- #1
arhzz
- 268
- 52
- Homework Statement
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- Relevant Equations
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Hello!
Consider this transfer function $$ G^\#(q) = q \cdot \frac{q^2 + 2q - 3}{q^2 - 25} $$
a) For which values of Ta > 0 is the G(q) step response capable?
b) For which values of Ta > 0 is the G(q) realizible?
c) Is it possible to find a sampling time Ta > 0 so that the G(q) is BIBO stable?
Now we have a table that defines all 3 of these for the G(q)
$$ \text{Realizability:} \quad \lim_{q \to \Omega_0} |G^\#(q)| < \infty $$
$$ \text{Jump response:} \quad \lim_{q \to \Omega_0} G^\#(q) \neq 0 $$
$$ \text{BIBO Stability:} \quad \text{only poles } q_i \text{ with } \Re(q_i) < 0 \text{ or } q_i = \Omega_0 \\ $$
Where ## \Omega_0## is defined as 2/Ta
So for c) I was able to get a solution. I looked at the denominator and saw that one Pole was -5 and the other 5. For the 5 we need ## \Omega_0 ## to be 5. We can achieve that when Ta = 2/5.
So my solution for c) is Ta = 2/5
a) I did the following. I pluged in ##\Omega_0 ## for q and look at the denominator again. Since we need the limit to not go to infinity, the only case where that happens is when the denominator is 0. I know that division with 0 is not defined, but in this class (control theory) we would say that division to zero (really really small number) would go to infinity.
Now this is what the denominator looks like when I plug in ##\Omega_0## and square it.
$$ \frac{4}{T^2} - 25 $$. Now we need it not to be equal to zero so I find for what values it is 0. And I get that it is 0 when Ta = 2/5
So to complete c) no it is not possible since the function cannot be realized for the sampling time T = 2/5
Now I cannot solve b). When I try plugging in ## \Omega_0 ## for q I just get a really complicated term that does now tell me much. Any advice?, or comments on my solution so far.
Many thanks!
Consider this transfer function $$ G^\#(q) = q \cdot \frac{q^2 + 2q - 3}{q^2 - 25} $$
a) For which values of Ta > 0 is the G(q) step response capable?
b) For which values of Ta > 0 is the G(q) realizible?
c) Is it possible to find a sampling time Ta > 0 so that the G(q) is BIBO stable?
Now we have a table that defines all 3 of these for the G(q)
$$ \text{Realizability:} \quad \lim_{q \to \Omega_0} |G^\#(q)| < \infty $$
$$ \text{Jump response:} \quad \lim_{q \to \Omega_0} G^\#(q) \neq 0 $$
$$ \text{BIBO Stability:} \quad \text{only poles } q_i \text{ with } \Re(q_i) < 0 \text{ or } q_i = \Omega_0 \\ $$
Where ## \Omega_0## is defined as 2/Ta
So for c) I was able to get a solution. I looked at the denominator and saw that one Pole was -5 and the other 5. For the 5 we need ## \Omega_0 ## to be 5. We can achieve that when Ta = 2/5.
So my solution for c) is Ta = 2/5
a) I did the following. I pluged in ##\Omega_0 ## for q and look at the denominator again. Since we need the limit to not go to infinity, the only case where that happens is when the denominator is 0. I know that division with 0 is not defined, but in this class (control theory) we would say that division to zero (really really small number) would go to infinity.
Now this is what the denominator looks like when I plug in ##\Omega_0## and square it.
$$ \frac{4}{T^2} - 25 $$. Now we need it not to be equal to zero so I find for what values it is 0. And I get that it is 0 when Ta = 2/5
So to complete c) no it is not possible since the function cannot be realized for the sampling time T = 2/5
Now I cannot solve b). When I try plugging in ## \Omega_0 ## for q I just get a really complicated term that does now tell me much. Any advice?, or comments on my solution so far.
Many thanks!