Signals and Systems Test 1

Concept:

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

Also, according to the sampling property, we have:

\(\mathop \smallint \nolimits_{ - \infty }^\infty x\left( t \right)\delta \left( {t - {t_0}} \right)dt = x\left( {{t_0}} \right)\) 

Analysis:

Given:

\(x\left( t \right) = \mathop \smallint \nolimits_{ - \infty }^\infty {e^{ - t}}\delta \left( {2t - 2} \right)dt\) 

This can also be written as:

\(x\left( t \right) = \mathop \smallint \nolimits_{ - \infty }^\infty {e^{ - t}}.\frac{1}{2}\delta \left( {t - 1} \right)dt\) 

\(x\left( t \right) = \frac{1}{2}\mathop \smallint \nolimits_{ - \infty }^\infty {e^{ - t}}.\delta \left( {t - 1} \right)dt\) 

Using the sampling property, we can write:

\(x\left( t \right) = \frac{1}{2}{e^{ - t}}{\left. \right|_{at\;t = 1}}\) 

\(x\left( t \right) = \frac{1}{2}{e^{ - 1}}\) 

\(x\left( t \right) = \frac{1}{2} \times 0.367 = 0.184\) 

Concept:

Discrete signal is periodic if

\(\frac{N}{K}=\frac{2\pi }{{{\omega }_{0}}}\) is integer

Application

^{\(\cos 15~t\underset{Sampling}{\mathop{\to }}\,\cos \left( 15\dot{n}{{T}_{s}} \right)\)}

Period:

\(\frac{N}{K}=\frac{2\pi }{15\left( {{T}_{s}} \right)}\)

\({{T}_{s}}=\frac{2\pi }{15}\times \frac{K}{N}\)

\({{T}_{s}}=\frac{2\pi }{15}\times \frac{K}{4}\)

\({{T}_{s}}=\frac{\pi }{30}K\)

Concept:

The energy of a signal x(t) is calculated as:

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

Analysis:

The given raised cosine pulse is defined as:

\(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{2}(\cos \omega t + 1),}\\{0,}\end{array}\begin{array}{*{20}{c}}{ - \frac{\pi }{\omega } \le t \le \frac{\pi }{\omega }}\\{otherwise}\end{array}} \right.\) 

For the defined range, the energy will be:

\(E = \mathop \smallint \limits_{ - \frac{\pi }{\omega }}^{\frac{\pi }{\omega }} \frac{1}{4}{\left( {\cos \omega t + 1} \right)^2}dt\)

Since the function inside the integration is an even function, we can write:

\(E = \frac{2}{4}\mathop \smallint \nolimits_0^{\frac{\pi }{\omega }} {\left( {\cos \omega t + 1} \right)^2}dt\) 

\( = \frac{1}{2}\mathop \smallint \nolimits_0^{\frac{\pi }{\omega }} \left( {{{\cos }^2}\omega t + 1 + 2\cos \omega t} \right)dt\) 

\(= \frac{1}{2}\mathop \smallint \nolimits_0^{\frac{\pi }{\omega }} \left( {\frac{1}{2}\cos 2\omega t + \frac{1}{2} + 2\cos \omega t + 1} \right)dt\) 

\( = \frac{1}{2}\left[ {\left. {\frac{1}{2}\frac{{\sin 2\omega t}}{{2\omega }}} \right|_0^{\frac{\pi }{\omega }} + \left. {\frac{3}{2}t} \right|_0^{\frac{\pi }{\omega }} + \left. {\frac{{2\sin \omega t}}{\omega }} \right|_0^{\frac{\pi }{\omega }}} \right]\) 

\(= \frac{1}{2} \times \frac{3}{2}\left( {\frac{\pi }{\omega } - 0} \right)\) 

\(= \frac{{3\pi }}{{4\omega }}\) 

From the given problem, we have:

\({E_x} = \frac{{k\pi }}{\omega }\) 

Comparing the two, we get:

\(k = \frac{3}{4} = 0.75\) 

All bounded periodic signals are power signals, because they do not converge to a finite value so their energy is infinite and their power is finite.

Time reversal property of Z-transform

X[n] ↔ X(z)

X[-n] ↔ X(z^-1)

Given that x[n] = x[-n]

⇒ X(z) = X(z^-1)

Zero of X(z) = 1 + j2

Then another zero will be:

\(\frac{1}{1+j2}=\frac{1-j2}{5}\)

\(=0.2-0.4j\)

Use the properties:

• An LTI system is stable if and only if the ROC includes |z| = 1 (unit circle)

• A causal LTI system with rational system function H(z) is stable is and only if all the poles of H(z) lie inside the unit circle.
• On changing z by \(\frac{1}{z}\), the poles of H(z) will lie outside the unit circle.
• But ROC of X(z^-1) is opposite to ROC of X(Z) i.e. inside the innermost pole.

For the given IIR filter that is both causal and stable, the poles will lie inside the unit circle and the ROC will include the |z| = 1 circle.

By replacing z with 1/z, the poles will shift to the outside of the unit circle and the ROC will now be the inside portion which will still include the |z| = 1 circle.
Thus changing z by \(\frac{1}{z},~H\left( z \right)\) becomes non-causal but stable.

Note that y[n] =  x[n] + x[n + N/2 ] has a period of N/2 and N has been assumed to be even,

The rectangular window is the most common windowing technique to design a finite impulse response (FIR) filter.

