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

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

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

which is the amplitude response, and

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

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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