Fourier Transform of Continuous Aperiodic Signals

Aperiodic Signal

Discrete Fourier Analysis

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

11.4.2 DFT of Aperiodic Discrete-Time Signals

We obtain the DFT of an aperiodic signal $y[n]$ by sampling in frequency its DTFT $Y(e^{j\omega })$ . Suppose we choose $\{{\omega }_k=2\pi k/L$ , $k=0,\cdots ,L-1\}$ as the sampling frequencies, where an appropriate value for the integer $L>0$ needs to be determined. Analogous to the sampling in time in Chapter 8, sampling in frequency generates a periodic signal in time

(11.50) $\tilde{y}[n]=\sum _{r=-\infty }^{\infty }y[n+rL].$

Now, if $y[n]$ is of finite length N, then when $L\ge N$ the periodic extension $\tilde{y}[n]$ clearly displays a first period equal to the given signal $y[n]$ (with some zeros attached at the end when $L>N$ ). On the other hand, if the length $L<N$ the first period of $\tilde{y}[n]$ does not coincide with $y[n]$ because of superposition of shifted versions of it—this corresponds to time aliasing, the dual of frequency-aliasing which occurs in time-sampling. Thus, for $y[n]$ of finite length N, and letting $L\ge N$ , we have

(11.51) $\tilde{y}[n]=\sum _{r=-\infty }^{\infty }y[n+rL]\iff Y[k]=Y(e^{j2\pi k/L})=\sum _{n=0}^{N-1}y[n]e^{-j2\pi nk/L}0\le k\le L-1.$

The equation on the right above is the DFT of $y[n]$ . The inverse DFT is the Fourier series representation of $\tilde{y}[n]$ (normalized with respect to L) or its first period

(11.52) $y[n]=\frac{1}{L}\sum _{k=0}^{L-1}Y[k]e^{j2\pi nk/L}0\le n\le L-1$

where $Y[k]=Y(e^{j2\pi k/L})$ .

In practice, the generation of the periodic extension $\tilde{y}[n]$ is not needed, we just need to generate a period that either coincides with $y[n]$ when $L=N$ , or that is $y[n]$ with a sequence of $L-N$ zeros attached to it (i.e., $y[n]$ is padded with zeros) when $L>N$ . To avoid time aliasing we do not consider choosing $L<N$ .

If the signal $y[n]$ is a very long signal, in particular if $N\to \infty$ , it does not make sense to compute its DFT, even if we could. Such a DFT would give the frequency content of the whole signal and since a large support signal could have all types of frequencies its DFT would just give no valuable information. A possible approach to obtain the frequency content of a signal with a large time support is to window it and compute the DFT of each of these segments. Thus when $y[n]$ is of infinite length, or its length is much larger than the desired or feasible length L, we use a window $W_L[n]$ of length L, and represent $y[n]$ as the superposition

(11.53) $y[n]=\sum _m{y}_m[n]\text{where}{y}_m[n]=y[n]W_L[n-mL],$

so that by the linearity of the DFT we have the DFT of $y[n]$ is

(11.54) $Y[k]=\sum _m\text{DFT}(y_m[n])=\sum _m{Y}_m[k]$

where each $Y_m[k]$ provides a frequency characterization of the windowed signal or the local frequency content of the signal. Practically, this would be more meaningful than finding the DFT of the whole signal. Now we have frequency information corresponding to segments of the signal and possibly evolving over time.

The DFT of an aperiodic signal $y[n]$ of finite length N, is found as follows:

•

Choose an integer $L\ge N$ , the length of the DFT, to be the fundamental period of a periodic extension $\tilde{y}[n]$ having $y[n]$ as a period with padded zeros if necessary.

•

Find the DFT of $\tilde{y}[n]$ ,

$\tilde{y}[n]=\frac{1}{L}\sum _{k=0}^{L-1}\tilde{Y}[k]e^{j2\pi nk/L}0\le n\le L-1$

and the inverse DFT

$\tilde{Y}[k]=\sum _{n=0}^{L-1}\tilde{y}[n]e^{-j2\pi nk/L}0\le k\le L-1.$

•

Then,

$\underset{\_}{\text{DFT of}y[n]:}$ $Y[k]=\tilde{Y}[k]$ for $0\le k\le L-1$ , and

$\underset{\_}{\text{Inverse DFT or IDFT of}Y[k]:}$ $y[n]=\tilde{y}[n]W[n]$ , $0\le n\le L-1$ , where $W[n]=u[n]-u[n-L]$ is a rectangular window of length L.

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Fourier Analysis of Discrete-Time Signals and Systems

Luis F. Chaparro , in Signals and Systems using MATLAB, 2011

10.4.2 DFT of Aperiodic Discrete-Time Signals

We obtain the DFT of an aperiodic signal y[n] by sampling its DTFT, Y(e ^jω ), in frequency. Suppose we choose $\mathfrak{\{}{\omega }_k=2\pi k∕L$ , $k=0,\dots ,L-1\mathfrak{\}}$ as the sampling frequencies, where an appropriate value for the integer L > 0 needs to be determined. Analogous to the sampling-in-time we did before, sampling-in-frequency generates a periodic signal in time:

(10.47) $\begin{array}{llll}ỹ\left[n\right]=\sum _{r=-\mathfrak{\infty }}^{\mathfrak{\infty }}y\left[n+rL\right]& & & \end{array}$

Now, if y[n] is of finite length N, then when L ≥ N the periodic expansion $ỹ\left[n\right]$ clearly displays a first period equal to the given signal y[n] (with some zeros attached at the end when L > N). On the other hand, if the length L < N the first period of $ỹ\left[n\right]$ does not coincide with y[n] because of superposition of shifted versions of it (this corresponds to time aliasing, the dual of frequency aliasing, which occurs in time sampling).

Assuming y[n] is of finite length N and that L ≥ N, as the dual of sampling in time we then have that

(10.48) $\begin{array}{llll}ỹ\left[n\right]=\sum _{r=-\mathfrak{\infty }}^{\mathfrak{\infty }}y\left[n+rL\right]\iff Y\left[k\right]=Y(e^{j2\pi k∕L})=\sum _{n=0}^{N-1}y\left[n\right]e^{-j2\pi nk∕L}k=0,\dots ,L-1& & & \end{array}$

The equation on the right is the DFT of y[n]. The inverse DFT is the Fourier series representation of $ỹ\left[n\right]$ (normalized with respect to L) or its first period

(10.49) $\begin{array}{llll}y\left[n\right]=\frac{1}{L}\sum _{k=0}^{L-1}Y\left[k\right]e^{j2\pi nk∕L}0\le n\le L-1& & & \end{array}$

where $Y\left[k\right]=Y(e^{j2\pi k∕L})$ .

