為什麼傳遞函數分母中s的階數n必不小於分子中s的階數m?
這個問題涉及到一些信號與系統的知識。我數學底子不好,就只談理解了,歡迎各位指證錯誤。首先,我們要了解什麼叫因果系統,因果系統就是系統當前的輸出只決定於當前以及過去的輸入,而與未來的輸入無關——因果系統才是在物理上可以實現的。@Garic Chuh貼出的答案其實把問題簡化在離散域,可以在離散z域根據z變換性質推出,當分子階數高於分母時,輸出會與未來的離散時刻點輸入相關,系統非因果,我們是找不到這樣的實際系統的。那麼我們回到連續的s域來看這個問題,也是一樣的。很多答案只說了結論,非因果系統不可實現卻沒說為什麼,這樣題主可能還是不明白的。我們來看最簡單理想的一階微分環節s,假設系統的傳遞函數H(s)=s,那麼意味著這個系統的單位衝擊響應h(t)=衝擊偶函數——這個系統的單位衝激響應在t&>0時,h(t)=0,該系統是反因果的;同樣,這個系統的單位階躍響應是什麼呢?e(t)=衝擊函數——系統的響應與t&>0之後的輸入都產生了關聯,這個系統顯然是一個非因果的系統。
那麼有同學產生了疑惑,我們學的電感的電壓對電流的傳遞函數不就是「sL」嗎?這不是一個非因果的系統嗎?為什麼物理上實現了?我覺得這個問題可以這樣解釋,理想電感在現實中是不存在的,線圈必然存在等效串聯電阻ESR,同時電感的線圈之間存在寄生電容,嚴格來說電感的實際模型並不是一個sL而已。
事實上,一般的系統總有摩擦,阻尼等損耗因素,使系統的響應產生衰減,因此建立起來的s域傳遞函數模型的分母階數一定會大於分子階數,這種系統函數也叫「慣性系統」。否則,系統會產生能量的自激振蕩。最後,有答主提到了微分環節。沒錯,在控制系統中理想微分環節也是物理上不可實現的。因此,我們常常做一些近似,通常的做法是對理想微分串聯一個一階慣性環節(一階低通濾波器),使得在低頻時表現出良好的近似微分特性。微分環節不可單獨存在道理是類似的
類似的問題Quora上有詳細的回答------------------------直接把答案貼出來吧-------------------------------------------------------「1) n &< m
To give a
concrete example, let us say n = 2 and m = 3, i.e., we have 2 poles and 3
zeroes. Then, the corresponding transfer function would be
G(s)=Y(s)/X(s)=(b0s^3+b1s^2+b2^s+b3)/(s^2+a1s+a2)
To see
things more clearly, convert G(s) to discrete time:
G(z)=Y(z)/X(z)=(bd0z^3+bd1z^2+bd2z+bd3)/(z^2+ad1z+ad2)
The
difference equation describing the system dynamics would be
y(k+2)=bd0
x(k+3)+bd1 x(k+2)+bd2 x(k+1)+bd3 x(k)?ad1 y(k+1)?ad2 y(k)
where k is
the discrete time. Here, we can clearly see that such a system is physically
impossible to construct: The system output y(k+2) now (assume that we are now
at the point "k+2" in time, with an arbitrary k) depends on a future
value of the system input, namely x(k+3). In other words, this system can
"foresee" into the future; it knows that the input x will have a
particular value 1 discrete time step ahead (x(k+3)) and behave in a certain
way now (y(k+2)) depending on that input value which did not happen yet.
Obviously, such a system is physically impossible.
2) n ≥ m
This time
say that n = m = 2. With similar derivations as above, the difference equation
would now be
y(k+2)=bd0
x(k+2)+bd1 x(k+1)+bd2
x(k)?ad1 y(k+1)?ad2 y(k)
Here, the
output y(k+2) reacts to the input values that happen now (x(k+2)) and to those
that happened in the past (x(k+1), x(k)). Thus, this system is physically
possible (this result would be the same for the case with n &> m).
control structures enable the controller to react to input values that will
happen at some point in the future, but these require that the reference signal
is known up to that certain point. In such cases the controller can have more
zeroes than poles or "negative" time delays (i.e., eθs with θ&>0),
but there is still no foreseeing since the future reference values are known.
Outside this very special case, the number of poles of a physical system are at
least equal to the number of its zeroes.」源自:https://www.quora.com/Why-is-it-said-that-for-control-systems-the-number-of-poles-should-be-more-than-the-number-of-zeroes
其實總的來說,正如前面有答主寫的,「超了就反因果了」,這樣就不存在物理意義的系統了。
因為要寫作業,我對上面的過程做了一個整理。
寫的很爛,也很亂。
寫了很久,就不整理成電子稿。
(這個1的題號,是我作業的題號。沒有其他意思,大家不要誤解。)
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更新:
首先聲明,上面的答案和下面的更新都是給予知乎上面的內容,和quora的內容總結、再加上我自己的計算的。
I??k ?lber S?rmatel, PhD student in EE 在quora的回答說他移除了上面 @Garic Chuh 的回答。給出了一個更精美的回答。
我只是語言的翻譯者把他搬運了過來。
才知道學了那麼久的casual,中文是「因果「,我都一直瞎叫成「常規「
搭車同問,電感不是s么?
來答這個有軟性門檻的題…超了就反因果了…
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