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 f_{c} can always be represented as

s(t)=a(t) cos(2πf_{c}t+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

s_{l}(t)=a(t)e^{jθ(t)}

Conversely, you may get s(t) back from s_{l}(t) (and f_{c}) using the formula

s(t)=Re[s_{l}(t) e^{j2πfct}].

You can easily verify this using Euler's formula.

Alternatively, if s(t) is represented as s(t)=x(t) cos(2πf_{c}t)-y(t) sin(2πf_{c}t), then the corresponding equivalent lowpass signal is s_{l}(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

s_{l}(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(2πf_{c}t+θ(t)), it can be shown that its Hilbert transform is s'(t)=a(t) sin(2πf_{c}t+θ(t)), so (s(t)+js'(t))
e^{-j2πfct} is just a(t) e^{jθ(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 f_{c}, S(f) must have two peaks centered near f_{c} and -f_{c} respectively, so the analytic signal just contains the peak near f_{c}, and by the multiplication of e^{-j2πfct} (frequency shift by -f_{c}), 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 f*_{c} 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 f_{c}.

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 f
_{c}, then the modulated signal is s(t)=a(t)cos(2πf_{c}t+&theta_{0}), where &theta_{0} is the initial phase. The corresponding equivalent lowpass signal is s_{l}(t)=a(t)e^{jθ0}, 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 {a
_{n}}, and those on the quadrature carrier are {b_{n}}, 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)=Σ_{n} a_{n}g(t-nT)cos(2πf_{c}t)-b_{n}g(t-nT)sin(2πf_{c}t),
(Σ is the summation sign)

while the equivalent lowpass signal is simply

s_{l}(t)= Σ_{n} (a_{n}+jb_{n})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 s_{l}(t)=x(t)+jy(t), thus s(t)=x(t)cos(2πf_{c}t)-y(t)sin(2πf_{c}t).
Now, the signal energy E is the time-integral of s^{2}(t), so we try to simplify s^{2}(t), obtaining

s^{2}(t)=(1/2)(x^{2}(t)+y^{2}(t))+(1/2)(x^{2}(t)-y^{2}(t))cos(4πf_{c}t)-x(t)y(t)sin(4πf_{c}t)

*With the assumption that f*_{c} is much larger than the signal bandwidth, x(t) and y(t) should change much more slowly than cos(4πf_{c}t) or sin(4πf_{c}t), 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 = ∫s^{2}(t)dt ≅ (1/2) ∫(x^{2}(t)+y^{2}(t))dt = (1/2) ∫|s_{l}(t)|^{2}dt,

where all integrations are done on (-∞,+∞). In other words, the signal energy can be approximately calculated with s_{l}(t) directly, using the same formula except that you will need to take the modulus of s_{l}(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, r_{l}(t), s_{l}(t) and h_{l}(t) denote the respective equivalent lowpass signals, and "*" denotes convolution, then

r_{l}(t) = (1/2) s_{l}(t) * h_{l}(t) = (1/2) ∫_{-∞}^{+∞} s_{l}(t-τ)h_{l}(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.