Signals and Systems Test 2

Concept:

Properties of unit impulse:

1) \(\delta \left( {at} \right) = \frac{1}{{\left| a \right|}}\delta \left( t \right)\)

2) \(\mathop \smallint \nolimits_{ - \infty }^\infty \delta \left( {t - a} \right)dt = 1\), for t = a; otherwise 0.

Analysis:

For an impulse function, we have:

\(\mathop \smallint \nolimits_{ - 1}^8 \left[ {\delta \left( {t + 3} \right) - 2\delta \left( {4t} \right)} \right]dt\)

\( = \mathop \smallint \nolimits_{ - 1}^8 \delta \left( {t + 3} \right)dt - 2\mathop \smallint \nolimits_{ - 1}^8 \delta \left( {4t} \right)dt\)

\( = 0 - 2\mathop \smallint \nolimits_1^8 \delta \left( {4t} \right)\)

\(\mathop \smallint \nolimits_{ - 1}^8 \delta \left( {t + 3} \right)dt = 0\) because t = -3 does not exist in the given interval -1 < t < 8

Since \(\delta \left( {at} \right) = \frac{1}{a}\delta \left( t \right)\), the above can be written as:

\(= - \frac{2}{4}\mathop \smallint \nolimits_1^8 \delta \left( t \right) = - \frac{1}{2} = - 0.5\)

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

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

Concept:

1) Fourier Series of Even Signals Contain only Cosine terms

2) Fourier Series of odd signals contain only sine terms.

Application:

Given f(t) is Even Periodic

Fourier series of f(t) will contain only cosine terms

For y(t) let us take an example

Let \(f\left( t \right)=Cos\left( \frac{2\pi }{T}t \right)\)

Time Period = T

Let us shift this signal by T/4

\(y\left( t \right)=cos\left( \frac{2\pi }{T}\left( t+\frac{T}{4} \right) \right)\)

\(y\left( t \right)=\cos \left( \frac{2\pi t}{\tau }+\frac{\pi }{2} \right)\)

\(y\left( t \right)-\sin \left( \frac{2\pi t}{T}\right)\)

y(t) is an Odd function

\(x\left[ n \right] = cos\left[ {\frac{{3\pi }}{4}n} \right] = even\;singal\)

\(y\left[ n \right] = sin\left[ {7\pi n} \right] = odd\;singal\)

x[n] y[n] = odd signal

\(\mathop \sum \limits_{n = - \infty }^{ + \infty } x\left[ n \right]y\left[ n \right] = 0\)

Concept:

\(If\;f\left( n \right)\mathop \leftrightarrow \limits^{DFT} F\left( k \right)\)

\(then\;f\left( {n - {n_0}} \right)\mathop \leftrightarrow \limits^{DFT} F\left( k \right){e^{ - jk.\frac{{2\pi }}{N}.{n_o}}}\)

Calculation:

Given f(n) ↔ F(k)

\(f\left( {n - 3} \right) \leftrightarrow F\left( k \right).{e^{ - jk.\frac{{2\pi }}{N}.3}}\)

The given DFT is a 4-point DFT. So, N = 4

\(\therefore f\left( {n - 3} \right) \leftrightarrow F\left( k \right){e^{ - jk.\frac{{2\pi }}{4}.3}}\)

\(F\left( k \right){e^{ - j\frac{{3\pi k}}{2}}} = F\left( k \right).{j^k}\;\;\left( {{e^{ - j\frac{{3\pi }}{2}}} = j} \right)\)

F(k) will now become: { 5.j⁰, (3 + 6j).(j¹), -1(j²), (3 – 6j).(j³)}

Causality:

A linear time-invariant discrete-time system is said to be causal if the impulse response h[n] = 0, for n < 0 and it is therefore right-sided.

The ROC of such a system H(z) is the exterior of a circle. If H(z) is rational, then the system is said to be causal if:

1) The ROC is the exterior of a circle outside the outermost pole; and

2) The degree of the numerator polynomial of H(z) should be less than or equal to the degree of the denominator polynomial.

Stability:

A discrete-time LTI system is said to be BIBO stable if the impulse response h[n] is summable, i.e.