Thus, instead of the frequency aliasing that sampling-in-time causes, we have time-aliasing whenever the length N of y[n] is greater than the chosen L in the sampling-in-frequency. In practice, the generation of the periodic extension $ỹ\left[n\right]$ is not needed—we just need to generate a period that either coincides with y[n] when L = N, or when L > N that coincides with y[n] with a sequence of L − N zeros attached to it (i.e., y[n] is padded with zeros). To avoid time aliasing we do not consider choosing L < N.

If the signal y[n] is a very long signal, in particular if N → ∞, it does not make sense to compute its DFT, even if we could. Such a DFT would give the frequency content of the whole signal and since an infinite-length signal could have all types of frequencies its DFT would just give no valuable information. A possible approach to obtain, over time, the frequency content of a signal with a large time support is to window it and compute the DFT of each of these segments. Thus, when y[n] is of infinite length, or its length is much larger than the desired or feasible length L, we use a window W _L [n] of length L, and represent y[n] as the superposition

(10.50) $\begin{array}{llll}y\left[n\right]=\sum _m{y}_m\left[n\right]\text{where}{y}_m\left[n\right]=y\left[n\right]W_L\left[n-mL\right]& & & \end{array}$

Therefore, by the linearity of the DFT, we have the DFT of y[n] is

(10.51) $\begin{array}{llll}Y\left[k\right]=\sum _m\text{DFT}(y_m\left[n\right])=\sum _m{Y}_m\left[k\right]& & & \end{array}$

where each Y _m [k] provides a frequency characterization of the windowed signal or the local frequency content of the signal. Practically, this would be more meaningful than finding the DFT of the whole signal. Now we have frequency information corresponding to segments of the signal and possibly evolving over time.

The DFT of an aperiodic signal x[n] of finite length N is found as follows:

■

Choose an integer L ≥ N that is the length of the DFT to be the period of a periodic extension $\tilde{x}\left[n\right]$ having x[n] as a period with padded zeros if necessary.

■

Find the normalized Fourier series representation of $\tilde{x}\left[n\right]$ ,

(10.52) $\begin{array}{llll}\tilde{x}\left[n\right]=\frac{1}{L}\sum _{k=0}^{L-1}\tilde{X}\left[k\right]e^{j2\pi nk∕L}0\le n\le L-1& & & \end{array}$

where

(10.53) $\begin{array}{llll}\tilde{X}\left[k\right]=\sum _{n=0}^{L-1}\tilde{x}\left[n\right]e^{-j2\pi nk∕L}0\le k\le L-1& & & \end{array}$

■

Then,

■

$X\left[k\right]=\tilde{X}\left[k\right]$ for 0 ≤ k ≤ L − 1 is the DFT of x[n].

■

$x\left[n\right]=\tilde{x}\left[n\right]W\left[n\right]$ where $W\left[n\right]=u\left[n\right]-u\left[n-L\right]$ is a rectangular window of length N, is the IDFT of X[k]. The IDFT x[n] is defined for 0 ≤ n ≤ L − 1.

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Frequency Analysis: The Fourier Transform

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

5.6.4 Symmetry of Spectral Representations

Now that the Fourier representation of aperiodic and periodic signals is unified, we can think of just one spectrum that accommodates both finite and infinite energy signals. The word "spectrum" is loosely used to mean different aspects of the frequency representation. The following provides definitions and the symmetry characteristic of the spectrum of real-valued signals.

If $X(\mathrm{\Omega })$ is the Fourier transform of a real-valued signal $x(t)$ , periodic or aperiodic, the magnitude $|X(\mathrm{\Omega })|$ and the real part $\mathcal{R}e[X(\mathrm{\Omega })]$ are even functions of Ω:

(5.19) $\begin{array}{ccc}\hfill & \hfill \hfill & |X(\mathrm{\Omega })|=|X(-\mathrm{\Omega })|,\hfill \\ \hfill & \hfill \hfill & \mathcal{R}e[X(\mathrm{\Omega })]=\mathcal{R}e[X(-\mathrm{\Omega })],\hfill \end{array}$

and the phase $\angle X(\mathrm{\Omega })$ and the imaginary part $\mathcal{I}m[X(\mathrm{\Omega })]$ are odd functions of Ω:

(5.20) $\begin{array}{ccc}\hfill & \hfill \hfill & \angle X(\mathrm{\Omega })=-\angle X(-\mathrm{\Omega }),\hfill \\ \hfill & \hfill \hfill & \mathcal{I}m[X(\mathrm{\Omega })]=-\mathcal{I}m[X(-\mathrm{\Omega })].\hfill \end{array}$

We then call the plots

$\begin{array}{ccc}\hfill & \hfill \hfill & |X(\mathrm{\Omega })|\text{ vs.}\mathrm{\Omega }\mathbf{magnitude\; spectrum}\hfill \\ \hfill & \hfill \hfill & \angle X(\mathrm{\Omega })\text{ vs.}\mathrm{\Omega }\mathbf{phase\; spectrum}\hfill \\ \hfill & \hfill \hfill & |X(\mathrm{\Omega }){|}^2\text{ vs.}\mathrm{\Omega }\mathbf{energy/power\; spectrum.}\hfill \end{array}$

To show this consider the inverse Fourier transform of a real-valued signal $x(t)$ ,

$x(t)={\int }_{-\infty }^{\infty }X(\mathrm{\Omega })e^{j\mathrm{\Omega }t}d\mathrm{\Omega },$

which is identical, because of being real, to

$x^{⁎}(t)={\int }_{-\infty }^{\infty }{X}^{⁎}(\mathrm{\Omega })e^{-j\mathrm{\Omega }t}d\mathrm{\Omega }={\int }_{-\infty }^{\infty }{X}^{⁎}(-{\mathrm{\Omega }}^{\prime })e^{j{\mathrm{\Omega }}^{\prime }t}d{\mathrm{\Omega }}^{\prime },$

since the integral can be thought of as an infinite sum of complex values and by letting ${\mathrm{\Omega }}^{\prime }=-\mathrm{\Omega }$ . Comparing the two integrals, we have the following identities:

$\begin{array}{ccc}\hfill & \hfill \hfill & X(\mathrm{\Omega })=X^{⁎}(-\mathrm{\Omega }),\hfill \\ \hfill & \hfill \hfill & |X(\mathrm{\Omega })|e^{j\angle X(\mathrm{\Omega })}=|X(-\mathrm{\Omega })|e^{-j\angle X(-\mathrm{\Omega })},\hfill \\ \hfill & \hfill \hfill & \mathcal{R}e[X(\mathrm{\Omega })]+j\mathcal{I}m[X(\mathrm{\Omega })]=\mathcal{R}e[X(-\mathrm{\Omega })]-j\mathcal{I}m[X(-\mathrm{\Omega })],\hfill \end{array}$

or that the magnitude is an even function of Ω and the phase an odd function of Ω. And that the real part of the Fourier transform is an even function and the imaginary part of the Fourier transform is an odd function of Ω.

