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I have a question regarding the equivalent lowpass signal $x_{LP}$ of a bandpass signal $x(t)$ and the quadrature component.

I'd like to find a condition on $x_{LP}(t)$ and $X_{LP}(f)$ such that the quadrature component of $x(t)$ is zero. (for example $x(t)=a(t)cos(2 \pi f_c t)$).

I know that $x(t)=Re\{(x(t)+j\hat{x}(t))e^{jw_0t}\}\\ = x(t)cos(w_0t)-\hat{x}(t)sin(w_0t)$

So my idea was that the hilbert transform of x(t) should be zero, in order that the signal has no quadrature component. That would mean that $X_{LP}(f)$ has even symmetry?

I hope that someone can help me with this question!

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Your idea with the Hilbert transform doesn't work. The only signal (apart from $x(t)=0$) for which the Hilbert transform is zero, is a constant signal.

A band pass signal $x(t)$ can be written in terms of its complex envelope $x_{LP}(t)$:

$$\begin{align}x(t)&=\textrm{Re}\left\{x_{LP}(t)e^{j\omega_0t}\right\}\\&=\textrm{Re}\{x_{LP}(t)\}\cos(\omega_0t))-\textrm{Im}\{x_{LP}(t)\}\sin(\omega_0t))\tag{1}\end{align}$$

From $(1)$ it is clear that the quadrature component is zero if the imaginary part of the complex envelope $x_{LP}(t)$ is zero, i.e., if $x_{LP}(t)$ is real-valued. In the frequency domain, this is equivalent with $X_{LP}(f)$ being conjugate symmetric: $X_{LP}(f)=X_{LP}^*(-f)$

Note that your equation involving the Hilbert transform is wrong. The complex band pass signal

$$x_a(t)=x_{LP}(t)e^{j\omega_0t}\tag{2}$$

is called the analytic signal. Its real part equals the (real-valued) band pass signal $x(t)$, and its imaginary part is the Hilbert transform of its real part:

$$x_a(t)=x(t)+j\hat{x}(t)\tag{3}$$

where $\hat{}$ denotes the Hilbert transform. So we simply have

$$x(t)=\textrm{Re}\{x_a(t)\}\tag{4}$$

which is the same as Eq. $(1)$.

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