Relating Fourier series coefficients through transformation

Question

Problem

A continuous, periodic function $x_1(t)$ has a fundamental frequency $\omega_1$ and Fourier coefficients $a_k$. Find the frequency $\omega_2$ of $x_2(t)$ in terms of $\omega_1$ and the Fourier coefficients $b_k$ of the same in terms of $a_k$.

$$x_2(t) = x_1(t - 1) + x_1(1 - t)$$

For the imaginary unit, use the symbol $j$.

Officially Provided Answer

It must be $e^{-jk\omega_1}[a_{-k}+a_k]$.

Officially Provided Transformation Rules.

• Linearity: If $y(t)=Ax_1(t)+Bx_2(t)$, and $x_1(t)$ has coefficients $a_k$ and $x_2(t)$ has $b_k$, then $y(t)$ has coefficients $Aa_k + Bb_k$.
• Time Shift: If $y(t)=x(t - t_0)$, and $x(t)$ has coefficients $a_k$, then $y(t)$ has $a_{-k}$.
• Time Reversal: If $y(t)=x(-t)$, and $x(t)$ has coefficients $a_k$, then $y(t)$ has $e^{-jkt_0\omega}a_k$, where $\omega$ is the fundamental frequency of $x(t)$.

My Solution

To Determine the coefficients of $h_1(t)=x_1(t-1)$.

Use the time shift rule with $t_0=1$. The coefficients $p_k$ of $h_1(t)$ are $e^{-j\omega_1k}a_k$.

To determine the coefficients of $h_2(t)=x_1(1-t)$.

Notice that $h_2(t)=h_1(-t)$. So, use the time reversal rule. The coefficients $q_k$ of $h_2(t)$ are $e^{j\omega_1k}a_{-k}$.

To determine the coefficients of $x_2(t) = h_1(t) + h_2(t)$.

Use the linearity property. The coefficients $b_k$ of $x_2(t)$ are $e^{-j\omega_1k}a_k+e^{j\omega_1k}a_{-k}$.

Question

My answer $e^{-j\omega_1k}a_k+e^{j\omega_1k}a_{-k}$ doesn't look like it can be reduced to $e^{-jk\omega_1}[a_{-k}+a_k]$. Where did I get it wrong?

• Is it one of the rules that are misremembered or misinterpreted?
• Is there a lapse in logic in my solution?

Auxiliary Information

The problem and the rules have been adapted from Oppenheim, A. V., Willsky, A. S., & Nawab, S. H. (1997). Signals and Systems (2nd ed., pp. 202, 251). London etc.: Prentice-Hall.

Update: Solution

I was pointed out that $h_2(t) \ne h_1(-t)$. Thus, the term was incorrectly derived.

$h_2(t)$ was instead correctly derived by first applying a shift of $t_0=-1$ to $x_1(t)$, and then applying a time reversal.

The newly calculated term is $e^{-j\omega_1kt}a_{-k}$.

With this term, the sum is now $e^{-j\omega_1kt}[a_k+a_{-k}]$, which is the same as the officially provided answer.

Answer 1

Your claim that $h_2(t)=h_1(-t)$ does not hold. $h_1(-t)=x_1(-t-1)\neq h_2(t)$. That is when your time reversing you only change the sign of the independent variable $t$ in the function argument and only that.

It is easier to proceed by relating $h_2(t)$ to $x_1(t)$ as follows:

• first, $x_1(t)$ is shifted by $t_0=-1$ resulting in the function $\tilde{x}_1(t)\triangleq x_1(t-(-1))=x_1(t+1)$
• then, $\tilde{x}_1(t)$ is time reversed to give $h_2(t) \triangleq \tilde{x}_1(-t)=x_1(-t+1)$

Note again that the time reversal only changes the sign of $t$ in the argument of ${x}_1$.

You should be able to compute the Fourier coefficients of $h_2$ from this.