Signals and Systems Test 2

The overall output y(n) is the same as the input x(n) with a one-unit delay.

y(n) = x(n - 1)

By applying z-transform,

Y(z) = z^-1X(z)

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

⇒ H₁(z) H₂(z) = z^-1

\( \Rightarrow \left( {\frac{{1 - 0.2{z^{ - 1}}}}{{1 - 0.4{z^{ - 1}}}}} \right){H_2}\left( z \right) = {z^{ - 1}}\) 

\(\Rightarrow {H_2}\left( z \right) = \frac{{{z^{ - 1}}\left( {1 - 0.4{z^{ - 1}}} \right)}}{{1 - 0.2{z^{ - 1}}}}\) 

For even function, f(t) = f(-t)

For odd function, f(t) = -f(-t)

Convolution in the time domain implies multiplication in S(or frequency) domain

The Laplace transform of any signal h(t) is given by

So we observe that the H(s) = 1/s³, corresponds to a unit step signal convoluted with itself thrice.

Therefore the correct answer is option 1

y(t) = tx(t)

y₁(t) = t.x₁(t) = r₁(t)

y₂(t) = tx₂(t) = r₂(t)

y₁(t) + y₂(t) = t(x₁(t) + x₂(t))

= r₁(t) + r₂(t)    ∴ linear

y(t) = t.x(t)

y( t - t_o) = (t - t_o) x ( t - t_o)

and for delayed input signal,

y(t) = t x (t - t_o)

y(t) ≠ y( t-t_o)

∴ Time varying signal

F(ω) is the Fourier transform of f(t).

Concept:

The Z-transform of standard signals with their ROC are:

\({a^n}u\left( n \right) \leftrightarrow \frac{z}{{z - a}};\left| z \right| > \left| a \right|\)

\( - {a^n}u\left( { - n - 1} \right) \leftrightarrow \frac{z}{{z - a}};\left| z \right| < \left| a \right|\)

Application:

\(X\left( z \right) = \frac{{6{z^2}}}{{\left( {2z - 1} \right)\left( {3z - 1} \right)}}\)

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

\(= z[\frac{z}{{\left( {z - \frac{1}{2}} \right)\left( {z - \frac{1}{3}} \right)}}\)

\(= z\left[ {\frac{3}{{\left( {z - \frac{1}{2}} \right)}} - \frac{2}{{\left( {z - \frac{1}{3}} \right)}}} \right]\)

\(= \frac{{3z}}{{\left( {z - \frac{1}{2}} \right)}} - \frac{{2z}}{{\left( {z - \frac{1}{3}} \right)}}\)

For \(\left| z \right| < \frac{1}{2},\;\frac{{3z}}{{\left( {z - \frac{1}{2}} \right)}} \leftrightarrow \; - 3{\left( {\frac{1}{2}} \right)^n}u\left[ { - n - 1} \right]\) 

For \(\left| z \right| > \frac{1}{3},\;\frac{{2z}}{{\left( {z - \frac{1}{3}} \right)}} \leftrightarrow 2{\left( {\frac{1}{3}} \right)^n}u\left[ n \right]\) 

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

\(X\left( {{e^{j\omega }}} \right) = 10 - 4{e^{ - j2\omega }}\)

y(t) = x(t - 2)

By applying Fourier transform,

\( \Rightarrow Y\left( {{e^{j\omega }}} \right) = {e^{ - j2\omega }}X\left( {{e^{j\omega }}} \right)\)

\(= {e^{ - j2\omega }}\left( {10 - 4\;{e^{ - j2\omega }}} \right)\) 

\(= 10{e^{ - j2\omega }} - 4{e^{ - j4\omega }}\) 

\(Y\left( {{e^{j\omega }}} \right){\left. \right|_{\omega = \pi }} = Y\left( {{e^{j\pi }}} \right) = 10{e^{ - j2\left( \pi \right)}} - 4{e^{ - j4\pi }}\) 

= 10 (cos 2π - j sin 2π) - 4(cos 4 π - j sin 4π)

= 10(1) - 4(1) = 6

Concept:

Aliased frequency component is given by:

f_a = |f_s – f_m|

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

Since f(t) u(t) = f(t) for t > 0 also we know

u ( t - t_o) = 1, for t > t_o

Here in right side shifting that means t_o > 0

by property on shifting right side,

The Discrete-Time Fourier transform of a signal of infinite duration x[n] is given by:

\(X\left( \omega \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{e^{ - j\omega n}}\)

For a signal x(n) = aⁿ u(n), the DTFT will be:

\(X\left( \omega \right) = \mathop \sum \limits_{n = - \infty }^\infty a^nu\left[ n \right]{e^{ - j\omega n}}\)

Since u[n] is zero for n < 0 and a constant 1 for n > 0, the above summation becomes:

\(X\left( \omega \right) = \mathop \sum \limits_{n = 0 }^\infty a^n{e^{ - j\omega n}}\)

