Signals and Systems Test 1

Given signal is, x[n] = r[n] (u[n] – u[n – 11])

Energy of x[n] is

\(E = \mathop \sum \limits_{n = 0}^{10} {n^2}\)

By using the formula \(\mathop \sum \limits_{k = 1}^N {k^2} = \frac{{N\left( {N + 1} \right)\left( {2N + 1} \right)}}{6}\)

\(= \frac{{10\; \times \;11\; \times\; 21}}{6} = 385\)

1. \(\frac{{dy}}{{dx}} + 6y\;\left( t \right) = 4\;x\left( t \right)\)

This is an ordinary differential equation with constant co-efficient, therefore it is linear and time invariant. It is casual.

 2. y[n] + 2y [n - 1] = x [n + 1]

This is a difference equation with the constant co-efficient. Therefore, it is linear and time-invariant. It is non-casual since the output depends on future values of x.

3. y[n + 1 ] + 4y[n] = 3x [n + 1] – x[n]

Rewriting the above equation by shifting one unit.

Y[n] + 4y [n - 1] = 3x [n] – x[n - 1]

This is a difference equation with constant co-efficient. So, it is linear and time invariant. The output does not depend on future values of the input, so it is causal.

4. \(\frac{{dy}}{{dt}} + {y^2}\left( t \right) = x\left( t \right)\)

The co-efficient of y means that this is non-linear and it does not depend on t. so, it is time invariant. It is causal.

y [n] is the first difference of x [n]

y(n) = x(n) – x(n - 1)

By applying z-transform.

⇒ Y(z) = X(z) – z^-1 X(z)

⇒ Y(z) = 3 + 4z – 6z^-1 + 2z^-2 – z^-1 (3 + 4z – 6z^-1 + 2z^-2)

= 3 + 4z – 6z^-1 + 2z^-2 – 3z^-1 – 4 + 6z^-2 – 2z^-3

= -1 + 4z – 9z^-1 + 8z^-2 – 2z^-3

Concept:

z-transform of a discrete signal can be written as;

\(X\left( z \right) = \mathop \sum \limits_{n = - \infty }^{ + \infty } x\left[ n \right]{z^{ - n}}\)      …1)

Calculation:

We have to calculate:

\(\mathop \sum \limits_{n = 0}^\infty \left( n \right){\left( {\frac{1}{2}} \right)^n} = \mathop \sum \limits_{n = - \infty }^{ + \infty } nu\left[ n \right].{\left( 2 \right)^{ - n}}\)      …2)

On comparing equation (1) & (2) we find that;

\(\mathop \sum \limits_{n = 0}^\infty \left( n \right){\left( 2 \right)^{ - n}} = X\left( z \right) = X\left( 2 \right)\)

Where X(z) is z-transform of nu[n] which is:

\(X\left( z \right) = \frac{{{z^{ - 1}}}}{{{{\left( {1 - - {z^{ - 1}}} \right)}^2}}}\)

So, \(X\left( z \right) = \frac{{{z^{ - 1}}}}{{{{\left( {1 - {z^{ - 1}}} \right)}^2}}}\) 

\(X\left( 2 \right) = \frac{{{2^{ - 1}}}}{{{{\left( {1 - {2^{ - 1}}} \right)}^2}}} = \frac{{{Y_2}}}{{{Y_4}}} = 2\)

Hence our required summation is

\(\mathop \sum \limits_{n = 0}^\infty \left( n \right){\left( 2 \right)^{ - n}} = X\left( 2 \right) = 2\)

Concept:

If the DTFT of a signal f(n) ↔ F(ω), then from the frequency shifting property

 \({{e}^{j{{\omega }_{o}}n}}f\left( n \right)\leftrightarrow F\left( \omega -{{\omega }_{o}} \right)\)

Calculation:

Given, f(n) = {a, b, c, d}

If f(n) ↔ F(ω)

Then, \({{e}^{j\pi n}}f\left( n \right)\leftrightarrow F\left( \omega -\pi \right)\)

So, the inverse DTFT of F(ω - π) is

e^jπn f(n), i.e

\(\Rightarrow {{\left( -1 \right)}^{n}}f\left( n \right)=\left\{ f\left( 0 \right),~\left( -1 \right)f\left( 1 \right),~f\left( 2 \right),~\left( -1 \right)f\left( 3 \right) \right\}\)

Concept:

The Fourier series representation of the discrete-time periodic sequence is given by:

\(x\left( n \right) = \mathop \sum \limits_{k = 0}^{N - 1} {a_k}{e^{jk{\omega _0}n}}\)

\(x\left( n \right) = \ldots + {a_{ - 1}}{e^{ - j{\omega _0}n}} + {a_1}{e^{j{\omega _0}n}} + \ldots \)

ak = Fourier series coefficient periodic by N.

