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Two linear time-invariant networks with individual delay times t_d1 and t_d2 are cascaded through an ideal unity-gain buffer (no loading). What is the overall delay time of the cascade?

Difficulty: Easy

Correct Answer: t_d,total = t_d1 + t_d2

Explanation:


Introduction / Context:
Delay time is a key metric for filters, amplifiers, and control systems. When blocks are cascaded, understanding how delays combine allows engineers to predict end-to-end latency and phase behavior. An ideal buffer between stages eliminates loading, so the cascade behaves like a pure series concatenation of time shifts.



Given Data / Assumptions:

  • Two linear time-invariant (LTI) networks characterized by delays t_d1 and t_d2.
  • Networks are cascaded via an ideal unity-gain buffer (infinite input impedance, zero output impedance).
  • Small-signal regime; no slew-rate or saturation effects.


Concept / Approach:
In LTI systems, a delay corresponds to multiplication by e^{-jωt_d} in the frequency domain, or equivalently a shift by t_d in the time domain. Cascading two delays results in multiplication of their frequency responses: e^{-jωt_d1} * e^{-jωt_d2} = e^{-jω(t_d1 + t_d2)}. Thus the net effect is a delay equal to the sum of the individual delays.



Step-by-Step Solution:

Model each stage as H_1(ω) = |H_1(ω)| e^{-jωt_d1} and H_2(ω) = |H_2(ω)| e^{-jωt_d2}.With an ideal buffer, no additional phase or loading is introduced.Overall transfer function H(ω) = H_1(ω) H_2(ω) = |H_1||H_2| e^{-jω(t_d1 + t_d2)}.Therefore the overall delay time is t_d1 + t_d2.


Verification / Alternative check:

Impulse response viewpoint: y(t) = h_2(t) * h_1(t) * x(t). For pure delays, convolution of δ(t − t_d1) and δ(t − t_d2) yields δ(t − (t_d1 + t_d2)).


Why Other Options Are Wrong:

Products or max/min do not model time shifts; delay adds along a signal path.The extra t_d1 t_d2 term has no basis in LTI cascades.


Common Pitfalls:

Confusing group delay with phase delay in non-minimum-phase systems; with an ideal buffer and simple delays, the sum rule holds.


Final Answer:

t_d,total = t_d1 + t_d2
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