The Exponential Fourier Series

Posted on 2024-05-02 Edited on 2024-09-17 In Electrical Engineering Views:

Sources:

B. P. Lathi & Roger Green. (2021). Chapter 3: Signal Representation by Fourier Series. Signal Processing and Linear Systems (2nd ed., pp. 286-301). Oxford University Press.

Exponential Fourier Series

For a signal \(x(t)\) whose Fourier series exists, its trigonometric Fourier series is \[ \color{red} {x(t)= {a_0}+\sum_{n=1}^{\infty} a_n \cos n \omega_0 t+ \sum_{n=1}^{\infty} b_n \sin n \omega_0 t}, \quad t_1<t<t_1+T_0 . \] We know each sinusoid of frequency \(n \omega_0\) can be expressed as a sum of two exponentials \(e^{j n \omega_0 t}\) and \(e^{-j n \omega_0 t}\). This is because \[ \begin{aligned} & \cos \theta=\frac{e^{i \theta}+e^{-i \theta}}{2} \\ & \sin \theta=\frac{e^{i \theta}-e^{-i \theta}}{2 i} \end{aligned} \] Meanwhile, we can show that (see my post) the set of exponentials \(e^{j n \omega_0 t}(n=0, \pm 1, \pm 2, \ldots)\) is complete over any interval of duration \(T_0=2 \pi / \omega_0\).

Thus, a signal \(x(t)\) whose Fourier series exists can be expressed by exponential Fourier Series: \[ \color{pink}{x(t)=\sum_{n=-\infty}^{\infty} D_n e^{j n \omega_0 t} } . \] where \[ \begin{equation} \label{eq_6_19_2} \color{blue} {D_n=\frac{1}{T_0} \int_{T_0} x(t) e^{-j n \omega_0 t} d t} . \end{equation} \]

Note that \(n\) is an integer and \(-\infty < n < \infty\).

Derivation of the coefficients \(D_n\)

To derive the coefficients \(D_n\), we multiply both sides of this equation by \(e^{-j m \omega_0 t}\) ( \(m\) integer) and integrate over one period. This yields \[ \int_{T_0} x(t) e^{-j m \omega_0 t} d t=\sum_{n=-\infty}^{\infty} D_n \int_{T_0} e^{j(n-m) \omega_0 t} d t \]

To simplify this expression, we use the orthogonality property of exponentials, which states that \[ \int_{T_0} e^{j n \omega_0 t} e^{-j m \omega_0 t} d t= \begin{cases}0 & m \neq n \\ T_0 & m=n\end{cases} \]

Thus, \[ \int_{T_0} x(t) e^{-j m \omega_0 t} d t= D_m\int_{T_0} e^{0} d t = D_m T_0 \] from which we obtain \[ D_m=\frac{1}{T_0} \int_{T_0} x(t) e^{-j m \omega_0 t} d t . \]

We can now relate \(D_n\) to trigonometric series coefficients \(a_n\) and \(b_n\). Setting \(n=0\) in \(\eqref{eq_6_19_2}\), we obtain \[ D_0= \frac{1}{T_0} \int_{T_0} x(t) d t = a_0 \]

Moreover, for \(n \neq 0\), \[ D_n= \frac{1}{T_0} \int_{T_0} x(t) e^{-j n \omega_0 t} d t = \frac{1}{T_0} \int_{T_0} x(t) \cos n \omega_0 t d t-\frac{j}{T_0} \int_{T_0} x(t) \sin n \omega_0 t d t=\frac{1}{2}\left(a_n-j b_n\right) \] and \[ D_{-n}=\frac{1}{T_0} \int_{T_0} x(t) \cos n \omega_0 t d t+\frac{j}{T_0} \int_{T_0} x(t) \sin n \omega_0 t d t=\frac{1}{2}\left(a_n+j b_n\right) \]

These results are valid for general \(x(t)\), real or complex.

