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<h2 class="hd hd-2 unit-title">Introduction to Transmission of AM Signals</h2>
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<p>
As previously emphasized, our objective is to learn how to send a signal [mathjaxinline]f(t)[/mathjaxinline] in a non-distorted way, through a dispersive medium. </p><p>
One way is through <i class="itshape">amplitude modulation</i>, or AM for short. Instead of sending any kind of signal, the specific form [mathjaxinline]f(t)=f_{s}(t)\cos (\omega _{0}t)[/mathjaxinline] is sent, where [mathjaxinline]\cos (\omega _{0}t)[/mathjaxinline] modulates the signal [mathjaxinline]f_{s}(t)[/mathjaxinline] at a very high frequency. Note, we are now defining [mathjaxinline]f_{s}(t)[/mathjaxinline] as the signal that contains "information," but [mathjaxinline]f(t)[/mathjaxinline] is the signal that is sent through the medium. </p><p>
We will show that this AM "trick" will yield a traveling wave composed of frequency components within a very narrow frequency window. Within this very narrow frequency window, the dispersion of any medium is approximately linear, and we will show why this is important for preserving the signal [mathjaxinline]f_{s}(t)[/mathjaxinline] as it travels. </p>
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<h2 class="hd hd-2 unit-title">L25v1: Idea for Transmission of AM Signals</h2>
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<h3 class="hd hd-2">L25v1: Idea for Transmission of AM Signals</h3>
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<h2 class="hd hd-2 unit-title">L25Q1: Consequences of a Narrow Frequency Range</h2>
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Consequences of a Narrow Frequency Range
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<p>
We have stated, but have not yet shown, that an amplitude modulated signal will have frequency components that are confined to a very narrow range of frequencies, between [mathjaxinline]\omega _{0}+\omega _{s}[/mathjaxinline] and [mathjaxinline]\omega _{0}-\omega _{s}[/mathjaxinline]. </p>
<p>
Let's assume this is the case. We have not quite explained why this will be useful for transmitting the signal [mathjaxinline]f_{s}(t)[/mathjaxinline]&#8212;regardless, let's consider the consequences. </p>
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Select ALL of the choices that are TRUE if the transmitted signal is confined to a narrow range of frequencies. </p>
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<text>a) The dispersion relation [mathjaxinline]\omega (k)[/mathjaxinline] is approximately linear.</text>
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<text>b) The group velocity of each frequency component is the same.</text>
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<text>c) The phase velocity of each frequency component is the same.</text>
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<h2 class="hd hd-2 unit-title">L25Q2: Ensuring a Narrow Frequency Range</h2>
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Ensuring a Narrow Frequency Range
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<p>
Again, we have stated, but have not yet proven, that an amplitude modulated signal will have frequency components that are confined to a very narrow range of frequencies, between [mathjaxinline]\omega _{0}+\omega _{s}[/mathjaxinline] and [mathjaxinline]\omega _{0}-\omega _{s}[/mathjaxinline]. </p>
<p><b class="bfseries">(Part a)</b> Let's define the frequency range to be [mathjaxinline]\Delta \omega[/mathjaxinline]. What is the ratio [mathjaxinline]\dfrac {\Delta \omega }{\omega _{0}}[/mathjaxinline]? Express your answer in terms of <code>omega_s</code> for [mathjaxinline]\omega _{s}[/mathjaxinline] and <code>omega_0</code> for [mathjaxinline]\omega _{0}[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\frac{\Delta \omega }{\omega _{0}}=[/mathjaxinline] </p>
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<td class="formulainput">fractions</td>
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<th class="formulainput" scope="row" rowspan="4">Operators</th>
<td class="formulainput"><code>+ - * /</code> (add, subtract, multiply, divide)</td>
<td class="formulainput">enter <code> (x+2*y)/(x-1)</code> for [mathjaxinline] \displaystyle \frac{x+2y}{x-1} [/mathjaxinline] </td>
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<td class="formulainput"><code>^</code> (raise to a power)</td>
