**Transcription for Educational Website**---**Analyzing Fourier Series Components for a Given Waveform**For the waveforms in Figure 26.32, determine whether the following components will be present in the Fourier series representation:a. dc term b. cosine terms c. sine terms d. even-ordered harmonics e. odd-ordered harmonics **Waveform Description**The graph depicted shows a piecewise linear waveform \( f(t) \) that resembles a triangular wave. The waveform has the following characteristics:- It is periodic with a period denoted as \( T \).- The waveform symmetrically oscillates between two amplitudes, \( A_m \) and \(-A_m\).- Key points on the time axis are labeled as \( -\frac{T}{2} \), \( 0 \), and \( \frac{T}{2} \).- The waveform is continuous and linear between these points, forming a triangular shape.**Graph Explanation**- The x-axis represents time \( t \).- The y-axis represents the waveform amplitude \( f(t) \).- The waveform starts at \( t = -T \), rises linearly to reach \( A_m \) at \( t = -\frac{T}{2} \), drops linearly to \( -A_m \) at \( t = 0 \), and then rises again to \( A_m \) at \( t = \frac{T}{2} \).- This pattern repeats every period \( T \).In conclusion, this waveform will have specific Fourier series components based on its symmetry and periodicity, which observers can analyze for applications in signal processing and communications.

According to the property of fouries series, for odd function all the coefficients of cosine will be zero. Let's prove this property below for a periodic function fx having period 2π. So, the Fourier series can be given by equation : f(x)=a02+∑n=1∞ an cos (nx) +∑n=1∞ bn sin (nx)where, a0=1π∫-ππ fx dx an=1π∫-ππ fx cos nx dx bn=1π∫-ππ fx sin nx dxnow, putting, x=2πT t , we have the fourier series as : ft=a02+∑n=1∞ an cos 2πntT + ∑n=1∞ bn sin 2πntT where, a0=2T∫-T2T2 ft dt an=2T∫-T2T2 ft cos 2πntT dt bn=2T∫-T2T2 ft sin 2πntT dtnow considering, a0 : a0=2T∫-T2T2 ft dt=2T∫-T20 ft dt +2T∫0T2 ft dt in the first integral putting, y=-t a0=-2T∫T20 f-y dy +2T∫0T2 ft dt=2T∫0T2 f-y dy +2T∫0T2 ft dt By using the properties of definite integral : a0=2T∫0T2 f-t dt +2T∫0T2 ft dt=0 as, the odd function in integrand f-t=-ftsimilarly, we can write : an=2T∫0T2 f-t cos -2πntT dt + 2T∫0T2 ft cos 2πntT dt=0the f-t cos 2πntT is the result of multiplication of odd and even functions. So, resultant function will be of odd nature and an=0 similarly bn=0 bn=2T∫0T2 f-t sin -2πntT dt + 2T∫0T2 ft sin 2πntT dt=0 the integrand will be the result of the multiplication of two odd functions so it is even so bn will be : bn=4T∫0T2 ft sin 2πntT dtit can be seen that there are only sine terms in the Fourier series expansion of the given waveform. So, Option (c) is correct.