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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.

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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.

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  • $\begingroup$ (+1) great answer. Much simpler than my approach. Just one question: could I have concluded after the last line in my derivation that, because the inner product is conjugate symmetric and invariant to frequency reversal and shifting, then... $\endgroup$
    – mhdadk
    Jan 28 at 15:37
  • $\begingroup$ ...$$\begin{align} \int_{-\infty}^\infty X(f)Y^*(f) \ \text{d}f &= \frac{1}{4} \int_{-\infty}^\infty X_l^*(-f - f_0) Y_l(-f - f_0) \ \text{d}f + \frac{1}{4} \int_{-\infty}^\infty X_l(f - f_0)Y_l^*(f - f_0) \ \text{d}f \\ &= \frac{1}{4} \left[\int_{-\infty}^\infty X_l(f - f_0)Y_l^*(f - f_0) \ \text{d}f\right]^* + \frac{1}{4} \int_{-\infty}^\infty X_l(f - f_0)Y_l^*(f - f_0) \ \text{d}f \\ &= \frac{1}{2} \text{Re}\left[\int_{-\infty}^\infty X_l(f - f_0) Y_l^*(f - f_0) \ \text{d}f\right] \\ &= \frac{1}{2} \text{Re}\left[\int_{-\infty}^\infty x_l(t) y_l^*(t) \ \text{d}t\right] \end{align}$$? $\endgroup$
    – mhdadk
    Jan 28 at 15:38

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