Hi,

I asked AI if there is a difference in a phase response of a system with and without a pole. For example if my system is defined as A(s) = x + s; where s = jw and then I have a second system defined as A_2(s) = 1/(x + s); assuming I found a frequency for which both generate the same phase difference of -90 degrees, is there any difference in a system behaviour?
And chat GPT says yes there is a difference:


Note at the very end of the response it says that only in case of a system with a pole we have "True dynamic lag (time delay)". I asked further what it means by "a true delay" and it says "true delay — the output literally happens later". So I am confused because how can a system without a pole that generates the same phase shift behave differently at the output?
Thank you.

 

Hi,

I asked AI if there is a difference in a phase response of a system with and without a pole. For example if my system is defined as A(s) = x + s; where s = jw and then I have a second system defined as A_2(s) = 1/(x + s); assuming I found a frequency for which both generate the same phase difference of -90 degrees, is there any difference in a system behaviour?
And chat GPT says yes there is a difference:
View attachment 355010

Note at the very end of the response it says that only in case of a system with a pole we have "True dynamic lag (time delay)". I asked further what it means by "a true delay" and it says "true delay — the output literally happens later". So I am confused because how can a system without a pole that generates the same phase shift behave differently at the output?
Thank you.

Hi,

Since you are quoting a lone numerator in one case and a proper transfer function in the other case, I would think you are referring to the effect these two have on an existing system.
You can find this out by looking at Bode representations, the approximate Bode representations.
I think we will find that one introduces a positive phase shift while the other introduces a negative phase shift.

What we would see with a system like 1/(s+c) is the one with the pole would cause an additional negative phase shift while the other would roughly cancel the phase shift.

The one without the pole would be a lead compensator, while the other would be a lag compensator.


USING ASSISTANCE FROM AI

Try this again but ask 'ai' some more specific questions like the ones outlined above. For example:
start with system:
C=1/(s+c)

and two different compensators:
A=a+s
B=1/(b+s)

then ask for the individual effects both A and B have on the phase of C.

 

@mondo90!! I am impressed! You have asked a simple question that AI can not find an answer for. GoodJob!!

 

The computer inferred that your question was ambiguous.
It replied:
"Phase is the angular displacement between two periodic signals and polarity is orientation either positive or negative.
Phase lag just means one signal is late compared to another. It doesn’t matter if you call the signal polar or non‑polar—the lag is about timing, not polarity.”
The Data center's trained electrical model puts the priority on the cause of lag. The root cause is energy storage in reactive elements (inductors, capacitors) and propagation delays in real systems.

 

@mondo90!! I am impressed! You have asked a simple question that AI can not find an answer for. GoodJob!!

I asked one if it could come up with a question that it would have to answer as true and false at the same time and it referred to the Liars Paradox so it seemed to be familiar with this.
I asked about updating on the fly and it said that it was not allowed to update it's internal workings which it referred to as its "system prompt".

 

Grok has more detailed response:

is there a difference in a phase response of a system with and without a pole. For example if my system is defined as A(s) = x + s; where s = jw and then I have a second system defined as A_2(s) = 1/(x + s); assuming I found a frequency for which both generate the same phase difference of -90 degrees, is there any difference in a system behaviour?


To analyze the phase response and system behavior of the two systems
A(s)=x+sA(s) = x + sA(s) = x + s
and
A2(s)=1x+sA_2(s) = \frac{1}{x + s}A_2(s) = \frac{1}{x + s}
, where
s=jωs = j\omegas = j\omega
, and to determine if there is any difference in system behavior when both produce a phase shift of
−90∘-90^\circ-90^\circ
, let’s proceed step-by-step.Step 1: Phase Response AnalysisThe phase response of a system is determined by the angle of its frequency response, obtained by substituting
s=jωs = j\omegas = j\omega
.System 1:
A(s)=x+sA(s) = x + sA(s) = x + s

