Note: the square root of -1 is denoted as "j" (rather than "i" in mathematics books) in the following text, according to the tradition in engineering books.

The equivalent lowpass representation is an extremely handy way to represent a real-valued bandpass signal (a signal whose a bandwidth much smaller than its center frequency, such as an AM signal) by a complex-valued lowpass signal (a signal with its spectrum concentrated around zero frequency). It is very similar to phasors, except that phasors are usually constants with respect to time, so they represent sinusoid signals, while equivalent lowpass signals are functions of time, representing the instantaneous amplitude and carrier phase of the corresponding bandpass signals, so they can actually represent any bandpass signal.

Specifically, a real bandpass signal s(t) with center frequency fc can always be represented as

s(t)=a(t) cos(2πfct+θ(t))

where a(t) can be viewed as the amplitude and θ(t) the phase of the carrier. In this case, we define the corresponding equivalent lowpass signal as

sl(t)=a(t)ejθ(t)

Conversely, you may get s(t) back from sl(t) (and fc) using the formula

s(t)=Re[sl(t) ej2πfct].

You can easily verify this using Euler's formula.

Alternatively, if s(t) is represented as s(t)=x(t) cos(2πfct)-y(t) sin(2πfct), then the corresponding equivalent lowpass signal is sl(t)=x(t)+jy(t). This can be derived from the above definition by expanding the cosine term.

If the bandpass signal s(t) is not represented as a sinusoid with time-varying amplitude and phase as above, its equivalent lowpass signal can be calculated using Hilbert transform. First calculate the Hilbert transform of s(t), denoted as s'(t) (this is not s(t)'s derivative!), then

sl(t)=(s(t)+js'(t)) e-j2πfct.

s(t)+js'(t) is usually called the analytic signal of s(t). For s(t)=a(t) cos(2πfct+θ(t)), it can be shown that its Hilbert transform is s'(t)=a(t) sin(2πfct+θ(t)), so (s(t)+js'(t)) e-j2πfct is just a(t) ejθ(t), consistent with the above definition.

In the frequency domain, the meanings of analytic signals and equivalent lowpass signals are intuitive: suppose s(t) becomes S(f) after a Fourier transform, then s'(t) becomes -j sgn(f)S(f), and s(t)+js'(t) becomes (1+sgn(f))S(f), or 2S(f) for f>0 and zero otherwise. Since s(t) is a bandpass signal with center frequency fc, S(f) must have two peaks centered near fc and -fc respectively, so the analytic signal just contains the peak near fc, and by the multiplication of e-j2πfct (frequency shift by -fc), the equivalent lowpass signal has that peak moved to around f=0.

The advantages of using the equivalent lowpass representation of bandpass signals are basically (1) things get simpler by getting rid of the carrier, and (2) when fc is much larger than the signal bandwidth (which is commonly the case), you can make use of the equivalent lowpass signal directly to calculate signal energy, filter the signal, etc., without caring about the actual value of fc.

To explain the first advantage, let's look at the following two examples:

• Double sideband AM with suppressed carrier: Suppose the modulating lowpass signal is a(t), the carrier frequency is fc, then the modulated signal is s(t)=a(t)cos(2πfct+&theta0), where &theta0 is the initial phase. The corresponding equivalent lowpass signal is sl(t)=a(t)e0, which is simpler since it has only one time-dependent part.
• Quadrature phase-shift keying (QPSK): Suppose the bits on the in-phase carrier are {an}, and those on the quadrature carrier are {bn}, T is the symbol interval, g(t) is the pulse shape (such as a rectangular pulse on [0,T]), and we ignore the initial phase, then the modulated signal becomes

s(t)=Σn ang(t-nT)cos(2πfct)-bng(t-nT)sin(2πfct), (Σ is the summation sign)

while the equivalent lowpass signal is simply

sl(t)= Σn (an+jbn)g(t-nT).

As for the second advantage, we first take note that the addition of two signals and the multiplication of a signal with a constant works the same way whether bandpass or equivalent lowpass representations are used. When calculating the energy of a bandpass signal s(t), suppose the equivalent lowpass form is sl(t)=x(t)+jy(t), thus s(t)=x(t)cos(2πfct)-y(t)sin(2πfct). Now, the signal energy E is the time-integral of s2(t), so we try to simplify s2(t), obtaining

s2(t)=(1/2)(x2(t)+y2(t))+(1/2)(x2(t)-y2(t))cos(4πfct)-x(t)y(t)sin(4πfct)

With the assumption that fc is much larger than the signal bandwidth, x(t) and y(t) should change much more slowly than cos(4πfct) or sin(4πfct), thus the latter two terms will be much smaller than the first one after integration, since these two terms change sign rapidly and cancel itself after integration. Therefore,

E = ∫s2(t)dt ≅ (1/2) ∫(x2(t)+y2(t))dt = (1/2) ∫|sl(t)|2dt,

where all integrations are done on (-∞,+∞). In other words, the signal energy can be approximately calculated with sl(t) directly, using the same formula except that you will need to take the modulus of sl(t) before squaring it (not surprising considering the non-negativity of energy), and the result must be multiplied by 1/2.

The convolution rule for equivalent lowpass signals can be obtained likewise. Suppose r(t) = s(t) * h(t), where r(t), s(t) and h(t) are bandpass signals, rl(t), sl(t) and hl(t) denote the respective equivalent lowpass signals, and "*" denotes convolution, then

rl(t) = (1/2) sl(t) * hl(t) = (1/2) ∫-∞+∞ sl(t-τ)hl(τ) dτ

In other words, when filtering a bandpass signal with a bandpass filter by convoluting the equivalent lowpass forms of the signal and the impulse response directly, the result must be multiplied by 1/2.

FIXME: Should also contain something about equivalent lowpass forms of bandpass random signals, but I'm currently too tired to write it.