Control System Doubt: Voltage Transfer Function of Resistor

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SUMMARY

The voltage transfer function of a resistor in the given circuit with Vs=1V, R=1Ω, and C=1F is defined as Vr(s)/Vs(s)=s/(s+1). At s=∞, the transfer function equals unity, indicating that the capacitor behaves as a short circuit. However, at s=-1, the transfer function approaches infinity, which signifies an infinite gain in the circuit due to the exponential decay of current represented by I=exp(-t). The real part σ in the Laplace transform indicates the rate of exponential decay or growth in the system.

PREREQUISITES
  • Understanding of Laplace transforms and their applications in control systems
  • Familiarity with voltage transfer functions and circuit analysis
  • Knowledge of complex frequency and its significance in system behavior
  • Basic concepts of resistors and capacitors in electrical circuits
NEXT STEPS
  • Study the implications of infinite gain in control systems and its practical limitations
  • Learn about the significance of the real part σ in Laplace transforms
  • Explore the behavior of circuits with different configurations using voltage transfer functions
  • Investigate the effects of varying resistor and capacitor values on transfer functions
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Electrical engineers, control system designers, and students studying circuit analysis and Laplace transforms will benefit from this discussion.

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))
 
Last edited:

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