\(X\left( \omega \right) = \mathop \sum \limits_{n = 0 }^\infty (a{e^{ - j\omega n})}\)   ---(1)

The geometric series states that:

\(\mathop \sum \limits_{n = 0 }^\infty r^n=\frac{1}{1-r}\), for 0 < r < 1

Equation (1) can now be written as:

\(X\left( \omega \right) =\frac{1}{1-ae^{-j\omega}}\)    ---(2)

Application:

Given x[n] = 0.5n u[n], i.e. a = 0.5

The DTFT of the above given sequence using Equation (2) will be:

\(X\left( \omega \right) =\frac{1}{1-0.5e^{-j\omega}}\)

Given, \(x\left[ n \right]\mathop \leftrightarrow \limits^{DFS\;12} X\left( k \right)\)

\(\Rightarrow {\rm{\Omega }} = \frac{{2\pi }}{{12}} = \frac{\pi }{6}\)

And Y(k) = X(k + 4) + X(k - 4)

\(\begin{array}{l} \Rightarrow y\left[ n \right] = {e^{ - \frac{{j4\pi }}{6}n}}x\left[ n \right] + {e^{\frac{{j4\pi }}{6}n}}x\left[ n \right]\\ = 2\cos \left( {\frac{{2\pi }}{3}n} \right)x\left[ n \right] \end{array}\)

Thus, we conclude that the output of the system is zero everywhere except for the time interval.

1 < t < 5

y(t) = x₁(t) * x₂(t)

x₁(t) = (2 cos 4t) u(t)

x₂(t) = (sin 2t) u(t)

We know convolution in time domain is multiplication in frequency domain

So, y(t) = x₁(t) * x₂(t)

y(t) = x₁(s) ⋅ x₂(s)

\({x_1}\left( s \right) = \frac{{2s}}{{{s^2} + 16}},\;{x_2}\left( s \right) = \frac{2}{{{s^2} + 4}}\) 

\(Y\left( s \right) = \frac{{4s}}{{\left( {{s^2} + 16} \right)\left( {{s^2} + 4} \right)}} = \frac{{As + B}}{{{s^2} + 16}} + \frac{{Cs + D}}{{{s^2} + 4}}\) 

4s = (As + B) (s² + 4) (Cs + D) (s² + 16)

Put s = 0 ⇒ 0 = 4B + 16D → 1)

Put s = 1 ⇒ 4 = (A + B) (5) + (C + D) (17) → 2)

Put s = 2 ⇒ 8 = (2A + B) (8) + (2C + D) (20) → 3)

Put s = 3 ⇒ 12 = (3A + B) (13) + (3C + D) (25) → 4)

Solving 1), 2), 3) and 4) we get

A = B = 1 / 3

C = D = -1 / 3

\(Y\left( s \right) = \frac{1}{3}\left( {\frac{s}{{{s^2} + 4}} - \frac{s}{{{s^2} + 16}}} \right)\) 

Applying inverse Laplace transform,

\(y\left( t \right) = \frac{1}{3}\left( {\cos 2t-\cos 4t} \right)u\left( t \right)\) 

\(\frac{{\left( {y\left( t \right)} \right)}}{t} = \frac{\pi }{2} = \frac{1}{3}\left( {\cos \pi - \cos 2\pi } \right)u\left( {\frac{\pi }{2}} \right)\) 

\(= \frac{1}{3}\left( { - 1 - 1} \right) = - \frac{2}{3}\) 

= -0.666

Given signal is \(x\left( t \right) = 10\;sinc\;\left( {\frac{{t + 4}}{7}} \right)\)

Area \( = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)dt = \mathop \smallint \limits_{ - \infty }^\infty 10\;sinc\;\left( {\frac{{t + 4}}{7}} \right)dt\)

\(= \mathop \smallint \limits_{ - \infty }^\infty 10\frac{{\sin \left( {\frac{{\pi \left( {t + 4} \right)}}{7}} \right)}}{{\frac{{\pi \left( {t + 4} \right)}}{7}}}dt\)

To solve this, we can use the relation

\(X\left( 0 \right) = {\left[ {\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - j2\pi ft}}dt} \right]_{f \to 0}} = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)dt\)

The Fourier transform of the given signal x(t) is,

\(X\left( f \right) = 70\;rect\left( {7f} \right){e^{j8\pi f}}\)

The area = X(0) = 70

The problem involves a signal f(t) = cos8πt + 0.5cos4πt. To find the maximum allowable sampling interval, we utilise the Nyquist-Shannon sampling theorem, which states that the sampling rate must be at least twice the highest frequency present in the signal.

