Continuous Time Fourier Series Exponential Form Vs Plar Form

Signals, Systems, and Spectral Analysis

Ali Grami , in Introduction to Digital Communications, 2016

3.6.5 Complex Exponential Fourier Series

The quadrature and polar forms of the Fourier series are one-sided spectral components, meaning the spectrum can exist for DC and positive frequencies, but on the other hand, the complex exponential Fourier series has two-sided spectral components. The complex exponential Fourier series is a simple form, in which the orthogonal functions are the complex exponential functions. Using (3.17), (3.34a) can thus be transformed into the following:

(3.37a) $g\left(t\right)=\sum _{n=-\infty }^{\infty }{c}_{n}exp\left(\frac{j2\pi nt}{T_0}\right),-\frac{T_0}{2}\le t\le \frac{T_0}{2}$

where c _n is defined as follows:

(3.37b) $c_n=\left(\frac{1}{T_0}\right){\int }_{-T_0/2}^{T_0/2}g\left(t\right)exp\left(-\frac{j2\pi nt}{T_0}\right)dt,n=0,\pm 1,\pm 2,\dots$

The coefficient c _n is, in general, a complex number. It is important to note that the presence of negative frequencies and complex-valued basis functions has no physical meaning, and from a system analysis standpoint they just provide a compact mathematical expression. When g(t) is real, then $c_{-n}=c_n^{*}$ , where the asterisk * denotes the complex conjugate operation. If g(t) is real and even, c _n is then a real number (i.e., its imaginary part is zero), and if g(t) is real and odd, c _n is then a pure imaginary number (i.e., its real part is zero).

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Biosignal Processing

Monty Escabí PhD , in Introduction to Biomedical Engineering (Third Edition), 2012

11.5.3 Exponential Fourier Series

The main result from the Fourier series analysis is that an arbitrary periodic signal can approximate by summing individual cosine terms with specified amplitudes and phases. This result serves as much of the conceptual and theoretical framework for the field of signal analysis. In practice, the Fourier series is a useful tool for modeling various types of quasi-periodic signals.

An alternative and somewhat more convenient form of this result is obtained by noting that complex exponential functions are directly related to sinusoids and cosines through Euler's identities: $cos(\theta )=\left(e^{j\theta }+e^{-j\theta }\right)/2$ and $sin(\theta )=\left(e^{j\theta }-e^{-j\theta }\right)/2j$ , where $j=\sqrt{-1}$ . By applying Euler's identity to the compact trigonometric Fourier series, an arbitrary periodic signal can be expressed as a sum of complex exponential functions:

(11.5a) $x(t)=\sum _{m=-\infty }^{+\infty }{c}_m{e}^{\mathit{jk}{\omega }_{o}t}$

This equation represents the exponential Fourier series of a periodic signal. The coefficients c_m are complex numbers that are related to the trigonometric Fourier coefficients

(11.5b) $c_m=\frac{a_m-j{b}_m}{2}=\frac{A_m}{2}{e}^{j{\varphi }_m}$

The proof for this result is beyond the scope of this text, but it is important to realize that the trigonometric and exponential Fourier series are intimately related, as can be seen by comparing their coefficients. The exponential coefficients can also be obtained directly by integrating x(t),

(11.5c) $c_m=\frac{1}{\text{T}}\underset{\text{T}}{\int }x(t)e^{-\mathit{jm}{\omega }_{o}t}\mathit{dt}$

over one cycle of the periodic signal. As for the trigonometric Fourier series, the exponential form allows us to approximate a periodic signal to any degree of accuracy by adding a sufficient number of complex exponential functions. A distinct advantage of the exponential Fourier series, however, is that it requires only a single integral (Eq. (11.5c)), compared to the trigonometric form, which requires three separate integrations.

Example Problem 11.8

Find the exponential Fourier series for the square wave of Figure 11.7a and implement in MATLAB for the first ten terms. Plot the time waveform and the Fourier series coefficients.

Solution

Like Example Problem 11.6, the Fourier coefficients are obtained by integrating from −1 to 1. Because a single cycle of the square wave signal has nonzero values between −1/2 and +1/2, the integral can be simplified by evaluating it between these limits:

$c_m=\frac{1}{\text{T}}\underset{\text{T}}{\int }x(t)e^{-\mathit{jm}{\omega }_{o}t}\mathit{dt}=\frac{1}{2}{\int }_{-1/2}^{1/2}5e^{-\mathit{jm}\pi t}\mathit{dt}=\frac{5}{2}\cdot {\frac{e^{-\mathit{jm}\pi t}}{-\mathit{jm}\pi }|}_{-1/2}^{1/2}$

$=\frac{5}{2}\cdot \frac{e^{+\mathit{jm}\pi /2}-e^{-\mathit{jm}\pi /2}}{\mathit{jm}\pi }=\frac{5}{2}\cdot \frac{sin\left(m\pi /2\right)}{m\pi /2}.$

Therefore,

$x(t)=\sum _{m=-\infty }^{+\infty }{c}_m{e}^{\mathit{jk}{\omega }_{o}t}=\sum _{m=-\infty }^{\infty }\frac{5}{2}\cdot \frac{sin\left(m\pi /2\right)}{m\pi /2}\cdot e^{\mathit{jm}\pi t}$

MATLAB implementation:

%Plotting Fourier Series Approximation

subplot(211)

time=-2:0.01:2; %Time Axis

x=0; %Initialize Signal

for m=-10:10

  if m==0

  x=x+5/2; %Term for m=0

  else

  x=x+5/2*sin(m*pi/2)/m/pi*2*exp(j*m*pi*time);

  end

end

plot(time,x,'k') %Plotting and Labels

xlabel('Time (sec)')

ylabel('Amplitude')

set(gca,'Xtick',[-2:2])

set(gca,'Ytick',[0 5])

set(gca,'Box','off')

%Plotting Fourier Magnitudes

subplot(212)

m=(-10:10)+1E-10;

A=[5/2*sin(m*pi/2)./m/pi*2]; %Fourier Magnitudes

Faxis=(-10:10)*.5; %Frequency Axis

plot(Faxis,A,'k.') %Plotting

axis([-5 5 -2 4])

set(gca,'Box','off')

xlabel('Frequency (Hz)')

ylabel('Fourier Amplitudes')

Note that we now require positive and negative frequencies in the approximation. Results showing the MATLAB output are shown in Figure 11.9.

