Signals and Systems Test 5

Conventional Method:

The Laplace transform of a function x(t) is defined as:

\(X\left( s \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - st}}dt\)

Given:

x(t) = (P + Qe^-bt) u(t)

x(t) = Pu(t) + Qe^-bt u(t)

Taking the Laplace Transform of the above, we get:

\(X\left( s \right) = \mathop \smallint \limits_0^\infty P{e^{ - st}}dt + \mathop \smallint \limits_0^\infty Q{e^{ - bt}} \cdot {e^{ - st}}dt\)

\(X\left( s \right) = \underbrace {P\mathop \smallint \limits_0^\infty {e^{ - st}}dt}_I + \underbrace {Q\mathop \smallint \limits_0^\infty {e^{ - \left( {b + s} \right)t}}dt}_{II}\)

Integral I converges for s > 0, and integral II converges for b + s > 0, or for Re(s) > -b.

∴ ROC is the intersection of Re(s) > 0 and Re(s) > - b, i.e. the resultant ROC will be:

ROC: Re(s) > max (-b, 0)

Conceptual Approach:

Since u(t) is a right sided signal, its ROC will also be right sided, i.e.

\(u\left( t \right)\mathop \leftrightarrow \limits^{LT} \frac{1}{s};Re\left( s \right) > 0\)

Similarly,

\({e^{ - bt}}u\left( t \right)\mathop \leftrightarrow \limits^{L.T.} \frac{1}{{s + b}};Re\left( s \right) > - b\)

The ROC of the sum of the two will be the intersection of the two, i.e.

ROC; Re(s) > max (-b, 0)

Concept:

\(\cos at\;u\left( t \right) \leftrightarrow \frac{s}{{{s^2} + {a^2}}}\) 

Since cos at u(t) is a right-sided signal, the ROC will be:

Re{s} > 0

Also,

\(- \cos at\;u\left( { - t} \right) \leftrightarrow \frac{s}{{{s^2} + {a^2}}}\) 

Re{s} < 0

Application:

Given \(X\left( s \right) = \frac{s}{{{s^2} + 9}}\;;Re\left( s \right) < 0\)

The inverse Laplace transform of the above will be:

\(X\left( s \right)\mathop \leftrightarrow \limits^{I.L.T} - \cos \left( {3t} \right)\;u\left( { - t} \right)\) 

∴ x(t) = - cos (3t) u(-t)

At \(t = - \frac{{\pi \;}}{3}\), we get

x(t) = -cos (-π) = 1 

\(Y\left( s \right) = \frac{{s + 4}}{{s + 8}}\)

\(= 1 - \frac{4}{{s + 8}}\)

By applying inverse Laplace transform

y(t) = δ(t) – 4 e^-8t u(t)

At t = 0.15 sec,

Y (0.15) = δ (0.15) – 4e^-1.2 u (0.15)

\(u(t)\leftrightarrow \frac{1}{s}; Re(s)>0\)

Given step response is y(t), i.e.

\(Y\left( s \right) = H\left( s \right)U\left( s \right) \)

\(Y(s)= \frac{{b\left( {s + a} \right)}}{{s\left( {s + b} \right)}}\)

From the initial value theorem, we can write:

\( y\left( 0 \right) = \mathop {{\rm{lt}}}\limits_{s \to \infty } sY\left( s \right)\)

\(\\ = \mathop {{\rm{lt}}}\limits_{s \to \infty } \frac{{sb\left( {s + a} \right)}}{{s\left( {s + b} \right)}} = 4 \)

b = 4

From final value theorem:

\(y\left( \infty \right) = \mathop {{\rm{lt}}}\limits_{s \to 0} sY\left( s \right)\)

\(\\ = \mathop {{\rm{lt}}}\limits_{s \to 0} \frac{{sb\left( {s + a} \right)}}{{s\left( {s + b} \right)}} = 20 \)

a = 20

\(\therefore \frac{b}{a} = \frac{4}{{20}} = 0.2\)