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So I'm given a pass band filter with specific transfer function $H_p(f)$, I want to implement this via baseband processing. I already know how to take the input signal $u(t)$ and process it such that I get the I/Q components $u_I(t), u_Q(t)$. I also have figured out that the following convolution relation holds for the filter output $y$'s I/Q components $y_I=(u_I * h_I) - (u_Q*h_Q)$ and $y_Q = (u_I*h_Q) + (u_Q*h_I)$ where $h_Q$ and $h_I$ are the filters I/Q components.

I have run into an issue where I can't figure out a simple way to determine $h_Q$ and $h_I$. I know for the complex envelope of the filter I have: $h(t) = h_I(t) + j\cdot h_Q(t)$. I can take the fourier transform of this to obtain $H(f) = H_I(f) + j\cdot H_Q(f)$.

So essentially my issue is with how to obtain $H_I$ and $H_Q$. My book basically gives the answer but I must be missing something since I can't figure out the reasoning behind it. They state that $H_I(f) = (H(f) + H^*(-f))/2$ and $j\cdot H_Q(f) = (H(f)-H^*(-f))/2$.

I'm guessing its some property to do with the fourier that I've forgotten but I'm hoping someone can explain the reasoning for it. Thanks!

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Since $h_I(t)$ is the real part of $h(t)$ you have

$$h_I(t)=\frac12[h(t)+h^*(t)]\tag{1}$$

where $h^*(t)$ is the complex conjugate of $h(t)$. For the imaginary part you have

$$h_Q(t)=\frac{1}{2j}[h(t)-h^*(t)]\tag{2}$$

Since the Fourier transform of $h^*(t)$ is $H^*(-f)$1, the Fourier transforms of (1) and (2) are

$$H_I(f)=\frac12[H(f)+H^*(-f)]\\ H_Q(f)=\frac{1}{2j}[H(f)-H^*(-f)]$$


1The fact that $H^*(-f)$ is the Fourier transform of $h^*(t)$ can be easily seen as follows:

With $$H(f)=\int_{-\infty}^{\infty}h(t)e^{-j2\pi ft}dt$$ the Fourier transform of $h^*(t)$ is $$\mathcal{F}\{h^*(t)\}=\int_{-\infty}^{\infty}h^*(t)e^{-j2\pi ft}dt= \left[\int_{-\infty}^{\infty}h(t)e^{j2\pi ft}dt\right]^*=H^*(-f)$$

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  • $\begingroup$ Thanks exactly what I needed, totally forgot about the definition of Re() and Im() through conjugation! $\endgroup$ – user67081 Jan 12 '15 at 19:49

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