Question

A first-order low-pass filter of time constant T is excited with different input signals (with zero initial conditions up to t = 0). Match the excitation signals X, Y, Z with the corresponding time responses for t ≥ 0:

X: Impulse      P: 1 − e^−t/T

Y: Unit step     Q: t − T(1 − e^−t/T)

Z: Ramp          R: e^−t/T

1. X→R, Y→Q, Z→P
2. X→Q, Y→P, Z→R
3. X→R, Y→P, Z→Q
4. X→P, Y→R, Z→Q

Answer 1

Correct Answer - Option 3 : X→R, Y→P, Z→Q

Let open-loop transfer function is:

\({\rm{G}}\left( {\rm{s}} \right) = \frac{1}{{{\rm{ST}}}}\)

then,

Closed-loop transfer function is:

\({\rm{H}}\left( {\rm{s}} \right) = \frac{1}{{1 + {\rm{sT}}}}\) (assume unity negative feedback)

\({\rm{Y}}\left( {\rm{s}} \right){\rm{\;}} = {\rm{\;X}}\left( {\rm{s}} \right).{\rm{\;H}}\left( {\rm{s}} \right)\)

Case i :

if input = impulse

\({\rm{X}}\left( {\rm{s}} \right){\rm{\;}} = {\rm{\;}}1\)

\(\begin{array}{l} {\rm{Y}}\left( {\rm{s}} \right) = \frac{1}{{1 + {\rm{sT}}}}.\frac{1}{{\rm{T}}}\left[ {\frac{1}{{\frac{1}{{\rm{T}}} + {\rm{s}}}}} \right]\\ {\rm{y}}\left( {\rm{t}} \right) = \frac{1}{{\rm{T}}}{{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{T}}}}} \end{array}\)

Case ii:

If input = unit step

\( {\rm{\;X}}\left( {\rm{s}} \right) = \frac{1}{{\rm{s}}}\)

\(\begin{array}{l} {\rm{Y}}\left( {\rm{s}} \right) = \frac{1}{{1 + {\rm{sT\;}}}}.\frac{1}{{\rm{s}}} \\= \frac{1}{{\rm{s}}} + \frac{{ - {\rm{T}}}}{{\left( {1 + {\rm{sT}}} \right)}} \\= \frac{1}{{\rm{s}}} - \frac{1}{{\frac{1}{{\rm{T}}} + {\rm{s}}}}\\ {\rm{y}}\left( {\rm{t}} \right) = 1 - {{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{T}}}}} \end{array}\)

Case iii:

 If input = Ramp \(\Rightarrow {\rm{\;x}}\left( {\rm{s}} \right) = \frac{1}{{{{\rm{s}}^2}}}\)

\(\begin{array}{l} {\rm{Y}}\left( {\rm{s}} \right) = \frac{1}{{1 + {\rm{sT}}}}.\frac{1}{{{{\rm{s}}^2}}}\\ = \frac{1}{{{{\rm{s}}^2}}} + \frac{{{{\rm{T}}^2}}}{{\left( {1 + {\rm{sT}}} \right)}} + \frac{{\left( { - {\rm{T}}} \right)}}{{\rm{s}}}\\ {\rm{y}}\left( {\rm{t}} \right) = {\rm{t}} + {\rm{T}}{{\rm{e}}^{ - {\rm{t}}/{\rm{T}}}} - {\rm{T}} \\= {\rm{t}} - {\rm{T}}\left( {1 - {{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{T}}}}}} \right) \end{array}\)