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!


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


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


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

  • $\begingroup$ 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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