### Rederiving the Frequency Response

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

*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*,

It is now easy to see that

and

*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 cut-off 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

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