Finding out modulation index and dc offset

Question

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

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

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

fourier transform $s_{AM-DSB-C}(t)$

We can get $S_{AM-DSB-C}(f)=\frac{A}{2} [S(f-f_c)+S(f+f_c)] + \frac {Ac}{2} [\delta (f-f_c) + \delta (f+f_c)]$

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

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

Help plz.

Answer 1

The sinusoid test signal is given by $$ s(t) = V_0cos(2\pi f_0) $$ and its Fourier transform by $$ S(f) = \frac{V_0}{2}[\delta(f-f_0) + \delta(f+f_0)]. $$ Inserted in $S_\mathrm{AM-DSB-C}$ it yields $$ S_\mathrm{AM-DSB-C}(f) = \frac{AV_0}{4}\left[\delta(f-f_c-f_0) + \delta(f-f_c+f_0) + \delta(f+f_c-f_0) + \delta(f+f_c+f_0) \right] + \frac {Ac}{2} \left[\delta (f-f_c) + \delta (f+f_c)\right]. $$ 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 $$ m = 1.5, c = \frac{V_0}{m}=4\text{ V}. $$