• Substitute
  s=jωs = j\omegas = j\omega
  :
  A(jω)=x+jωA(j\omega) = x + j\omegaA(j\omega) = x + j\omega
• The phase is the angle of the complex number
  x+jωx + j\omegax + j\omega
  :
  ϕ1=∠(x+jω)=tan⁡−1(ωx)\phi_1 = \angle(x + j\omega) = \tan^{-1}\left(\frac{\omega}{x}\right)\phi_1 = \angle(x + j\omega) = \tan^{-1}\left(\frac{\omega}{x}\right)
• For the phase to be
  −90∘-90^\circ-90^\circ
  , we need:
  tan⁡−1(ωx)=−90∘\tan^{-1}\left(\frac{\omega}{x}\right) = -90^\circ\tan^{-1}\left(\frac{\omega}{x}\right) = -90^\circ
  However,
  tan⁡−1(ω/x)\tan^{-1}(\omega/x)\tan^{-1}(\omega/x)
  ranges from
  0∘0^\circ0^\circ
  to
  90∘90^\circ90^\circ
  for positive
  ω\omega\omega
  and (x), or
  −90∘-90^\circ-90^\circ
  to
  0∘0^\circ0^\circ
  if
  ω/x\omega/x\omega/x
  is negative. A phase of exactly
  −90∘-90^\circ-90^\circ
  occurs only in the limit as
  ω→−∞\omega \to -\infty\omega \to -\infty
  (since
  ω\omega\omega
  is typically positive in frequency response analysis, this suggests a negative frequency or a misinterpretation). For positive
  ω\omega\omega
  , the phase is positive, so achieving
  −90∘-90^\circ-90^\circ
  is not possible unless (x) is negative or we consider the system differently. Let’s assume the question implies a condition where the phase approaches
  −90∘-90^\circ-90^\circ
  asymptotically or reconsider the setup later.

System 2:
A2(s)=1x+sA_2(s) = \frac{1}{x + s}A_2(s) = \frac{1}{x + s}

• Substitute
  s=jωs = j\omegas = j\omega
  :
  A2(jω)=1x+jωA_2(j\omega) = \frac{1}{x + j\omega}A_2(j\omega) = \frac{1}{x + j\omega}
• The phase is:
  ϕ2=∠(1x+jω)=−∠(x+jω)=−tan⁡−1(ωx)\phi_2 = \angle\left(\frac{1}{x + j\omega}\right) = -\angle(x + j\omega) = -\tan^{-1}\left(\frac{\omega}{x}\right)\phi_2 = \angle\left(\frac{1}{x + j\omega}\right) = -\angle(x + j\omega) = -\tan^{-1}\left(\frac{\omega}{x}\right)
• For the phase to be
  −90∘-90^\circ-90^\circ
  :
  −tan⁡−1(ωx)=−90∘  ⟹  tan⁡−1(ωx)=90∘-\tan^{-1}\left(\frac{\omega}{x}\right) = -90^\circ \implies \tan^{-1}\left(\frac{\omega}{x}\right) = 90^\circ-\tan^{-1}\left(\frac{\omega}{x}\right) = -90^\circ \implies \tan^{-1}\left(\frac{\omega}{x}\right) = 90^\circ
• This occurs when
  ω/x→∞\omega/x \to \infty\omega/x \to \infty
  , i.e.,
  ω≫x\omega \gg x\omega \gg x
  . For large
  ω\omega\omega
  ,
  tan⁡−1(ω/x)≈90∘\tan^{-1}(\omega/x) \approx 90^\circ\tan^{-1}(\omega/x) \approx 90^\circ
  , so
  ϕ2≈−90∘\phi_2 \approx -90^\circ\phi_2 \approx -90^\circ
  .