ω0 = Fundamental frequency.

Application:

Given sequence is: \(x\left( n \right) = \cos \frac{\pi }{4}n\)

The fundamental frequency of the sequence \({\omega _0} = \frac{\pi }{4}\)

We know that, \(\cos \theta = \frac{{{e^{j\theta }} + {e^{ - j\theta }}}}{2}\)

Now, we can rewrite the given sequence as

\(x\left( n \right) = \frac{{{e^{\frac{{j\pi }}{4}n}} + {e^{ - \frac{{j\pi }}{4}n}}}}{2}\)

\( = \frac{1}{2}{e^{\frac{{j\pi }}{4}n}} + \frac{1}{2}{e^{\frac{{ - j\pi }}{4}n}}\)

We can write \({e^{\frac{{ - j\pi }}{4}n}} = {e^{\frac{{j7\pi }}{4}n}}\)

Now, x(n) becomes

\(x\left( n \right) = \frac{1}{2}{e^{\frac{{j\pi }}{4}n}} + \frac{1}{2}{e^{\frac{{j7\pi }}{4}n}}\)

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.

\(x\left( t \right) = j\left( {{e^{\frac{{j\pi t}}{4}}} - {e^{\frac{{ - j\pi t}}{4}}}} \right) + 2\left( {{e^{\frac{{j5\pi t}}{4}}} - {e^{\frac{{ - j5\pi t}}{4}}}} \right)\)

\( = - 2\sin \left( {\frac{{\pi t}}{4}} \right) + 4\cos \left( {\frac{{5\pi t}}{4}} \right)\)

The steady state response of the output is

\(y\left( t \right) = 4\left| {H\left( {\frac{{j5\pi }}{4}} \right)} \right|\cos \left( {\frac{{5\pi }}{4} + \angle H\left( {\frac{{j5\pi }}{4}} \right)} \right)\)

\( = 4\cos \left( {\frac{{5\pi t}}{4} - \frac{{5\pi }}{4}} \right)\)

Given that f(t) = v(t) = e^-2t u(t)

\(F\left( \omega \right) = \frac{1}{{2 + j\omega }}\)

\({\left| {F\left( \omega \right)} \right|^2} = \frac{1}{{4 + {\omega ^2}}}\)

The total energy dissipated by the resistor is

\(W = \frac{1}{\pi }\mathop \smallint \limits_0^\infty {\left| {F\left( \omega \right)} \right|^2}d\omega \)

\( = \frac{1}{\pi }\mathop \smallint \limits_0^\infty \frac{1}{{4 + {\omega ^2}}}d\omega \)

\(= \frac{1}{\pi }\left[ {\frac{1}{2}{{\tan }^{ - 1}}\frac{\omega }{2}} \right]_0^\infty = 0.25\;J\)

The energy in the frequency band 0 < ω < 10 is,

\(W = \frac{1}{\pi }\mathop \smallint \limits_0^{10} {\left| {F\left( \omega \right)} \right|^2}d\omega \)

\( = \frac{1}{\pi }\mathop \smallint \limits_0^{10} \frac{1}{{4 + {\omega ^2}}}d\omega \)

\( = \frac{1}{\pi }\left[ {\frac{1}{2}{{\tan }^{ - 1}}\frac{\omega }{2}} \right]_0^{10} = 0.218\;J\)

The percentage of the total energy is \(= \frac{{0.218}}{{0.25}} \times 100 = 87.43\)

y[n] + 3y[n – 1] + 2y[n – 2] = 2x[n] – x[n – 1]

By applying Z-transform,

Y[z] + 3(z^-1Y(z) + y[-1]) + 2(z^-2Y(z) + y[-2] + z^-1y[-1]) = 2X(z) – z^-1X(z) – x[-1]

\(\Rightarrow Y\left( z \right)\left( {1 + 3{z^{ - 1}} + 2{z^{ - 2}}} \right) + 2 = \frac{{2z}}{{z - 1}} + \frac{1}{{z - 1}}\)