Figure 11.9. (a) MATLAB result showing the first ten terms of exponential Fourier series approximation for the square wave. (b) The compact Fourier coefficients are shown as a function of the harmonic frequency. Note that both negative and positive frequencies are now necessary to approximate the square wave signal.

In practice, many periodic or quasi-periodic biological signals can be accurately approximated with only a few harmonic components. Figures 11.10 and 11.11 illustrate a harmonic reconstruction of an aortic pressure waveform obtained by applying a Fourier series approximation. Figure 11.10 plots the coefficients for the cosine series representation as a function of the harmonic number. Note that the low-frequency coefficients are large in amplitude, whereas the high-frequency coefficients contain little energy and do not contribute substantially to the reconstruction. The amplitude coefficients, A _m , are plotted on a log₁₀ scale so the smaller values are magnified and are therefore visible. Figure 11.11 shows several levels of harmonic reconstruction. The mean plus the first and second harmonics provide the basis for the general systolic and diastolic shape, since the amplitudes of these harmonics are large and contribute substantially to the reconstructed waveform. Additional harmonics add fine details but do not contribute significantly to the raw waveform.

Figure 11.10. Harmonic coefficients of the aortic pressure waveform shown in Figure 11.2.

Figure 11.11. Harmonic reconstruction of the aortic pressure waveform shown in Figure 11.2.

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Frequency Analysis: The Fourier Series

Luis F. Chaparro , Aydin Akan , in Signals and Systems Using MATLAB (Third Edition), 2019

4.3.2 Trigonometric Fourier Series

In this section we develop an equivalent expression for the Fourier series using sinusoids. First we need to show that sinusoids of harmonic frequencies are orthonormal. Orthogonality of the complex exponential Fourier basis indicates that an equivalent basis can be obtained from cosine and sine functions. Choosing $t_0=-T_0/2$ in Equation (4.16) which displays the orthogonality of the exponentials, we have, for $k\ne \ell$ :

(4.22) $\begin{array}{ccc}\hfill 0& \hfill =\hfill & \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}{e}^{jk{\mathrm{\Omega }}_{0}t}{[e^{j\ell {\mathrm{\Omega }}_{0}t}]}^{⁎}dt\hfill \\ \hfill & \hfill =\hfill & \underset{0}{\underbrace{\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\cos ((k-\ell ){\mathrm{\Omega }}_{0}t)dt}}+j\underset{0}{\underbrace{\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\sin ((k-\ell ){\mathrm{\Omega }}_{0}t)dt}}\hfill \end{array}$

or that the real and imaginary parts are zero. Expanding $\cos ((k-\ell ){\mathrm{\Omega }}_{0}t)$ and $\sin ((k-\ell ){\mathrm{\Omega }}_{0}t)$ in the integrals on the right in Equation (4.22) we obtain

$\begin{array}{ccc}\hfill 0& \hfill =\hfill & \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\cos (k{\mathrm{\Omega }}_{0}t)\cos (\ell {\mathrm{\Omega }}_{0}t)dt+\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\sin (k{\mathrm{\Omega }}_{0}t)\sin (\ell {\mathrm{\Omega }}_{0}t)dt,\hfill \\ \hfill 0& \hfill =\hfill & \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\sin (k{\mathrm{\Omega }}_{0}t)\cos (\ell {\mathrm{\Omega }}_{0}t)dt-\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\cos (k{\mathrm{\Omega }}_{0}t)\sin (\ell {\mathrm{\Omega }}_{0}t)dt.\hfill \end{array}$

Using the trigonometric identities

$\sin (\alpha )\sin (\beta )=0.5[\cos (\alpha -\beta )-\cos (\alpha -\beta )],\cos (\alpha )\cos (\beta )=0.5[\cos (\alpha -\beta )+\cos (\alpha -\beta )],$

and the fact that the argument of the sinusoids is a nonzero integer times ${\mathrm{\Omega }}_0$ the top integrals are zero as the integrals are over one or more periods. Thus, $\cos (k\mathrm{\Omega }t)$ is orthogonal to $\cos (\ell \mathrm{\Omega }t)$ for any $k\ne \ell$ , and $\sin (k\mathrm{\Omega }t)$ is orthogonal to $\sin (\ell \mathrm{\Omega }t)$ for $k\ne \ell$ . Likewise, the second set of integrals are zero since the integrands are odd functions. Thus, $\cos (k\mathrm{\Omega }t)$ is orthogonal to $\sin (\ell \mathrm{\Omega }t)$ when $k\ne \ell$ . Thus for different frequencies the cosine and the sine functions are orthogonal.

To normalize the sinusoidal basis, we find that

$\begin{array}{ccc}\hfill \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}{\cos }^2(k{\mathrm{\Omega }}_{0}t)dt& \hfill =\hfill & \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}{\sin }^2(k{\mathrm{\Omega }}_{0}t)dt\hfill \\ \hfill & \hfill =\hfill & \frac{1}{T_0}[{\int }_{-T_0/2}^{T_0/2}0.5dt\pm {\int }_{-T_0/2}^{T_0/2}0.5\cos (2k{\mathrm{\Omega }}_{0}t)dt]=0.5\hfill \end{array}$

by using that ${\cos }^2(\theta )=0.5(1+\cos (2\theta ))$ and ${\sin }^2(\theta )=0.5(1-\cos (2\theta ))$ . Therefore, if we choose as basis $\{\sqrt{2}\cos (k{\mathrm{\Omega }}_{0}t),\sqrt{2}\sin (k{\mathrm{\Omega }}_{0}t)\}$ we obtain the trigonometric Fourier series, equivalent to the exponential Fourier series presented before.