Remarks

1.

Clearly if the signal is complex, the above symmetry will not hold. For instance, if $x(t)=e^{j{\mathrm{\Omega }}_{0}t}=\cos ({\mathrm{\Omega }}_{0}t)+j\sin ({\mathrm{\Omega }}_{0}t)$ , using the frequency shift property its Fourier transform is

$X(\mathrm{\Omega })=2\pi \delta (\mathrm{\Omega }-{\mathrm{\Omega }}_o),$

which occurs at $\mathrm{\Omega }={\mathrm{\Omega }}_0$ only, so the symmetry in the magnitude and phase does not exist.

2.

It is important to recognize the meaning of "negative" frequencies. In reality, only positive frequencies exist and can be measured, but as shown the spectrum, magnitude or phase, of a real-valued signal requires negative frequencies. It is only under this context that negative frequencies should be understood—as necessary to generate "real-valued" signals.

Example 5.12

Use MATLAB to compute the Fourier transform of the following signals:

$\begin{array}{cc}(a)x_1(t)=u(t)-u(t-1),\hfill & (b)x_2(t)=e^{-t}u(t).\hfill \end{array}$

Plot their magnitude and phase spectra.

Solution: Three possible ways to compute the Fourier transforms of these signals using MATLAB are: (i) find their Laplace transforms, as in Chapter 3, using the symbolic function laplace and compute the magnitude and phase functions by letting $s=j\mathrm{\Omega }$ , (ii) use the symbolic function fourier, and (iii) sample the signals and approximate their Fourier transforms (this requires sampling theory to be discussed in Chapter 8, and the Fourier representation of sampled signals to be considered in Chapter 11).

The Fourier transform of $x_1(t)=u(t)-u(t-1)$ can be found by considering the advanced signal $z(t)=x_1(t+0.5)=u(t+0.5)-u(t-0.5)$ with Fourier transform

$Z(\mathrm{\Omega })={\int }_{-0.5}^{0.5}{e}^{-j\mathrm{\Omega }t}dt=\frac{\sin (\mathrm{\Omega }/2)}{\mathrm{\Omega }/2}.$

Since $z(t)=x_1(t+0.5)$ we have $Z(\mathrm{\Omega })=X_1(\mathrm{\Omega })e^{j0.5\mathrm{\Omega }}$ so that

$X_1(\mathrm{\Omega })=e^{-j0.5\mathrm{\Omega }}Z(\mathrm{\Omega })\text{and}|X_1(\mathrm{\Omega })|=\left|\frac{\sin (\mathrm{\Omega }/2)}{\mathrm{\Omega }/2}\right|.$

Given that $Z(\mathrm{\Omega })$ is real, its phase is either zero when $Z(\mathrm{\Omega })\ge 0$ or ±π when $Z(\mathrm{\Omega })<0$ (using these values so that the phase is an odd function of Ω) and as such the phase of $X_1(\mathrm{\Omega })$ is

$\angle X_1(\mathrm{\Omega })=\angle Z(\mathrm{\Omega })-0.5\mathrm{\Omega }=\{\begin{array}{cc}-0.5\mathrm{\Omega }\hfill & Z(\mathrm{\Omega })\ge 0,\hfill \\ \pm \pi -0.5\mathrm{\Omega }\hfill & Z(\mathrm{\Omega })<0.\hfill \end{array}$

The Fourier transform of $x_2(t)=e^{-t}u(t)$ is

$X_2(\mathrm{\Omega })=\frac{1}{1+j\mathrm{\Omega }}.$

The magnitude and phase are given by

$|X_2(\mathrm{\Omega })|=\frac{1}{\sqrt{1+{\mathrm{\Omega }}^2}},\angle (X_2(\mathrm{\Omega }))=-{\tan }^{-1}\mathrm{\Omega }.$

Computing for different values of Ω we have

$\begin{array}{ccc}\hfill \mathrm{\Omega }& \hfill |X_2(\mathrm{\Omega })|\hfill & \angle (X_2(\mathrm{\Omega }))\hfill \\ \hfill 0& \hfill 1\hfill & 0\hfill \\ \hfill 1& \hfill \frac{1}{\sqrt{2}}=0.707\hfill & -\pi /4\hfill \\ \hfill \infty & \hfill 0\hfill & -\pi /2,\hfill \end{array}$

i.e., the magnitude spectrum decays as Ω increases. The following script gives the necessary instructions to compute and plot the signal $x_1(t)$ and the magnitude and phase of its Fourier transform using symbolic MATLAB. Similar calculations are done for $x_2(t)$ .

Notice the way the magnitude and the phase are computed. The computation of the phase is complicated by the lack of the function atan2 in symbolic MATLAB (atan2 extends the principal values of the inverse tangent to $(-\pi ,\pi ]$ by considering the sign of the real part of the complex function). The phase computation can be done by using the log function:

$\log (X_1(\mathrm{\Omega }))=\log [|X_1(\mathrm{\Omega })|e^{j\angle X_1(\mathrm{\Omega })}]=\log (|X_1(\mathrm{\Omega })|)+j\angle X_1(\mathrm{\Omega }),$

so that

$\angle X_1(\mathrm{\Omega })=\mathcal{I}m[\log (X_1(\mathrm{\Omega }))].$

Changing the above script we can find the magnitude and phase of $X_2(\mathrm{\Omega })$ . See Fig. 5.7 for results. □

Figure 5.7. Top three figures: pulse x ₁(t)=u(t)−u(t − 1) and its magnitude and phase spectra. Bottom three figures: decaying exponential x ₂(t)=e ^−t u(t) and its magnitude and phase spectra.

Example 5.13

It is not always the case that the Fourier transform is a complex-valued function. Consider the signals

$\begin{array}{cc}(a)x(t)=0.5e^{-|t|},\hfill & (b)y(t)=e^{-|t|}\cos ({\mathrm{\Omega }}_{0}t).\hfill \end{array}$

Find their Fourier transforms. Let ${\mathrm{\Omega }}_0=1$ ; use the magnitudes $|X(\mathrm{\Omega })|$ and $|Y(\mathrm{\Omega })|$ to discuss the smoothness of the signals $x(t)$ and $y(t)$ .

