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.

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Nyquist Sampling Theorem: 

A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.
f_s ≥ 2f_m

Therefore when we want to convert continuous signals to discrete signals, the sampling must be done at the Nyquist rate.

Calculation:

Given that,
f_m = 3.1 kHz 
⇒ f_s ≥ 2f_m
⇒ f_s ≥ 2 × 3.1 = 6.4 kHz

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:
Multiplication in the time domain is the convolution in the frequency domain and the maximum frequency of convolution of the signal is the sum of individual frequencies, i.e.
If z(t) = x(t) y(t)

The maximum frequency of Z(ω) = ω_x + ω_y
where ω_x is the maximum frequency of signal x(t)
ω_y is the maximum frequency of signal y(t)

Application:
The maximum frequency present in x(t) is:

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)]

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

Calculation:
f_s = Sampling Frequency = 56 MHz
The frequencies present at the input are:
f_m1 = 10 MHz, f_m2 = 50 MHz and, f_m3 = 70 MHz

Once sampled, the frequency content at the output will be: f_m ± f_s
For, f_m1 = 10 MHz, the output will contain frequencies of,
= f_m1, f_m1 ±, f_m1 ± 2f_s ….
= 10 MHz, 66 MHz, 122 MHz

For, f_m2 = 50 MHz, the output will contain frequencies of
= f_m2, f_m2 ± f_s, f_m2 ± 2f_s ….
= 50 MHz, 6 MHz, 106 MHz, 162 MHz

For, f_m3 = 70 MHz, the output will contain frequencies of
= f_m3, f_m3 ± f_s, f_m3 ± 2f_s ….
= 70 MHz, 14 MHz, 182 MHz ….

When these frequencies are passed through a low pass filter with a cutoff frequency of 15 MHz, the only remaining frequencies at its output will be 6MHz, 10 MHz and 14 MHz.

Period of sampling train, T_s = 20 ms

If frequency of x(t) is fx, then after sampling the signal, the sampled signal has the frequency,
f_s - f_x = 50 - f_x and f_s + f_x = 50 + f_s
Now, the sampled signal is applied to and ideal low pass filter with cut off frequency.
f_c = 25 Hz
Now, the o/p of filter carried a single frequency component of 20 Hz
∴, only (f_s−f_x) component passes through the filter, ie.
f_s - f_x < 25
and f_s - f_x = 20
50 - f_x­ = 20
f_x = 50 - 20 = 30 Hz

Given:

• Input signal: x(t) = 2 + 5sin(100πt).
• Sampling frequency: f_s = 400 Hz.
• Transfer function: Y(z)/X(z) = (1/N) * ((1 - z^-N) / (1 - z^-1)), where N = number of samples per cycle.

Step 1: Determine the frequency of the input signal

• The frequency of the sinusoidal component in x(t) is f = 50 Hz (from sin(100πt), where ω = 2πf).

Step 2: Calculate the number of samples per cycle (N)

• N = f_s / f.
• Substitute values: N = 400 / 50 = 8.

Step 3: Analyze the transfer function

• The transfer function is a moving average system over N samples.
• For a steady-state sinusoidal input, the moving average system passes the DC component (mean) and filters out the sinusoidal component.

Step 4: Compute the steady-state output

• The DC component of the input signal x(t) is 2 (the constant term).
• The sinusoidal component is filtered out by the system.
• Thus, the steady-state output y[n] is 2.

Answer: The steady-state output y[n] is 2.

Concept:

Duty-Cycle is defined as the ratio of time for which the waveform is High (1) to the total period of the waveform, i.e.

\(Duty\;Cycle = \frac{{{t_{on}}}}{{{t_{on}} + {t_{off}}}} = \frac{{{t_{on}}}}{T}\)

Where T is the Time period of the rectangular wave.

Also, for a periodic waveform, the Fourier series coefficients are defined as:

\({c_k} = \frac{1}{T}\smallint x\left( t \right){e^{ - jk{\omega _0}t}}dt\)

Calculation:

Given Duty cycle of the rectangular wave = 30 % = 3/10

\(So,\frac{{{t_{on}}}}{T} = \frac{3}{{10}}\)

\({t_{on}} = \frac{{3T}}{{10}}.\)

The given rectangular pulse with amplitude A(let) is now drawn as shown below:

To find the frequency content of a periodic wave, we can do Fourier analysis and evaluate the Fourier Series coefficients as:

\( \Rightarrow {c_k} = \frac{1}{T}\mathop \smallint \nolimits_0^T x\left( t \right){e^{ - jk{\omega _0}t}}dt\)

For the given rectangular wave,

\( \Rightarrow {c_k} = \frac{1}{T}\mathop \smallint \nolimits_0^{\frac{{3T}}{{10}}} A.{e^{ - jk{\omega _0}t}}dt\)

\( \Rightarrow {c_k} = \frac{A}{T}\mathop \smallint \nolimits_0^{\frac{{3T}}{{10}}} {e^{ - jk{\omega _0}t}}dt\)

\( = \frac{A}{T}\left. {\left[ {\frac{{{e^{ - jk{\omega _0}t}}}}{{ - jk{\omega _0}}}} \right]} \right|_0^{\frac{{3T}}{{10}}}\)

\( = - \frac{A}{{T\left( {jk{\omega _0}} \right)}}\left[ {{e^{ - jk{\omega _0}.\frac{{3T}}{{10}}}} - 1} \right]\)

\( = \frac{A}{{jTk{\omega _0}}}\left[ {1 - {e^{ - jk.{\omega _0}.\frac{{3T}}{{10}}}}} \right]\)

\({\rm{Replacing\;}}{\omega _0} = \frac{{2\pi }}{T},\;we\;get,\)

\( = \frac{{A.T}}{{jT.k.2\pi }}\left[ {1 - {e^{ - jk.\frac{{2\pi }}{T}.\frac{{3T}}{{10}}}}} \right]\)

\( = \frac{A}{{jk2\pi }}\left[ {1 - {e^{ - jk.\frac{{3\pi }}{5}}}} \right]\)

From the above equation, we observe that when k is a multiple of 10, the coefficient c is 0. i.e.

for k = 10n, the coefficients are zero.

\( \Rightarrow {c_k} = {c_{10n}} = \frac{A}{{j.10n.2\pi }}\left[ {1 - {e^{ - j\left( {10n} \right).\frac{{3\pi }}{5}}}} \right]\)

\( = \frac{A}{{jn \times 20\pi }}\left( {1 - {e^{ - j6n\pi }}} \right) = 0\)  (∵ e^-j6nπ = cos 6nπ – j sin 6nπ = cos 6nπ = 1)

So, we conclude that when k is a multiple of 10, there are no frequency components for the given rectangular pulse.

So, Option (1) is correct.

Concept:
Aliased frequency component is given by:
fa = |fs – fm|
f_s = sampling frequency
f_m = message signal frequency

Calculation:
For a sampling rate of 140 Hz, since f_s > f_m:
Aliased Frequency = f_s - f_m,
= 140 – 100 = 40 Hz
For a sampling rate of 90 Hz, since f_s < f_m:
Aliased Frequency = f_m – f_s,
= 100 – 90 = 10 Hz
For a sampling rate of 30 Hz, since f_s < f_m:
Aliased Frequency = f_m – f_s
= 100 – 30 = 70 Hz

Aliasing is explained with the help of the spectrum as shown: