Effects of interchanging sine terms with cosine terms

Question

Suppose we have a real signal $x(t)$. Now, we know that $x(t)$ can be represented as a sum of sines and cosines. w be the angular frequency.

• If $a(\omega)$ be the coefficients of the cosine terms, and $b(\omega)$ be the coefficients of the sine terms, then if we perform a transformation on this signal such that now $b(\omega)$ are the coefficients of not sines but cosines and similarly, $a(\omega)$ the coefficients of the sines, then how will the waveform transform$[x(t)]$ look like?

  • What will be the similarity between this transformed waveform and the original waveform?

  • What is the name for this transformation (is it Hilbert transformation? )

Please attach graphs for an example $x(t)$ to clarify.

Answer 1

If you have a $T$-periodic signal $x(t)$ with a Fourier series expansion

$$x(t)=a_0+\sum_{n=1}^{\infty}\left[a_n\cos(n\omega_0t)+b_n\sin(n\omega_0t)\right],\quad \omega_0=\frac{2\pi}{T}\tag{1}$$

then its Hilbert transform is given by

$$\mathcal{H}\{x(t)\}=\hat{x}(t)=\sum_{n=1}^{\infty}\left[a_n\sin(n\omega_0t)-b_n\cos(n\omega_0t)\right]\tag{2}$$

because $\mathcal{H}\{\cos(t)\}=\sin(t)$, $\mathcal{H}\{\sin(t)\}=-\cos(t)$, and $\mathcal{H}\{c\}=0$ for constant $c$.

This is not exactly the same as just swapping the coefficients $a_n$ and $b_n$. The signal with swapped coefficients (assuming $a_0=0$) is

$$y(t)=\sum_{n=1}^{\infty}\left[a_n\sin(n\omega_0t)+b_n\cos(n\omega_0t)\right]$$

which is related to $\hat{x}(t)$ by

$$y(t)=-\hat{x}(-t)\tag{3}$$

It is also equal to the Hilbert transform of $x(-t)$:

$$y(t)=\mathcal{H}\{x(-t)\}\tag{4}$$

So swapping the coefficients is equivalent to applying a linear time-varying system to the signal. It cannot be achieved by a Hilbert transform alone (which is a time-invariant operation).

Note that for even signals ($b_n=0$) and for odd signals ($a_n=0$) one does indeed obtain the Hilbert transform (for odd signals with a negative sign) by swapping coefficients. This is shown in the example in hops's answer.

Answer 2

I will base my answer on the following definitions until I hear otherwise from you. I will use $T$ as the fundamental period of the periodic signal $x(t)$. So, $\omega$ takes on discrete values at $\omega = \frac{2 \pi}{T} k$ where $k$ varies over all non-negative integers ($0, 1, 2, \cdots, \infty$). The underlying signal may then be decomposed into $$ x(t) = a_0 + \sum_{k=1}^{\infty} a_k \cos\left(\frac{2\pi}{T} k t\right) + b_k \sin\left(\frac{2\pi}{T} k t\right).$$ I'm changing notations to something more natural for me, $a_k = a\left(\frac{2\pi}{T}k\right)$ and similarly $b_k = b\left(\frac{2\pi}{T}k\right)$.

Example 1: Let's consider a simple example. Perhaps the simplest non-trivial example (definitions of trivial may vary). Take $$x(t) = \left\{\begin{array}{cc}A&x \geq 0\\ -A & x < 0\end{array}\right.$$ The Fourier Series coefficients for this function (given as a sum of sines and cosines) with a fundamental period $T$ is given as $$a_k = \frac{1}{T} \int_{-T/2}^{T/2} x(t) \cos \left( \frac{2\pi}{T} k t\right)dt$$ and $$b_k = \frac{1}{T} \int_{-T/2}^{T/2} x(t) \sin \left( \frac{2\pi}{T} k t\right)dt.$$ All of the $a_k = 0$ since the integrand is odd (an even function times an odd function is odd). So, we only need to consider the $b_k$. Cranking through the calculus, we find that $$ b_k = \left\{\begin{array}{cc}0 & \mbox{$k$ is even}\\ \frac{4 A}{\pi k} & \mbox{$k$ is odd}\end{array}\right.$$

The MATLAB code below can be used to construct an approximation of both signals (using $a_k$ and $b_k$ conventionally and using them swapped).

A = 1;
T = 1;
N = 1000;

t = (-0.5*T:0.0001*T:0.5*T).';
x = zeros(size(t));
y = zeros(size(t));

for k = 1:2:N
    x = x + 4 * A / (pi*k) * cos(2*pi*k/T*t);
    y = y + 4 * A / (pi*k) * sin(2*pi*k/T*t);
end

figure()
plot(t, x)
xlabel('time')
ylabel('signal')
title('Using a_k and b_k')

figure()
plot(t, y)
xlabel('time')
ylabel('signal')
title('Swapping a_k and b_k')

The images below show that indeed, for this particular example, swapping the $a_k$ and $b_k$ yields (if not exactly then a scalar multiple of) the Hilbert Transform. It doesn't seem like a very practical way to obtain it though, and I haven't proven that it works in general for all periodic signals.

Answer 3

Let us just run a quick simulation and try ourselves. Assume we have a signal $s[n] = A \cos{(2 \pi \frac{f}{Fs} n )} $

A = 1000;
x = 0:1000;
f = 2;
Fs = 50;
s = A*cos(2*pi*f/Fs.*x);
F = fit(x(:),s(:),'fourier2');

I am using fourier series fitting upto 2 coefficients i.e., $(a_0, a_1, a_2)$ for the cosine term and $(b_1, b_2)$ for the sine term. $F$ will have a general model:

$$F(x) = a_0 + a_1 \cos{(\omega x)} + b_1 \sin{(\omega x)} + a_2 \cos{(2\omega x)} + b_2 \sin{(2 \omega x)}$$

Let us now flip the coefficients and see what happens to the phase of the new signal.

G = @(x) F.b1*cos(x.*F.w) + F.a1*sin(x.*F.w) + F.b2*cos(2.*x.*F.w) + F.a2*sin(2.*x.*F.w)
plot(x,s,x,G(x)); % To see the difference visually

h1 = hilbert(s);
h2 = hilbert(G(x));

phase_offset = mean(angle(h1./h2));

You will see the phase offset is about $1.57$ i.e. $\pi/2$. Basically switching the coefficients changes the phase of the signal.

You can also see this mathematically:

$$ s(t) = a_0 + \sum\limits_{n=1}^{\infty} a_n \cos{(n\omega_0 t)} + b_n \sin{(n\omega_0 t)} $$

Now if you were to reverse the signal in time and change the phase of these sine and cosine terms by $\pi/2$, i.e.,

$$ s^{'}(t) = a_0 + \sum\limits_{n=1}^{\infty} a_n \cos{(\pi/2 - n\omega_0 t)} + b_n \sin{(\pi/2 - n\omega_0 t)} $$

which is nothing but

$$ s^{'}(t) = a_0 + \sum\limits_{n=1}^{\infty} b_n \cos{(n\omega_0 t)} + a_n \sin{(n\omega_0 t)}$$