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Inside Collection (Textbook):

Textbook by: Carlos E. Davila. E-mail the author

# Transfer Functions and Frequency Response

Module by: Carlos E. Davila. E-mail the author

Summary: Looks at the relationship between the transfer function and the frequency response.

We saw in (Reference) that the transfer function of a linear time-invariant system is given by

H ( s ) = Y ( s ) X ( s ) H ( s ) = Y ( s ) X ( s )
(1)

If we assume that H(s)H(s) is a rational function of ss then we can write

H ( s ) = ( s - β 1 ) ( s - β 2 ) ( s - β q ) ( s - α 1 ) ( s - α 2 ) ( s - α p ) H ( s ) = ( s - β 1 ) ( s - β 2 ) ( s - β q ) ( s - α 1 ) ( s - α 2 ) ( s - α p )
(2)

where β1,β2,...,βqβ1,β2,...,βq are the zeros, and α1,α2,...,αpα1,α2,...,αp are the poles of H(s)H(s). The poles and zeros are points in the ss-plane where the transfer function is either non-existent (for a pole) or zero (for a zero). These points can be plotted in the ss-plane with “××" representing the location of a pole and a “" representing the location of a zero. Since

H j Ω = H ( s ) s = j Ω H j Ω = H ( s ) s = j Ω
(3)

we have

H j Ω = j Ω - β 1 j Ω - β 2 j Ω - β q j Ω - α 1 j Ω - α 2 j Ω - α p H j Ω = j Ω - β 1 j Ω - β 2 j Ω - β q j Ω - α 1 j Ω - α 2 j Ω - α p
(4)

and

H j Ω = k = 1 q j Ω k - β k - l = 1 p j Ω l - α l H j Ω = k = 1 q j Ω k - β k - l = 1 p j Ω l - α l
(5)

Each of the quantities jΩ-βkjΩ-βk have the same magnitude and phase as a vector from the zero βkβk to the jΩjΩ axis in the complex ss-plane. Likewise, the quantities jΩ-αkjΩ-αk have the same magnitude and phase as vectors from the pole αkαk to the jΩjΩ axis.

Example 3.1 Consider a first-order lowpass filter with transfer function

H ( s ) = 1 s + 2 H ( s ) = 1 s + 2
(6)

Then

H j Ω = 1 j Ω + 2 H j Ω = 1 j Ω + 2
(7)

The quantity jΩ+2jΩ+2 has the same magnitude and phase as a vector from the pole at s=-2s=-2 to the jΩjΩ axis in the complex plane. The magnitude of the frequency response is the inverse of the magnitude of this vector. The length of jΩ+2jΩ+2 increases as ΩΩ increases thereby making HjΩHjΩ decrease as one would expect of a lowpass filter.

Example 3.2 Consider a first-order highpass filter with transfer function

H ( s ) = s s + 2 H ( s ) = s s + 2
(8)

Then

H j Ω = j Ω j Ω + 2 H j Ω = j Ω j Ω + 2
(9)

When Ω=0Ω=0, HjΩ=0HjΩ=0, but as ΩΩ approaches infinity, the two vectors jΩjΩ and jΩ+2jΩ+2 have equal lengths, so the magnitude of the frequency response approaches unity.

Example 3.3 Let's now look at the transfer function corresponding to a second-order filter

H ( s ) = s + 1 ( s + 1 + j 5 ) ( s + 1 - j 5 ) H ( s ) = s + 1 ( s + 1 + j 5 ) ( s + 1 - j 5 )
(10)

Or

H j Ω = j Ω + 1 j Ω + 1 + j 5 j Ω + 1 - j 5 H j Ω = j Ω + 1 j Ω + 1 + j 5 j Ω + 1 - j 5
(11)

The pole-zero plot for this transfer function is shown in Figure 1. The corresponding magnitude and phase of the frequency response are shown in Figure 2.

The two poles are at s=-3±j5s=-3±j5 and the zero is at s=-1s=-1. When the frequency gets close to either one of the poles, the frequency response magnitude increases since the lengths of one of the vectors jΩ±j5jΩ±j5 is small.

In the previous example we saw that as the poles get close to the jΩjΩ axis in the ss-plane, the frequency response magnitude increases at frequencies that are close to the poles. In fact, when a pole is located directly on the jΩjΩ axis, the filter's frequency response magnitude becomes infinite, and will begin to oscillate. If a zero is located on the jΩjΩ axis, the filters frequency response magnitude will be zero.

Example 3.4 Suppose a filter has transfer function

H ( s ) = s s 2 + 25 H ( s ) = s s 2 + 25
(12)

The two poles are at s=±j5s=±j5 and there is a zero at s=0s=0. From (Reference), the impulse response of the filter is h(t)=cos(5t)u(t)h(t)=cos(5t)u(t). This impulse response doesn't die out like most of the impulse responses we've seen. Instead, it oscillates at a fixed frequency. The filter is called an oscillator. Oscillators are useful for generating high-frequency sinusoids used in wireless communications.

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