# Quadrature component - condition on equivalent lowpass signal

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!

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. $$(1)$$.