Real signal case

When \(x(t)\) is real, \(a_n\) and \(b_n\) are real, and \(D_n=\frac{1}{2}\left(a_n-j b_n\right)\) and \(D_{-n}=\frac{1}{2}\left(a_n+j b_n\right)\) are conjugates: \[ D_{-n}=D_n^* , \]

since any complex number \(z = a + bj\) can be converted to polar form¹, we have \[ a_n-j b_n=\sqrt{a_n^2+b_n^2} e^{j \arctan \left(\frac{-b_n}{a_n}\right)} , \]

let \(C_n = \sqrt{a_n^2+b_n^2}\), \(\theta_n = \arctan \left(\frac{-b_n}{a_n}\right)\), we have \[ a_n-j b_n=\sqrt{a_n^2+b_n^2} e^{j \arctan \left(\frac{-b_n}{a_n}\right)}=C_n e^{j \theta_n} \] where \(n \neq 0\).

Recall that for \(n \neq 0\), \(D_n=\frac{1}{2}\left(a_n-j b_n\right)\) and \(D_{-n}=\frac{1}{2}\left(a_n+j b_n\right)\), hence we obtain \[ D_n=\frac{1}{2} C_n e^{j \theta_n} \quad D_{-n}=\frac{1}{2} C_n e^{-j \theta_n} . \]

Meanwhile, we also have \[ D_0=a_0=C_0 . \]

Therefore, for \(n \neq 0\), \[ \left|D_n\right|=\left|D_{-n}\right|=\frac{1}{2} C_n, \quad \angle D_n=\theta_n, \quad \text { and } \angle D_{-n}=-\theta_n \]

Note that \(\left|D_n\right|\) are the amplitudes and \(\angle D_n\) are the angles of various exponential components. From this equation, it follows that

When \(x(t)\) is real, the amplitude spectrum \(\left(\left|D_n\right|\right.\) versus \(\omega\) ) is an even function of \(\omega\) and the angle spectrum ( \(\angle D_n\) versus \(\omega\) ) is an odd function of \(\omega\).
For complex \(x(t)\), \(D_n\) and \(D_{-n}\) are generally not conjugates.
\(\left|D_n\right|\) is an even function of \(n\) and \(\angle D_n\) is an odd function of \(n\).

What is a negative frequency?

The existence of the spectrum at negative frequencies is somewhat disturbing because, by definition, the frequency (number of repetitions per second) is a positive quantity. How do we interpret a negative frequency? We can use a trigonometric identity to express a sinusoid of a negative frequency \(-\omega_0\) as \[ \cos \left(-\omega_0 t+\theta\right)=\cos \left(\omega_0 t-\theta\right) \]

This equation clearly shows that the frequency of a sinusoid \(\cos \left(\omega_0 t+\theta\right)\) is \(\left|\omega_0\right|\), which is a positive quantity. The same conclusion is reached by observing that \[ e^{ \pm j \omega_0 t}=\cos \omega_0 t \pm j \sin \omega_0 t \]

Thus, the frequency of exponentials \(e^{ \pm j \omega_0 t}\) is indeed \(\left|\omega_0\right|\). How do we then interpret the spectral plots for negative values of \(\omega\)? A more satisfying way of looking at the situation is to say that exponential spectra are a graphical representation of coefficients \(D_n\) as a function of \(\omega\). Existence of the spectrum at \(\omega=-n \omega_0\) is merely an indication that an exponential component \(e^{-j n \omega_0 t}\) exists in the series. We know that a sinusoid of frequency \(n \omega_0\) can be expressed in terms of a pair of exponentials \(e^{j n \omega_0 t}\) and \(e^{-j n \omega_0 t}\).

Example: Relating Exponential to Trigonometric Fourier Series Spectra

The trigonometric Fourier spectra of a certain periodic signal \(x(t)\) are shown in Fig. 6.14a. After inspecting these spectra, sketch the corresponding exponential Fourier spectra and verify your results analytically.

Solution

The trigonometric spectral components exist at frequencies 0, 3, 6, and 9. The exponential spectral components exist at 0, 3, 6, 9, and -3, -6, -9.