Step 2: Finding a Common FrequencyThe question assumes there exists a frequency
ω\omega\omega
where both systems produce a phase shift of
−90∘-90^\circ-90^\circ
. For
A2(s)A_2(s)A_2(s)
, this is achievable at high frequencies (
ω≫x\omega \gg x\omega \gg x
). For ( A(s) ), achieving exactly
−90∘-90^\circ-90^\circ
is problematic since
ϕ1=tan⁡−1(ω/x)\phi_1 = \tan^{-1}(\omega/x)\phi_1 = \tan^{-1}(\omega/x)
is positive for positive
ω\omega\omega
and (x). Let’s hypothesize that the question may involve a misinterpretation or a specific condition (e.g., negative
ω\omega\omega
or a modified system). To proceed, assume
A2(s)A_2(s)A_2(s)
reaches
−90∘-90^\circ-90^\circ
at high
ω\omega\omega
, and check if ( A(s) ) can match this under certain conditions or if the question implies a comparison at a specific phase.Let’s try to find a frequency where
A2(s)A_2(s)A_2(s)
has a phase of
−90∘-90^\circ-90^\circ
. At
ω≫x\omega \gg x\omega \gg x
, the phase of
A2(jω)≈−90∘A_2(j\omega) \approx -90^\circA_2(j\omega) \approx -90^\circ
. For ( A(s) ), if we consider negative frequencies (
ω<0\omega < 0\omega < 0
):
s=j(−ω)=−jω,A(j(−ω))=x−jω,ϕ1=tan⁡−1(−ωx)=−tan⁡−1(ωx)s = j(-\omega) = -j\omega, \quad A(j(-\omega)) = x - j\omega, \quad \phi_1 = \tan^{-1}\left(\frac{-\omega}{x}\right) = -\tan^{-1}\left(\frac{\omega}{x}\right)s = j(-\omega) = -j\omega, \quad A(j(-\omega)) = x - j\omega, \quad \phi_1 = \tan^{-1}\left(\frac{-\omega}{x}\right) = -\tan^{-1}\left(\frac{\omega}{x}\right)
This can approach
−90∘-90^\circ-90^\circ
for large
ω\omega\omega
. Let’s assume the question allows negative frequencies or a system modification, but typically, frequency response considers
ω≥0\omega \geq 0\omega \geq 0
. If we assume a typo or a specific condition, let’s compare at a frequency where
A2(s)A_2(s)A_2(s)
is at
−90∘-90^\circ-90^\circ
(e.g.,
ω→∞\omega \to \infty\omega \to \infty
).Step 3: System Behavior ComparisonAssuming we find a frequency where both systems have a phase of
−90∘-90^\circ-90^\circ
(or approximately so), we need to compare their behavior. The phase alone doesn’t fully describe system behavior; we must consider:

1. Magnitude Response:
   • For
     A(s)=x+jωA(s) = x + j\omegaA(s) = x + j\omega
     :
     ∣A(jω)∣=x2+ω2|A(j\omega)| = \sqrt{x^2 + \omega^2}|A(j\omega)| = \sqrt{x^2 + \omega^2}
     This increases with
     ω\omega\omega
     , indicating a high-pass behavior.
   • For
     A2(s)=1x+jωA_2(s) = \frac{1}{x + j\omega}A_2(s) = \frac{1}{x + j\omega}
     :
     ∣A2(jω)∣=1x2+ω2|A_2(j\omega)| = \frac{1}{\sqrt{x^2 + \omega^2}}|A_2(j\omega)| = \frac{1}{\sqrt{x^2 + \omega^2}}
     This decreases with
     ω\omega\omega
     , indicating a low-pass behavior.
   • At high
     ω\omega\omega
     ,
     ∣A(jω)∣→∞|A(j\omega)| \to \infty|A(j\omega)| \to \infty
     , while
     ∣A2(jω)∣→0|A_2(j\omega)| \to 0|A_2(j\omega)| \to 0
     . This is a significant difference in gain.
2. System Dynamics:
   • A(s)=x+sA(s) = x + sA(s) = x + s
     is a high-pass system (or a differentiator-like system) with no poles, only a zero at
     s=−xs = -xs = -x
     . It amplifies high frequencies and has no stability concerns since it’s a simple gain function.
   • A2(s)=1x+sA_2(s) = \frac{1}{x + s}A_2(s) = \frac{1}{x + s}
     has a pole at
     s=−xs = -xs = -x
     , characteristic of a first-order low-pass system. It attenuates high frequencies and has a stable, causal response with a time constant
     τ=1/x\tau = 1/x\tau = 1/x
     .
   • The presence of a pole in
     A2(s)A_2(s)A_2(s)
     introduces a memory effect (exponential decay in the time domain), while ( A(s) ) is memoryless in terms of poles, behaving more like a proportional-derivative system.
3. Time-Domain Behavior:
   • For ( A(s) ), the impulse response is:
     h(t)=xδ(t)+δ′(t)h(t) = x\delta(t) + \delta'(t)h(t) = x\delta(t) + \delta'(t)
     This includes an impulse and its derivative, indicating instantaneous and differential response.
   • For
     A2(s)A_2(s)A_2(s)
     , the impulse response is:
     h2(t)=e−xtu(t)h_2(t) = e^{-xt}u(t)h_2(t) = e^{-xt}u(t)
     This is an exponentially decaying response, typical of a low-pass filter.
4. Phase at
   −90∘-90^\circ-90^\circ
   :
   • For
     A2(s)A_2(s)A_2(s)
     , the phase reaches
     −90∘-90^\circ-90^\circ
     as
     ω→∞\omega \to \infty\omega \to \infty
     , where the magnitude is near zero, meaning the output is heavily attenuated.
   • For ( A(s) ), achieving
     −90∘-90^\circ-90^\circ
     requires negative
     ω\omega\omega
     or a reinterpretation (e.g., a system modification like
     A(s)=−s+xA(s) = -s + xA(s) = -s + x
     ). If we assume a typo or specific condition, the magnitude at high frequencies is large, amplifying the input.

Step 4: Differences in BehaviorEven if both systems achieve a
−90∘-90^\circ-90^\circ
phase shift at some frequency (e.g.,
A2(s)A_2(s)A_2(s)
at high
ω\omega\omega
, and ( A(s) ) at negative
ω\omega\omega
or under a modified definition), their behaviors differ significantly:

• Magnitude: ( A(s) ) amplifies high frequencies, while
  A2(s)A_2(s)A_2(s)
  attenuates them. This leads to vastly different output amplitudes for the same input.
• Frequency Response: ( A(s) ) is high-pass, emphasizing rapid changes, while
  A2(s)A_2(s)A_2(s)
  is low-pass, smoothing signals.
• Stability and Dynamics: ( A(s) ) has no poles, so no dynamic memory, while
  A2(s)A_2(s)A_2(s)
  has a pole, introducing a time constant and filtering behavior.
• Application: ( A(s) ) might represent a differentiator or high-pass filter, while
  A2(s)A_2(s)A_2(s)
  is a typical low-pass filter used for noise reduction or signal smoothing.

ConclusionEven if a frequency exists where both systems produce a
−90∘-90^\circ-90^\circ
phase shift, their behaviors differ due to:

• Opposite magnitude responses (high-pass vs. low-pass).
• Different time-domain responses (impulsive vs. exponential decay).
• The presence of a pole in
  A2(s)A_2(s)A_2(s)
  versus none in ( A(s) ), affecting stability and dynamics. Thus, the systems process inputs differently, with ( A(s) ) amplifying high frequencies and
  A2(s)A_2(s)A_2(s)
  attenuating them, leading to distinct signal outputs despite the same phase at a specific frequency.

 

Grok has more detailed response:

is there a difference in a phase response of a system with and without a pole. For example if my system is defined as A(s) = x + s; where s = jw and then I have a second system defined as A_2(s) = 1/(x + s); assuming I found a frequency for which both generate the same phase difference of -90 degrees, is there any difference in a system behaviour?