For the given rectangular sequence, the Fourier transform will be a sinc sequence given by:

f(t) = cos²t + cos 2t

\(= \frac{{1 + \cos 2t}}{2} + \cos 2t\) 

\( = \frac{1}{2} + \frac{3}{2}\cos 2t\) 

The function has a DC term and a cosine function. The frequency of cosine terms is

ω = 2

⇒ 2πf = 2

\(\Rightarrow f = \frac{1}{\pi }Hz\) 

1. Write the Laplace-domain relation for an inductor
For an inductor of inductance L with zero initial current:
V(s) = L s I(s)

2. Express I(s) in terms of the given voltage transform
Given V(s) = (4 s² + 3 s + 2) / (7 s² + 6 s + 5) and L = 2 H,
I(s) = V(s) / (L s)
= (4 s² + 3 s + 2) / [2 s (7 s² + 6 s + 5)]

3. Apply the Final-Value Theorem to find the steady current
The final-value theorem states that i(∞) = limₜ→∞ i(t) = limₛ→0 [s I(s)], provided all poles of s I(s) lie in the left half-plane.
So
i(∞) = limₛ→0 [s · (4 s² + 3 s + 2) / (2 s (7 s² + 6 s + 5))]
= limₛ→0 [(4 s² + 3 s + 2) / (2 (7 s² + 6 s + 5))]
= 2 / 10
= 0.2 A

Choose the correct option: The steady-state current is 0.2 A (Option B).

Given difference equation is:

y(n) – ay(n - 1) = bx(n) + x(n - 1)

Taking the fourier transform on both the sides, we get:

Y(e^jω) – ae^-jω ⋅ y(e^jω) = bX(e^jω) + e^-jω X(e^jω)

Y(e^jω) (1 – ae^-jω) = X(e^jω) [b + e^-jω]

\(\frac{{Y\left( {{e^{j\omega }}} \right)}}{{X\left( {{e^{j\omega }}} \right)}} = H\left( {{e^{j\omega }}} \right)\)

\(H\left( {{e^{j\omega }}} \right) = \frac{{b + {e^{ - j\omega }}}}{{1 - a{e^{ - j\omega }}}}\)

For an all-pass system, the magnitude of the transfer function is always unity, i.e.

|H(e^jω)| = 1 for all ω.

|b + e^-jω| = |1 – ae^-jω|

|b + cosω  – j sin ω| = |1 – (a cos ω – aj sin ω)|

(b + cos ω)² + (-sin ω)² = |(1 – a cos ω)² + (-a sin ω)²|

b² + cos² ω + 2b cos ω + sin² ω  = 1 + a² cos² ω  – 2a cos ω  + a² sin² ω

b² + 1 + 2b cos ω  = 1 + a² – 2a cos ω

The above equality will hold true only when:

b = -a, i.e.

\(\frac{b}{a} = - 1\)

Concept:

The z- transform of an impulse function is:

\(\delta \left( n \right)\overset{z-transfrom}{\longleftrightarrow} 1,\)

with ROC the entire z – plane.

Also according to the time-shifting property of the z-transform,

If δ(n) ↔ 1

δ(n- z₀) ↔ z^-zo

Calculation:

Given,

\(X\left( z \right)=\mathop \sum \limits_{k = 5}^{10}\frac{1}{k}{{z}^{-k}},\left| \text{z} \right|>0\)

Expanding the above, we get;

\(\Rightarrow X\left( z \right)=\frac{1}{5}.{{z}^{-5}}+\frac{1}{6}{{z}^{-6}}+\frac{1}{7}{{z}^{-7}}+\frac{1}{8}{{z}^{-8}}+~---+\frac{1}{10}{{z}^{-10}}\)

Taking the Inverse- z transform of the above, we get;

\(x\left( n \right)=\frac{1}{5}\delta \left( n-5 \right)+\frac{1}{6}\delta \left( n-6 \right)+~----\frac{1}{10}\delta \left( n-10 \right)\)

The above can be written as;

\(x\left( n \right)=\mathop \sum \limits_{k = 5}^{10}\frac{1}{k}.\delta \left( n-k \right)\)

So, Option (4) is correct.

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:

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

H(z) = pz + q + pz^-1

Z = e^jωT

\(f = \frac{1}{4}\;Hz\) 

\(\omega = 2\pi \times \frac{1}{4} = \frac{\pi }{2}\) 

\(f = \frac{1}{8}\;Hz\)

\(\omega = 2\pi \times \frac{1}{8} = \frac{\pi }{4}\) 

\(H\left( {{e^{j\omega }}} \right) = p\;{e^{j\omega }} + q + p\;{e^{ - j\omega }}\) 

At \(\omega = \frac{\pi }{2}\)

\(\left| {H\left( {{e^{j\omega }}} \right)} \right| = 0\)

\(0 = p\;{e^{j\frac{\pi }{2}}} + q + p\;{e^{ - j\frac{\pi }{2}}}\)

\(0 = p\left( {0 + j} \right) + q + p\left( {0 - j} \right)\) 

0 = q

At \(\omega = \frac{\pi }{4}\)

\(H\left( {{e^{j\omega }}} \right) = p\;{e^{j\frac{\pi }{4}}} + q + p\;{e^{ - j\frac{\pi }{4}}}\) 

q = 0

\(= p\left( {\frac{1}{{\sqrt 2 }} - j\frac{1}{{\sqrt 2 }}} \right) + p\left( {\frac{1}{{\sqrt 2 }}\frac{{ - j}}{{\sqrt 2 }}} \right)\) 

\(\frac{{2p}}{{\sqrt 2 }} = H\left( {{e^{j\omega }}} \right)\) 

\(p = \frac{1}{{\sqrt 2 }}\)