(-0.25,0)
% out to y
\rput(3.2,0){$y(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(0.25,0)(2.7,0)
% up arrow
\rput(0,-2){$\cos(\omega_0 t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(0,-1.65)(0,-0.25)

% multiplier
\pscircle(0,0){0.25}
\psline(-0.175,0.175)(0.175,-0.175)
\psline(0.175,0.175)(-0.175,-0.175)

% box
\pspolygon(-1.5,-2.65)(-1.5,1)(1.5,1)(1.5,-2.65)

\end{pspicture}
\end{center}

Where $x(t)$ is the \emph{information-bearing} signal (a.k.a. the modulating signal), and $h(t) = \cos(\omega_0 t)$ is the \emph{carrier} signal (a.k.a. the modulated signal). The output signal $y(t)$ is the product of these two signals, $x(t)\cos(\omega_0 t)$. We can begin solving for the spectrum of $y$ by first finding the spectrum of $h$. 

\begin{align*}
h(t) &= \cos(\omega_0 t) \\
h(t) &= \frac{1}{2}e^{i\omega_0 t} + \frac{1}{2}e^{-i\omega_0 t} \\
\ftrans{h(t)} &= \frac{1}{2}\ftrans{e^{i\omega_0 t}} + \frac{1}{2}\ftrans{e^{-i\omega_0 t}} \\
H(\omega) &= \frac{1}{2}2\pi\delta(\omega-\omega_0) + \frac{1}{2}2\pi\delta(\omega+\omega_0) \\
H(\omega) &= \pi\delta(\omega-\omega_0) + \pi\delta(\omega+\omega_0) \\
\end{align*}

$H(\omega)$ is two dirac deltas symmetrically shifted each by $\omega_0$, 

% \img{images/dirac/doublefreq.ps}

\begin{center}
\begin{pspicture}(-3,-1)(3,3)

\psline{->}(-2,0)(2,0)
\rput(2,-0.25){$\omega$}
\rput(2,1){$H(\omega)$}

\psline{->}(-1,0)(-1,1)
\rput(-1.3,1){$(\pi)$}
\rput(-1,-0.25){$-\omega_0$}

\psline{->}(1,0)(1,1)
\rput(1.3,1){$(\pi)$}
\rput(1,-0.25){$\omega_0$}

\end{pspicture}
\end{center}

which gives us an output signal of shifted versions of the input signal with half of the original height:

% \img{images/triangles/double.ps}

\begin{center}
\begin{pspicture}(-5,-2)(5,3)

% axis
\psline{->}(-5,0)(5,0)


\psline(-3,0)(-2,1)
\psline(-2,1)(-1,0)

\psline(1,0)(2,1)
\psline(2,1)(3,0)


\rput(-3,-0.25){$-\omega_0-A$}
\rput(-2,-1.25){$-\omega_0$}
\rput(-1,-0.25){$-\omega_0+A$}

\rput(3,-0.25){$\omega_0+A$}
\rput(2,-1.25){$\omega_0$}
\rput(1,-0.25){$\omega_0-A$}

\psline[linestyle=dashed](-2,-1)(-2,1)
\psline[linestyle=dashed](2,-1)(2,1)

\rput(5.25,-0.25){$\omega$}
\rput(-3.5,1){$Y(\omega)$}

\rput(2.6,1){$(1/2)$}
\rput(-1.4,1){$(1/2)$}


\end{pspicture}
\end{center}

where we must require that $-\omega_0+A \lt \omega_0-A$, so that $A \lt \omega_0$. This saves us a few complications when $\omega_0 \gt\gt A$. How can we demodulate this signal? Once way would be to pass it through a amplitude modulator using a complex exponential $e^{-i\omega_0 t}$, and then we would get the signal shifted over to the left. We can then pass it through a low-pass filter with a pass band amplification of 2 to get the original height back. But, what if we had an oscillator using a cosine function at the recieving end instead of a complex exponential?



The following scheme is called \emph{synchronous demodulation}, and is when we have an identical cosine at the receiving end. Looking at what happened upon transmission, the signal was shifted $\pm \omega_0$ and the amplitude was cut in half. Applying this again, we would get a signal $q(t) = y(t)\cos(\omega_0 t)$, where we shift the shifted signal $\pm \omega_0$. This gives us three disjoint triangular components in the frequency spectrum. The first shift to the left $\omega_0$ gives us two triangles with a height of $1/4$ centered at $-2\omega_0$ and $0$. The shift to the right gives us two triangles with a height of $1/4$ centered at $2\omega_0$ and $0$. The two triangles centered at 0 add to form the original triangle halfed. So we can now put this signal $q(t)$ through a low-pass filter and amplify the signal by 2 to retrieve the original signal.

% \img{images/triangles/three.ps}

\begin{center}
\begin{pspicture}(-5,-1)(5,3)

% axis
\psline{->}(-5,0)(5,0)


\psline(-4,0)(-3,0.5)
\psline(-3,0.5)(-2,0)

\psline(2,0)(3,0.5)
\psline(3,0.5)(4,0)

\psline(-1,0)(0,1)
\psline(0,1)(1,0)


\rput(-3,-0.25){$-2\omega_0$}
\rput(0,-0.25){$0$}
\rput(3,-0.25){$2\omega_0$}


\rput(5.25,-0.25){$\omega$}
\rput(-3.5,1){$Q(\omega)$}

\rput(0.5,1){$(1/2)$}
\rput(3.7,0.5){$(1/4)$}
\rput(-2.3,0.5){$(1/4)$}


\end{pspicture}
\end{center}


The reason this is called amplitude modulation is because given an input signal $x(t)$, that signal multiplied with a $\cos(\omega_0 t)$ will give us a new signal $y(t)$, which is a cosine wave oscillating between $\pm x(t)$, where $x(t)$ serves as an envelope.


What happens if at the reciever we have $q(t) = y(t)\cos(\omega_0t + \theta)$?
\begin{align*}
y(t) &= x(t)\cos(\omega_0 t) \\
q(t) &= y(t)\cos(\omega_0 t + \theta) \\
q(t) &= x(t)\cos(\omega_0 t) \cos(\omega_0 t + \theta) \\
\end{align*}

We can use the trigonometric identity:

$$ \cos\alpha \cos\beta = \frac{1}{2}\lr{\cos\lr{\alpha+\beta}+\cos\lr{\alpha-\beta}} $$

\begin{align*}
q(t) &= x(t)\cos(\omega_0 t) \cos(\omega_0 t + \theta) \\
q(t) &= \frac{x(t)}{2}\lr{\cos(2\omega_0 t + \theta) + \cos(-\theta)} \\
q(t) &= x(t)\lr{\frac{1}{2}\cos(2\omega_0 t + \theta) + \frac{1}{2}\cos(\theta)} \\
\end{align*}

If $\theta \approx \frac{\pi}{2}$, you get an extremely weak baseband. When $\theta = 0$, you have a synchronous demodulation scheme:

$$ q(t) = \frac{x(t)}{2} + \frac{1}{2}\cos(2\omega_0 t)x(t) $$

14.2 Amplitude Modulation