Question 17.14: If the sawtooth waveform in Fig. 17.45(a) is applied to an i......
If the sawtooth waveform in Fig. 17.45(a) is applied to an ideal lowpass filter with the transfer function shown in Fig. 17.45(b), determine the output.
The input signal in Fig. 17.45(a) is the same as the signal in Fig. 17.9. From Practice Prob. 17.2, we know that the Fourier series expansion is
x(t) = \frac{1}{2} – \frac{1}{π} \sin ω_{0} t – \frac{1}{2π} \sin 2ω_{0}t – \frac{1}{3π} \sin 3ω_{0} t – . . .
where the period is T = 1 s and the fundamental frequency is ω_{0} = 2π rad / s . Since the corner frequency of the filter is ω_{c} = 10 rad/s , only the dc component and harmonics with nω_{0} < 10 will be passed. For n = 2 , nω_{0} = 4 π = 12.566 rad/s , which is higher than 10 rad/s , meaning that second and higher harmonics will be rejected. Thus, only the dc and fundamental components will be passed. Hence, the output of the filter is
y (t) = \frac{1}{2} – \frac{1}{π} \sin 2πt