The trigonometric Fourier Series of a real-valued, periodic signal $x(t)$ , of fundamental period $T_0$ , is an equivalent representation that uses sinusoids rather than complex exponentials as the basis functions. It is given by

(4.23) $\begin{array}{ccc}\hfill x(t)& \hfill =\hfill & X_0+2\sum _{k=1}^{\infty }|X_k|\cos (k{\mathrm{\Omega }}_{0}t+{\theta }_k)\hfill \\ \hfill & \hfill =\hfill & c_0+2\sum _{k=1}^{\infty }[c_k\cos (k{\mathrm{\Omega }}_{0}t)+d_k\sin (k{\mathrm{\Omega }}_{0}t)]{\mathrm{\Omega }}_0=\frac{2\pi }{T_0}\hfill \end{array}$

where $X_0=c_0$ is called the dc component, and $\{2|X_k|\cos (k{\mathrm{\Omega }}_{0}t+{\theta }_k)\}$ are the kth harmonics for $k=1,2\cdots$ . The coefficients $\{c_k,d_k\}$ are obtained from $x(t)$ as follows:

(4.24) $\begin{array}{ccc}\hfill c_k& \hfill =\hfill & \frac{1}{T_0}{\int }_{t_0}^{t_0+T_0}x(t)\cos (k{\mathrm{\Omega }}_{0}t)dtk=0,1,\cdots \hfill \\ \hfill d_k& \hfill =\hfill & \frac{1}{T_0}{\int }_{t_0}^{t_0+T_0}x(t)\sin (k{\mathrm{\Omega }}_{0}t)dtk=1,2,\cdots .\hfill \end{array}$

The coefficients $X_k=|X_k|e^{j{\theta }_k}$ are connected with the coefficients $c_k$ and $d_k$ by

$\begin{array}{ccc}\hfill & \hfill \hfill & |X_k|=\sqrt{c_k^2+d_k^2}\hfill \\ \hfill & \hfill \hfill & {\theta }_k=-{\tan }^{-1}\left[\frac{d_k}{c_k}\right].\hfill \end{array}$

The sinusoidal basis functions $\{\sqrt{2}\cos (k{\mathrm{\Omega }}_{0}t),\sqrt{2}\sin (k{\mathrm{\Omega }}_{0}t)\}$ , $k=0,\pm 1,\cdots$ , are orthonormal in $[0,T_0]$ .

Using the relation $X_k=X_{-k}^{⁎}$ , obtained in the previous section, we express the exponential Fourier series of a real-valued periodic signal $x(t)$ as

$\begin{array}{ccc}\hfill x(t)& \hfill =\hfill & X_0+\sum _{k=1}^{\infty }[X_k{e}^{jk{\mathrm{\Omega }}_{0}t}+X_{-k}{e}^{-jk{\mathrm{\Omega }}_{0}t}]=X_0+\sum _{k=1}^{\infty }[|X_k|e^{j(k{\mathrm{\Omega }}_{0}t+{\theta }_k)}+|X_k|e^{-j(k{\mathrm{\Omega }}_{0}t+{\theta }_k)}]\hfill \\ \hfill & \hfill =\hfill & X_0+2\sum _{k=1}^{\infty }|X_k|\cos (k{\mathrm{\Omega }}_{0}t+{\theta }_k),\hfill \end{array}$

which is the top equation in (4.23).

Let us then show how the coefficients $c_k$ and $d_k$ can be obtained directly from the signal. Using the relation $X_k=X_{-k}^{⁎}$ and the fact that for a complex number $z=a+jb$ , then $z+z^{⁎}=(a+jb)+(a-jb)=2a=2\mathcal{R}e(z)$ we have

$\begin{array}{ccc}\hfill x(t)& \hfill =\hfill & X_0+\sum _{k=1}^{\infty }[X_k{e}^{jk{\mathrm{\Omega }}_{0}t}+X_{-k}{e}^{-jk{\mathrm{\Omega }}_{0}t}]=X_0+\sum _{k=1}^{\infty }[X_k{e}^{jk{\mathrm{\Omega }}_{0}t}+X_k^{⁎}{e}^{-jk{\mathrm{\Omega }}_{0}t}]\hfill \\ \hfill & \hfill =\hfill & X_0+\sum _{k=1}^{\infty }2\mathcal{R}e[X_k{e}^{jk{\mathrm{\Omega }}_{0}t}].\hfill \end{array}$

Since $X_k$ is complex,

$\begin{array}{ccc}\hfill 2\mathcal{R}e[X_k{e}^{jk{\mathrm{\Omega }}_{0}t}]& \hfill =\hfill & 2\mathcal{R}e[(\mathcal{R}e[X_k]+j\mathcal{I}m[X_k])(\cos (k{\mathrm{\Omega }}_{0}t)+j\sin (k{\mathrm{\Omega }}_{0}t))]\hfill \\ \hfill & \hfill =\hfill & 2\mathcal{R}e[X_k]\cos (k{\mathrm{\Omega }}_{0}t)-2\mathcal{I}m[X_k]\sin (k{\mathrm{\Omega }}_{0}t).\hfill \end{array}$

