Question 17.6: Let the function f(t) in Example 17.1 be the voltage source......

From Example 17.1,

v_{s} (t) = \frac{1}{2}  + \frac{2}{π} \sum\limits^{∞}_{k = 1}{\frac{1}{n} \sin nπt}       ,         n = 2k  –  1 

where ω_{n} = n ω_{0} = n π rad / s .   Using phasors, we obtain the response V_{0}   in the circuit of Fig. 17.20 by voltage division:

V_{0} = \frac{jω_{n} L}{R  +  jω_{n} L} V_{s} = \frac{j2nπ}{5  +  j2nπ} V_{s}

For the dc component (ω_{n}  = 0   or  n = 0 )  

V_{s} = \frac{1}{2}      ⇒          V_{o} = 0

This is expected, since the inductor is a short circuit to dc. For the nth harmonic,

V_{s} = \frac{2}{nπ} \underline{/ – 90°}                      (17.6.1)

and the corresponding response is

V_{o} = \frac{2nπ \underline{/  90°} }{\sqrt{25  +  4n^{2} π² } \underline{/ \tan^{-1}  2nπ / 5}} (\frac{2}{nπ} \underline{/ – 90°})                           (17.6.2)

= \frac{4\underline{/ – \tan^{-1}  2nπ / 5} }{\sqrt{25  +  4n^{2} π² }}

In the time domain,

v_{o} (t) = \sum\limits^{∞}_{k = 1}{\frac{4}{\sqrt{25  +  4n^{2} π² }}} \cos (nπt  –  \tan^{-1} \frac{2nπ}{5}) ,                       n = 2k  –  1

The first three terms ( k = 1 , 2 , 3 or n = 1 , 3 , 5) of the odd harmonics in the summation give us

v_{o}(t) = 0.4981  \cos (πt  –  51.49°)  + 0.2051  \cos (3πt  –  75.14°)

+ 0.1257 \cos (5πt  –  80.96°) +  . . .  V

Figure 17.21 shows the amplitude spectrum for output voltage v_{o}(t) ,  while that of the input voltage v_{s}(t)   is in Fig. 17.4(a). Notice that the two spectra are close. Why? We observe that the circuit in Fig. 17.20 is a highpass filter with the corner frequency ω_{c} = R/L = 2.5  rad/s ,   which is less than the fundamental frequency ω_{0} = π  rad/s .   The dc component is not passed and the first harmonic is slightly attenuated, but higher harmonics are passed. In fact, from Eqs. (17.6.1) and (17.6.2), V_{o}   is identical to  V_{s} for large n, which is characteristic of a highpass filter.