Solution: (a) The Fourier transform of $x(t)$ is

$X(\mathrm{\Omega })=\frac{0.5}{s+1}+\frac{0.5}{-s+1}{|}_{s=j\mathrm{\Omega }}=\frac{1}{{\mathrm{\Omega }}^2+1},$

a positive and real-valued function of Ω. Thus $\angle (X(\mathrm{\Omega }))=0$ ; $x(t)$ is a "low-pass" signal like $e^{-t}u(t)$ as its magnitude spectrum decreases with frequency:

$\begin{array}{cc}\hfill \mathrm{\Omega }& \hfill |X(\mathrm{\Omega })|=X(\mathrm{\Omega })\hfill \\ \hfill 0& \hfill 1\hfill \\ \hfill 1& \hfill 0.5\hfill \\ \hfill \infty & \hfill 0\hfill \end{array}$

but $0.5e^{-|t|}$ is "smoother" than $e^{-t}u(t)$ because the magnitude response is more concentrated in the low frequencies. Compare the values of the magnitude responses at $\mathrm{\Omega }=0$ and 1 to verify this.

(b) The signal $y(t)=x(t)\cos ({\mathrm{\Omega }}_{0}t)$ is a "band-pass" signal. It is not as smooth as $x(t)$ given that the energy concentration of

$Y(\mathrm{\Omega })=0.5[X(\mathrm{\Omega }-{\mathrm{\Omega }}_0)+X(\mathrm{\Omega }+{\mathrm{\Omega }}_0)]=\frac{0.5}{{(\mathrm{\Omega }-{\mathrm{\Omega }}_0)}^2+1}+\frac{0.5}{{(\mathrm{\Omega }+{\mathrm{\Omega }}_0)}^2+1}$

is around the frequency ${\mathrm{\Omega }}_0$ and not the zero frequency as for $x(t)$ . Since $Y(\mathrm{\Omega })$ is real-valued and positive the corresponding phase is zero. The magnitude of $Y(\mathrm{\Omega })$ for ${\mathrm{\Omega }}_0=1$ is

$\begin{array}{cc}\hfill \mathrm{\Omega }& \hfill |Y(\mathrm{\Omega })|=Y(\mathrm{\Omega })\hfill \\ \hfill 0& \hfill 0.5\hfill \\ \hfill 1& \hfill 1\hfill \\ \hfill 2& \hfill 0.5\hfill \\ \hfill \infty & \hfill 0\hfill \end{array}$

The higher the frequency ${\mathrm{\Omega }}_0$ the more variation is displayed by the signal. □

The bandwidth of a signal $x(t)$ is the support—on the positive frequencies—of its Fourier transform $X(\mathrm{\Omega })$ . There are different definitions of the bandwidth of a signal depending on how the support of its Fourier transform is measured. We will discuss some of the bandwidth measures used in filtering and in communications in Chapter 7.

The concept of bandwidth of a filter that you learned in circuit theory is one of its possible definitions, other possible definitions will be introduced later. The bandwidth together with the information about whether the energy or power of the signal is concentrated in the low, middle, high frequencies or a combination of them provides a good characterization of the signal. The spectrum analyzer, a device used to measure the spectral characteristics of a signal, will be presented after the section on filtering.

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Frequency Analysis: The Fourier Transform

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

5.6.4 Symmetry of Spectral Representations

Now that the Fourier representation of aperiodic and periodic signals is unified, we can think of just one spectrum that accommodates both finite as well as infinite energy signals. The word "spectrum" is loosely used to mean different aspects of the frequency representation. The following provides definitions and the symmetry characteristic of the spectrum of real-valued signals.

If $X(\mathrm{\Omega })$ is the Fourier transform of a real-valued signal x(t), periodic or aperiodic, the magnitude $|X(\mathrm{\Omega })|$ and the real part $\mathcal{R}e[X(\mathrm{\Omega })]$ are even functions of $\mathrm{\Omega }$ :

(5.19) $\begin{array}{cc}\hfill & |X(\mathrm{\Omega })|=|X(-\mathrm{\Omega })|\hfill \\ \hfill & \mathcal{R}e[X(\mathrm{\Omega })]=\mathcal{R}e[X(-\mathrm{\Omega })]\hfill \end{array}$

and the phase $\angle X(\mathrm{\Omega })$ and the imaginary part $\mathcal{I}m[X(\mathrm{\Omega })]$ are odd functions of $\mathrm{\Omega }$ :

(5.20) $\begin{array}{cc}\hfill & \angle X(\mathrm{\Omega })=-\angle X(-\mathrm{\Omega })\hfill \\ \hfill & \mathcal{I}m[X(\mathrm{\Omega })]=-\mathcal{I}m[X(-\mathrm{\Omega })]\hfill \end{array}$

We then call the plots

$\begin{array}{cc}\hfill & |X(\mathrm{\Omega })|\mathit{vs}\mathrm{\Omega }\mathbf{Magnitude}\mathbf{Spectrum}\hfill \\ \hfill & \angle X(\mathrm{\Omega })\mathit{vs}\mathrm{\Omega }\mathbf{Phase}\mathbf{Spectrum}\hfill \\ \hfill & |X(\mathrm{\Omega }){|}^2\mathit{vs}\mathrm{\Omega }\mathbf{Energy}/\mathbf{Power}\mathbf{Spectrum}\hfill \end{array}$

To show this consider the inverse Fourier transform of a real-valued signal x(t),

$x(t)={\int }_{-\infty }^{\infty }X(\mathrm{\Omega })e^{j\mathrm{\Omega }t}d\mathrm{\Omega }$

which is identical, because of being real, to

$x^{\ast }(t)={\int }_{-\infty }^{\infty }{X}^{\ast }(\mathrm{\Omega })e^{-j\mathrm{\Omega }t}d\mathrm{\Omega }={\int }_{-\infty }^{\infty }{X}^{\ast }(-\mathrm{\Omega }\text{'})e^{j\mathrm{\Omega }\text{'}t}d\mathrm{\Omega }\text{'}$

since the integral can be thought of as an infinite sum of complex values and by letting ${\mathrm{\Omega }}^{\prime }=-\mathrm{\Omega }$ . Comparing the two integrals, we have the following identities:

$\begin{array}{cc}\hfill & X(\mathrm{\Omega })=X^{\ast }(-\mathrm{\Omega })\hfill \\ \hfill & |X(\mathrm{\Omega })|e^{j\angle X(\mathrm{\Omega })}=|X(-\mathrm{\Omega })|e^{-j\angle X(-\mathrm{\Omega })}\hfill \\ \hfill & \mathcal{R}e[X(\mathrm{\Omega })]+j\mathcal{I}m[X(\mathrm{\Omega })]=\mathcal{R}e[X(-\mathrm{\Omega })]-j\mathcal{I}m[X(-\mathrm{\Omega })]\hfill \end{array}$

or that the magnitude is an even function of $\mathrm{\Omega }$ and the phase is an odd function of $\mathrm{\Omega }$ . And that the real part of the Fourier transform is an even function and the imaginary part of the Fourier transform is an odd function of $\mathrm{\Omega }$ .