<td class="formulainput">enter <code> x^(n+1) </code> for [mathjaxinline] x^{n+1} [/mathjaxinline]</td>
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<td class="formulainput"><code>_</code> (add a subscript)</td>
<td class="formulainput">enter <code> v_0 </code> for [mathjaxinline] v_0 [/mathjaxinline] </td>
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<td class="formulainput">use <code>( )</code> to clarify order of operations</td>
<td class="formulainput"> enter <code>(2+3)*2 </code> for 10 <br/>
enter <code> 2+3*2 </code> for 8 </td>
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<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">enter (english) name of letter</td>
<td class="formulainput">enter <code>alpha </code> for [mathjaxinline] \alpha [/mathjaxinline]<br/>
enter <code>lambda </code> for [mathjaxinline]\lambda [/mathjaxinline]
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<th class="formulainput" scope="row">Mathematical <br/> constants</th>
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<code>e, pi</code>
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<td class="formulainput">enter <code>e^x </code> for [mathjaxinline] e^x [/mathjaxinline]<br/>
enter <code>2*pi </code> for [mathjaxinline] 2\pi [/mathjaxinline]
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<code>abs, ln, sqrt</code>
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<td class="formulainput">enter <code>abs(x+y) </code> for [mathjaxinline] \left|x+y \right| [/mathjaxinline]<br/>
enter <code>sqrt(x^2-y) </code> for [mathjaxinline] \sqrt{x^2-y} [/mathjaxinline]
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<code>sin, cos, tan, sec, csc, cot</code>
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<td class="formulainput">enter <code>sin(4*x+y)^2 </code> for [mathjaxinline]\sin^2(4x+y) [/mathjaxinline]</td>
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<td class="formulainput"><code>arcsin, arccos, arctan</code>, etc.</td>
<td class="formulainput">enter <code>arctan(x^2/3) </code> for [mathjaxinline]\tan^{-1}\left(\frac{x^2}{3}\right) [/mathjaxinline]</td>
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<td class="formulainput"><code>sinh, cosh, arcsinh</code>, etc.</td>
<td class="formulainput">enter <code>cosh(4*x+y) </code> for [mathjaxinline]\cosh(4x+y) [/mathjaxinline]</td>
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<th class="formulainput" scope="row" rowspan="3">Matrices<br/>&amp; Vectors</th>
<td class="formulainput">matrix</td>
<td class="formulainput">enter <code>[[1,0],[0,-1]]</code> for [mathjaxinline]\begin{pmatrix} 1 &amp; &amp; 0 \\ 0 &amp; &amp; -1 \end{pmatrix}[/mathjaxinline]</td>
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<td class="formulainput">column vector</td>
<td class="formulainput">enter <code>[[1],[2],[3]]</code> for [mathjaxinline]\begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}[/mathjaxinline]</td>
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<td class="formulainput">enter <code>[[1,2,3]]</code> for [mathjaxinline]\begin{pmatrix} 1 &amp; &amp; 2 &amp; &amp; 3 \end{pmatrix}[/mathjaxinline]</td>
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<p><b class="bfseries">(Part b)</b> What should be the relation between [mathjaxinline]\omega _{s}[/mathjaxinline] and [mathjaxinline]\omega _{0}[/mathjaxinline] such that the "relative" range [mathjaxinline]\Delta \omega /\omega _{0}[/mathjaxinline] is very small? </p>
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<text> a) [mathjaxinline]\omega _{s} \ll \omega _{0}[/mathjaxinline]</text>
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<text> b) [mathjaxinline]\omega _{s} \gg \omega _{0}[/mathjaxinline]</text>
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<text> c) [mathjaxinline]\omega _{s} \approx \omega _{0}[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">Review of AM Derivation</h2>
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<p><h2>REVIEW OF OBJECTIVE</h2></p><p>
Recall, our objective has been to show that an AM signal enables us to send information in a non-distorted way, through a dispersive medium. </p><p>
If the information that we want to send is represented by the signal [mathjaxinline]f_{s}(t)[/mathjaxinline], then we can successfully transmit this information if the traveling waveform [mathjaxinline]\psi (x,t)[/mathjaxinline] can be expressed in terms of the original signal traveling at a constant velocity—therefore, the signal should not disperse. </p><p>
