The block diagrams of an ideal system and a real system with their impulse responses are shown below. An auxiliary path is added to the delayed impulse response in the real system. For a unit impulse input ($x(t)=\delta(t)$) to both systems, gain $\beta$ is chosen such that $y(4T)$ is the same for both systems. The value of $\beta$ is _________.
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When matching delayed system outputs, always evaluate unit step activation intervals carefully. For impulse inputs, the system output is simply the impulse response shifted and scaled.
To determine the value of $\beta$, we need to match the output of the ideal system at $t=4T$ with the output of the real system at the same instant.
Step 1: Output of the ideal system at $t=4T$.
The impulse response of the ideal system is
\[
h_1(t) = e^{-\alpha t} u(t).
\]
For a unit impulse input $x(t)=\delta(t)$, the output is simply the impulse response evaluated at $t=4T$:
\[
y_{\text{ideal}}(4T) = e^{-4\alpha T}.
\]
Step 2: Output of the real system at $t=4T$.
The real system has two contributions:
1. A delayed impulse response
\[
h_1(t) = e^{-\alpha (t-T)} u(t-T).
\]
At $t=4T$, this contributes
\[
e^{-\alpha(4T - T)} = e^{-3\alpha T}.
\]
2. An auxiliary path
\[
h_{\text{aux}}(t) = \beta [u(t) - u(t - 5T)].
\]
At $t = 4T$, this term is active because
- $u(4T)=1$,
- $u(4T-5T)=u(-T)=0$.
Thus, its contribution is exactly $\beta$.
Therefore, the total real-system output is:
\[
y_{\text{real}}(4T) = e^{-3\alpha T} + \beta.
\]
Step 3: Equate outputs of ideal and real systems.
\[
y_{\text{ideal}}(4T) = y_{\text{real}}(4T)
\]
\[
e^{-4\alpha T} = e^{-3\alpha T} + \beta.
\]
Solve for $\beta$:
\[
\beta = e^{-4\alpha T} - e^{-3\alpha T}.
\]
Factor out $e^{-3\alpha T}$:
\[
\beta = e^{-3\alpha T}(e^{-\alpha T} - 1)
= - e^{-3\alpha T}(1 - e^{-\alpha T}).
\]
This matches option (C).