Remarks

1.

Clearly if the signal is complex, the above symmetry will not hold. For instance, if $x(t)=e^{j{\mathrm{\Omega }}_{0}t}=\cos ({\mathrm{\Omega }}_{0}t)+j\sin ({\mathrm{\Omega }}_{0}t)$ , using the frequency shift property its Fourier transform is

$X(\mathrm{\Omega })=2\pi \delta (\mathrm{\Omega }-{\mathrm{\Omega }}_0)$

which occurs at $\mathrm{\Omega }={\mathrm{\Omega }}_0$ only, so the symmetry in the magnitude and phase does not exist.

2.

It is important to recognize the meaning of "negative" frequencies. In reality, only positive frequencies exist and can be measured, but as shown the spectrum, magnitude, or phase, of a real-valued signal requires negative frequencies. It is only under this context that negative frequencies should be understood—as necessary to generate "real-valued" signals.

Example 5.11

Use MATLAB to compute the Fourier transform of the following signals

$(a)x_1(t)=u(t)-u(t-1)\text{,}(b)x_2(t)=e^{-t}u(t)$

Plot their magnitude and phase spectra.

Solution

Three possible ways to compute the Fourier transforms of these signals using MATLAB are: (i) find their Laplace transforms, as in Chapter 3, using the symbolic function laplace and compute the magnitude and phase functions by letting $s=j\mathrm{\Omega }$ , (ii) use the symbolic function fourier, and (iii) sample the signals and approximate their Fourier transforms (this requires sampling theory to be discussed in Chapter 8, and the Fourier representation of sampled signals to be considered in Chapter 11).

The Fourier transform of x ₁(t)   = u(t)   − u(t  −   1) can be found by considering the advanced signal z(t)   = x ₁(t  +   0.5)   = u(t  +   0.5)   − u(t  −   0.5) with Fourier transform

$Z(\mathrm{\Omega })=\frac{\sin (\mathrm{\Omega }/2)}{\mathrm{\Omega }/2}$

Since z(t)   = x ₁(t  +   0.5) then $Z(\mathrm{\Omega })=X_1(\mathrm{\Omega })e^{j0.5\mathrm{\Omega }}$ so that

$X_1(\mathrm{\Omega })=e^{-j0.5\mathrm{\Omega }}Z(\mathrm{\Omega })\text{and}|X_1(\mathrm{\Omega })|=\left|\frac{\sin (\mathrm{\Omega }/2)}{\mathrm{\Omega }/2}\right|$

Given that $Z(\mathrm{\Omega })$ is real, its phase is either zero when $Z(\mathrm{\Omega })\ge 0$ or ±π when $Z(\mathrm{\Omega })<0$ (using these values so that the phase is an odd function of $\mathrm{\Omega }$ ) and as such the phase of $X_1(\mathrm{\Omega })$ is

$\angle X_1(\mathrm{\Omega })=\angle Z(\mathrm{\Omega })-0.5\mathrm{\Omega }=\left\{\begin{array}{cc}-0.5\mathrm{\Omega }\hfill & Z(\mathrm{\Omega })\ge 0\hfill \\ \pm \pi -0.5\mathrm{\Omega }\hfill & Z(\mathrm{\Omega })<0\hfill \end{array}\right.$

The Fourier transform of x ₂(t)   = e ^−t u(t) is

$X_2(\mathrm{\Omega })=\frac{1}{1+j\mathrm{\Omega }}$

The magnitude and phase are given by

$|X_2(\mathrm{\Omega })|=\frac{1}{\sqrt{1+{\mathrm{\Omega }}^2}}\text{,}\angle (X_2(\mathrm{\Omega }))=-{\tan }^{-1}\mathrm{\Omega }$

Computing for different values of $\mathrm{\Omega }$ we have

$\begin{array}{cc}\hfill & \mathrm{\Omega }|X_2(\mathrm{\Omega })|\angle (X_2(\mathrm{\Omega }))\hfill \\ \hfill & 010\hfill \\ \hfill & 1\frac{1}{\sqrt{2}}=0.707-\pi /4\hfill \\ \hfill & \infty 0-\pi /2\hfill \end{array}$

i.e., the magnitude spectrum decays as $\mathrm{\Omega }$ increases. The following script gives the necessary instructions to compute and plot the signal x ₂(t)   = e ^−t u(t) and the magnitude and phase of its Fourier transform using symbolic MATLAB.

%%

% Example 5.11

%%

x2=heaviside(t)*exp(−t)

X2=fourier(x2);

X2m=sqrt((real(X2))ˆ+(imag(X2))ˆ2);% magnitude

X2a=imag(log(X2));              % phase

Notice the way the magnitude and the phase are computed. The computation of the phase is complicated by the lack of the function atan2 in symbolic MATLAB (atan2 extends the principal values of the inverse tangent to (−π,π] by considering the sign of the real part of the complex function). The phase computation can be done using the inverse tangent function, or by using the log function:

$\log (X_2(\mathrm{\Omega }))=\log [|X_2(\mathrm{\Omega })|e^{j\angle X_2(\mathrm{\Omega })}]=\log (|X_2(\mathrm{\Omega })|)+j\angle X_2(\mathrm{\Omega })$

so that

$\angle X_2(\mathrm{\Omega })=\mathcal{I}m[\log (X_2(\mathrm{\Omega }))]$

Changing the above script we can find the magnitude and phase of $X_1(\mathrm{\Omega })$ . See Figure 5.7 for results.   ▪

Example 5.12

It is not always the case that the Fourier transform is a complex-valued function. Consider the signals

$(a)x(t)=0.5e^{-|t|}\text{,}(b)y(t)=e^{-|t|}\cos ({\mathrm{\Omega }}_{0}t)$

Find their Fourier transforms. Let ${\mathrm{\Omega }}_0=1$ , use the magnitudes $|X(\mathrm{\Omega })|$ and $|Y(\mathrm{\Omega })|$ to discuss the smoothness of the signals x(t) and y(t).