Now, if we let

$\begin{array}{ccc}\hfill c_k& \hfill =\hfill & \mathcal{R}e[X_k]=\frac{1}{T_0}{\int }_{t_0}^{t_0+T_0}x(t)\cos (k{\mathrm{\Omega }}_{0}t)dtk=1,2,\cdots ,\hfill \\ \hfill d_k& \hfill =\hfill & -\mathcal{I}m[X_k]=\frac{1}{T_0}{\int }_{t_0}^{t_0+T_0}x(t)\sin (k{\mathrm{\Omega }}_{0}t)dtk=1,2,\cdots ,\hfill \end{array}$

we then have

$\begin{array}{ccc}\hfill x(t)& \hfill =\hfill & X_0+\sum _{k=1}^{\infty }(2\mathcal{R}e[X_k]\cos (k{\mathrm{\Omega }}_{0}t)-2\mathcal{I}m[X_k]\sin (k{\mathrm{\Omega }}_{0}t))\hfill \\ \hfill & \hfill =\hfill & X_0+2\sum _{k=1}^{\infty }(c_k\cos (k{\mathrm{\Omega }}_{0}t)+d_k\sin (k{\mathrm{\Omega }}_{0}t)),\hfill \end{array}$

and since the average $X_0=c_0$ we obtain the second form of the trigonometric Fourier series. Notice that $d_0=0$ and so it is not necessary to define it.

The coefficients $X_k=|X_k|e^{j{\theta }_k}$ are connected with the coefficients $c_k$ and $d_k$ by

$|X_k|=\sqrt{c_k^2+d_k^2},{\theta }_k=-{\tan }^{-1}\left[\frac{d_k}{c_k}\right].$

This can be shown by adding the phasors corresponding to $c_k\cos (k{\mathrm{\Omega }}_{0}t)$ and $d_k\sin (k{\mathrm{\Omega }}_{0}t)$ and finding the magnitude and phase of the resulting phasor.

Example 4.4

Find the exponential Fourier series of a raised cosine signal ( $B\ge A$ ),

$x(t)=B+A\cos ({\mathrm{\Omega }}_{0}t+\theta ),$

which is periodic of fundamental period $T_0$ and fundamental frequency ${\mathrm{\Omega }}_0=2\pi /T_0$ . Let $y(t)=B+A\sin ({\mathrm{\Omega }}_{0}t)$ , find its Fourier series coefficients and compare them to those for $x(t)$ . Use symbolic MATLAB to compute the Fourier series of $y(t)=1+\sin (100t)$ , find and plot its magnitude and phase line spectra.

Solution: We do not need to compute the Fourier coefficients in this case since $x(t)$ is in the Fourier series trigonometric form. According to Equation (4.23) the dc value of $x(t)$ is B, and A is the coefficient of the first harmonic in the trigonometric Fourier series. Thus $X_0=B$ and $|X_1|=A/2$ and $\angle X_1=\theta$ . Likewise, using Euler's identity we obtain

$x(t)=B+\frac{A}{2}[e^{j({\mathrm{\Omega }}_{0}t+\theta )}+e^{-j({\mathrm{\Omega }}_{0}t+\theta )}]=B+\frac{A{e}^{j\theta }}{2}{e}^{j{\mathrm{\Omega }}_{0}t}+\frac{A{e}^{-j\theta }}{2}{e}^{-j{\mathrm{\Omega }}_{0}t},$

which gives

$X_0=B,X_1=\frac{A{e}^{j\theta }}{2},X_{-1}=X_1^{⁎}=\frac{A{e}^{-j\theta }}{2}.$

If we let $\theta =-\pi /2$ in $x(t)$ we get

$y(t)=B+A\sin ({\mathrm{\Omega }}_{0}t)$

and from the above results its Fourier series coefficients are

$\begin{array}{ccc}\hfill & \hfill \hfill & Y_0=B,Y_1=\frac{A}{2}{e}^{-j\pi /2}=Y_{-1}^{⁎}\text{so that}\hfill \\ \hfill & \hfill \hfill & |Y_1|=|Y_{-1}|=\frac{A}{2}\text{and}\angle Y_1=-\angle Y_{-1}=-\frac{\pi }{2}.\hfill \end{array}$

The line spectrum of the raised cosine ( $\theta =0$ ) and of the raised sine ( $\theta =-\pi /2$ ) are shown in Fig. 4.2. For both $x(t)$ and $y(t)$ there are only two frequencies, the dc frequency and ${\mathrm{\Omega }}_0$ , and as such the power of the signal is concentrated at those two frequencies. The difference between the line spectrum of the raised cosine and sine is in the phase.

Figure 4.2. Line spectrum of raised cosine (left) and of raised sine (right).

We find the Fourier series coefficients with our symbolic MATLAB function fourierseries. The corresponding magnitude and phase are then plotted using stem to obtain the line spectra. The magnitude and phase line spectra corresponding to $y(t)=1+\sin (100t)$ are shown in Fig. 4.3. □

Figure 4.3. Line spectrum of Fourier series of $y(t)=1+\sin (100t)$ . Notice the even and the odd symmetries of the magnitude and phase spectra. The phase is −π/2 at Ω = 100 (rad/s).

Remark

Just because a signal is a sum of sinusoids, each of which is periodic, is not enough for it to have a Fourier series. The whole signal should be periodic. The signal $x(t)=\cos (t)-\sin (\pi t)$ has components with fundamental periods $T_1=2\pi$ and $T_2=2$ so that the ratio $T_1/T_2=\pi$ is not a rational number. Thus $x(t)$ is not periodic and no Fourier series for $x(t)$ is possible.

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BIOSIGNAL PROCESSING

Monty A. Escabí PhD , in Introduction to Biomedical Engineering (Second Edition), 2005

EXERCISES

1.

What types of biosignals would the nerves in your legs produce during a sprint across the street?

2.

What types of biosignals can be recorded with an EEG? Describe in terms of both origins and characteristics of the signal.

3.

Describe the biosignal that the electrical activity of a normal heart would generate during a bicycle race.

4.

A 16-bit A/D converter is used to convert an analog biosignal with a minimum voltage of −30 mV and a maximum voltage of 90 mV. What is the sensitivity?

5.