We HOPE that the traveling wave has the following functional form (in complex notation): </p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\psi (x,t)=f_{s}(t-x/v_{g})e^{i(k_{0}x-\omega _{0}t)}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where the carrier [mathjaxinline]e^{i(k_{0}x-\omega _{0}t)}[/mathjaxinline] may have a different velocity than the traveling signal [mathjaxinline]f_{s}(t-x/v_{g})[/mathjaxinline]. </p><p>
Ultimately, we showed that [mathjaxinline]\psi (x,t)[/mathjaxinline] yields this result if the transmitting signal has the following form (in complex notation): </p><table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(t)=f_{s}(t)e^{i\omega _{0} t}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
To produce this result, we also had to assume that the dispersion of the medium was approximately linear in the relevant frequency range. </p><p><h2>ALTERNATE DERIVATION</h2></p><p>
The derivation carried out in the lesson may seem a little backwards. We started with the solution, i.e., [mathjaxinline]f(t)=f_{s}(t)e^{i\omega _{0} t}[/mathjaxinline], and showed that this yields the expected traveling waveform [mathjaxinline]\psi (x,t)=f_{s}(t-x/v_{g})e^{i(k_{0}x-\omega _{0}t)}[/mathjaxinline]. </p><p>
It's worthwhile to work the other way around, without stating the form of [mathjaxinline]f(t)[/mathjaxinline], and see where our assumptions lead. <i class="itshape">The assumption that we will make is that the dispersion is approximately linear over the relevant frequency range.</i> Note, this is more of a requirement rather than an assumption, and our result will tell us what type of signal satisfies this condition. </p><p>
So, let's consider a traveling waveform [mathjaxinline]\psi (x,t)[/mathjaxinline] that is produced by the time-varying signal [mathjaxinline]f(t)[/mathjaxinline]: </p><table id="a0000000004" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\psi (x,t)=\int _{-\infty }^{\infty } d\omega \, c(\omega ) e^{-i(\omega t - k(\omega )x)}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where [mathjaxinline]k(\omega )[/mathjaxinline] is frequency dependent wavenumber, which is responsible for dispersion. Now, let's assume that the dispersion is linear around some wavenumber [mathjaxinline]k_{0}[/mathjaxinline], such that </p><table id="a0000000005" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\omega (k) \approx \omega _{0} + (k-k_{0})v_{g}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Then, we can express [mathjaxinline]k(\omega )[/mathjaxinline] as </p><table id="a0000000006" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]k(\omega )=\frac{\omega - \omega _{0}}{v_{g}} + k_{0}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Note, this is equivalent to the requirement that the frequency range of the signal is very narrow, centered around the very high frequency [mathjaxinline]\omega _{0}=k_{0}v_{p}[/mathjaxinline], as discussed in the videos. </p><p>
With this requirement, the waveform is </p><table id="a0000000007" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000008"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \psi (x,t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle =\int _{-\infty }^{\infty } d\omega \, c(\omega ) e^{-i\omega t} e^{i\left[\frac{\omega - \omega _{0}}{v_{g}} + k_{0}\right]x}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000009"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = e^{ik_{0}x} \int _{-\infty }^{\infty } d\omega \, c(\omega ) e^{-i\omega \left(t-\frac{x}{v_{g}}\right)} e^{-i\frac{\omega _{0}}{v_{g}}x}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
Now, we can shift the integral with [mathjaxinline]\omega \rightarrow \omega + \omega _{0}[/mathjaxinline], then </p><table id="a0000000010" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000011"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \psi (x,t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = e^{ik_{0}x} \int _{-\infty }^{\infty } d\omega \, c(\omega + \omega _{0}) e^{-i(\omega +\omega _{0})\left(t-\frac{x}{v_{g}}\right)} e^{-i\frac{\omega _{0}}{v_{g}}x}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000012"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = e^{i(k_{0}x-\omega _{0}t)} \int _{-\infty }^{\infty } d\omega \, c(\omega + \omega _{0}) e^{-i\omega \left(t-\frac{x}{v_{g}}\right)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