\(\mathop \sum \nolimits_{n = - \infty }^\infty \left| {h\left[ n \right]} \right| < \infty \)

z-transform of h[h] is given as:

\(H\left( z \right) = \mathop \sum \nolimits_{n = - \infty }^\infty h\left[ h \right]{z^{ - n}}\)

Let z = e^jΩ (which describes a unit circle in the z-plane), then

\(\left| {H\left[ {{e^{j{\rm{\Omega }}}}} \right]} \right| = \left| {\mathop \sum \nolimits_{n = - \infty }^\infty h\left[ n \right]{e^{ - j{\rm{\Omega }}n}}} \right|\)

The above equality can be written as:

\( \le \;\mathop \sum \nolimits_{n = - \infty }^\infty \left| {h\left[ n \right]{e^{ - j{\rm{\Omega }}n}}} \right|\)

\(= \mathop \sum \nolimits_{n = - \infty }^\infty \left| {h\left[ n \right]} \right| < \infty \)

This is the condition for stability. Thus we can conclude that an LTI system is stable if the ROC of the system function H(z) contains the unit circle |z| = 1

Causality and Stability:

For a system with rational transfer function H(z) to be causal, the ROC should lie outside the outermost pole, and for BIBO stability, the ROC should include the unit circle |z| = 1. 

∴ An LTI discrete-time causal system with the rational system function H(z) is said to be stable if all the poles of H(z) lie inside the unit circle.

Concept:

The energy of the signal is defined as:

\(E= \mathop \smallint \limits_{ - \infty }^\infty {x^2}\left( {t} \right)dt\)

Calculation:

Given

\(\frac{E}{4} = \mathop \smallint \limits_{ - \infty }^\infty {x^2}\left( t \right)dt\)

The signal is shrunk (compressed) by a factor of 3, i.e.

x'(t) = x(3t)

The new energy will be:

\(E' = \mathop \smallint \limits_{ - \infty }^\infty {x^2}\left( {3t} \right)dt\) 

Taking 3t = u

3 dt = du

\(E' = \mathop \smallint \limits_{ - \infty }^\infty {x^2}\left( u \right)\frac{{du}}{3}\)

\(= \frac{1}{3}\left( {\mathop \smallint \limits_{ - \infty }^\infty {x^2}\left( u \right)du} \right)\)

\(E'= \frac{1}{3}.\frac{E}{4}=\frac{E}{12}\)

The z-transform of a sequence x(n) is defined as:

\(X (Z) = \sum_{n =- \infty}^{+ \infty} x(n) \cdot z^{-n}\)

Now, \(Y(z) = X (z^3) = \sum_{n=- \infty}^\infty x(n) (z^3)^{-n}\)

\(\Rightarrow Y(z) = X(z^3) = \sum_{n=-\infty}^\infty x(n)z^{-3n}\)

Let 3n = k

\(\Rightarrow n=\frac{k}{3}\)

\(Y(z) =\sum_{k=-\infty}^\infty x\left(\frac{k}{3}\right) z^{-k}\)

Thus, \(y(n) = x \left(\frac{n}{3}\right)\)

\(y\left( n \right)=\left\{ \begin{matrix} ~x\left( n \right), \\ 0, \\\end{matrix} \right.\begin{matrix} ~n=0,3,6,-- \\ otherwise \\\end{matrix}\)

So, y(n) at n = 4 is 0.

Concept:

The energy of a continuous-time signal is defined as:

\({E_{x\left( t \right)}} = \mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_{ - T}^T {\left| {x\left( t \right)} \right|^2}dt\)

Also, the Parseval's Power theorem states that:

\(\mathop \smallint \limits_{ - \infty }^\infty {\left| {x\left( t \right)} \right|^2}dt = \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \infty }^\infty {\left| {X\left( \omega \right)} \right|^2}d\omega \)

And, If x(t) having Fourier transform as X(f)

\(\mathop \smallint \limits_{ - \infty }^\infty {\left| {x\left( t \right)} \right|^2}dt = \mathop \smallint \limits_{ - \infty }^\infty {\left| {X\left( f \right)} \right|^2}df\)

