Lowpass Equivalent Representation

Bandpass and Lowpass Signals

Definition. A bandpass signal is a real signal whose frequency content is located around some non-zero frequency $f_c$.

$$ X(f) = \begin{cases} \text{nonzero}, & \left[ f_c - \frac{W}{2}, f_c+\frac{W}{2} \right] \\ 0, & {\text{elsewhere}} \end{cases}$$

 

Then, the spectrum of a bandpass signal is composed of two components such that

$$ \begin{align*} \text{positive spectrum }: X_{+}(f) &= \begin{cases} X(f), & f>0 \\ \frac{1}{2}X(0), & f = 0 \\ 0, & f<0 \end{cases} = X(f) \cdot u_{-1}(f) \\ \text{negative spectrum }: X_{-}(f) &= \begin{cases} X(f), & f<0 \\ \frac{1}{2}X(0), & f = 0 \\ 0, & f>0 \end{cases} = X(f)\cdot u_{-1}(-f) \end{align*} $$

 

Definition. A lowpass, or baseband, signal is a signal whose spectrum is located around the zero frequency. 

 

Hermitian Symmetry

$$ \begin{align*} \overline{X(f)} &= \overline{ \left( \int_{-\infty}^\infty e^{-2\pi i ft} x(t) dt \right) } \\ &= \int_{-\infty}^\infty e^{2\pi i ft} \overline{x(t)} dt \\ &= \int_{-\infty}^\infty e^{-2\pi i (-f)t} x(t)dt \\ &= X(-f) \end{align*} $$

Therefore, the Fourier transform of a real signal $x(t)$ has Hermitian Symmetry:

$$ X(-f) = \overline{X(f)} $$

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For a real signal $x(t)$,

$$ \begin{align*} x_+ (t) &= \mathcal{F}^{-1} \left( X_+ (f)\right) (t) \\ &= \mathcal{F}^{-1} \{ X(f) \cdot u_{-1}(f) \} \\ &= \mathcal{F}^{-1} \left( X(f) \right) (t) * \mathcal{F}^{-1} \left( u_{-1}(f) \right) (t) \\ &= x(t) * \left[ \frac{1}{2} \delta (t) - \frac{1}{2i\pi t} \right] = \frac{1}{2}x(t) + \frac{i}{2} \left( x(t) * \frac{1}{\pi t} \right) \\ &= \frac{1}{2}x(t) + i\frac{1}{2} \hat x (t), \end{align*} $$

where $ \hat x(t) = x(t) * \frac{1}{\pi t} = \int_{-\infty}^\infty \frac{x(\tau)}{\pi (t-\tau)} d\tau $ is the Hilbert transform.

 

Then, we have

$$ X_+ (f) = \mathcal{F} \left( \frac{1}{2}x(t) + i\frac{1}{2}\hat x (t) \right)(f) = \frac{1}{2}X(f) + i\frac{1}{2} \mathcal{F} \hat x (f) \left( \text{by linearity} \right) $$

 

Similarily, $x_{-}(t)$ and $X_-(f)$ can be written as

$$ \begin{align*} x_-(t) &= \mathcal{F}^{-1} (X_-(f))(t) \\ &= \mathcal{F}^{-1} \left( X(f) \right) (t) * \mathcal{F}^{-1} \left( u_{-1}(f) \right)(t) \\ &= x(t) * \left( \frac{1}{2}x(t) - i\frac{i}{2\pi t} \right) \\ &= \frac{1}{2}x(t) - j\frac{1}{2}\hat x (t) \\ X_-(f) &= \mathcal{F} \left( \frac{1}{2}x(t) - i\frac{1}{2}\hat x (t) \right) (f) \\ &= \frac{1}{2} X(f) - i\frac{1}{2} \mathcal{F} \hat x (f) \end{align*} $$

 

Leveraging the Hermitian Symmetry of $x(t)$ and $\hat x (t)$ and denoting the Fourier transform of $\hat x (t)$ as $\hat X (f)$, we obtain the following relationship

$$ \left( \text{Note that in reality, } \mathcal{F} \hat x (f) = -i \text{sgn}(f) X(f) \right) $$

$$ \begin{align*} \overline{X_+ (-f)} &= \overline{ \left( \frac{1}{2}X(-f) + i\frac{1}{2} \mathcal{F} \hat x (-f) \right) } \\ &= \frac{1}{2}\overline{ X(-f) } - i\frac{1}{2} \overline{ \hat X (-f) } \\ &= \frac{1}{2}X(f) - i\frac{1}{2}\hat X (f) \\ & \hspace{2cm} \left( \text{ Since } x(t) \text{ and } \hat x (t) \text{ are real,} \overline{X(-f)} = X(f) \text{ and } \overline{\hat X (-f)} = \hat X (f) \right) \\ &= X_-(f) \end{align*} $$

$$ \therefore \overline{X_+(-f)} = X_- (f) $$ 

 

Therefore, for the real bandpass signal $x(t)$,

$$ X(f) = X_+(f) + X_-(f) = X_+(f) + \overline{X_+(-f)}. $$

This implies that knowledge of the positive spectrum $X_+(f)$ only is sufficient to reconstruct $X(f)$.