• The signal consists of two cosine terms: cos8πt and 0.5cos4πt.
• The frequencies of these terms are 4 Hz and 2 Hz respectively.
• The highest frequency component is 4 Hz.
• According to the sampling theorem, the minimum sampling rate is 2 × 4 Hz = 8 Hz.
• The maximum allowable sampling interval T_s is the reciprocal of the sampling rate, thus 1/8 sec.

y(n) = x(n) * x(n)

By taking z-transform both sides

Y(z) = X(z) ⋅ X(z)

x(n) = (0.7)ⁿ u(n)

\( \Rightarrow X\left( z \right) = \left[ {\frac{1}{{1 - 0.7\;{z^{ - 1}}}}} \right]\)

\(Y\left( z \right) = {\left( {\frac{1}{{1 - 0.7{z^{ - 1}}}}} \right)^2}\)

We know that; \(Y\left( z \right) = \mathop \sum \nolimits_{ - \infty }^\infty y\left( n \right){z^{ - n}}\)

\(z = 1\;;Y\left( 1 \right) = \mathop \sum \nolimits_{ - \infty }^\infty y\left( n \right)\)

\(\mathop \sum \nolimits_{ - \infty }^\infty y\left( n \right) = Y\left( z \right){\left. \right|_{z = 1}} = {\left( {\frac{1}{{1 - 0.7}}} \right)^2} = \frac{{100}}{9} = 11.11\)

Frequency response of \(H\left( {j\omega } \right) = \frac{{j\omega + 2}}{{\left( {j\omega + 1} \right)\left( {j\omega + 3} \right)}}\)

Input to the system is x(t) = e^-t u(t)

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

The output in frequency domain is given by

\(Y\left( {j\omega } \right) = H\left( {j\omega } \right)X\left( {j\omega } \right) = \frac{{j\omega + 2}}{{\left( {j\omega + 1} \right)\left( {j\omega + 3} \right)}} \times \frac{1}{{j\omega + 1}}\)

\( = \frac{{j\omega + 2}}{{{{\left( {j\omega + 1} \right)}^2}\left( {j\omega + 3} \right)}}\)

By applying partial fraction method,

\(Y\left( {j\omega } \right) = \frac{1}{4}\frac{1}{{\left( {j\omega + 1} \right)}} + \frac{1}{2}\frac{1}{{{{\left( {j\omega + 1} \right)}^2}}} + \frac{1}{4}\frac{1}{{\left( {j\omega + 3} \right)}}\)

By applying the inverse Fourier transform,

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

At t = 1, y(t) = 0.263

Concept:

The DTFT of a discrete sequence is a complex quantity, which can be defined as:

X(e^jω) = Real {X(e^jω)} + Img. {X(e^jω)}

Where the Real {X(e^jω)} is nothing but the DTFT of the even part of the signal.

Proof:

A signal can be written as the sum of the even part and odd part, is:

\({x_e}\left( n \right) = \frac{{x\left( n \right) + x\left( { - n} \right)}}{2}\)

\({x_0}\left( n \right) = \frac{{x\left( n \right) - x\left( { - n} \right)}}{2}\)

\(x\left( -n \right)\mathop \leftrightarrow \limits^{DTFT} X\left( {{e^{ - jω }}} \right)\)

Taking the Fourier transform of the even part, we get:

\({X_e}\left( {{e^{jω }}} \right) = \frac{{X\left( {{e^{jω }}} \right) + X\left( {{e^{ - jω }}} \right)}}{2}\)

This can be written as:

\({X_e}\left( {{e^{jω }}} \right) = \frac{{R\dot e\left\{ {X\left( {{e^{jω }}} \right)} \right\} + jIm\left\{ {X\left( {{e^{jω }}} \right)} \right\} + Re\left\{ {X\left( {{e^{jω }}} \right)} \right\} - jIm\left\{ {X\left( {{e^{jω }}} \right)} \right\}}}{2}\)  

X_e(e^jω) = Re{X(e^jω)}

Application:

We are required to find F^-1{Re{X(e^jω)}} at n = 0

The inverse DTFT of Re{X(e^jω)} is x_e(n)

Given:

\(x\left( n \right) = \left\{ { - 1,0,1,\begin{array}{*{20}{c}} 2\\ \uparrow \end{array},1,0,1,2,1,0, - 1} \right\}\)

\(x\left( { - n} \right) = \left\{ { - 1,0,1,2,1,0,1,\begin{array}{*{20}{c}} 2\\ \uparrow \end{array},1,0, - 1} \right\}\)

x_e(n) will be:

\(\frac{{x\left( n \right) + x\left( { - n} \right)}}{2} = \left\{ {\frac{{ - 1}}{2},0,\frac{1}{2},1,0,0,1,\begin{array}{*{20}{c}} 2\\ \uparrow \end{array},1,0,0,1,\frac{1}{2},0,\frac{{ - 1}}{2}} \right\}\)

∴ At n = 0, we have:

\({F^{ - 1}}\left\{ {Re\left\{ {X\left( {{e^{jw}}} \right)} \right\}} \right\} = {x_e}\left( 0 \right) = 2\)