\(\Rightarrow Y\left( z \right) = \frac{{{z^2}}}{{{z^2} + 3z + 2}}\left( {\frac{{2z - 1 - 2z + 2}}{{z - 1}}} \right)\)

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

By applying partial fractions,

\(zy\left( z \right) = - \frac{1}{2}\frac{z}{{\left( {z + 1} \right)}} + \frac{4}{3}\frac{z}{{\left( {z + 2} \right)}} + \frac{1}{6}\frac{z}{{\left( {z - 1} \right)}}\)

By applying the inverse z-transform,

\(y\left[ n \right] = \left[ { - \frac{1}{2}{{\left( 1 \right)}^{n - 1}} + \frac{4}{3}{{\left( { - 2} \right)}^{n - 1}} + \frac{1}{6}} \right]u\left( {n - 1} \right)\)

2 y [n] = α y [n - 2] – 2 x [n] + β x [n - 1]

By applying z – transform,

2Y(z) = α Y(z) z^-2 – 2 X(z) + β X(z) z^-1

\(\Rightarrow \frac{{Y\left( z \right)}}{{X\left( z \right)}} = \frac{{(\beta {z^{ - 1}} - 2}}{{\left( {2 - \alpha {z^{ - 2}}} \right)}}\)

\(\Rightarrow H\left( z \right) = \frac{{z\left( {\beta - 2z} \right)}}{{\left( {2{z^2} - \alpha } \right)}}\)

H(z) has poles at \(z \pm \sqrt {\frac{\alpha }{2}}\) and zeros at \(z = 0,\frac{\beta }{2}\).

For a stable system, poles must be lie inside the unit circle of z plane.

\(\Rightarrow \left| {\sqrt {\frac{\alpha }{z}} } \right| < 1 \Rightarrow \left| \alpha \right| < 2\)

Zero can lie anywhere in plane.

The given system is stable for |α | < 2, and any value of β.

Concept:

The Fourier Series Coefficient of a discrete time Periodic sequence is given by:

\({{c}_{K}}=\frac{1}{N} \sum_{n=0}^{N-1}x\left( n \right){{e}^{\frac{-j2\pi kn}{N}}}\)

Where N = Period of the sequence.

Calculation:

Given, x (n) = {1, 1, 0, 0} and N = 4

So, \({{c}_{k}}=\frac{1}{N}~\sum_{n=0}^{N-1}x\left( n \right){{e}^{-jk\omega n}}=\frac{1}{N}\sum_{n=0}^{N-1}x\left( n \right){{e}^{\frac{-jK2\pi n}{N}~}}\)

With N = 4,

\(\Rightarrow {{c}_{k}}=\frac{1}{4}\sum_{n=0}^{N-1}x\left( n \right){{e}^{\frac{-j2\pi kn}{4}}}=\frac{1}{4}\sum_{n=0}^{N-1}x\left( n \right){{e}^{\frac{-j\pi kn}{2}}}\)

\({{c}_{k}}=\frac{1}{4}\left[ x\left( 0 \right)+x\left( 1 \right){{e}^{\frac{-j\pi k}{2}}}+x\left( 2 \right){{e}^{\frac{-j\pi k\left( 2 \right)}{2}}}+x\left( 3 \right){{e}^{\frac{-j\pi k\left( 3 \right)}{2}}} \right]\)

With, x(0) = 1, x(1) = 1, x(2) = 0 and x(3) = 0.

\({{c}_{k}}=\frac{1}{4}\left[ 1+1.{{e}^{\frac{-j\pi k}{2}}}+0+0 \right]\)

\({{c}_{k}}=\frac{1}{4}\left[ 1+{{e}^{\frac{-j\pi k}{2}}} \right]\)

We are asked to calculate \(c_3^*\)

Putting k = 3.

\({{c}_{3}}=\frac{1}{4}\left[ 1+{{e}^{\frac{-j3\pi }{2}}} \right]\)

\(=\frac{1}{4}\left( 1+\cos \frac{3\pi }{2}-jsin\frac{3\pi }{2} \right)\)

\(=\frac{1}{4}\left( 1+j \right)\)

So, \(c_{3}^{*}=\frac{1}{4}\left( 1-j \right)\)

Option (3) is correct.