An EMG recording of skeletal muscle activity has been sampled at 200–250 Hz and correctly digitized. What is the highest frequency of interest in the original EMG signal?

6.

Two signals, x ₁(t) and x ₂(t), have the magnitude spectrum shown in Figure 10.36. Find the Nyquist rate for:

a)

x ₁(t)

b)

x ₂(t)

c)

x(t) = x₁(t)* x ₂(t) (Hint: apply the convolution theorem)

Figure 10.36.

7.

Consider the signal

$x(t)=3+\text{sin}(2\pi 100t)+\text{cos}(2\pi 250t+\pi /3)$

Find the Nyquist frequency.

8.

A sinusoid with the frequency of 125 kHz is sampled at 70,000 samples per second. What is the apparent frequency of the sampled signal?

9.

An electroencephalographic (EEG) signal has a maximum frequency of 300 Hz. The signal is sampled and quantized into a binary sequence by an A/D converter.

a)

Determine the sampling rate if the signal is sampled at a rate 50% higher than the Nyquist rate.

b)

The samples are quantized into 2048 levels. How many binary bits are required for each sample

10.

Find the exponential Fourier series for the signal shown in Fig. 10.37a.

Figure 10.37.

11.

Find the exponential Fourier series for the signal shown in Fig. 10.37b.

12.

f(t) is a periodic signal shown in Figure 10.38. Find its trigonometric Fourier series.

Figure 10.38.

13.

Consider the following trigonometric Fourier series:

$f(t)=3+\text{cos}(t)+2\text{cos}(2t)+4\text{sin}(2t)-4(\frac{e^{j4t}+e^{-j4t}}{2})$

Write f(t) in its compact trigonometric Fourier series form.

14.

Explain why the exponential Fourier series requires negative frequencies.

15.

Find the Fourier transform of

a)

u(t)

b)

e ^−at u(t)

c)

cos(at)u(t)

16.

Find the Fourier transform of f ₁(t) = e ^−3t u(t).

17.

Find the Fourier transform of f(t) = e^−3|t| and sketch its time and frequency domain representations. Hint: Find a few points on the curve by substituting values for the variable.

18.

Prove the shift property of the Fourier transform.

19.

Given x(t) = e ^−at u(t) and h(t) = e ^−bt u(t) where a and b are constants greater than zero, explain why it would be easier to evaluate the convolution x(t)*h(t) in the frequency domain.

20.

A brief current pulse of duration 50 ms and amplitude 1 mA is presented to a cell membrane with time constant 10 ms. Find the cell membrane voltage output.

21.

The ion exchange process of a cell is estimated to have the following impulse response: h(t) = e ^−4t u(t).

a)

Explain what type of general information would be available to the researcher if this estimation of h(t) were accurate.

b)

If sodium ions are injected into the system for two seconds in the form of a brief pulse approximated by the following equation, x(t) = 3u(t) — 3u(t — 2), how would the cell respond to (e.g., pump out ions) such input? Find the answer using time-domain procedures. Hint: Convolve the input and the impulse response.

22.

An ECG recording of the electrical activity of the heart during ventricular fibrillation is digitized and the signal begins with the following data sequence: −90.0, 10.0, −12.0, −63.0, 7.0, −22.0. The units of the data sequence are given in mV. What is the z transform of this data sequence of the biosignal?

23.

For the systems described by the following equations, determine which of the systems is linear and which is not.

a)

$\frac{dy}{dt}+2y(t)=f^2(t)$

b)

$\frac{dy}{dt}+3ty(t)=t^{2}f(t)$

c)

$\frac{dy}{dt}+2y(t)=f(t)\frac{df}{dt}$

d)

$y(t)={\int }_{-\infty }^{t}f(\tau )d\tau$

24.

Examine the characteristics of the digital filter

$y(k)=\frac{1}{4}x(k)+\frac{1}{4}x(k-1)+\frac{1}{2}y(k-1)$

Find the impulse response, H(z) and |H′(Ω). Use MATLAB to calculate and plot |H′(Ω)| for 0 < Ω < π Observe the difference between this filter and the filter in Example Problem 10.12. Why is this a better low-pass filter? What is the output if the input sequence is x(k)= 100 sin $(\frac{\pi }{2}k+\frac{\pi }{8})$ ? What is the output if the input sequence is x(k) = 100u(k)?

25.

Find the z transform of

a)

x(k) = u(k)

b)

x(k) = a^ku(k)

c)

x(k) = cos(b ·k)u(k)

26.

Find the z transform of the following:

a)

$x[k]={(\frac{1}{2})}^{k}u(k)$

b)

x[k] = (cos Ωk)u[k]

27.

Find the first four outputs of the following discrete system

$\begin{array}{l}y[k]-3y[k-1]+2y[k-2]=f[k-1]\hfill \\ \text{if}y[-1]=2,y[-2]=3,\text{and}f[k]=3^{k}u[k].\hfill \end{array}$

28.

Find the first four outputs of the following discrete system

$\begin{array}{l}y[k]-2y[k-1]+2y[k-2]=f[k-2]\hfill \\ \text{if}y[-1]=1,y[-2]=0,\text{and}f[k]=u[k].\hfill \end{array}$

29.

In MATLAB, design a routine to show that averaging random noise across many trials approaches zero as the number of trials increases.

30.

Accurate measurements of blood glucose levels are needed for the proper treatment of diabetes. Glucose is a primary carbohydrate, which circulates throughout the body and serves as an energy source for cells. In normal individuals the hormone insulin regulates the levels of glucose in the blood by promoting glucose transport out of the blood to skeletal muscle and fat tissues. Diabetics suffer from improper management of glucose levels, and the levels of glucose in the blood can become too high. Describe how fuzzy logic might be used in the control of a system for measuring blood glucose levels. What advantages would the fuzzy logic system have over a more conventional system?

31.

