Fourier series of a bandwidth limited periodic function

[hotvette]

Homework Statement

Find Fourier coefficients of the periodic function whose template is x(t) where the Fourier Transform of x(t) is $$X(f) = (1-f^2)^2$$ where $\left|f\right|<1$ and period $T_0= 4$.

Homework Equations

$$FC=\hat x_T(k,T_0)=\sum_{k=-\infty}^\infty\frac{1}{T_0}X\left(k/T_0\right)$$

The Attempt at a Solution

I know the expression for FC only applies with the Fourier transform X(f) has no limit on bandwidth, so something different needs to be done. We've had nothing in lectures about finding the Fourier series for this situation (i.e. where FT is zero outside of a specific frequency range), so I'm not sure where to go from here.

There is a short section in the book about waveform sampling stating that for $T=1/(2f_c)$:
$$x(t) = T\sum_{k=-\infty}^\infty x(kT) \frac{sin 2 \pi f_c (t-kT)}{\pi(t-kT)}$$ which I think is just saying:
$$T\sum_{k=-\infty}^\infty x(kT) \frac{sin 2 \pi f_c (t-kT)}{\pi(t-kT)} <==> (1-f^2)^2$$ where $|f|<1$ and $f_c=1$, but I don't immediately see how that's useful. Any tips or suggestions?

 

[mighty2000]

Thank you for your question. The Fourier coefficients can still be found using the formula you provided, even if the Fourier transform has a limited bandwidth. In this case, the Fourier coefficients will correspond to the discrete frequencies within the bandwidth.

To find the Fourier coefficients for your function x(t), we can use the formula FC=\hat x_T(k,T_0)=\sum_{k=-\infty}^\infty\frac{1}{T_0}X\left(k/T_0\right). Since the Fourier transform, X(f), is only defined for frequencies within the range of |f|<1, we only need to consider the discrete frequencies within this range. These frequencies can be found by dividing the sampling frequency, f_c=1, by the period T_0=4. This gives us the discrete frequencies of f=-1/4, -1/2, 0, 1/4, and 1/2.

Now, we can plug these frequencies into the formula for the Fourier coefficients to find their values. For example, for k=0, we have FC=\hat x_T(0,T_0)=\frac{1}{T_0}X(0)=\frac{1}{4}(1-0)^2=1/4. Similarly, for k=-1/4, we have FC=\hat x_T(-1/4,T_0)=\frac{1}{T_0}X(-1/4)=\frac{1}{4}(1-(-1/4)^2)^2=\frac{15}{256}. The other coefficients can be found in a similar manner.

I hope this helps you with your problem. If you have any further questions, please don't hesitate to ask.