But since the w-plane can be regarded an analogue of the s-plane, i.e. s = σ + j w in the s-plane and w = σw + j ww in the z-plane, for the frequency response we can set
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w = j ww |
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G( j w |
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1.2 j ww − 2.8w2w |
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The magnitude and the phase are then given by
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G( j w |
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1 + ww |
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1.2 |
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G( j ww ) = tan−1 |
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where ww is related to w by the expression |
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ww = tan |
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wT |
8.6 BODE DIAGRAMS
The Bode diagrams used in the analysis of continuous-time systems are not very practical when used directly in the z-plane. This is because of the e j ωT term present in the sampled data system transfer functions when the frequency response is to be obtained. However, it is possible to draw the Bode diagrams of sampled data systems by transforming the system into the w-plane by making the substitution
where the frequency in the w-plane (ww ) is related to the frequency in the s-plane (w) by the expression
It is common in practice to use a similar transformation to the one given above, known as the w -plane transformation, which gives a closer analogy between the frequency in the s-plane and the w -plane. The w -plane transformation defined as
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(8.9) |
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= T z |
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or |
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and the frequencies in the two planes are related by the expression |
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(8.11) |
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206 System Stability
Note that for small values of the real frequency (s-plane frequency) such that wT is small, (8.11) reduces to
w = T tan |
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(8.12) |
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Thus, the w -plane frequency is approximately equal to the s-plane frequency. This approximation is only valid for small values of wT such that tan(wT /2) ≈ wT , i.e.
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which can also be written as |
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w ≤ |
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10T |
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w ≤ |
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where ws is the sampling frequency in radians per second. The interpretation of this is that the w -plane and the s-plane frequencies will be approximately equal when the frequency is less than one-tenth of the sampling frequency.
We can use the transformations given in (8.10) and (8.11) to transform a sampled data system into the w -plane and then use the standard continuous system Bode diagram analysis.
Some example Bode plots for sampled data systems are given below.
Example 8.13
Consider the closed-loop sampled data system given in Figure 8.11. Draw the Bode diagram and determine the stability of this system. Assume that T = 0.1 s.
Solution |
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From Figure 8.11, |
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s 5 |
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s T |
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0.5 |
or |
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− − |
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G(z) = |
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1 − e−sT |
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y(s) |
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s + 5 |
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Figure 8.11 Closed-loop system
Transforming the system into the w -plane gives
G(w ) = 0.393 − 0.0196w , 0.08w − 0.393
or
G(w ) = 4.9(1 − 0.05w ) . w + 4.9
The magnitude of the frequency response and the phase can now be calculated if we set w = j v where v is the analogue of true frequency ω. Thus,
G( j v) = 4.9(1 − 0.05 j v) . j v + 4.9
The magnitude is
|G( j v)| = 4.9√1 + (0.25v)2 v2 + 4.92
and
G( j v) = − tan−1(0.05) − tan−1 v . 4.9
The Bode diagram of the system is shown in Figure 8.12. The system is stable.
Figure 8.12 Bode diagram of the system
208 System Stability
Example 8.14
The loop transfer function of a unity feedback sampled data system is given by
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0.368z + 0.264 |
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Draw the Bode diagram and analyse the stability of the system. Assume that T = 1 s.
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Solution |
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Using the transformation |
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(T /2)w = |
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we get |
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G(w) |
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0.368(1 + 0.5w/1 − 0.5w) + 0.264 |
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− 0.5w)2 − 1.368(1 + 0.5w/1 − 0.5w) + 0.368 |
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G(w) |
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0.0381(w − 2)(w + 12.14) |
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w(w |
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To obtain the frequency response, we can replace w with j v, giving |
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G( j v) |
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0.0381( j v − 2)( j v + 12.14) |
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j v( j v |
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The magnitude is then
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√ |
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G( j v) |
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v√ |
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and |
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v2 + 0.9242 |
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G( j v) |
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tan−1 |
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tan−1 |
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90 |
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tan−1 |
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12.14 |
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8.7 EXERCISES
1.Given below are the characteristic equations of some sampled data systems. Using Jury’s test, determine if the systems are stable.
(a)z2 − 1.8z + 0.72 = 0
(b)z2 − 0.5z + 1.2 = 0
(c)z3 − 2.1z2 + 2.0z − 0.5 = 0
(d)z3 − 2.3z2 + 1.61z − 0.32 = 0
2.The characteristic equation of a sampled data system is given by
(z − 0.5)(z2 − 0.5z + 1.2) = 0.
Determine the stability of the system.
