A Design Tip

A Note on the Frequency Response for an Example 3^rdOrder System

There is a simple system that can be used to show many of the properties of Nyquist plots and how they behave. The system's transfer function is:

KG(s) = K/[s(st₁ + 1)(st₂ + 1)]

This system has the following characteristics:

The system is third order.
The system is type one, i.e. it has a single pole at the origin (s = 0).
The system has two real poles (at the reciprocal values of the time constants).

The Nyquist plot for this system is drawn below for the following parameters.

K = 100
t₁ = .01 sec
t₂ = .002 sec

Note the features in this Nyquist plot.

It asymptotically approaches -270^o.
It starts from infinity at an angle of -90^o.
It seems to have an asymptote at around -1.2. For very low frequencies, it seems to
It crosses the negative real axis as frequency increases.

The questions that arise from this mainly concern how to calculate the last two points. We'll tackle them one at a time.

The Asymptote

The transfer function - with jw substituted for s - is:

G(jw) = K/[jw(jwt₁ + 1)(jwt₂ +1)]

We can expand this with an eye to isolating real and imaginary parts.

G(jw) = K/[jw(-w²t₁t₂ + jw(t₁+ t₂) + 1)]

G(jw) = K/[jw(1 -w²t₁t₂ + jw(t₁+ t₂)]

Multiply through by the complex conjugate of the quadratic factor only, and you get:

G(jw) = K(1 -w²t₁t₂ - jw(t₁+ t₂)/[jw((1 -w²t₁t₂)² + w²(t₁+ t₂)²]

The real part of this (taking into accoun the common jw factor) is:

Re[G(jw)] = -K(t₁+ t₂)/[(1 -w²t₁t₂)²]

When the frequency is very small, this becomes:

Re[G(j0)] = -K(t₁+ t₂)

For the example used in the illustration above, we had:

K = 100
t₁ = .01 sec
t₂ = .002 sec

So, the asymptote for the real part should be Re[G(j0)] = -K(t₁+ t₂) = -100(.012) = -1.2. Here is a rescaled graph of that Nyquist plot showing the asymptote more clearly - and you can see pretty well that it matches our calculation.

Note that many texts show incorrect graphs that asymptotically approach the vertical axis.

Real Axis Intercept

The real axis intercept can be obtained by noting when the imaginary part of the frequency response is zero. Using our expression from above, we have:

G(jw) = K/[jw(1 -w²t₁t₂ + jw(t₁+ t₂)]

That expression will be real (have no imaginary part) when:

1 -w²t₁t₂= 0

or when:

w² = 1/(t₁t₂)

With a little algebra, you should be able to show that the frequency response is then:

-Kt₁t₂/(t₁+ t₂)

For the example used in the illustration above, we had:

K = 100
t₁ = .01 sec
t₂ = .002 sec

So the real part is equal to:

-100(.01)(.002)/(.01 + .002)
= -.002/.012
= -0.167

Again, examine a rescaled plot, and you can see that the real intercept is at -0.167. Here's the plot.