Describe three different biosignal processing applications for which artificial neural networks might be used. Give at least two advantages of artificial neural networks over traditional biosignal processing methods for the applications you listed.

32.

The fuzzy sets in Example Problem 10.26 have been calibrated so that a person with a resting heart rate of 95 beats per minute has a 75% degree of membership in the normal category and a 25% degree of membership in the tachycardia category. A resting heart rate of 65 beats per minute indicates a 95% degree of membership in the normal category. Draw a graph of the fuzzy sets.

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Frequency Analysis: The Fourier Series

Luis Chaparro , in Signals and Systems Using MATLAB (Second Edition), 2015

4.3.2 Trigonometric Fourier Series

In this section we develop an equivalent expression for the Fourier series using sinusoids. First we need to show that sinusoids of harmonic frequencies are orthonormal. Orthogonality of the complex exponential Fourier basis indicates that an equivalent basis can be obtained from cosine and sine functions. Choosing $t_0=-T_0/2$ in Equation (4.16) which displays the orthogonality of the exponentials, we have for $k\ne \ell$ :

(4.22) $\begin{array}{cc}\hfill 0=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}{e}^{\mathit{jk}{\mathrm{\Omega }}_{0}t}{[e^{j\ell {\mathrm{\Omega }}_{0}t}]}^{\ast }\mathit{dt}=& {\displaystyle \underset{0}{\underbrace{\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\cos ((k-\ell ){\mathrm{\Omega }}_{0}t)\mathit{dt}}}}\hfill \\ \hfill & +j{\displaystyle \underset{0}{\underbrace{\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\sin ((k-\ell ){\mathrm{\Omega }}_{0}t)\mathit{dt}}}}\hfill \end{array}$

or that the real and imaginary parts are zero. Expanding $\cos ((k-\ell ){\mathrm{\Omega }}_{0}t)$ and $\sin ((k-\ell ){\mathrm{\Omega }}_{0}t)$ in the integrals on the right in equation (4.22) we obtain

$\begin{array}{cc}\hfill 0=& \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\cos (k{\mathrm{\Omega }}_{0}t)\cos (\ell {\mathrm{\Omega }}_{0}t)\mathit{dt}+\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\sin (k{\mathrm{\Omega }}_{0}t)\sin (\ell {\mathrm{\Omega }}_{0}t)\mathit{dt}\hfill \\ \hfill 0=& \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\sin (k{\mathrm{\Omega }}_{0}t)\cos (\ell {\mathrm{\Omega }}_{0}t)\mathit{dt}-\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\cos (k{\mathrm{\Omega }}_{0}t)\sin (\ell {\mathrm{\Omega }}_{0}t)\mathit{dt}\hfill \end{array}$

Using the trigonometric identities

$\begin{array}{cc}\hfill & \sin (\alpha )\sin (\beta )=0.5[\cos (\alpha -\beta )-\cos (\alpha -\beta )]\text{,}\hfill \\ \hfill & \cos (\alpha )\cos (\beta )=0.5[\cos (\alpha -\beta )+\cos (\alpha -\beta )]\hfill \end{array}$

and that the argument of the sinusoids is a non-zero integer times ${\mathrm{\Omega }}_0$ the top integrals are zero as the integrals are over one or more periods. Thus, $\cos (k\mathrm{\Omega }t)$ is orthogonal to $\cos (\ell \mathrm{\Omega }t)$ for any $k\ne \ell$ , and $\sin (k\mathrm{\Omega }t)$ is orthogonal to $\sin (\ell \mathrm{\Omega }t)$ for $k\ne \ell$ . Likewise, the second set of integrals are zero since the integrands are odd functions. Thus, $\cos (k\mathrm{\Omega }t)$ is orthogonal to $\sin (\ell \mathrm{\Omega }t)$ when $k\ne \ell$ . Thus for different frequencies the cosine and the sine functions are orthogonal.

To normalize the sinusoidal basis, we find that

$\begin{array}{cc}\hfill \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}{\cos }^2(k{\mathrm{\Omega }}_{0}t)\mathit{dt}& =\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}{\sin }^2(k{\mathrm{\Omega }}_{0}t)\mathit{dt}\hfill \\ \hfill & =\frac{1}{T_0}\left[{\int }_{-T_0/2}^{T_0/2}0.5\mathit{dt}\pm {\int }_{-T_0/2}^{T_0/2}0.5\cos (2k{\mathrm{\Omega }}_{0}t)\mathit{dt}\right]=0.5\hfill \end{array}$

by using that ${\cos }^2(\theta )=0.5(1+\cos (2\theta ))$ and ${\sin }^2(\theta )=0.5(1-\cos (2\theta ))$ . Therefore, if we choose as basis $\{\sqrt{2}\cos (k{\mathrm{\Omega }}_{0}t)\text{,}\sqrt{2}\sin (k{\mathrm{\Omega }}_{0}t)\}$ we obtain the trigonometric Fourier series, equivalent to the exponential Fourier series presented before.

The trigonometric Fourier Series of a real-valued, periodic signal $x(t)$ , of fundamental period $T_0$ , is an equivalent representation that uses sinusoids rather than complex exponentials as the basis functions. It is given by

(4.23) $\begin{array}{cc}\hfill x(t)=& X_0+2{\displaystyle \sum _{k=1}^{\infty }}|X_k|\cos (k{\mathrm{\Omega }}_{0}t+{\theta }_k)\hfill \\ \hfill =& c_0+2{\displaystyle \sum _{k=1}^{\infty }}[c_k\cos (k{\mathrm{\Omega }}_{0}t)+d_k\sin (k{\mathrm{\Omega }}_{0}t)]{\mathrm{\Omega }}_0=\frac{2\pi }{T_0}\hfill \end{array}$

where $X_0=c_0$ is called the dc-component, and $\{2|X_k|\cos (k{\mathrm{\Omega }}_{0}t+{\theta }_k)\}$ are the $k$ th harmonics for $k=1\text{,}2\text{,}\dots$ . The coefficients $\{c_k\text{,}{d}_k\}$ are obtained from $x(t)$ as follows