Calculation:

f(t) = e^-t u(t)

\({E_f}\left( t \right) = \mathop \smallint \limits_{ - \infty }^\infty {\left| {f\left( t \right)} \right|^2}dt = \mathop \smallint \limits_0^\infty {\left( {{e^{ - t}}} \right)^2}dt\)

\(= \frac{1}{2} = 0.5\)

The energy (E_f) for |ω| ≤ 5 will be:

\( = \frac{1}{{2\pi }}\;\mathop \smallint \limits_{ - 5}^5 \frac{1}{{1 + {ω ^2}}}dω \)

\( = \frac{1}{{2\pi }}\left( {{{\tan }^{ - 1}}ω } \right)_{ - 5}^5 \)

\(= \frac{2}{{2\pi }}\left( {{{\tan }^{ - 1}}ω } \right)_0^5\)

\( = \frac{1}{\pi }\left( {{{\tan }^{ - 1}}5 - {{\tan }^{ - 1}}0} \right)\)

\(= \frac{{1.37}}{\pi } = 0.437\)

Percentage of energy contained will be:

\( = \frac{{0.437}}{{0.5}} \times 100\)

= 87.433%

Concept:

Properties of Fourier series coefficient:

1) If x(t) is real, then the Fourier series coefficient satisfies the condition:

\({c_n} = c_{ - n}^*\) (* denotes conjugate)

2) If x(t) is even, then:

cn = c-n

3) Fourier series coefficient of \(\frac{{dx\left( t \right)}}{{dt}}\) is given by:

\(\frac{{dx\left( t \right)}}{{dt}}\mathop \leftrightarrow \limits^{C.T.F.S} jn\frac{{2\pi }}{{{T_0}}} \cdot {c_n}\)

Application:

Given:

\({c_n} = \left\{ {\begin{array}{*{20}{c}} {2,\;\;\;\;\;\;\;,\;\;\;\;\;\;\;\;\;~n = 0}\\ {j{{\left( {\frac{1}{2}} \right)}^{\left| n \right|}},\;\;\;\;otherwise} \end{array}} \right.\)

c-n will be defined as:

\(c_{ - n}^* = \left\{ {\begin{array}{*{20}{c}} {2\;\;\;\;\;\;\;\;\;,\;\;\;\;\;\;n = 0}\\ { - j{{\left( {\frac{1}{2}} \right)}^{\left| n \right|}},\;\;\;\;otherwise} \end{array}} \right.\)

Since \({c_n} \ne c_{ - n}^*\), x(t) is not real.

Also, we observe that cn = c-n.

∴ The function x(t) is even.

\(x\left( t \right) \leftrightarrow {c_n}\)

\(\frac{{dx\left( t \right)}}{{dt}}\mathop \leftrightarrow \limits^{C.T.F.S} c_n' = jn\frac{{2\pi }}{{{T_0}}} \cdot {c_n}\)

\(c_n' = \left\{ {\begin{array}{*{20}{c}} {0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;,\;\;\;\;\;\;n = 0}\\ { - n{{\left( {\frac{1}{2}} \right)}^{\left| n \right|}}.\frac{{2\pi }}{{{T_0}}}\;\;,\;\;\;\;otherwise} \end{array}} \right.\)

We observe that \(c_n' \ne c_{ - n}'\)

\(\therefore y\left( t \right) = \frac{{dx\left( t \right)}}{{dt}}\) is not even.

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

Put S = j2

\(\begin{array}{l} \left| {H\left( {2j} \right)} \right| = \frac{1}{{ - 4 + 6j + 1}} = \frac{1}{{ - 3 + 6j}} = \frac{1}{{\sqrt {9 + 36} }} = \frac{1}{{\sqrt {45} }}\\ \angle H\left( {j2} \right) = -180+ ta{n^{ - 1}}\left( {\frac{6}{{ 3}}} \right) = -116.56^\circ \end{array}\)  

y_ss(t) = 2 |H(2j)| cos [2t + 25 + ∠H(j2)]

\(= \frac{2}{{\sqrt {45} }}\cos \left( {2t + 25 -116.56^\circ } \right)\)