Signals and Systems Test 4

Concept:

If X(s) is Laplace transform of x(t), then

\(x\left( t \right) = {e^{ - at}}u\left( t \right) \leftrightarrow X\left( s \right) = \frac{1}{{\left( {s + a} \right)}};ROC:s > - a\)

\(x\left( t \right) = - {e^{ - at}}u\left( { - t} \right) \leftrightarrow X\left( s \right) = \frac{1}{{\left( {s + a} \right)}};ROC:s < - a\)

\(x\left( t \right) = {e^{at}}u\left( t \right) \leftrightarrow X\left( s \right) = \frac{1}{{\left( {s - a} \right)}};ROC:s > a\)

\(x\left( t \right) = - {e^{at}}u\left( t \right) \leftrightarrow X\left( s \right) = \frac{1}{{\left( {s - a} \right)}};ROC:s < a\)

Calculation:

\(X\left( s \right) = \frac{{s + 2}}{{{s^2} + 7s + 12}}\)

\(\frac{{s + 2}}{{\left( {s + 3} \right)\left( {s + 4} \right)}}\)

\(= \frac{2}{{\left( {s + 4} \right)}} - \frac{1}{{\left( {s + 3} \right)}}\)

For \(Re\left( s \right) > - 4,\;\frac{2}{{\left( {s + 4} \right)}} \leftrightarrow 2{e^{ - 4t}}u\left( t \right)\) 

For \(Re\left( s \right) < - 3,\;\frac{1}{{\left( {s + 3} \right)}} \leftrightarrow \; - {e^{ - 3t}}u\left( { - t} \right)\) 

x(t) = 2e^-4t u(t) + e^-3t u(-t)

When the given system is fed with the input u(t), the output produced is (1 – e^-t – te^-t) u(t).

Hence, the system function H(s), is the ratio of the Laplace transform of the output divided by the Laplace transform of the input.

\(H\left( s \right) = L\left( {h\left( t \right)} \right) = \frac{{L\left[ {\left( {1 - {e^{ - t}} - t{e^{ - t}}} \right)u\left( t \right)} \right]}}{{L\left( {u\left( t \right)} \right)}}\) 

\(H\left( s \right) = \frac{{\frac{1}{s} - \frac{1}{{s + 1}} - \frac{1}{{{{\left( {s + 1} \right)}^2}}}}}{{\frac{1}{s}}} = \frac{1}{{{{\left( {s + 1} \right)}^2}}}\) 

For the given input x(t), the observed output

y(t) = (2 – 3e^-t + e^-3t) u(t)

\(Y\left( s \right) = \frac{2}{s} - \frac{3}{{\left( {s + 1} \right)}} + \frac{1}{{\left( {s + 3} \right)}}\) 

\( = \frac{{2\left( {s + 1} \right)\left( {s + 3} \right) - 3\left( s \right)\left( {s + 3} \right) + s\left( {s + 1} \right)}}{{s\left( {s + 1} \right)\left( {s + 3} \right)}}\) 

\( = \frac{{2\left( {{s^2} + 4s + 3} \right) - 3\left( {{s^2} + 3s} \right) + {s^2} + s}}{{s\left( {s + 1} \right)\left( {s + 3} \right)}}\) 

\(= \frac{6}{{s\left( {s + 1} \right)\left( {s + 3} \right)}}\) 

\(X\left( s \right) = \frac{{Y\left( s \right)}}{{H\left( s \right)}}\) 

\( = \frac{{\frac{6}{{s\left( {s + 1} \right)\left( {s + 3} \right)}}}}{{\frac{1}{{{{\left( {s + 1} \right)}^2}}}}}\) 

\(= \frac{{6\left( {s + 1} \right)}}{{s\left( {s + 3} \right)}}\) 

\(= \frac{4}{{\left( {s + 3} \right)}} + \frac{2}{s}\) 

⇒ x(t) = (4e^-3t + 2) u(t)

1. X(s) converges to \(\frac{2}{{{s^3}}}\) with ROC: Re (s) > 0.

For limit \(t \to + \infty ,\;\;{t^2}{e^{ - st}} = {t^2}{e^{ - \left( {\sigma + j\omega } \right)t}}\) tends to zero for σ = Re(s) > 0 as the exponent term dominates and its integral from 0 to +∞ converges.

Therefore, the given statement is false.

2. \({e^{\left( {{t^2} - st} \right)}}\) does not tend to zero as t → +∞, which means the integral from 0 to +∞ is unbounded for all s. Thus the Laplace transform cannot converge for any value of s.

Therefore, the given statement is true.

3. For \(Re\left( s \right) > 0,\;\;{e^{j{\omega _0}t - st}}\) does not tend to zero for t → -∞.

For Re(s) < 0, it does not converge for t → +∞.

Hence for no value of s does the Laplace transform converge.

Therefore, the given statement is true.

4. |t| = t u(t) – t u(-t).

For t u(t), the ROC is Re(s) > 0

For t u(-t), the ROC is Re(s) < 0

Hence, there is no value of s common to both the ROCs.