Consider first the amplitude spectrum. The dc component remains unchanged: that is, \(D_0=C_0=16\).

Now \(\left|D_n\right|\) is an even function of \(\omega\) and \(\left|D_n\right|=\left|D_{-n}\right|=C_n / 2\). Thus, all the remaining spectrum \(\left|D_n\right|\) for positive \(n\) is half the trigonometric amplitude spectrum \(C_n\), and the spectrum \(\left|D_n\right|\) for negative \(n\) is a reflection about the vertical axis of the spectrum for positive \(n\), as shown in Fig. 6.14b.

The angle spectrum is \(\angle D_n=\theta_n\) for positive \(n\) and is \(-\theta_n\) for negative \(n\), as depicted in Fig. 6.14b.

Verify

We shall now verify that both sets of spectra represent the same signal.

For the trigonometric spectra shown in Fig. 6.14a, signal \(x(t)\) has four spectral components of frequencies 0, 3, 6, and 9. The dc component is 16. The amplitude and the phase of the component of frequency 3 are 12 and \(-\pi / 4\), respectively. Therefore, this component can be expressed as \(12 \cos (3 t-\pi / 4)\). Proceeding in this manner, we can write the Fourier series for \(x(t)\) as \[ x(t)=16+12 \cos \left(3 t-\frac{\pi}{4}\right)+8 \cos \left(6 t-\frac{\pi}{2}\right)+4 \cos \left(9 t-\frac{\pi}{4}\right) \]

For the exponential spectra shown in Fig. 6.14b, they contain components of frequencies 0 (dc), \(\pm 3, \pm 6\), and \(\pm 9\). The dc component is \(D_0=16\). The component \(e^{j 3 t}\) (frequency 3) has magnitude 6 and angle \(-\pi / 4\). Therefore, this component strength is \(6 e^{-j \pi / 4}\), and it can be expressed as \(\left(6 e^{-j \pi / 4}\right) e^{j 3 t}\). Similarly, the component of frequency -3 is \(\left(6 e^{j \pi / 4}\right) e^{-j 3 t}\). Proceeding in this manner, \(\hat{x}(t)\), the signal corresponding to the spectra in Fig. 6.14b, is \[ \begin{aligned} \hat{x}(t) & =16+\left[6 e^{-j \pi / 4} e^{j 3 t}+6 e^{j \pi / 4} e^{-j 3 t}\right]+\left[4 e^{-j \pi / 2} e^{j 6 t}+4 e^{j \pi / 2} e^{-j 6 t}\right] \\ & +\left[2 e^{-j \pi / 4} e^{j 9 t}+2 e^{j \pi / 4} e^{-j 9 t}\right] \\ & =16+6\left[e^{j(3 t-\pi / 4)}+e^{-j(3 t-\pi / 4)}\right]+4\left[e^{j(6 t-\pi / 2)}+e^{-j(6 t-\pi / 2)}\right] \\ & +2\left[e^{j(9 t-\pi / 4)}+e^{-j(9 t-\pi / 4)}\right] \\ & =16+12 \cos \left(3 t-\frac{\pi}{4}\right)+8 \cos \left(6 t-\frac{\pi}{2}\right)+4 \cos \left(9 t-\frac{\pi}{4}\right) \end{aligned} \]

Clearly both sets of spectra represent the same periodic signal.

BANDWIDTH OF A SIGNAL

The difference between the highest and the lowest frequencies of the spectral components of a signal is the bandwidth of the signal.

The bandwidth of the signal whose exponential spectra are shown in Fig. 6.14b is 9 (in radians).

The highest and lowest frequencies are 9 and 0, respectively. Note that the component of frequency 12 has zero amplitude and is nonexistent.

Moreover, the lowest frequency is 0 , not -9 . Recall that the frequencies (in the conventional sense) of the spectral components at \(\omega\) = -3, -6, and -9 in reality are 3,6 and 9. The bandwidth can be more readily seen from the trigonometric spectra in Fig. 6.14a.