Review of Analog Signal Processing Basics - Digital Signal Processing: Fundamentals and Applications - page 779

Applying Euler's formula yields

c n ¼ 10 cos 0:5p n j sin ð0:5p n Þ ½cos ð0:5p n Þ þ j sin ð0:5p n Þ
jn2p

¼ 5 sin ð0:5p n Þ
0:5pn

Second, using the relationship between the sine-cosine form and the complex exponential form, it follows that

a n jb n
2

a n

2 ¼ 5 sin ð0:5npÞ

c n ¼ ¼

ð0:5npÞ
Certainly, the result is identical to the one obtained directly from the formula. Note that c 0 cannot be evaluated
directly by substituting n ¼ 0, since we have the indeterminate term 0

0 . Using L'Hospital's rule, described in

Appendix G, leads to

dðsin ð0:5npÞÞ
dn
dð0:5npÞ
dn

n / 0 5 sin ð0:5 n pÞ

c 0 ¼ lim ¼ lim

n / 0 5 ð0:5npÞ

n / 0 5 0:5p cos ð0:5 n pÞ

¼ lim ¼ 5

0:5p

Finally, the Fourier expansion in terms of the complex exponential form is shown as follows:
x t ¼ / þ 10
p

þ 5 þ 10
p

10
3p

þ 2
p

10
7p

e j2pt e j2pt e j6pt e j10pt e j14pt

þ /

B.1.4 Spectral Plots
As previously discussed, the magnitude-phase form can provide information to create a one-sided
spectral plot. The amplitude spectrum is obtained by plotting A n versus the harmonic frequency nu 0 ,
and the phase spectrum is obtained by plotting f n versus nu 0 , both for n 0. Similarly, if the complex
exponential form is used, the two-sided amplitude and phase spectral plots of jc n j and f n versus nu 0
for N < n < N can be achieved, respectively. We illustrate this by the following example.

EXAMPLE B.2
Based on the solution to Example B.1, plot the one-sided amplitude spectrum and two-sided amplitude spectrum,
respectively.
Solution:
Based on the solution for A n , the one-sided amplitude spectrum is shown in Figure B.2 .

FIGURE B.2
One-sided spectrum of the square waveform in Example B.2.