Signals and Systems Test 1

Given that, x(t) = e^-3t u(t)

h(t) = e^-6t u(t)

We know x(t) * h(t) = x(s) ⋅ H(s) i.e. convolution in one domain is multiplication in another domain

x(t) → x(s)

\({e^{ - 3t}}u\left( t \right) \to \frac{1}{{s + 3}}\) 

\({e^{ - 6t}}u\left( t \right) \to \frac{1}{{s + 6}}\) 

\({e^{ - 3t}}u\left( t \right)*{e^{ - 6t}}u\left( t \right) = \frac{1}{{s + 3}} \cdot \frac{1}{{s + 6}}\) 

\( \Rightarrow \frac{1}{{\left( {s + 3} \right)\left( {s + 6} \right)}} = \frac{A}{{s + 3}} \cdot \frac{B}{{s + 6}}\) 

\(A = {\left. {\frac{1}{{s + 6}}} \right|_{s = - 3}} = \frac{1}{3}\) 

\(B = {\left. {\frac{1}{{s + 3}}} \right|_{s = - 6}} = - \frac{1}{3}\) 

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

By applying inverse Laplace transform,

\( = \frac{1}{3}{e^{ - 3t}}u\left( t \right) - \frac{1}{3}{e^{ - 6t}}u\left( t \right)\) 

\(= \left( {\frac{{{e^{ - 3t}} - {e^{ - 6t}}}}{3}} \right)u\left( t \right)\) 

We know power \(\left( P \right) = \mathop {{\rm{lt}}}\limits_{T \to 0} \frac{1}{T}\mathop \smallint \nolimits_0^T {\left[ {x\left( t \right)} \right]^2}dt\) 

\(\left( P \right) = \mathop {{\rm{lt}}}\limits_{T \to 0} \frac{1}{T}\mathop \smallint \nolimits_0^T {\left( {4 + 3\sin t} \right)^2}dt\) 

The time period of sine signal is 2π.

So, \(P = \mathop {{\rm{lt}}}\limits_{T \to 0} \left( {\frac{1}{{2\pi }}} \right)\mathop \smallint \nolimits_0^{2\pi } \left( {16 + 9{{\sin }^2}t + 24\sin t} \right)dt\) 

\(P = \mathop {{\rm{lt}}}\limits_{T \to 0} \frac{1}{{2\pi }}\left[ {16\left( t \right)_0^{2\pi } + 9\mathop \smallint \nolimits_0^{2\pi } \frac{{1 - \cos 2t}}{2} + 24\mathop \smallint \nolimits_0^{2\pi } \sin t\;dt} \right]\) 

\(= \mathop {{\rm{lt}}}\limits_{T \to 0} \frac{1}{{2\pi }}\left[ {\left( {32\pi } \right) + \frac{9}{2}\left( {2\pi - \frac{1}{2}\left( {\sin 2t} \right)_0^{2\pi }} \right) + 24\left( { - \cos t} \right)_0^{2\pi }} \right]\) 

\(= \mathop {{\rm{lt}}}\limits_{T \to 0} \frac{1}{{2\pi }}\left( {32\pi + 9\pi + 0} \right)\) 

\( = \frac{{41\pi }}{{2\pi }} = 20.5\;W\) 

Given integral is \(\mathop \smallint \nolimits_{ - 4}^6 \left( {3{t^3} + 6{t^2} + \cos \pi t} \right)\frac{{{d^3}}}{{d{t^3}}}\delta \left( {t - 2} \right)dt\) 

δ (t - 2) indicates the unit impulse function shifted by t = 2, but still in limit of (-4, 6)

So function value is non zero.

We know that \(\mathop \smallint \nolimits_a^b x\left( t \right)\frac{{{d^n}}}{{d{t^n}}}\delta \left( {t - {t_0}} \right) = {\left( { - 1} \right)^n}{\left. {\frac{{{d^n}}}{{d{t^n}}}x\left( t \right)} \right|_{t = {t_0}}}\) 

\(\mathop \smallint \nolimits_{ - 4}^6 \left( {3{t^3} + 6{t^2} + \cos \pi t} \right)\frac{{{d^3}}}{{d{t^3}}}\delta \left( {t - 2} \right)dt\) 

\(= {\left( { - 1} \right)^3}{\left. {\left[ {\frac{{{d^3}}}{{d{t^3}}}\left( {3{t^3} + 6{t^2} + \cos \pi t} \right)} \right]} \right|_{t = 2}}\) 

\(= \left( { - 1} \right){\left[ {3 \times 3\frac{{{d^2}}}{{d{t^2}}}\left( {{t^2}} \right) + 6 \times 2\frac{{{d^2}}}{{d{t^2}}}t - \pi \sin \pi t} \right]_{t = 2}}\) 

\(= \left( { - 1} \right){\left( {18\frac{d}{{dt}}\left( t \right) + 0 - {\pi ^2}\frac{d}{{dt}}\cos \pi t} \right)_{t = 2}}\) 

= (-1) (18 + π³ sin πt)_{t = 2}

= (-1) (18 + π³ sin 2π)

= -18

\(h\left( n \right) = \left\{ {\begin{array}{*{20}{c}}{{a^n},}&{n \ge 0}\\{{b^n},}&{n < 0}\end{array}} \right.\) 

For stability, h(n) must be absolute summable.

\(\mathop \sum \limits_{ - \infty }^\infty {\left| {h\left( n \right)} \right|^2} < \infty \) 

\(\Rightarrow \mathop \sum \limits_{n = 0}^\infty {\left| {{a^n}} \right|^2} + \mathop \sum \limits_{ - \infty }^{ - 1} {\left| {{b^n}} \right|^2} < \infty \) 

The above equation valid for |a|< 1 and |b|> 1.

\(\omega \;\left( {Angular\;frequency} \right) = \frac{{2\pi }}{T}\)

So, \(T= \frac{{2\pi }}{\omega }\)      ---- (1)

The given signal can be represented as:

\( \Rightarrow x\left( n \right) = 1 + \sin \left( {{\omega _1}n} \right) + 3\cos \left( {{\omega _2}n} \right) + \cos \left( {{\omega _3}n + \frac{\pi }{2}} \right)\)

Where \({\omega _1} = \frac{{2\pi }}{5},\;{\omega _2} = \frac{{2\pi }}{7}\;and\;{\omega _3} = \frac{{4\pi }}{{22}}\) 

Using Equation (1)

 \({T_1} = \frac{{2\pi }}{{2\pi }} \times 5 = 5\)

\(\;{T_2} = \frac{{2\pi }}{{2\pi }} \times 7 = 7\)

\({T_3} = \frac{{2\pi }}{{4\pi }} \times 22 = 11\)

The period of the combination of the given signals is therefore, the LCM [T₁, T₂, T₃], which is = 385