Solution

(a)

The Fourier transform of x(t) is

$X(\mathrm{\Omega })=\frac{0.5}{s+1}+\frac{0.5}{-s+1}{|}_{s=j\mathrm{\Omega }}=\frac{1}{{\mathrm{\Omega }}^2+1}$

a positive and real-valued function of $\mathrm{\Omega }$ . Thus $\angle (X(\mathrm{\Omega }))=0$ ; x(t) is a "low-pass" signal like e ^−t u(t) (in Example 5.11) as its magnitude spectrum decreases with frequency:

$\begin{array}{cc}\hfill & \mathrm{\Omega }|X(\mathrm{\Omega })|=X(\mathrm{\Omega })\hfill \\ \hfill & 01\hfill \\ \hfill & 10.5\hfill \\ \hfill & \infty 0\hfill \end{array}$

but 0.5e ^−∣t∣ is "smoother" than e ^−t u(t) because the magnitude response is more concentrated in the low frequencies. Compare the values of the magnitude responses at $\mathrm{\Omega }=0$ and 1 to verify this.

(b)

The signal $y(t)=x(t)\cos ({\mathrm{\Omega }}_{0}t)$ is a "band-pass" signal. It is not as smooth as x(t) given that the energy concentration of

$Y(\mathrm{\Omega })=0.5[X(\mathrm{\Omega }-{\mathrm{\Omega }}_0)+X(\mathrm{\Omega }+{\mathrm{\Omega }}_0)]=\frac{0.5}{{(\mathrm{\Omega }-{\mathrm{\Omega }}_0)}^2+1}+\frac{0.5}{{(\mathrm{\Omega }+{\mathrm{\Omega }}_0)}^2+1}$

is around the frequency ${\mathrm{\Omega }}_0$ and not the zero frequency as for x(t). Since $Y(\mathrm{\Omega })$ is real-valued and positive the corresponding phase is zero. The magnitude of $Y(\mathrm{\Omega })$ for ${\mathrm{\Omega }}_0=1$ is

$\begin{array}{cc}\hfill & \mathrm{\Omega }|Y(\mathrm{\Omega })|=Y(\mathrm{\Omega })\hfill \\ \hfill & 00.5\hfill \\ \hfill & 10.6\hfill \\ \hfill & 20.3\hfill \\ \hfill & \infty 0\hfill \end{array}$

The higher the frequency ${\mathrm{\Omega }}_0$ the more variation is displayed by the signal.   ▪

Figure 5.7. Top three figures: pulse x ₁(t)   = u(t)   − u(t  −   1) and its magnitude and phase spectra. Bottom three figures: decaying exponential x ₂(t)   = e ^−t u(t) and its magnitude and phase spectra.

The bandwidth of a signal x(t) is the support—on the positive frequencies—of its Fourier transform $X(\mathrm{\Omega })$ . There are different definitions of the bandwidth of a signal depending on how the support of its Fourier transform is measured. We will discuss some of the bandwidth measures used in filtering and in communications in Chapter 7.

The concept of bandwidth of a filter that you learned in Circuit Theory is one of its possible definitions; other possible definitions will be introduced later. The bandwidth together with the information about whether the energy or power of the signal is concentrated in the low-, middle-, high-frequencies or a combination of them provide a good characterization of the signal. The spectrum analyzer, a device used to measure the spectral characteristics of a signal, will be presented after the section on filtering.

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Frequency Analysis

Luis F. Chaparro , in Signals and Systems using MATLAB, 2011

Publisher Summary

This chapter discusses the frequency analysis of the Fourier series for periodic signals. The frequency representation of periodic and aperiodic signals indicates how their power or energy is allocated to different frequencies. The Fourier representation of periodic signals is fundamental in finding a representation for non-periodic signals. Complex exponentials or sinusoids are used in the Fourier representation of periodic as well as aperiodic signals by taking advantage of the Eigen-function property of linear time invariant (LTI) systems. Fourier analysis considers the steady state, while Laplace analysis considers both transient and steady state. The Fourier representation is also useful in finding the frequency response of linear time-invariant systems that is related to the transfer function obtained with the Laplace transform. The frequency response of a system indicates how an LTI system responds to sinusoids of different frequencies. Such a response characterizes the system and permits easy computation of its steady-state response, and will be equally important in the synthesis of systems. It is important to understand the driving force behind the representation of signals in terms of basic signals when applied to LTI systems. The Laplace transform is seen as the representation of signals in terms of general Eigen-functions. The frequency representation of signals and systems is extremely important in signal processing and in communications.

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Frequency Analysis: The Fourier Series

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

Abstract

In this chapter we consider the frequency representation of periodic signals by means of the Fourier series, an infinite expansion in terms of orthonormal complex exponentials or sinusoids. Representing aperiodic signals in terms of periodic signals will permit us to extend the Fourier series representation to the Fourier transform valid for periodic and aperiodic signals. The Fourier series coefficients are obtained using the orthonormality of complex exponentials or sinusoidal bases and efficiently computed using the Laplace transform of a period. The line spectrum, obtained from the Fourier series coefficients, indicates how the power of the signal is distributed to harmonic frequency components in the series. Properties of the Fourier series allow visualization of the power distribution over frequency, the symmetry of the spectrum, and the nature of the Fourier coefficients depending on the symmetry of the signal. Taking advantage of the eigenfunction property of linear time-invariant (LTI) systems, the steady-state response of these systems to periodic signals is easily obtained. MATLAB is used to represent and process periodic continuous-time signals.

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Frequency Analysis

Luis F. Chaparro , in Signals and Systems using MATLAB, 2011

5.1 Introduction

In this chapter we continue the frequency analysis of signals. In particular, we will concentrate in the following issues:

■

Generalization of the Fourier series —The frequency representation of signals as well as the frequency response of systems are tools of great significance in signal processing, communications, and control theory. In this chapter we will complete the Fourier representation of signals by extending it to aperiodic signals. By a limiting process the harmonic representation of periodic signals is extended to the Fourier transform, a frequency-dense representation for nonperiodic signals. The concept of spectrum introduced for periodic signals is generalized for both finite-power and finite-energy signals. Thus, the Fourier transform measures the frequency content of a signal, and unifies the representation of periodic and aperiodic signals.

■

Laplace and Fourier transform—In this chapter the connection between the Laplace and the Fourier transforms will be highlighted for computational and analytical reasons. The Fourier transform turns out to be a very important case of the Laplace transform for signals of which the region of convergence includes the jΩ axis. There are, however, signals where the Fourier transform cannot be obtained from the Laplace transform; for those cases, properties of the Fourier transform will be used. The duality of the direct and inverse transforms is of special interest in computing the Fourier transform.

■

Basics of filtering—Filtering is an important application of the Fourier transform. The Fourier representation of signals and the eigenfunction property of LTI systems provide the tools to change the frequency content of a signal by processing it with an LTI system with a desired frequency response.

