# Fourier coefficients of two discrete-time signals of different periods

I'm trying to understand the Fourier series coefficients of the sum of two discrete-time periodic signals.

Consider two discrete-time periodic signals $$x[n]$$ and $$y[n]$$. $$x[n]$$ has period $$N$$, its Fourier series coefficients $$a_k$$. $$y[n]$$ has period $$M$$, its Fourier series coefficients $$b_k$$.

According to this problem solution (S10.10), the Fourier series coefficients $$c_k$$ of the summation signal $$x[n] + y[n]$$ can be represented by $$a_k$$, $$b_k$$ as follows:

$$c_k = \begin{cases} \frac{1}{N}a_{k/M} + \frac{1}{M}b_{k/N} &k \text{ a multiple of }M \text{ and } N\\ \frac{1}{N}a_{k/M}&k \text{ a multiple of }M\\ \frac{1}{M}b_{k/N}&k \text{ a multiple of }N\\ 0 &\text{otherwise} \end{cases}$$

I don't see where $$\frac{1}{N}$$ and $$\frac{1}{M}$$ come from. With a concrete example, say $$N=3$$ and $$M=2$$.

\begin{align} x[n] & = a_0 + 0 + a_1 e^{j\frac{4\pi}{6}n} + 0 + a_2 e^{j\frac{8\pi}{6}n} + 0\\ y[n] & = b_0 + 0 + 0 + b_1 e^{j\frac{6\pi}{6}n} + 0 + 0 \\ x[n]+y[n] &= c_0 + 0 + c_2 e^{j\frac{4\pi}{6}n} + c_3 e^{j\frac{6\pi}{6}n} + c_4 e^{j\frac{8\pi}{6}n} + 0 \end{align}

In this example，$$c_0 = a_0 + b_0$$, $$c_1=0$$, $$c_2 = a_1$$, $$c_3 = b_1$$, $$c_4 = a_2$$, $$c_5 = 0$$

To extrapolate, I think the correct formula to compute $$c_k$$ is

$$c_k = \begin{cases} a_{k/M} + b_{k/N} &k \text{ a multiple of }M \text{ and } N\\ a_{k/M}&k \text{ a multiple of }M\\ b_{k/N}&k \text{ a multiple of }N\\ 0 &\text{otherwise} \end{cases}$$

Also the formula doesn't make sense in special cases like $$x[n] = y[n]$$ and $$y[n] = 0$$.

If $$y[n] = x[n]$$, apparently $$a_k = b_k$$ and $$c_k = 2a_k$$. However, according to the formula above, $$c_k = \frac{2}{N} a_{k/N}$$ for $$k$$ a multiple of $$N$$ and $$c_k = 0$$ otherwise.

If $$y[n] = 0$$, $$c_k = a_k$$. But by the formula, $$c_k = \frac{1}{N}a_{k/M}$$. I don't see how the two representations of $$c_k$$ can be the same thing.

The original problem (P10.10)

– Jdip
Dec 15, 2022 at 9:03

• At first I thought the scaling factors $$1/N$$ and $$1/M$$ were coming from one possible definition that conserves energy in both domains (some prefer applying the scaling factor $$1/\sqrt{N}$$ to both), but from the links you provided, clearly the authors use the following definition:

A real, $$N$$-periodic, discrete-time signal $$x[n]$$ can be represented by a linear combination of complex exponential signals: $$x[n] = \sum_{k=0}^{N-1}a_ke^{j2\pi kn/N}$$ The complex coefficients $$a_k$$ can be calculated with: $$a_k = \frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-j2\pi kn/N}$$

So, I must agree with you, I don't understand where those scaling factors come from for the coefficients. Not sure why the coefficients themselves would be scaled, since the authors apply the scaling as per the above definition, and not on the coefficients. If they applied the factors on the 1st equation (synthesis) and not the 2nd (analysis), then it would make sense. As such, either it is an error in the given solution, or we're both missing something...

• Second, kudos to you for not taking the solution at face value and trying out special cases yourself!

As far as your concerns, the authors could probably have been a little more mathematically rigorous. However, without actually going through the math (what you wrote makes sense), I'm willing to make a couple assumptions derived from how they formulated the problem.

1. The problem does state

two specific periodic sequences $$x[n]$$ and $$y[n]$$

I'm assuming that excludes the special case $$x[n] = y[n]$$

2. $$y[n] = 0$$ is a constant function and so is indeed periodic, but can take on any period
(i.e. $$y[n+M] = y[n]$$ for any $$M$$), and the problem does imply a single possible value for $$M$$, so I'm assuming that case is excluded as well.