Hints for Midterm

Problem I.

Use the following facts:

• Plug in the formulae for $a_0$, $a_k$ and $b_k$ in the equation for $s_K(t)$.
• $cos(a)cos(b) + sin(a)sin(b) = cos(a-b)$.
• Integration and summation can be interchanged.

II. Properties of Fourier Series

Hints for various parts.

• P 2. The definition of periodicity: there exists a $T$ such that $s(t) = s(t+MT)$ for all $M$. The smallest such $T$ is the period.
• P. 3. The proof is similar to that for the multiplicative property of Fourier Series. The multiplication property for periodic signals is as follows: Given two signals $x(t)$ and $y(t)$ with period $T$, and given that the Fourier Series coefficients of $x(t)$ and $y(t)$ are $X_k$ and $Y_k$ respectively, the FS coefficients of $z(t) = x(t)y(t)$ are $\sum_{l=-\infty}^\infty Z_k = X_l Y_{k-l}$. The proof is simple: $$Z_k = \frac{1}{T}\int_0^T z(t) \exp\left({\frac{-j2\pi k t}{T}}\right)dt \\ = \frac{1}{T}\int_0^T x(t)y(t) \exp\left({\frac{-j2\pi k t}{T}}\right)dt$$ Writing $x(t)$ in its Fourier Series form: $x(t) = \sum_{k=-\infty}^\infty X_k exp\left(\frac{j2\pi kt}{T}\right)$, we get $$Z_k = \frac{1}{T}\int_0^T \sum_{l=-\infty}^\infty X_l \exp\left(\frac{j2\pi lt}{T}\right)~ y(t) \exp\left({\frac{-j2\pi k t}{T}}\right)dt$$ Since $X_l$ is not a function of $t$ and integration and summation can be interchanged, we can write $$Z_k = \sum_{l=-\infty}^\infty X_l \frac{1}{T}\int_0^T \exp\left(\frac{j2\pi lt}{T}\right)~ y(t) \exp\left({\frac{-j2\pi k t}{T}}\right)dt \\ = \sum_{l=-\infty}^\infty X_l \frac{1}{T}\int_0^T y(t) \exp\left({\frac{-j2\pi (k-l) t}{T}}\right)dt \\ = \sum_{l=-\infty}^\infty X_l Y_{l-k}$$
• Remember that if $x(t)$ is even, $x(t) = x(-t)$. If $X_k$ is real, $X_k = X_k^*$ (The $^*$ represents complex conjugation).
• Similarly, if $x(t)$ is odd, $x(t) = -x(-t)$. If $X_k$ is imaginary, $X_k = -X_k^*$

P. III. Examples of Fourier Series

The following properties will be useful:

• Time shift: if $x(t) \rightarrow X_k$, $x(t-\tau) \rightarrow X_k e^{\frac{-j 2\pi k\tau}{T}}$