■

Modulation and communications—The idea of changing the frequency content of a signal via modulation is basic in analog communications. Modulation allows us to send signals over the airwaves using antennas of reasonable sizes. Voice and music are relative low-frequency signals that cannot be easily radiated without the help of modulation. Continuous-wave modulation changes the amplitude, the frequency, or the phase of a sinusoidal carrier of frequency much greater than the frequencies present in the message we wish to transmit.

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Discrete-Time Signals and Systems

Luis F. Chaparro , in Signals and Systems using MATLAB, 2011

Publisher Summary

This chapter discusses the theory of discrete-time signals and systems, whose basic theory is very much like that for continuous-time signals and systems. However, there are significant differences that need to be understood. Characteristics such as energy, power, and symmetry of continuous-time signals are conceptually the same for discrete-time signals. Integrals are replaced by sums, derivatives by finite differences, and differential equations by difference equations. The discrete-time signals such as periodic and aperiodic signals, finite-energy and finite-power discrete-time signals, even and odd signals, and basic discrete-time signals are discussed in the chapter. The discrete approximation of derivatives and integrals provides an approximation of differential equations, representing dynamic continuous-time systems by difference equations. A computationally significant difference with continuous-time systems is that the solution of difference equations can be recursively obtained and that the convolution sum provides a class of systems that do not have a counterpart in the analog domain. The chapter also discusses the basic structure for discrete-time signals and continues developing the theory of linear time-invariant discrete-time systems using transforms. The relation that exists between the Z-transform and the Fourier representations of discrete-time signals and systems, not only with each other but with the Laplace and Fourier transforms have also been discussed. There is a great deal of connection among all of these transforms and a clear understanding of this would help with the analysis and synthesis of discrete-time signals and systems. The chapter also discusses the discrete-time systems such as linearity time invariance, stability, and causality.

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Introduction to the Design of Discrete Filters

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

12.1 Introduction

Filtering is an important application of linear time-invariant (LTI) systems. According to the eigenfunction property of discrete-time LTI systems, the steady-state response of a discrete-time LTI system to a sinusoidal input is also a sinusoid of the same frequency as that of the input, but with magnitude and phase affected by the response of the system at the frequency of the input. Since periodic as well as aperiodic signals have Fourier representations consisting of sinusoids of different frequencies, these signal components can be modified by appropriately choosing the frequency response of a LTI system, or filter. Filtering can thus be seen as a way to change the frequency content of an input signal.

The appropriate filter is specified using the spectral characterization of the input and the desired spectral characteristics of the output of the filter. Once the specifications of the filter are set, the problem becomes one of approximation, either by a ratio of polynomials or by a polynomial (if possible). After establishing that the filter resulting from the approximation satisfies the given specifications, it is then necessary to check its stability (if not guaranteed by the design method)—in the case of the filter being a rational approximation—and if stable, we need to figure out what would be the best possible way to implement the filter in hardware or in software. If not stable, we need either to repeat the approximation or to stabilize the filter before its implementation.

In the continuous-time domain, filters are obtained by means of rational approximation. In the discrete-time domain, there are two possible types of filters: one that is the result of rational approximation—these filters are called recursive or infinite impulse response (IIR) filter. The other is the non-recursive or finite impulse response (FIR) filter that results from polynomial approximation. As we will see, the discrete filter specifications can be in the frequency or in the time domain. For recursive or IIR filters, the specifications are typically given in the form of magnitude and phase specifications, while the specifications for non-recursive or FIR filters can be in the time domain as a desired impulse response. The discrete filter design problem then consists in: Given the specifications of a filter we look for a polynomial or rational (ratio of polynomials) approximation to the specifications. The resulting filter should be realizable, which besides causality and stability requires that the filter coefficients be real-valued.

There are different ways to attain a rational approximation for discrete IIR filters: by transformation of analog filters, or by optimization methods that include stability as a constraint. We will see that the classical analog design methods (Butterworth, Chebyshev, Elliptic, etc.) can be used to design discrete filters by means of the bilinear transformation that maps the analog s-plane into the Z-plane. Given that the FIR filters are unique to the discrete domain, the approximation procedures for FIR filters are unique to that domain.

The difference between discrete and digital filters is in quantization and coding. For a discrete filter we assume that the input and the coefficients of the filter are represented with infinite precision, i.e., using an infinite number of quantization levels and thus no coding is performed. The coefficients of a digital filter are binary, and the input is quantized and coded. Quantization thus affects the performance of a digital filter, while it has no effect in discrete filters.

Considering continuous to discrete (C/D) and discrete to continuous (D/C) ideal converters simply as samplers and reconstruction filters, theoretically it is possible to implement the filtering of band-limited analog signals using discrete filters (Figure 12.1). In such an application, an additional specification for the filter design is the sampling period. In this process it is crucial that the sampling period in the C/D and D/C converters be synchronized. In practice, filtering of analog signals is done using analog to digital (A/D) and digital to analog (D/A) converters together with digital filters.

Figure 12.1. Discrete filtering of analog signals using ideal continuous to discrete (C/D), or a sampler, and discrete to continuous (D/C) converter, or a reconstruction filter.

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Continuous-Time Signals

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

1.3.3 Periodic and Aperiodic Signals

A useful characterization of signals is whether they are periodic or aperiodic (non-periodic).

A continuous-time signal $x(t)$ is periodic if the following two conditions are satisfied:

1.

it is defined for all possible values of t, $-\infty <t<\infty$ , and

2.

there is a positive real value $T_0$ , the fundamental period of $x(t)$ , such that

(1.7) $x(t+k{T}_0)=x(t)$

for any integer k.

The fundamental period of $x(t)$ is the smallest $T_0>0$ that makes the periodicity possible.

Remarks

1.

The infinite support and the uniqueness of the fundamental period make periodic signals non-existent in practical applications. Despite this, periodic signals are of great significance in the Fourier signal representation and processing as we will see later. Indeed, the representation of aperiodic signals is obtained from that of periodic signals, and the response of systems to sinusoids is fundamental in the theory of linear systems.

2.

Although seemingly redundant, the first part of the definition of a periodic signal indicates that it is not possible to have a nonzero periodic signal with a finite support (i.e., the signal is zero outside an interval $t\in [t_1,t_2]$ ) or any support different from the infinite support $-\infty <t<\infty$ . This first part of the definition is needed for the second part to make sense.

3.

