Filtering Passband Signals using Complex Baseband Filtering 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

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

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