To analyze the phase response and system behavior of the two systems
A(s)=x+sA(s) = x + sA(s) = x + s
and
A2(s)=1x+sA_2(s) = \frac{1}{x + s}A_2(s) = \frac{1}{x + s}
, where
s=jωs = j\omegas = j\omega
, and to determine if there is any difference in system behavior when both produce a phase shift of
−90∘-90^\circ-90^\circ
, let’s proceed step-by-step.Step 1: Phase Response AnalysisThe phase response of a system is determined by the angle of its frequency response, obtained by substituting
s=jωs = j\omegas = j\omega
.System 1:
A(s)=x+sA(s) = x + sA(s) = x + s

• Substitute
  s=jωs = j\omegas = j\omega
  :
  A(jω)=x+jωA(j\omega) = x + j\omegaA(j\omega) = x + j\omega
• The phase is the angle of the complex number
  x+jωx + j\omegax + j\omega
  :
  ϕ1=∠(x+jω)=tan⁡−1(ωx)\phi_1 = \angle(x + j\omega) = \tan^{-1}\left(\frac{\omega}{x}\right)\phi_1 = \angle(x + j\omega) = \tan^{-1}\left(\frac{\omega}{x}\right)
• For the phase to be
  −90∘-90^\circ-90^\circ
  , we need:
  tan⁡−1(ωx)=−90∘\tan^{-1}\left(\frac{\omega}{x}\right) = -90^\circ\tan^{-1}\left(\frac{\omega}{x}\right) = -90^\circ
  However,
  tan⁡−1(ω/x)\tan^{-1}(\omega/x)\tan^{-1}(\omega/x)
  ranges from
  0∘0^\circ0^\circ
  to
  90∘90^\circ90^\circ
  for positive
  ω\omega\omega
  and (x), or
  −90∘-90^\circ-90^\circ
  to
  0∘0^\circ0^\circ
  if
  ω/x\omega/x\omega/x
  is negative. A phase of exactly
  −90∘-90^\circ-90^\circ
  occurs only in the limit as
  ω→−∞\omega \to -\infty\omega \to -\infty
  (since
  ω\omega\omega
  is typically positive in frequency response analysis, this suggests a negative frequency or a misinterpretation). For positive
  ω\omega\omega
  , the phase is positive, so achieving
  −90∘-90^\circ-90^\circ
  is not possible unless (x) is negative or we consider the system differently. Let’s assume the question implies a condition where the phase approaches
  −90∘-90^\circ-90^\circ
  asymptotically or reconsider the setup later.

System 2:
A2(s)=1x+sA_2(s) = \frac{1}{x + s}A_2(s) = \frac{1}{x + s}

• Substitute
  s=jωs = j\omegas = j\omega
  :
  A2(jω)=1x+jωA_2(j\omega) = \frac{1}{x + j\omega}A_2(j\omega) = \frac{1}{x + j\omega}
• The phase is:
  ϕ2=∠(1x+jω)=−∠(x+jω)=−tan⁡−1(ωx)\phi_2 = \angle\left(\frac{1}{x + j\omega}\right) = -\angle(x + j\omega) = -\tan^{-1}\left(\frac{\omega}{x}\right)\phi_2 = \angle\left(\frac{1}{x + j\omega}\right) = -\angle(x + j\omega) = -\tan^{-1}\left(\frac{\omega}{x}\right)
• For the phase to be
  −90∘-90^\circ-90^\circ
  :
  −tan⁡−1(ωx)=−90∘  ⟹  tan⁡−1(ωx)=90∘-\tan^{-1}\left(\frac{\omega}{x}\right) = -90^\circ \implies \tan^{-1}\left(\frac{\omega}{x}\right) = 90^\circ-\tan^{-1}\left(\frac{\omega}{x}\right) = -90^\circ \implies \tan^{-1}\left(\frac{\omega}{x}\right) = 90^\circ
• This occurs when
  ω/x→∞\omega/x \to \infty\omega/x \to \infty
  , i.e.,
  ω≫x\omega \gg x\omega \gg x
  . For large
  ω\omega\omega
  ,
  tan⁡−1(ω/x)≈90∘\tan^{-1}(\omega/x) \approx 90^\circ\tan^{-1}(\omega/x) \approx 90^\circ
  , so
  ϕ2≈−90∘\phi_2 \approx -90^\circ\phi_2 \approx -90^\circ
  .

