Lect16EEE 2021 System Responses Dr. Holbert March 24, 2008

Lect16EEE 2022 Introduction Today, we explore in greater depth the three cases for second-order systems –Real and unequal poles –Real and equal poles –Complex conjugate pair This material is ripe with new terminology

Lect16EEE 2023 Second-Order ODE Recall: the second-order ODE has a form of For zero-initial conditions, the transfer function would be

Lect16EEE 2024 Second-Order ODE The denominator of the transfer function is known as the characteristic equation To find the poles, we solve : which has two roots: s 1 and s 2

Lect16EEE 2025 Damping Ratio (  ) and Natural Frequency (  0 ) The damping ratio is ζ The damping ratio determines what type of solution we will obtain: –Exponentially decreasing (  >1) –Exponentially decreasing sinusoid (  < 1) The undamped natural frequency is  0 –Determines how fast sinusoids wiggle –Approximately equal to resonance frequency

Lect16EEE 2026 Characteristic Equation Roots The roots of the characteristic equation determine whether the complementary (natural) solution wiggles

Lect16EEE Real and Unequal Roots If  > 1, s 1 and s 2 are real and not equal This solution is overdamped

Lect16EEE 2028 Overdamped Both of these graphs have a response of the form i(t) = K 1 exp(–t/τ 1 ) + K 2 exp(–t/ τ 2 )

Lect16EEE Complex Roots If  < 1, s 1 and s 2 are complex Define the following constants: This solution is underdamped

Lect16EEE Underdamped A curve having a response of the form i(t) = e –t/τ [K 1 cos(ωt) + K 2 sin(ωt)]

Lect16EEE Real and Equal Roots If  = 1, then s 1 and s 2 are real and equal This solution is critically damped

Lect16EEE IF Amplifier Example This is one possible implementation of the filter portion of an intermediate frequency (IF) amplifier 10  769pFvs(t)vs(t) i(t)i(t) 159  H +–+–

Lect16EEE IF Amplifier Example (cont’d.) The ODE describing the loop current is For this example, what are ζ and ω 0 ?

Lect16EEE IF Amplifier Example (cont’d.) Note that  0 = 2  f = 2  455,000 Hz) Is this system overdamped, underdamped, or critically damped? What will the current look like?

Lect16EEE IF Amplifier Example (cont’d.) The shape of the current depends on the initial capacitor voltage and inductor current

Lect16EEE Slightly Different Example Increase the resistor to 1k  Exercise: what are  and  0 ? 1k  769pFvs(t)vs(t) i(t)i(t) 159  H +–+–

Lect16EEE Different Example (cont’d.) The natural (resonance) frequency does not change:  0 = 2  455,000 Hz) But the damping ratio becomes  = 2.2 Is this system overdamped, underdamped, or critically damped? What will the current look like?

Lect16EEE Different Example (cont’d.) The shape of the current depends on the initial capacitor voltage and inductor current

Lect16EEE Damping Summary Damping Ratio Poles (s 1, s 2 )Damping ζ > 1Real and unequalOverdamped ζ = 1Real and equalCritically damped 0 < ζ < 1Complex conjugate pair set Underdamped ζ = 0Purely imaginary pairUndamped

Lect16EEE Transient and Steady-State Responses The steady-state response of a circuit is the waveform after a long time has passed, and depends on the source(s) in the circuit –Constant sources give DC steady-state responses DC steady-state if response approaches a constant –Sinusoidal sources give AC steady-state responses AC steady-state if response approaches a sinusoid The transient response is the circuit response minus the steady-state response

Lect16EEE Transient and Steady-State Responses Consider a time-domain response from an earlier example this semester Steady State Response Transient Response Transient Response Steady-State Response

Lect16EEE Class Examples Drill Problems P7-6, P7-7, P7-8