It is a well-known result that when two signals with the same period $T$ get multiplied in the time domain, the resulting signal's Fourier coefficients are given by the discrete convolution of individual Fourier coefficients. $$c_k = \sum_{n=-\infty}^{+\infty} a_nb_{k-n}$$

where $c_k$ is the resulting signal's Fourier coefficients and $a_n$ and $b_n$ are the constituent signal, Fourier coefficients. What I am interested in knowing is if the same is true for two signals with different frequencies.

To start off, the two frequencies should at least be rational multiples as explained here. So, if we assume $\omega_x = p\omega_0$ and $\omega_y = q\omega_0$ and follow the steps for inspecting the nature of the resulting signal's fourier coefficients by,

$$c_k = \frac{1}{T}\int_0^T \sum_{n=-\infty}^{+\infty} a_{n} e^{jpn\omega_0t}\sum_{l=-\infty}^{+\infty}b_{l}e^{jlq\omega_0t} e^{-jk\omega_0t}dt$$ Then by following the procedure shown here, we arrive at,

$$c_k = \frac{1}{T} \int_0^T \sum_{n=-\infty}^{+\infty}a_{n}\sum_{l=-\infty}^{+\infty} b_{l}\delta(k-(np+lq))$$

Normally, we would apply sifting property here if $p =1$ and $q=1$ and proceed to get, $$c_{k}= \frac{1}{T} \sum _{n}a_{n} b_{k-n} = \mathbf{a}*\mathbf{b}$$

However, I am unsure how to proceed with the case where we have $\delta(k-(np+lq))$. How do we go about applying time sifting property to the above delta function to arrive at a definitive result (if at all possible)?


Let's start with a generic solution where the both signals are periodic but with a different period. That means we need to use the continuous Fourier Transform. Let's say we have

$$a(t) = \sum_m a_k e^{jk\omega_at} \\ A(\omega) = \sum_k a_k \delta(\omega - k\omega_a)$$

If multiply the signal we get a double sum

$$c(t) = a(t) \cdot b(t) = \sum_k \sum_m a_k b_m e^{jk\omega_at} e^{jm\omega_bt} = \\ \sum_k \sum_m a_k b_m e^{j(k\omega_a + m\omega_b)t} $$

The spectrum is simply $$C(\omega) = \sum_k \sum_m a_k b_m \delta(\omega-k\omega_a - m\omega_b) $$

We could also have derived this by convolution in the frequency domain and would have gotten the same result. If the ratio of the frequency is real (and not rational), than the result is aperiodic and that's the end of the rope.

If it's rational the result is periodic with the largest common divisor

$$\omega_c = \frac{\omega_a}{A} = \frac{\omega_a}{B} \\ \omega_a = A\omega_c \\\omega_b = B\omega_c$$

So we can express this as a function of $\omega_c$

$$C(\omega) = \sum_k \sum_m a_k b_m \delta(\omega-\omega_c(kA+mB)) $$

Next step would be to reorder so that the inner sum sums over all the same frequencies, but I'm currently out of time. Your end result will be

$$C(\omega) = \sum_n c_n \delta(\omega-n \omega_c)$$

where the $c_n$ are the sum of $a_kb_m$ over all possible permutations of $n = Ak+ Bm$ and the $c_n$ are the coefficients of the Fourier series of the result.

  • $\begingroup$ Thank you for the reply. However, do you think we can apply sifting property to the above signal $C(\omega) = \sum_k \sum_m a_k b_m \delta(\omega-\omega_c(kA+mB)) $ when we can see that $k$ and $m$ are getting constantly scaled. Can we still write it as a convolution sum? $\endgroup$
    – user58865
    Aug 24 '21 at 15:33
  • $\begingroup$ Sort of. It's not a standard convolution sum. Determining the elements to sum over is the tricky part. You can do an outer sum over $n$ and the inner sum over $m$ and substitute $k = \frac{n-Bm}{A}$ . However you need to skip all elements where $k$ is not a whole number. $\endgroup$
    – Hilmar
    Aug 25 '21 at 16:14

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