Amplitude Modulation vs Ringmodulation¶

Both amplitude modulation and ringmodulation are based on the multiplication of two signals - a carrier $c$ and a modulator $x$ . The basic formula is the same for both:

$y(t) = x(t) \cdot c(t)$

However, the carrier signals are different:

For ringmodulation both the carrier and the modulator are symmetric (bipolar):

$y(t) = \sin\left[2 \pi f_c t\right] \cdot \left(\sin\left[2 \pi f_m t\right]\right)$

For amplitude modulation, the carrier ist asymmetric (unipolar):

$y(t) = \left( 1+ m \sin \left[ 2 \pi f_c t \right] \right) \cdot \left(\sin\left[2 \pi f_m t\right]\right)$

This differnce has an influence on the resulting spectrum and on the sound, as the following examples show.

AM Spectrum¶

In the discrete time case, the spectrum for amplitude modulation can be calculated as follows:

$Y[k] = DFT(y[n])$

The modulation index is set to $m=1$ :

$\begin{align} Y[k] = & \sum_{n=0}^{N-1} y[n] \cdot e^{-j 2 \pi k \frac{n}{N}}\\ =& \sum_{n=0}^{N-1} \sin\left(2 \pi f_c \frac{n}{f_m}\right) \cdot \left( 1+ \sin\left[2 \pi f_m \frac{n}{f_m}\right]\right) \cdot e^{-j 2 \pi k \frac{n}{N}} \\ &\\ & \textsf{\hspace{80pt} Use: } \sin(a) \cdot \sin(b) = 0.5 (\cos(a-b) - \cos(a+b) \\ &\\ =&\sum_{n=0}^{N-1} \left( \sin\left(2 \pi f_c \frac{n}{f_m}\right) + 0.5 \left( \cos\left(2 \pi (f_c - f_m)\frac{n}{f_m}\right) - \cos\left(2 \pi (f_1 + f_m)\frac{n}{f_m}\right) \right) \right) \cdot e^{-j 2 \pi k \frac{n}{N}} \\ &\\ & \textsf{\hspace{40pt} Sines/cosines appear as $\delta$-peaks at the relevant frequency bin.} \\ &\\ = & \boxed{\delta[f_c] + 0.5 \delta[f_c - f_m] + 0.5 \ \delta[f_c + f_m]} \end{align}$

AM creates a spectrum with a peak at the carrier frequency and two additional peaks - one below and one above the carrier. Their position is defined by the difference between carrier and modulator. For a modulation index of $m=1$ , the sidebands have half the carrier's amplitude.

The following example shows an AM signal with a carrier frequency $f_c = 10$ Hz and a modulator frequency of $f_m=200$ Hz.

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Ringmodulator Spectrum¶

The spectrum of a ringmodulator is calculated in the continous time domain for this example.

$\begin{align} \mathcal{F} [ y(t)] = & \int\limits_{-\inf}^{\inf} y(t) e^{-j 2 \pi f t} \mathrm{d}t\\ =& \int\limits_{-\inf}^{\inf} \left( \sin(2 \pi f_c t) \cdot \sin(2 \pi f_m t) \right) e^{-j 2 \pi f t} \mathrm{d}t\\ &\\ & \textsf{Use Euler to get complex representation:}\\ &\\ =& \frac{1}{2 j} \int\limits_{-\inf}^{\inf} \left( (-e^{-j 2 \pi f_c t} +e^{j 2 \pi f_c t}) (-e^{-j 2 \pi f_m t} +e^{j 2 \pi f_m t}) \right) \ e^{-j 2 \pi f t} \mathrm{d}t\\ =& \frac{1}{2 j} \int\limits_{-\inf}^{\inf} \left( e^{j 2 \pi (f_c+f_m) t} - e^{j 2 \pi (f_c-f_m) t} - e^{j 2 \pi (-f_c+f_m) t} + e^{j 2 \pi (-f_c-f_m) t} \right) e^{-j 2 \pi f t} \mathrm{d}t\\ &\\ & \textsf{Each exponential represents a peak in the spectrum (including positive and negative):}\\ &\\ =& \frac{1}{2 j} \left[ \delta(f_c+f_m) -\delta(f_c-f_m) - \delta(-f_c+f_m) + \delta(-f_c-f_m) \right] \end{align}$

Ringmodulation creates a spectrum with two peaks - one below and one above the carrier frequency. Their position is defined by the difference between carrier and modulator. The carrier is supressed, since it is a bipolar signal.