Finding out modulation index and DC offset 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:


  
  
  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\} &amp;=S_{AM-DSB-C}(f)\\
&amp;=\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 $\displaystyle \frac {Acm}{4}=3$.
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
$$
m = 1.5, c = \frac{V_0}{m}=4\text{ V}.
$$