We have seen that a periodic signal x(t) can be represented by its Fourier series (3.19)....

We have seen that a periodic signal x(t) can be represented by its Fourier series (3.19). It turns out that we can transform many operations on the signal into corresponding operations on the Fourier coefficients ak. For example, suppose that we want to consider a new periodic signal . What would the Fourier coefficients be for y(t)? The answer comes from differentiating the Fourier series representation

Thus, we see that y(t) is also in the Fourier series form

so the Fourier series coefficients of y(t) are related to the Fourier series coefficients of x(t) by bk = (jkω0). This is a nice result, because it allows us to find the Fourier coefficients without actually differentiating x(t), and without doing any tedious evaluation of integrals to obtain the Fourier coefficients bk. It is a general result that holds for every periodic signal and its derivative.

We can use this style of manipulation to obtain some other useful results for Fourier series. In each case below, use (3.19) as the starting point and express the the given definition for y(t) as a Fourier series and then manipulate the equation so that you can pick off an expression for the Fourier coefficients bk as a function of the original coefficients ak.

(a) Suppose that y(t) = Ax(t) where A is a real number; i.e., y(t) is just a scaled version of x(t). Show that the Fourier coefficients for y(t) are bk = Aak.

(b) Suppose that y(t) = x(t – td) where td is a real number; i.e., y(t) is just a delayed version of x(t). Show that the Fourier coefficients for y(t) in this case are .