I have a question form my teachers, and I cannot understand why I can find out the modulation index form the figure.

The question provide a Figure like this:

Figure 1

And the information signal is a sinusoidal test signal with peak amplitude $6\ \rm V$ and is applied to an AM-DSB-C modulator, the Fourier spectrum of the modulated signal is shown above.

The solution is like this:

As $\displaystyle \frac {\frac {Acm}{4}}{\frac{Ac}{2}}=\frac{3}{4}$, $m=1.5$, so the modulation index is $1.5$ Moreover as $\displaystyle m=\frac{x}{c}$, the peak amplitude of $s(t)$ is $6\ \rm V$, so the DC offset($c$) is $4\ \rm V$.

I know where is the $\displaystyle \frac {Ac}{2}=4$ come form

as $s_{AM-DSB-C}(t)=A\big(s(t)+c\big)\cos(2 \pi f_c t)$

\begin{align} \mathcal F\left\{s_{AM-DSB-C}(t)\right\} &=S_{AM-DSB-C}(f)\\ &=\frac{A}{2}\big[S(f-f_c)+S(f+f_c)\big] + \frac {Ac}{2} \big[\delta (f-f_c) + \delta (f+f_c)\big] \end{align}

so, form the second term we can get $\displaystyle\frac {Ac}{2}=4$

But, I cannot understand how the solution can get $\displaystyle \frac {Acm}{4}=3$.

  • $\begingroup$ and This question is made by my teacher, so no reference. $\endgroup$
    – Samuel
    Commented Feb 3, 2013 at 6:57
  • $\begingroup$ Is the solution only "like this" or is it definitely the correct solution? $\endgroup$
    – Deve
    Commented Feb 3, 2013 at 8:45
  • $\begingroup$ this is the provided solution. $\endgroup$
    – Samuel
    Commented Feb 3, 2013 at 9:06
  • $\begingroup$ From $Ac/2 = 4$ follows $c = 8/A$, not $c = 4$. Are you sure that $m = cm/(2c)$? And what do you mean by $cm$? Is this the amplitude $c_\mathrm{m}$ of the AC part of the test signal? $\endgroup$
    – Deve
    Commented Feb 3, 2013 at 9:19
  • $\begingroup$ The solution is actually write Ac/2 =4. and cm mean c times m, c stand for dc offset and m stand for modulation index. and the amplitude of the test signal is 6V. $\endgroup$
    – Samuel
    Commented Feb 3, 2013 at 9:30

1 Answer 1


The sinusoid test signal is given by $$ s(t) = V_0\cos(2\pi f_0) $$ and its Fourier transform by $$ S(f) = \frac{V_0}{2}\big[\delta(f-f_0) + \delta(f+f_0)\big]. $$ Inserted in $S_\mathrm{AM-DSB-C}$ it yields $$ S_\mathrm{AM-DSB-C}(f) = \frac{AV_0}{4}\big[\delta(f-f_c-f_0) + \delta(f-f_c+f_0) + \delta(f+f_c-f_0) + \delta(f+f_c+f_0) \big] + \frac {Ac}{2} \big[\delta (f-f_c) + \delta (f+f_c)\big]. $$ Comparing that with the figure given we see that $AV_0/4 = 3$ and $Ac/2 = 4$. Additionaly, $f_0 = 10\text{ kHz, } f_c = 100\text{ kHz}$. Using $m = V_0/c$ and $V_0 = 6\text{ V}$ we find $$ m = 1.5, c = \frac{V_0}{m}=4\text{ V}. $$

  • $\begingroup$ I think I forget $c=V_0/m$, so I have problems on understand the solutions. Thanks Deve, greats answer. $\endgroup$
    – Samuel
    Commented Feb 3, 2013 at 10:36

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