Behaviours Flashcards

1
Q

whats difference between steady state gain and steady state output?

A

s-s gain: –> K=lim s–>0 G(s)
G(0) –> asymptotic(s-s) ratio of the sytem o/p to i/p. G(s)=Y(s)/U(s). Assumes i/p is constant. It assumes that the o/p is convergent and thus dont need to check poles- we assume u(t) is such that limits exists and therefore works. therefore system gain always meaningfully defined. Could be infinite if G(s) has poles on origin aka p=0.

ss- output: –> asymptotic(basically at s-s) o/p of system. in practice thats what we are interested in. A=i/p magnitude.
We use FVT:
lim t–> infinity y(t) = lim s–>0 sY(s).
note: only applied if signal convergent, so MUST check first if poles are stable.

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2
Q

laplace tranforms are defined for a…

A

signal

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3
Q

what is a TF?

A

it is the system, G(s).
it gives the relationship btw the i/p and o/p of the system
U(s)—> G(s) —-> Y(s)
Y(s)=G(s)U(s)

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4
Q

what do the o/p Y(s) depend on

A

Y(s) contains poles from both i/p, u and the system,G and therfore the dynamics of the o/p depends on G(s) and U(s).
G(s) is the natural dynamics of the system(free response)
and
U(s) is the behaviour that enters the sytem(forced response)

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5
Q

what do we mean when we talk about instability?

A

When the signal Y(s) include RHP poles and therfore the signal Y(s) is divergent and goes to infinity.
As Y=GU…
… it can be unstable if i/p, U OR the inherent sytem dynamics, G OR both are divergent.

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6
Q

where does the instability originate from in an open and closed loop system?

A

but think about it. in open loop system you choose the input, why would put a divergent signal into sytem. Thats MAD! hence instability originates in the TF.
In a closed sytem, u is a target/ set point. instabilty willl typically arise from Gc(s) [closed loop TF] having RHP poles.
loop i/p is again assumed to be convergent so instability originates from TF.
In Summary, so even though Y depends on G and U, most of time instability originates from G(s). so if Y has RHP poles its cuz they originated from the RHP poles in G.
thats why convergence of Y(s) can be assessed solely by analysing G(s).

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7
Q

what is steady state gain?

A
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8
Q

what is steady-state output?

A

when U(s)=A/s
Only if G(s) has not integrators[when G(0)=infinity
Y(s) = G(s)U(s) = G(s)A/s
We use FVT:
lim t–> infinity y(t) = lim s–>0 sY(s).
sG(s)A/s =
G(0)A (as s in num and den cancel out)

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9
Q

The only converging signal with a non-zero asymptotic value is what?

A

when it is a contains a constant, aka a step function as L[u] will therefore give us A/s.
Therefore using FVT lim s–>0 (sF(s))
asymptotic value is A if all others converge to 0. The s at num and den cancel out that’s why so not timesing by 0/s anymore

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10
Q

if a transform does not contain a single ‘s’ in the denominator, what will Final Value be?

A

zero or infinite(except pure sinusoids)

FVT only applies if signal=converging.

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11
Q

what’s condition of FVT

A

FVT only applies if signal=converging.

if applied to Laplace transform of diverging signal, answer is meaningless

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12
Q

FVT w/ Matlab

A

y=impulse(tf([1 6],[1 4 3]),1000);

y(end)

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13
Q

Analysing behaviours, what are the things to look at. (4) SSSO

A
  1. Stability
  2. Settling time/ Speed of Response
  3. Shape of Response–> Smooth/Oscillitary
  4. Offset –> Do we reach the Target? How far off?
    If closed loop, we need the closed-loop gain/offset Gc(s)
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14
Q

How to calculate offset?

A

1/ 1 + G(0)M(0)

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15
Q

How to calculate s-s offset with a simple feedback loop?

A

Since e = 1/ 1 + GM
lim t–> ∞ e(t) = lim s–>0 ( s / 1 + G(0)M(0) ) * R(s)

note: only meaning full in step input R(s)=A/s
Therefore the s at num and den in e (error) cancel out

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16
Q

When considering offset and s-s, what must you ask yourself? Its only meaningful when…

A

Offset and s-s are only meaningful for STEP signal. Therefore we assume R(s)=A/s (as r=A)

Must check for stability first as well. ‘ iff closed-loop is stable’ If unstable, no offset as its diverging–> ∞

17
Q

What is the offset of a simple closed loop system?

A

e= ( 1 / 1 + GM ) * r

lim t–> ∞ e(t) = lim s–>0 ( s / 1 + GM) * R(s)

18
Q

With a stable closed loop system. G has an integrator. M=2 and r(s)= 3/s. What is the s-s offset?

A

lim t–> ∞ e(t) = lim s–>0 ( s / 1 + GM) * R(s)

G(0)–>∞
lim s–>0 3/∞ = 0 offset iff stable