(4.24) $\begin{array}{cc}\hfill c_k=& \frac{1}{T_0}{\int }_{t_0}^{t_0+T_0}x(t)\cos (k{\mathrm{\Omega }}_{0}t)\mathit{dt}k=0\text{,}1\text{,}\dots \hfill \\ \hfill d_k=& \frac{1}{T_0}{\int }_{t_0}^{t_0+T_0}x(t)\sin (k{\mathrm{\Omega }}_{0}t)\mathit{dt}k=1\text{,}2\text{,}\dots \hfill \end{array}$

The coefficients $X_k=|X_k|e^{j{\theta }_k}$ are connected with the coefficients $c_k$ and $d_k$ by

$\begin{array}{cc}\hfill & |X_k|=\sqrt{c_k^2+d_k^2}\hfill \\ \hfill & {\theta }_k=-{\tan }^{-1}\left[\frac{d_k}{c_k}\right]\hfill \end{array}$

The sinusoidal basis functions $\{\sqrt{2}\cos (k{\mathrm{\Omega }}_{0}t)\text{,}\sqrt{2}\sin (k{\mathrm{\Omega }}_{0}t)\}\text{,}k=0\text{,}\pm 1\text{,}\dots$ , are orthonormal in $[0\text{,}{T}_0]$ .

Using the relation $X_k=X_{-k}^{\ast }$ , obtained in the previous section, we express the exponential Fourier series of a real-valued periodic signal $x(t)$ as

$\begin{array}{cc}\hfill x(t)=& X_0+{\displaystyle \sum _{k=1}^{\infty }}[X_k{e}^{\mathit{jk}{\mathrm{\Omega }}_{0}t}+X_{-k}{e}^{-\mathit{jk}{\mathrm{\Omega }}_{0}t}]=X_0+{\displaystyle \sum _{k=1}^{\infty }}\left[|X_k|e^{j(k{\mathrm{\Omega }}_{0}t+{\theta }_k)}+|X_k|e^{-j(k{\mathrm{\Omega }}_{0}t+{\theta }_k)}\right]\hfill \\ \hfill =& X_0+2{\displaystyle \sum _{k=1}^{\infty }}|X_k|\cos (k{\mathrm{\Omega }}_{0}t+{\theta }_k)\hfill \end{array}$

which is the top equation in (4.23).

Let us then show how the coefficients $c_k$ and $d_k$ can be obtained directly from the signal. Using the relation $X_k=X_{-k}^{\ast }$ and the fact that for a complex number $z=a+\mathit{jb}$ , then $z+z^{\ast }=(a+\mathit{jb})+(a-\mathit{jb})=2a=2\mathcal{R}e(z)$ we have that

$\begin{array}{cc}\hfill x(t)=& X_0+{\displaystyle \sum _{k=1}^{\infty }}[X_k{e}^{\mathit{jk}{\mathrm{\Omega }}_{0}t}+X_{-k}{e}^{-\mathit{jk}{\mathrm{\Omega }}_{0}t}]=X_0+{\displaystyle \sum _{k=1}^{\infty }}[X_k{e}^{\mathit{jk}{\mathrm{\Omega }}_{0}t}+X_k^{\ast }{e}^{-\mathit{jk}{\mathrm{\Omega }}_{0}t}]\hfill \\ \hfill =& X_0+{\displaystyle \sum _{k=1}^{\infty }}2\mathcal{R}e[X_k{e}^{\mathit{jk}{\mathrm{\Omega }}_{0}t}]\hfill \end{array}$

Since $X_k$ is complex,

$\begin{array}{cc}\hfill 2\mathcal{R}e[X_k{e}^{\mathit{jk}{\mathrm{\Omega }}_{0}t}]=& 2\mathcal{R}e[(\mathcal{R}e[X_k]+j\mathcal{I}m[X_k])(\cos (k{\mathrm{\Omega }}_{0}t+j\sin (k{\mathrm{\Omega }}_{0}t))]\hfill \\ \hfill =& 2\mathcal{R}e[X_k]\cos (k{\mathrm{\Omega }}_{0}t)-2\mathcal{I}m[X_k]\sin (k{\mathrm{\Omega }}_{0}t)\hfill \end{array}$

Now, if we let

$\begin{array}{cc}\hfill c_k=& \mathcal{R}e[X_k]=\frac{1}{T_0}{\int }_{t_0}^{t_0+T_0}x(t)\cos (k{\mathrm{\Omega }}_{0}t)\mathit{dt}k=1\text{,}2\text{,}\cdots \hfill \\ \hfill d_k=& -\mathcal{I}m[X_k]=\frac{1}{T_0}{\int }_{t_0}^{t_0+T_0}x(t)\sin (k{\mathrm{\Omega }}_{0}t)\mathit{dt}k=1\text{,}2\text{,}\cdots \hfill \end{array}$

we then have

$\begin{array}{cc}\hfill x(t)=& X_0+{\displaystyle \sum _{k=1}^{\infty }}\left(2\mathcal{R}e[X_k]\cos (k{\mathrm{\Omega }}_{0}t)-2\mathcal{I}m[X_k]\sin (k{\mathrm{\Omega }}_{0}t)\right)\hfill \\ \hfill =& X_0+2{\displaystyle \sum _{k=1}^{\infty }}\left(c_k\cos (k{\mathrm{\Omega }}_{0}t)+d_k\sin (k{\mathrm{\Omega }}_{0}t)\right)\hfill \end{array}$

and since the average $X_0=c_0$ we obtain the second form of the trigonometric Fourier series. Notice that $d_0=0$ and so it is not necessary to define it.

