Performance in Sampled Systems - DC Gain, Decay Rate

System Performance in Sampled Systems

The questions that need to be answered are the following.

How do I compute DC gain?
How long does it take for a signal to die out?
Are there a lot of oscillations?

We will answer these questions in order.

Computing DC Gain in a Sampled System

In continuous systems the answer to this question really relies on the final value theorem. In linear systems we have:

In other words, the final value (as time becomes very large) can be computed from the value of the transform at s = 0. In sampled systems, the point z = 1 often plays the role that s = 0 plays in continuous systems. The final value theorem for sampled systems is:

Now, the transform of a sampled unit step is:

Transform of a Sampled Step = z/(z - 1)

And, that is the transform of this signal:

So, if you multiply the transform of the step times the transfer function and take the limit, you get the limit shown below.

And, that's the story on DC gain.

How Quickly Does a Response Die Out?

How quickly a system response is always an issue in control systems. Often the control problem is to speed up the response of a system to get it within some sort of specification. In continuous systems rise time and settling time are measures of response time. In sampled systems it is best to start by determining how long it takes the system to get within a certain percentage of the final response (settling time) but by measuring that as a number of sample periods. In ordere to get a handle on response time, we will need to examine the response of some typical, low order systems.

First Order Step Response

We will assume that we have a first order system with a step input.

The transfer function of the system is G[z] = G/(z - a)

That means that the DC gain of the system is G/(1 - a)

The transform of the step input is z/(z - 1)
The difference equation satisfied by the system is:

y_k+1 = ay_k + Gu_k

Now, to compute the response of the system, we can use Z-transform methods.

The output transform is the produce of the input transform and the transfer function.

Output transform = Y[z] = zG[z]/(z - 1)
Y[z] = zG/[(z - a)(z - 1)]

If we do a partial fraction expansion on this we have:

Y[z]/z = A/(z - a) + B/(z - 1)

We do Y[z]/z to get a standard form later.

A = -G/(1 - a)
B = G/(1 - a)

And, the output sequence is:

Y[z] = G/[(1 - a)(z - 1)] - G/[(1 - a)(z - a)]
y_k = G/(1 - a) - (a^k)[G/(1 - a)]

So, the response will asymptotically approach the steady state value of G/(1 - a) (which is the DC gain multiplied by the input, 1.0)

The distance from the steady state value decreases as (a^k)[G/(1 - a)]. The (a^k) factor is like a deacying exponential in a continuous system - and when you graph it, it looks like samples taken from a decaying exponential. Consider this:

(a^k) <=> (e^-kT/^t) =(e^-T/^t)^k
So, we can say: a = e^-T/^t
or, ln(a) = -T/t
or, and equivalent time constant in this system is:

t = -T/ln(a) (And, remember a is less than one, so the logarithm is negative.)
And, this equivalent time constant is peculiar because it changes if you change the sample period, T. If you sample faster, the system responds faster. That's because we assumed that a was constant. If you have a first order continuous system, and you change the sampling period, a would change - if you are driving the system with a zero order hold signal. So, be very careful applying this concept.

One question that arises is how long will it take for the response to die out to within a certain percentage of the final value. We can do that calculation.

Basically, you are asking "When does (a^k) become equal to FractionalValue?"
(a^k) = FractionalValue
kln(a) = ln(FractionalValue)
k = ln(FractionalValue)/ln(a)
And, remember that both the FractionalValue and a are numbers less than 1.0 so both ln() values are negative.

Example:

You want to design a system that settles out to within 5% in five sample periods. What is the value of a?

To solve for the value of, use k = 5, and FractionalValue = 5%. Then we have.

k = ln(FractionalValue)/ln(a), or
ln(a) = ln(FractionalValue)/k
a = FractionalValue^1/k
a = .05^1/5 = 0.549

What if a is negative?

If a is negative, the same argument holds, but the amount of decay each sample period is given by |a |. We need to adjust our result to:

k = ln(FractionalValue)/ln(|a|)

Now, the same question can be posed for a system with two (complex?) poles. How can we estimate the time it takes a second order sampled system to get within some given fractional value and stay there.

We assume that the the system will have a transfer function with a denominator of z² + a z+ b.

We know that a quadratic factor like this will contribute two terms to the response.

C₁z₁^k+ C₂z₂^k
Here, z₁ and z₁are the roots of the quadratic polynomial in z (above) and are given by:

z₁ = (b + jd) = Ae^j^f
z₂ = (b - jd) = Ae^-j^f

Here, the coefficients in the quadratic are:
a = 2b and b = b² + d² = |A|²

And, the response, is going to have two terms.

C₁A^ke^j^fk+ C₂A^ke^-j^fk

We can bound the response (taking advantage of the fact that C₁ = C₁^*, so |C₁| = |C₂|).

C₁A^ke^j^fk+ C₂A^ke^-j^fk < 2|C₁||A|^k

And this will decay to some fractional value when:

|A|^k = FractionalValue or when: kln(|A|) = ln(FractionalValue) or when: k = [FractionalValue]/ln(|A|)

Interpreting This Result

The result is really the same for both cases. We can rephrase the result in this manner.

The amount of decay in a single sample period is the distance of the pole from the origin of the z-plane.

For a single real pole, the distance from the origin is |a| and that is how much the response due to that pole decays in magnitude each sample period.
For complex poles, the distance from the origin is SQRT(b² + d²)= A. (Remember b and d are the real and the imaginary parts of the pole.)