Problem No. 3: Fourier Series

One period of a periodic signal is shown below:

a)	Using symmetry arguments, explain which Fourier coefficients of the trigonometric
Fourier series will be	zero ({an} and {bn}).

Because the waveform is periodic, it is defined for all time t:(-∞,+∞). This means that the exact same waveform will appear on the left-hand side of the x(t) axis. An analysis of this signal shows that f(t) = f(-t). Therefore, the waveform exhibits even function symmetry. Also, a further analysis of the waveform shows that

f(t) = -f(t-T0/2) centered around the average (dc) value of the signal. This is known as half-wave symmetry. With these two characteristics combined, the waveform is said to exhibit half-wave even symmetry. The coefficients of a waveform that has half-wave even symmetry are as follows:

bn = 0, for all n

(proven in the following manner:)

an = bn = 0, for n even

(proven in the following manner:)

b)	Compute the first term in the trigonometric Fourier series, a0.

As displayed earlier, the equation for finding a0 (the dc value of the input signal) is as follows:

Therefore,

And,

So

c)	Compute the remaining coefficients of the trigonometric Fourier series.

The remaining coefficients of the trigonometric Fourier series of x(t) are an with n odd. This is given just as before:

Which is:

d)	Compute the power of the output signal, y(t).

Parseval's theorem states that the average power of a signal is the same whether it be in the time domain or the frequency domain. The equations for evaluating average power in each domain are as follows:

Time domain:

Frequency domain:

In these equations, T is the fundamental period of the signal, y(t) is the signal, Yn is the magnitude of the Fourier coefficients of y(t), and Y0 is the average (dc) value of y(t).

Since we do not know what y(t) is, but we can find Yn and Y0, we will calculate the average power in the frequency domain.

For the given figure, the spectral content of the output (y(t)) will contain only the dc value, Y0, and the fundamental harmonic, Y1. This is due to the fact that the system is "a linear system that removes all frequencies above 30Hz", which means that 50Hz harmonics and above are filtered. (There are no harmonics between 25Hz and 50Hz). Plugging in the value 1 for n in the equation found in 3-c) gives the following:

So,

This gives PAV (from the aforementioned equation) as:

Or

PAV = 3.06 Watts