The coefficients $X_k=|X_k|e^{j{\theta }_k}$ are connected with the coefficients $c_k$ and $d_k$ by

$|X_k|=\sqrt{c_k^2+d_k^2}\text{,}{\theta }_k=-{\tan }^{-1}\left[\frac{d_k}{c_k}\right]$

This can be shown by adding the phasors corresponding to $c_k\cos (k{\mathrm{\Omega }}_{0}t)$ and $d_k\sin (k{\mathrm{\Omega }}_{0}t)$ and finding the magnitude and phase of the resulting phasor.

Example 4.4

Find the exponential Fourier series of a raised cosine signal ( $B\ge A$ ),

$x(t)=B+A\cos ({\mathrm{\Omega }}_{0}t+\theta )$

which is periodic of fundamental period $T_0$ and fundamental frequency ${\mathrm{\Omega }}_0=2\pi /T_0$ . Let $y(t)=B+A\sin ({\mathrm{\Omega }}_{0}t)$ , find its Fourier series coefficients and compare them to those for $x(t)$ . Use symbolic MATLAB to compute the Fourier series of $y(t)=1+\sin (100t)$ , find and plot its magnitude and phase line spectra.

Solution

We do not need to compute the Fourier coefficients in this case since $x(t)$ is in the Fourier series trigonometric form. According to equation (4.23) the dc value of $x(t)$ is B, and A is the coefficient of the first harmonic in the trigonometric Fourier series. Thus $X_0=B$ and $|X_1|=A/2$ and $\angle X_1=\theta$ . Likewise, using Euler's identity we obtain that

$x(t)=B+\frac{A}{2}\left[e^{j({\mathrm{\Omega }}_{0}t+\theta )}+e^{-j({\mathrm{\Omega }}_{0}t+\theta )}\right]=B+\frac{{\mathit{Ae}}^{j\theta }}{2}{e}^{j{\mathrm{\Omega }}_{0}t}+\frac{{\mathit{Ae}}^{-j\theta }}{2}{e}^{-j{\mathrm{\Omega }}_{0}t}$

which gives

$X_0=B\text{,}{X}_1=\frac{{\mathit{Ae}}^{j\theta }}{2}\text{,}{X}_{-1}=X_1^{\ast }=\frac{{\mathit{Ae}}^{-j\theta }}{2}$

If we let $\theta =-\pi /2$ in $x(t)$ we get

$y(t)=B+A\sin ({\mathrm{\Omega }}_{0}t)$

and from the above results its Fourier series coefficients are

$\begin{array}{cc}\hfill & Y_0=B\text{,}{Y}_1=\frac{A}{2}{e}^{-j\pi /2}=Y_{-1}^{\ast }\text{so that}\hfill \\ \hfill & |Y_1|=|Y_{-1}|=\frac{A}{2}\text{and}\angle Y_1=-\angle Y_{-1}=-\frac{\pi }{2}\hfill \end{array}$

The line spectrum of the raised cosine ( $\theta =0$ ) and of the raised sine ( $\theta =-\pi /2$ ) are shown in Figure 4.2. For both $x(t)$ and $y(t)$ there are only two frequencies, the dc frequency and ${\mathrm{\Omega }}_0$ , and as such the power of the signal is concentrated at those two frequencies. The difference between the line spectrum of the raised cosine and sine is in the phase.

We find the Fourier series coefficients with our symbolic MATLAB function fourierseries. The corresponding magnitude and phase are then plotted using stem to obtain the line spectra. The magnitude and phase line spectra corresponding to $y(t)=1+\sin (100t)$ are shown in Figure 4.3  ▪

Figure 4.2. Line spectrum of raised cosine (left) and of raised sine (right).

Figure 4.3. Line spectrum of Fourier series of $y(t)=1+\sin (100t)$ . Notice the even and the odd symmetries of the magnitude and phase spectra. The phase is $-\pi /2$ at $\mathrm{\Omega }=100$   (rad/sec).

%%

% Example 4.4

%%

clear all; syms t

% signal

y=1+sin(100*t); T0=2*pi/100; N=5;    % N harmonics

figure(1)

subplot(211)

ezplot(y,[0,0.25]); grid; xlabel('t (sec)');ylabel('y(t)')

% harmonics and harmonic frequencies

[Y1, w1]=fourierseries(y,T0,N);

Y=[conj(fliplr(Y1(2:N))) Y1];w=[−fliplr(w1(2:N)) w1];

subplot(223)

stem(w,abs(Y)); grid; axis([−400 400 −0.1 1.1])

xlabel('k\Omega_0 (rad/sec)'); ylabel('|Y_k|')

subplot(224)

stem(w,angle(Y)); grid; axis([−400 400 −2 2])

xlabel('k\Omega_0 (rad/sec)'); ylabel('\angle{Y_k}')

function [X, w]=fourierseries(x,T0,N)

% function fourierseries

% Computes harmonics of the Fourier series of a continuous-time signal

% symbolically

% input: periodic signal x(t), its period (T0), number of harmonics (N)

% output: harmonics X and corresponding harmonic frequency w

% use: [X, w]=fourier(x,T0,N)

syms t

for k=1:N,

X1(k)=int(x*exp(−j*2*pi*(k−1)*t/T0),t,0,T0)/T0;

X(k)=subs(X1(k));

w(k)=(k−1)*2*pi/T0;

end

Remark

Just because a signal is a sum of sinusoids, each of which is periodic, is not enough for it to have a Fourier series. The whole signal should be periodic. The signal $x(t)=\cos (t)-\sin (\pi t)$ has components with fundamental periods $T_1=2\pi$ and $T_2=2$ so that the ratio $T_1/T_2=\pi$ is not a rational number. Thus $x(t)$ is not periodic and no Fourier series for $x(t)$ is possible.

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wilkersoncasim1999.blogspot.com

Source: https://www.sciencedirect.com/topics/engineering/exponential-fourier-series

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