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Baseband(lowpass) Equivalent (Complex Envelope)

Definition. The baseband equivalent representation, or the complex envelope, of $x(t)$ is defined as the signal $x_b(t)$ whose spectrum is

$$\mathcal{F} x_b(f) = X_b(f)=\sqrt{2} X_+(f+f_c)=\sqrt{2} X(f+f_c) \cdot u_{-1}(f+f_c). $$

The spectrum is centered about the zero frequency; hence, in general, $x_b(t)$ is a complex baseband waveform.

 

Remark. $X_+ (f)$ only contains positive frequency components, so it is NOT Hermitian symmetric. It thus follows that $x_b(t)$ is a complex signal. 

 

Then, we can write $x_b(t)$ as 

$$ \begin{align*} x_b(t) &= \mathcal{F}^{-1} X_b (t) = \mathcal{F}^{-1} \left( \sqrt{2} X_+ (f+f_c) \right) (t) \\ &= \sqrt{2} \mathcal{F}^{-1} \left( X_{+} (f) * \delta (f + f_c) \right) (t) \\ &= \sqrt{2} \left( x_{+} (t) \cdot e^{-2 \pi i f_ct} \right) \\ &= \frac{1}{\sqrt{2}} \left( x(t) + i \hat x (t) \right) e^{-2\pi i f_ct} \end{align*} $$

 

Hence, we have

$$ \begin{align*} x(t) &= \sqrt{2} \Re{ \{ x_b(t) e^{2\pi i f_ct} \} } \tag{1} \\ &= \frac{1}{\sqrt{2}} \left( x_b(t)e^{2\pi i f_ct} \right) + \frac{1}{\sqrt{2}} \overline{ \left( x_b(t)e^{2\pi i f_ct} \right) } \\ X(f) &= \frac{1}{\sqrt{2}} \left[ \mathcal{F}x_b(f) * \delta (f-f_fc) + \mathcal{F} ( \overline{x_b} (t) ) (f) * \delta (f+f_c) \right] \\ &= \frac{1}{\sqrt{2}} \left[ X_b(f) * \delta (f-f_c) + \overline{ X_b(-f) } * \delta (f+f_c) \right] \\ &= \frac{1}{\sqrt{2}} \left[ X_b (f-f_c) + \overline{ X_b (- f - f_c) } \right] \end{align*} $$ 

 

$$ \therefore \begin{align*} x(t) &= \sqrt{2} \Re{ \{ x_b(t) e^{2\pi i f_ct} \} } \\ X(f) &= \frac{1}{\sqrt{2}} \left[ X_b (f-f_c) + \overline{ X_b (-f - f_c) } \right] \end{align*} $$

 

Exploting Euler's formula, Eq.(1) can be decomposed further into

$$ \begin{align*} x_b(t) &= \sqrt{2} \Re{ \{ x_b(t) e^{2 \pi i f_ct} \} } \\ &= \sqrt{2} \Re{ \{ \left( \Re{ \{ x_b(t) \} } + i\Im{ \{ x_b(t) \} } \right) \left( \cos{2\pi f_ct} + i\sin{2\pi f_ct} \right) \} } \\ &= \sqrt{2} \left( \Re{ \{ x_b(t) \} } \cos{2\pi f_ct} - \Im{ \{ x_b(t) \} } \sin{2\pi f_ct} \right) \end{align*} $$

 

Then, the real(in-phase) and imaginary(quadrature) parts of the baseband signal $x_b(t)$ can be obtained by modulating $x(t)$ by $\sqrt{2}\cos{2\pi f_ct}$ and $-\sqrt{2}\sin{2\pi f_ct}$, respectively, followed by ideal low-pass filtering at the baseband $[-W/2, W/2]$.

$$ \begin{align*} x_I(t) = \Re{ \{ x_b(t) \} } &= \text{LPF} \left( \sqrt{2} x(t) \cos{2\pi f_ct} \right) \\ &= \text{LPF} \left( 2x_I(t)\cos^2{2\pi f_ct} - 2x_Q(t)\cos{2\pi f_ct} \sin{2\pi f_ct} \right) \\ &= \text{LPF} \left( x_I(t) \left( 1+\cos{2\pi \cdot 2f_ct} \right) - x_Q(t) \left( \sin{2\pi \cdot 2 f_ct} - 0 \right) \right) \\ &= x_I(t) \end{align*} $$

 

$$ \begin{align*} x_Q(t) = \Im{ \{ x_b(t) \} } &= \text{LPF} \left( -\sqrt{2} x(t) \sin{2\pi f_ct} \right) \\ &= \text{LPF} \left( -2 x_I(t) \cos{2\pi f_ct} \sin{2\pi f_ct} + 2 x_Q(t) \sin^2{2\pi f_ct} \right) \\ &= \text{LPF} \left( -x_I(t) \left( \sin{2\pi \cdot 2 f_ct} - 0 \right) + x_Q(t) \left( 1 - \cos{2\pi \cdot 2 f_ct} \right) \right) \\ &= x_Q(t) \end{align*} $$