1
$\begingroup$

Let the complex exponential Fourier series coefficients of two periodic signals $x_1(t)$ and $x_2(t)$ be $C_{1n}$ and $C_{2n}$, respectively, with $T_0$ being the fundamental time period of both the signals.

Let the signal $x(t)$ be equal to $x_1(t) * x_2(t)$, where $*$ denotes the periodic convolution operation, and its complex exponential Fourier series coefficient be $C_n$.

The aim is to find a relationship between $C_n$, $C_{1n}$ and $C_{2n}$.

$$C_n = \frac1{T_0} \int_{T_0}\left(x(t)\cdot e^{-jnw_0t}\right)dt$$ $$= \frac{1}{T_0} \int_0^{T_0}\left((x_1(t) * x_2(t)) \cdot e^{-jnw_0t}\right)dt $$$$= \frac{1}{T_0} \int_0^{T_0}\left(\left(\int_0^{T_0}\left(x_1(\tau) \cdot x_2(t - \tau)\right)d\tau\right) \cdot e^{-jnw_0t}\right)dt$$

Since $e^{-jnw_0t}$ is constant with respect to the integral $\int_0^{T_0}\left(x_1(\tau) \cdot x_2(t - \tau)\right)d\tau$, therefore we can write

$$C_n = \frac{1}{T_0} \int_0^{T_0}\left(\int_0^{T_0}\left(e^{-jnw_0t} \cdot x_1(\tau) \cdot x_2(t - \tau)\right)d\tau\right)dt $$$$= \frac{1}{T_0} \int_0^{T_0}\left(\int_0^{T_0}\left(\left(x_1(\tau) \cdot e^{-jnw_0\tau}\right) \cdot \left(x_2(t-\tau) \cdot e^{-jnw_0(t-\tau)}\right)\right)d\tau\right)dt$$

Now, I would like to break this double integral into a product of two integrals, but since $x_2(t-\tau) \cdot e^{-jnw_0(t-\tau)}$ involves both $t$ and $\tau$, therefore I don't know how to proceed further.

Could someone help me with this?

$\endgroup$

1 Answer 1

1
$\begingroup$

It looks like you're almost there. Note that for any $T$-periodic function $f(t)$ you have

$$\int_{0}^Tf(t)dt=\int_{0}^Tf(t+\tau)dt$$

because as long as you integrate over one whole period (or an integer number of periods), it doesn't matter where you start integrating.

Hence you can split the last expression in your question into the product of two integrals, the first over $\tau$ and the second over $t$. This will give you the desired solution.

$\endgroup$
8
  • 1
    $\begingroup$ Lemme see: $$ \int\limits_{0}^{1} e^{j2\pi x} \, \mathrm{d}x = \int\limits_{0}^{1} e^{j2\pi (x+0.1)} \, \mathrm{d}x $$ ?? Are you sure about that? $\endgroup$ Commented May 29, 2023 at 17:12
  • $\begingroup$ I think what you need to do is just adjust the limits of the integral by $\tau$. $\endgroup$ Commented May 29, 2023 at 17:23
  • $\begingroup$ I guess it's true, because it's zero on both sides. But I would still express that differently. $\endgroup$ Commented May 29, 2023 at 17:30
  • $\begingroup$ @robertbristow-johnson: Sure it's true. Either you change the integration variable or the integration bounds, same thing, as long as you integrate over one full period. That's why people also write $$\int_{<T>}f(t)dt$$ meaning some interval of length $T$, because it doesn't matter where the interval starts. $\endgroup$
    – Matt L.
    Commented May 29, 2023 at 18:35
  • 1
    $\begingroup$ @KushagrJaiswal: The product of two $T$-periodic functions is also $T$-periodic. Try to show it, it's straightforward! $\endgroup$
    – Matt L.
    Commented May 30, 2023 at 5:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.