It is exasperating to find the period of a constant signal $x(t)=A$ . Visually $x(t)$ is periodic but its fundamental period is not clear. Any positive value could be considered the period, but none will be taken. The reason is that $x(t)=A\cos (0t)$ or a cosine of zero frequency, and as such its period ( $2\pi /\text{frequency}$ ) is not determined since we would have to divide by zero—which is not permitted. Thus, a constant signal is periodic of non-definable fundamental period!

Example 1.8

Consider the sinusoid $x(t)=A\cos ({\mathrm{\Omega }}_{0}t+\theta )$ , $-\infty <t<\infty$ . Determine the fundamental period of this signal, and indicate for what frequency ${\mathrm{\Omega }}_0$ the fundamental period of $x(t)$ is not defined.

Solution: The analog frequency is ${\mathrm{\Omega }}_0=2\pi /T_0$ so $T_0=2\pi /{\mathrm{\Omega }}_0$ is the fundamental period. Whenever ${\mathrm{\Omega }}_0>0$ these sinusoids are periodic. For instance, consider $x(t)=2\cos (2t-\pi /2)$ , $-\infty <t<\infty$ ; its fundamental period is found from its analog frequency ${\mathrm{\Omega }}_0=2=2\pi f_0$ (rad/s), or a Hz frequency of $f_0=1/\pi =1/T_0$ , so that $T_0=\pi$ is the fundamental period in s. That this is so can be seen using the fact that ${\mathrm{\Omega }}_0{T}_0=2T_0=2\pi$ and as such

$\begin{array}{ccc}\hfill x(t+N{T}_0)& \hfill =\hfill & 2\cos (2(t+N{T}_0)-\pi /2)=2\cos (2t+2\pi N-\pi /2)\hfill \\ \hfill & \hfill =\hfill & 2\cos (2t-\pi /2)=x(t)\text{integer}N,\hfill \end{array}$

since adding $2\pi N$ (a multiple of 2π) to the angle of the cosine gives the original angle. If ${\mathrm{\Omega }}_0=0$ , i.e., dc frequency, the fundamental period cannot be defined because of the division by zero when finding $T_0=2\pi /{\mathrm{\Omega }}_0$ . □

Example 1.9

Consider a periodic signal $x(t)$ of fundamental period $T_0$ , determine whether the following signals are periodic and if so find their corresponding fundamental periods:

1.

$y(t)=A+x(t)$ , where A is either a positive, negative or zero value,

2.

$z(t)=x(t)+v(t)$ where $v(t)$ is periodic of fundamental period $T_1=N{T}_0$ , where N is a positive integer, i.e., a multiple of $T_0$ ,

3.

$w(t)=x(t)+s(t)$ where $s(t)$ is periodic of fundamental period $T_1$ , not a multiple of $T_0$ .

Solution: (1) Adding a constant to a periodic signal does not change the periodicity. Thus, $y(t)$ is periodic of fundamental period $T_0$ , i.e., for an integer k, $y(t+k{T}_0)=A+x(t+k{T}_0)=A+x(t)$ since $x(t)$ is periodic of fundamental period $T_0$ .

(2) The fundamental period $T_1=N{T}_0$ of $v(t)$ is also a period of $x(t)$ and so $z(t)$ is periodic of fundamental period $T_1$ since for any integer k

$z(t+k{T}_1)=x(t+k{T}_1)+v(t+k{T}_1)=x(t+kN{T}_0)+v(t)=x(t)+v(t)$

given that $v(t+k{T}_1)=v(t)$ , and that kN is an integer so that $x(t+kN{T}_0)=x(t)$ . The periodicity can be visualized by considering that in one period of $v(t)$ we can include N periods of $x(t)$ .

(3) The condition for $w(t)$ to be periodic is that the ratio of the periods of $x(t)$ and of $s(t)$ be

$\frac{T_1}{T_0}=\frac{N}{M}$

where N and M are positive integers not divisible by each other, or that $M{T}_1$ periods of $s(t)$ can be exactly included into $N{T}_0$ periods of $x(t)$ . Thus, $M{T}_1=N{T}_0$ becomes the fundamental period of $w(t)$ , or

$w(t+M{T}_1)=x(t+M{T}_1)+s(t+M{T}_1)=x(t+N{T}_0)+s(t+M{T}_1)=x(t)+s(t).\text{\hspace{1em}□}$

Example 1.10

Let $x(t)=e^{j2t}$ and $y(t)=e^{j\pi t}$ , consider their sum $z(t)=x(t)+y(t)$ , and their product $w(t)=x(t)y(t)$ . Determine if $z(t)$ and $w(t)$ are periodic, and if so their fundamental periods. Is $p(t)=(1+x(t))(1+y(t))$ periodic?

Solution: According to Euler's identity

$x(t)=\cos (2t)+j\sin (2t),y(t)=\cos (\pi t)+j\sin (\pi t)$

indicating $x(t)$ is periodic of fundamental period $T_0=\pi$ (the frequency of $x(t)$ is ${\mathrm{\Omega }}_0=2=2\pi /T_0$ ) and $y(t)$ is periodic of fundamental period $T_1=2$ (the frequency of $y(t)$ is ${\mathrm{\Omega }}_1=\pi =2\pi /T_1$ ).

For $z(t)=x(t)+y(t)$ to be periodic $T_1/T_0$ must be a rational number, which is not the case as $T_1/T_0=2/\pi$ is irrational due to π. So $z(t)$ is not periodic.

The product, $w(t)=x(t)y(t)=e^{j(2+\pi )t}=\cos ({\mathrm{\Omega }}_{2}t)+j\sin ({\mathrm{\Omega }}_{2}t)$ where ${\mathrm{\Omega }}_2=2+\pi =2\pi /T_2$ . Thus, $w(t)$ is periodic of fundamental period $T_2=2\pi /(2+\pi )$ .

The terms $1+x(t)$ and $1+y(t)$ are periodic of fundamental period $T_0=\pi$ and $T_1=2$ , respectively, and from the case of the product above giving $w(t)$ , one would hope their product $p(t)=(1+x(t))(1+y(t))$ would be periodic. But since $p(t)=1+x(t)+y(t)+x(t)y(t)$ and as shown above $x(t)+y(t)=z(t)$ is not periodic, $p(t)$ is not periodic.  □

1.

A sinusoid of frequency ${\mathrm{\Omega }}_0>0$ is periodic of fundamental period $T_0=2\pi /{\mathrm{\Omega }}_0$ . If ${\mathrm{\Omega }}_0=0$ the fundamental period is not defined.

2.

The sum of two periodic signals $x(t)$ and $y(t)$ , of periods $T_1$ and $T_2$ , is periodic if the ratio of the periods $T_1/T_2$ is a rational number $N/M$ , with N and M non-divisible integers. The fundamental period of the sum is $M{T}_1=N{T}_2$ .

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