Control System Doubt: Voltage Transfer Function of Resistor

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The voltage transfer function for the given circuit is Vr(s)/Vs(s)=s/(s+1), indicating that at s=∞, the capacitor behaves like a short circuit, resulting in unity gain. At s=-1, the transfer function approaches infinity, which signifies an infinite gain under specific conditions, particularly when considering exponential inputs like I=exp(-t). The real part σ in the Laplace transform represents the decay rate of the system, while jω accounts for oscillations. This mathematical scenario suggests that the circuit can theoretically produce infinite gain, although such conditions are not physically realizable. Understanding these concepts is crucial for analyzing control systems and their behavior in response to different inputs.
cnh1995
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Consider the following circuit with Vs=1V,
R=1Ω and C=1F.
upload_2016-3-20_16-49-34.png

The voltage transfer function of the resistor can be written as
Vr(s)/Vs(s)=s/(s+1).
Now I understand s is complex frequency and at s=∞
the transfer function becomes unity since capacitor acts as a short and Vs=Vr. What I don't understand is at s=-1, the TF becomes infinity.
If s=-1 and s=σ+jω, σ=-1 gives TF=∞. What is the meaning of this? What is the general significance of σ? I know jω represents the oscillations in the system but what does the real part σ represent?
 
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σ is for an exponential voltage (or current) proportional to exp(σt).
Consider the current I=exp(-t).
It gives voltages exp(-t) and -exp(-t)+const passing through the resistor and capacitor - so the sum is constant (1 in your case) which means infinite "AC" gain of your circuit.

(but of course it's only in mathematics where a current can be infinite in the past))
 
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