So, we can identify the following term as a traveling wave: </p><table id="a0000000013" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\int _{-\infty }^{\infty } d\omega \, c(\omega + \omega _{0}) e^{-i\omega \left(t-\frac{x}{v_{g}}\right)} \equiv f(t-\frac{x}{v_{g}})[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where [mathjaxinline]f(t)[/mathjaxinline] is defined as </p><table id="a0000000014" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(t)=\int _{-\infty }^{\infty } d\omega \, c(\omega + \omega _{0}) e^{-i\omega t}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
We're almost there! We want to know what [mathjaxinline]f(t)[/mathjaxinline] is in terms of the signal that we are producing, which is: </p><table id="a0000000015" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f_{s}(t)=\int _{-\infty }^{\infty } d\omega \, c(\omega ) e^{-i\omega t}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
We can shift the frequency again to further simplify, letting [mathjaxinline]\omega \rightarrow \omega - \omega _{0}[/mathjaxinline]: </p><table id="a0000000016" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000017"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle f(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle =\int _{-\infty }^{\infty } d\omega \, c(\omega ) e^{-i(\omega - \omega _{0}) t}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000018"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = e^{i \omega _{0} t} \int _{-\infty }^{\infty } d\omega \, c(\omega ) e^{-i\omega t}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000019"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = f_{s}(t)e^{i \omega _{0} t}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
SO THERE IT IS! If we want to send a signal that produces our desired traveling waveform [mathjaxinline]\psi (x,t)=f_{s}(t-x/v_{g})e^{i(k_{0}x-\omega _{0}t)}[/mathjaxinline], the signal that we send should have the form [mathjaxinline]f(t)=f_{s}(t)e^{i \omega _{0} t}[/mathjaxinline]. </p><p>
The equivalence is valid only in a narrow frequency window, which enabled us to approximate the linear dispersion. The frequency window will be narrow if the carrier frequency [mathjaxinline]\omega _{0}[/mathjaxinline] is much higher than the characteristic frequency of the modulating signal [mathjaxinline]f_{s}(t)[/mathjaxinline]. </p>
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<h2 class="hd hd-2 unit-title">L25Q3: Physical Interpretation</h2>
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Physical Interpretation
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We have shown that a signal of the form: </p>
<table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
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<td class="equation" style="width:80%; border:none">[mathjax]f(t)=f_{s}(t)e^{i\omega _{0}t}[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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<p>
will produce a traveling wave of the form: </p>
<table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
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<td class="equation" style="width:80%; border:none">[mathjax]\psi (x,t)=f_{s}(t-\frac{x}{v_{g}})e^{i(k_{0}x-\omega _{0}t)}[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
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<p>
We concluded that this solution shows the signal [mathjaxinline]f_{s}(t)[/mathjaxinline] can be transmitted without being distorted. For the following, select ALL statements that accurately characterize this solution and its consequences. </p>
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<text>a) All frequency components of [mathjaxinline]f_{s}[/mathjaxinline] travel at approximately the same velocity, so they do not spread out.</text>
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<text>b) The traveling waveform has an envelope with the same "shape" as the signal [mathjaxinline]f_{s}(t)[/mathjaxinline].</text>
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<text>c) The carrier frequency [mathjaxinline]\omega _{0}[/mathjaxinline] can be any frequency, so this is always applicable.</text>
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<text>d) The carrier wave can travel faster or slower than the modulating signal.</text>
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<text>e) This result is not possible if the medium is dispersive.</text>
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