Let's repeat the mathematical sine-wave analysis of the simplest
low-pass filter, but this time using a complex sinusoid instead of a
real one. Thus, we will test the filter's response at frequency
by setting its input to

Again, because of time-invariance, the frequency response will not
depend on , so let . Similarly, owing to linearity, we
may normalize to 1. By virtue of Euler's relation Eq. (1.8) and
the linearity of the filter, setting the input to
is physically equivalent to putting
into one
copy of the filter and
into a separate copy of the
same filter. The signal path where the cosine goes in is the
real part of the signal, and the other signal path is simply called
the imaginary part. Thus, a complex signal in real life is
implemented as two real signals processed in parallel; in particular,
a complex sinusoid is implemented as two real sinusoids, side by side,
one-quarter cycle out of phase. When the filter itself is real, two
copies of it suffice to process a complex signal. If the filter is
complex, we must implement complex multiplies between the complex
signal samples and filter coefficients.

Using the normal rules for manipulating exponents, we find that the
output of the simple low-pass filter in response to the complex
sinusoid at frequency
Hz is given by

where we have defined
, which we
will show is in fact the frequency response of this filter at
frequency . This derivation is clearly easier than the
trigonometry approach. What may be puzzling at first, however, is
that the filter is expressed as a frequency-dependent complex
multiply (when the input signal is a complex sinusoid). What does
this mean? Well, the theory we are blindly trusting at this point
says it must somehow mean a gain scaling and a phase shift. This is
true and easy to see once the complex filter gain is expressed in
polar form,

where the gain versus frequency is given by
(the absolute value, or complex modulus of ), and the
phase shift in radians versus frequency is given by the phase angle
(complex argument)
. In
other words, we must find

which is the phase response. There is a trick we can call ``balancing
the exponents,'' which will work nicely for the simple low-pass of
Eq. (1.1).

It is now easy to see that

and

We have derived again the graph of Fig.1.7, which shows the
complete frequency response of Eq. (1.1). The gain of the simplest
low-pass filter varies, as cosine varies, from 2 to 0 as the
frequency of an input sinusoid goes from 0 to half the sampling
rate. In other words, the amplitude response of Eq. (1.1) goes
sinusoidally from 2 to 0 as goes from 0 to . It does
seem somewhat reasonable to consider it a low-pass, and it is a poor
one in the sense that it is hard to see which frequency should be
called the cutoff frequency. We see that the spectral ``roll-off'' is
very slow, as low-pass filters go, and this is what we pay for the
extreme simplicity of Eq. (1.1). The phase response
is linear in frequency, which gives rise to a constant
time delay irrespective of the signal frequency.

It deserves to be emphasized that all a linear time-invariant filter
can do to a sinusoid is scale its amplitude and change
its phase. Since a sinusoid is completely determined by its amplitude
, frequency , and phase , the constraint on the filter is
that the output must also be a sinusoid, and furthermore it must be at
the same frequency as the input sinusoid. More explicitly:

Mathematically, a sinusoid has no beginning and no end, so there really are
no start-up transients in the theoretical setting. However, in practice, we
must approximate eternal sinusoids with finite-time sinusoids whose starting
time was so long ago that the filter output is essentially the same as if the
input had been applied forever.

Tying it all together, the general output of a linear time-invariant
filter with a complex sinusoidal input may be expressed as