Step 2: Finding a Common FrequencyThe question assumes there exists a frequency
ω\omega\omega
where both systems produce a phase shift of
−90∘-90^\circ-90^\circ
. For
A2(s)A_2(s)A_2(s)
, this is achievable at high frequencies (
ω≫x\omega \gg x\omega \gg x
). For ( A(s) ), achieving exactly
−90∘-90^\circ-90^\circ
is problematic since
ϕ1=tan⁡−1(ω/x)\phi_1 = \tan^{-1}(\omega/x)\phi_1 = \tan^{-1}(\omega/x)
is positive for positive
ω\omega\omega
and (x). Let’s hypothesize that the question may involve a misinterpretation or a specific condition (e.g., negative
ω\omega\omega
or a modified system). To proceed, assume
A2(s)A_2(s)A_2(s)
reaches
−90∘-90^\circ-90^\circ
at high
ω\omega\omega
, and check if ( A(s) ) can match this under certain conditions or if the question implies a comparison at a specific phase.Let’s try to find a frequency where
A2(s)A_2(s)A_2(s)
has a phase of
−90∘-90^\circ-90^\circ
. At
ω≫x\omega \gg x\omega \gg x
, the phase of
A2(jω)≈−90∘A_2(j\omega) \approx -90^\circA_2(j\omega) \approx -90^\circ
. For ( A(s) ), if we consider negative frequencies (
ω<0\omega < 0\omega < 0
):
s=j(−ω)=−jω,A(j(−ω))=x−jω,ϕ1=tan⁡−1(−ωx)=−tan⁡−1(ωx)s = j(-\omega) = -j\omega, \quad A(j(-\omega)) = x - j\omega, \quad \phi_1 = \tan^{-1}\left(\frac{-\omega}{x}\right) = -\tan^{-1}\left(\frac{\omega}{x}\right)s = j(-\omega) = -j\omega, \quad A(j(-\omega)) = x - j\omega, \quad \phi_1 = \tan^{-1}\left(\frac{-\omega}{x}\right) = -\tan^{-1}\left(\frac{\omega}{x}\right)
This can approach
−90∘-90^\circ-90^\circ
for large
ω\omega\omega
. Let’s assume the question allows negative frequencies or a system modification, but typically, frequency response considers
ω≥0\omega \geq 0\omega \geq 0
. If we assume a typo or a specific condition, let’s compare at a frequency where
A2(s)A_2(s)A_2(s)
is at
−90∘-90^\circ-90^\circ
(e.g.,
ω→∞\omega \to \infty\omega \to \infty
).Step 3: System Behavior ComparisonAssuming we find a frequency where both systems have a phase of
−90∘-90^\circ-90^\circ
(or approximately so), we need to compare their behavior. The phase alone doesn’t fully describe system behavior; we must consider:

1. Magnitude Response:
   • For
     A(s)=x+jωA(s) = x + j\omegaA(s) = x + j\omega
     :
     ∣A(jω)∣=x2+ω2|A(j\omega)| = \sqrt{x^2 + \omega^2}|A(j\omega)| = \sqrt{x^2 + \omega^2}
     This increases with
     ω\omega\omega
     , indicating a high-pass behavior.
   • For
     A2(s)=1x+jωA_2(s) = \frac{1}{x + j\omega}A_2(s) = \frac{1}{x + j\omega}
     :
     ∣A2(jω)∣=1x2+ω2|A_2(j\omega)| = \frac{1}{\sqrt{x^2 + \omega^2}}|A_2(j\omega)| = \frac{1}{\sqrt{x^2 + \omega^2}}
     This decreases with
     ω\omega\omega
     , indicating a low-pass behavior.
   • At high
     ω\omega\omega
     ,
     ∣A(jω)∣→∞|A(j\omega)| \to \infty|A(j\omega)| \to \infty
     , while
     ∣A2(jω)∣→0|A_2(j\omega)| \to 0|A_2(j\omega)| \to 0
     . This is a significant difference in gain.
2. System Dynamics:
   • A(s)=x+sA(s) = x + sA(s) = x + s
     is a high-pass system (or a differentiator-like system) with no poles, only a zero at
     s=−xs = -xs = -x
     . It amplifies high frequencies and has no stability concerns since it’s a simple gain function.
   • A2(s)=1x+sA_2(s) = \frac{1}{x + s}A_2(s) = \frac{1}{x + s}
     has a pole at
     s=−xs = -xs = -x
     , characteristic of a first-order low-pass system. It attenuates high frequencies and has a stable, causal response with a time constant
     τ=1/x\tau = 1/x\tau = 1/x
     .
   • The presence of a pole in
     A2(s)A_2(s)A_2(s)
     introduces a memory effect (exponential decay in the time domain), while ( A(s) ) is memoryless in terms of poles, behaving more like a proportional-derivative system.
3. Time-Domain Behavior:
   • For ( A(s) ), the impulse response is:
     h(t)=xδ(t)+δ′(t)h(t) = x\delta(t) + \delta'(t)h(t) = x\delta(t) + \delta'(t)
     This includes an impulse and its derivative, indicating instantaneous and differential response.
   • For
     A2(s)A_2(s)A_2(s)
     , the impulse response is:
     h2(t)=e−xtu(t)h_2(t) = e^{-xt}u(t)h_2(t) = e^{-xt}u(t)
     This is an exponentially decaying response, typical of a low-pass filter.
4. Phase at
   −90∘-90^\circ-90^\circ
   :
   • For
     A2(s)A_2(s)A_2(s)
     , the phase reaches
     −90∘-90^\circ-90^\circ
     as
     ω→∞\omega \to \infty\omega \to \infty
     , where the magnitude is near zero, meaning the output is heavily attenuated.
   • For ( A(s) ), achieving
     −90∘-90^\circ-90^\circ
     requires negative
     ω\omega\omega
     or a reinterpretation (e.g., a system modification like
     A(s)=−s+xA(s) = -s + xA(s) = -s + x
     ). If we assume a typo or specific condition, the magnitude at high frequencies is large, amplifying the input.

Step 4: Differences in BehaviorEven if both systems achieve a
−90∘-90^\circ-90^\circ
phase shift at some frequency (e.g.,
A2(s)A_2(s)A_2(s)
at high
ω\omega\omega
, and ( A(s) ) at negative
ω\omega\omega
or under a modified definition), their behaviors differ significantly:

• Magnitude: ( A(s) ) amplifies high frequencies, while
  A2(s)A_2(s)A_2(s)
  attenuates them. This leads to vastly different output amplitudes for the same input.
• Frequency Response: ( A(s) ) is high-pass, emphasizing rapid changes, while
  A2(s)A_2(s)A_2(s)
  is low-pass, smoothing signals.
• Stability and Dynamics: ( A(s) ) has no poles, so no dynamic memory, while
  A2(s)A_2(s)A_2(s)
  has a pole, introducing a time constant and filtering behavior.
• Application: ( A(s) ) might represent a differentiator or high-pass filter, while
  A2(s)A_2(s)A_2(s)
  is a typical low-pass filter used for noise reduction or signal smoothing.

ConclusionEven if a frequency exists where both systems produce a
−90∘-90^\circ-90^\circ
phase shift, their behaviors differ due to:

• Opposite magnitude responses (high-pass vs. low-pass).
• Different time-domain responses (impulsive vs. exponential decay).
• The presence of a pole in
  A2(s)A_2(s)A_2(s)
  versus none in ( A(s) ), affecting stability and dynamics. Thus, the systems process inputs differently, with ( A(s) ) amplifying high frequencies and
  A2(s)A_2(s)A_2(s)
  attenuating them, leading to distinct signal outputs despite the same phase at a specific frequency.

Hi,

Oh that's great, I always wanted to learn how to read native Latex

 

oh well... i was just pasting something in another forum and got rendered perfectly. so i did not bother to check.
guess if i ever do it here, should take screenshots...

 

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