# How can I show this equality for the inner product of two bandpass signals?

Given the real bandpass signals $$x(t)$$ and $$y(t)$$ with center frequency $$f_0$$ and lowpass equivalents $$x_l(t)$$ and $$y_l(t)$$ respectively, I want to show that $$\int_{-\infty}^\infty x(t)y(t) \ dt= \frac{1}{2} \text{Re}\left[\int_{-\infty}^\infty x_l(t)y_l^*(t) \ dt \right]$$ but I am stuck in the following derivation. Using Parseval's theorem, \begin{align} \int_{-\infty}^\infty x(t)y^*(t) \text{d}t &= \int_{-\infty}^\infty X(f)Y^*(f) \ \text{d}f \end{align} and since $$y(t)$$ is real, \begin{align} \int_{-\infty}^\infty X(f)Y^*(f) \ \text{d}f &= \int_{-\infty}^\infty X(f)Y(-f) \ \text{d}f \\ &= \int_{-\infty}^\infty \left[ \frac{1}{2}\left(X_l(f - f_0) + X_l^*(-f - f_0)\right) \right]\left[ \frac{1}{2}\left(Y_l(-f - f_0) + Y_l^*(f - f_0)\right) \right] \ \text{d}f \\ &= \int_{-\infty}^\infty \frac{1}{4}X_l(f - f_0)Y_l(-f - f_0) + \frac{1}{4}X_l^*(-f - f_0) Y_l(-f - f_0) ~+ \\ &\frac{1}{4}X_l(f - f_0)Y_l^*(f - f_0) + \frac{1}{4}X_l^*(-f - f_0)Y_l^*(f - f_0) \ \text{d}f \\ &= \int_{-\infty}^\infty \frac{1}{4}X_l^*(-f - f_0) Y_l(-f - f_0) + \frac{1}{4}X_l(f - f_0)Y_l^*(f - f_0) \ \text{d}f \\ &= \frac{1}{4} \int_{-\infty}^\infty X_l^*(-f - f_0) Y_l(-f - f_0) + \frac{1}{4} \int_{-\infty}^\infty X_l(f - f_0)Y_l^*(f - f_0) \ \text{d}f \end{align} I'm not sure what do after this, so any help would be appreciated.

The product $$x(t)y(t)$$ can be written as
\begin{align}x(t)y(t)&=\textrm{Re}\left\{x_l(t)e^{j2\pi f_ct}\right\}\textrm{Re}\left\{y_l(t)e^{j2\pi f_ct}\right\}\\&=\textrm{Re}\left\{x_l(t)e^{j2\pi f_ct}\right\}\frac12\big[y_l(t)e^{j2\pi f_ct}+y_l^*(t)e^{-j2\pi f_ct}\big]\\&=\frac12\textrm{Re}\left\{x_l(t)y_l^*(t)+x_l(t)y_l(t)e^{j4\pi f_ct}\right\}\tag{1}\end{align}
Integrating over $$(1)$$ gives
$$\int_{-\infty}^{\infty}x(t)y(t)dt=\frac12\textrm{Re}\left\{\int_{-\infty}^{\infty}x_l(t)y_l^*(t)dt\right\}+\frac12\textrm{Re}\left\{\int_{-\infty}^{\infty}x_l(t)y_l(t)e^{j4\pi f_ct}dt\right\}\tag{2}$$
The last term in $$(2)$$ is the Fourier transform of $$x_l(t)y_l(t)$$ evaluated at $$-2f_c$$. Since $$x_l(t)$$ and $$y_l(t)$$ are lowpass signals, that term vanishes and the result follows.
• ...\begin{align} \int_{-\infty}^\infty X(f)Y^*(f) \ \text{d}f &= \frac{1}{4} \int_{-\infty}^\infty X_l^*(-f - f_0) Y_l(-f - f_0) \ \text{d}f + \frac{1}{4} \int_{-\infty}^\infty X_l(f - f_0)Y_l^*(f - f_0) \ \text{d}f \\ &= \frac{1}{4} \left[\int_{-\infty}^\infty X_l(f - f_0)Y_l^*(f - f_0) \ \text{d}f\right]^* + \frac{1}{4} \int_{-\infty}^\infty X_l(f - f_0)Y_l^*(f - f_0) \ \text{d}f \\ &= \frac{1}{2} \text{Re}\left[\int_{-\infty}^\infty X_l(f - f_0) Y_l^*(f - f_0) \ \text{d}f\right] \\ &= \frac{1}{2} \text{Re}\left[\int_{-\infty}^\infty x_l(t) y_l^*(t) \ \text{d}t\right] \end{align}? Jan 28, 2022 at 15:38