636 SIGNAL PROCESSING
(a) Rectangular pulse train
(b) Triangular wave
(c) Sawtooth wave
(d) Square wave
(e) Half-rectified sine waveIdentify the waveform symmetry and find expressions for the Fourier coefficients.SolutionWaveform Symmetry a 0 anor bn(a) Rectangular pulse train Even DA
T
an=^2 A
πn
sinπDn
T
n= 1 , 2 , 3 , ...(b) Triangular wave Even and half-wave 0 an=
8 A
π^2 n^2
n= 1 , 3 , 5 , ...(c) Sawtooth wave Odd 0 bn=
2 A
πn n=^1 ,^2 ,^3 , ...
(d) Square wave Odd and half-wave 0 bn=^4 A
πn
n= 1 , 3 , 5 , ...(e) Half-rectified sine wave None A
π
b 1 =A
2
an=−
2 A
π(n^2 − 1 )
n= 2 , 4 , 6 , ...Spectral Analysis and Signal Bandwidth
Spectral analysis is based on the fact that a sinusoidal waveform is completely characterized by
three quantities: amplitude, phase, and frequency. By plotting amplitude and phase as a function
of frequencyf(=ω/ 2 π= 1 /T ), the frequency-domain picture conveys all the information,
including the signal’s bandwidth and other significant properties about the signal that consists
entirely of sinusoids. The two plots together (amplitude versus frequency and phase versus
frequency) constitute theline spectrumof the signalx(t).
Ifx(t) happens to be a periodic signal whose Fourier coefficients are known, Equation
(14.1.11) can be rewritten asx(t)=a 0 +∑∞n= 1Ancos(nωt+φn) (14.1.15)whereAn=√
an^2 +b^2 n and φn=−arctan(
bn
an)The phase angles are referenced to the cosine function, in agreement with our phasor notation.
The corresponding phasor diagram of Fourier coefficients is shown in Figure 14.1.3.
Equation (14.1.15) reveals that the spectrum of a periodic signal contains lines at frequencies
of 0,f,2f, and all higher harmonics off, although some harmonics may be missing in particular
cases. The zero-frequency or dc component represents the average valuea 0 , the components
corresponding to the first few harmonics represent relatively slow time variations, and the higher
harmonics represent more rapid time variations.