Impulse Response MCQ Quiz - Objective Question with Answer for Impulse Response - Download Free PDF

Explanation:

Cascade Form Realisation of FIR System

Definition: A Finite Impulse Response (FIR) filter is a type of digital filter that responds to a finite number of input samples before settling to zero. In the cascade form realization of an FIR system, the overall transfer function is implemented as a series of second-order sections (biquads) and possibly one first-order section if the order of the filter is odd. This method is preferred for its numerical stability and ease of implementation.

Working Principle: In cascade form realization, the FIR filter of order \( M-1 \) is decomposed into a series of smaller sections, each of which can be implemented with fewer computational resources. Each section is either a first-order or a second-order filter. These sections are then connected in series (cascaded) to achieve the desired overall filter response.

Advantages:

• Improved numerical stability compared to direct form implementations.
• Easier to design and implement using second-order sections.
• Modular structure allows for flexible design and implementation.

Disadvantages:

• Potentially higher computational complexity compared to direct form for certain applications.
• Requires careful scaling to avoid overflow in fixed-point implementations.

Correct Option Analysis:

The correct option is:

Option 2: \((M - 1)\) adders and \( M \) multipliers are required for \((M - 1)\)th order FIR transfer function.

To understand why this is the correct option, let's analyze the requirements for implementing an \((M - 1)\)th order FIR filter. An FIR filter of order \((M - 1)\) has \( M \) coefficients, denoted as \( \{b_0, b_1, \ldots, b_{M-1}\} \). The general form of the FIR filter equation is:

\( y[n] = b_0 x[n] + b_1 x[n-1] + \ldots + b_{M-1} x[n-(M-1)] \)

In cascade form realization, each section typically involves two main operations:

• Multiplication of the input or intermediate signal by the filter coefficients.
• Addition of the resulting products to form the output or intermediate signal.

For an \((M - 1)\)th order FIR filter, we need \( M \) multipliers because each coefficient \( b_i \) is used to multiply the corresponding input or intermediate signal. Additionally, we need \((M - 1)\) adders because we have to sum \( M \) products to get the final output. This is why option 2 is correct.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: \((M + 1)\) adders and \((M - 1)\) multipliers.

This option is incorrect because it overestimates the number of adders and underestimates the number of multipliers. For an \((M - 1)\)th order FIR filter, we only need \((M - 1)\) adders, not \((M + 1)\). The number of multipliers should be \( M \), not \((M - 1)\).

Option 3: \((M - 1)\) adders and \((M + 1)\) multipliers.

This option is incorrect because it overestimates the number of multipliers needed. For an \((M - 1)\)th order FIR filter, we need exactly \( M \) multipliers, not \((M + 1)\). The number of adders \((M - 1)\) is correct, though.

Option 4: \( M \) adders and \((M - 1)\) multipliers.

This option is incorrect because it overestimates the number of adders needed. For an \((M - 1)\)th order FIR filter, we only need \((M - 1)\) adders. The number of multipliers should be \( M \), not \((M - 1)\).

Conclusion:

Understanding the cascade form realization of FIR filters is essential for correctly identifying the computational requirements for their implementation. An FIR filter of order \((M - 1)\) requires \((M - 1)\) adders and \( M \) multipliers. This modular and stable approach makes it suitable for various applications, despite its potential complexity compared to direct form implementations.

Concept:

The Laplace transform of a general exponential signal is given 

\(L[e^{-at}]\longleftrightarrow \frac{1}{s+a} \)

where 'a' is any positive integer.

Calculation:

Given, f (t) = e^-3t 

The Laplace transform is given as

F(s) = \(\frac{1}{s+3} \)

The impulse response of e^-3t is 

Concept :- 

A causal system is the one in which the output y(n) at time n depends only on the current input x(n) at time n, and its past input sample values such as x(n − 1), x(n − 2),…. Otherwise, if a system output depends on the future input values such as x(n + 1), x(n + 2),…, the system is non causal.

According to DTFT property Fourier transform of Sinc function in discrete time domain is rectangular pulse in frequency domain.

\(\rm x[n]=\frac{\sin (\hat {\omega}_bn)}{\pi n}\) \(\overset{DTFT}{\large\Leftrightarrow}\)

\(\rm X(e^{j\hat{\omega}})=\left\{\begin{matrix}1,&|\hat{\omega}|\le\hat{\omega}_b\\\ 0,&\hat{\omega}_b<|\hat{\omega}|\le \pi\end{matrix}\right.\)

Analysis:-

given function depends on the future input value such x[n+1] so that function is non causal function.

\(\rm {h[n] = \left\{ \begin{matrix} \frac{\omega_c}{\pi} & n = 0 \\\ \frac{\sin \omega_c n}{\pi n} & n \ne 0 \end{matrix} \right.}\)

this is Sinc function and for it , rectangular pulse generate in frequency domain which provide gain in only low frequency region so that, it is behave like low pass filter.

x (t) = u (t - 1)

\(\therefore X\left( s \right) = \frac{{{e^{ - s}}}}{s}\)

Impulse response, h (t) = t u (t)

\(\therefore H\left( s \right) = \frac{1}{{{s^2}}}\)

Y (s) = H (s) X (s)

\(= \frac{{{e^{ - s}}}}{{{s^3}}}\)

Taking inverse Laplace,

\(y\left( t \right) = \frac{{{{\left( {t - 1} \right)}^2}}}{2}u\left( {t - 1} \right)\)

Concept:

y[n] = x[n] * h[n]

When the two systems are cascaded, then the resultant response is the convolution of the individual responses.

Convolution in one domain is a multiplication in another domain.

An LTI system satisfies the following property:

x[n] * δ[n - n0] = x[n - n0]

Analysis:

The output of the cascaded system:

The impulse response of the cascaded system

hc (n) = h1(n) * h2(n)

hc (n) = δ (n - 1) * δ (n - 2)

= δ (n - 3)

Concept:

A transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, the transfer function is also known as the impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Analyses:-

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

\(h\left( t \right) = \frac{{ds\left( t \right)}}{{dt}}\)

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

Concept:

The Laplace transform of a general exponential signal is given 

\(L[e^{-at}]\longleftrightarrow \frac{1}{s+a} \)

where 'a' is any positive integer.

Calculation:

Given, f (t) = e^-3t 

The Laplace transform is given as

F(s) = \(\frac{1}{s+3} \)

The impulse response of e^-3t is 

Concept:

y[n] = x[n] * h[n]

When the two systems are cascaded, then the resultant response is the convolution of the individual responses.

Convolution in one domain is a multiplication in another domain.

An LTI system satisfies the following property:

x[n] * δ[n - n0] = x[n - n0]

Analysis:

The output of the cascaded system:

The impulse response of the cascaded system

hc (n) = h1(n) * h2(n)

hc (n) = δ (n - 1) * δ (n - 2)

= δ (n - 3)

Concept:

The transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

The closed-loop transfer function for the second-order system is:

\(\frac{{Y\left( s \right)}}{{X\left( s \right)}} = \frac{{ω _n^2}}{{{s^2} + 2ζ {ω _n}s + ω _n^2}}\)

Auxilliary Equation is given as:

s² + 2ξω_ns + ωn² = 0    ---(1)

ωn is the natural frequency of oscillations in rad/sec 

ξ  is the damping ratio

Calculation:

Impulse response h(t) as:

\(h(t)=\frac{1}{6}{e^{ - 0.8t}}\sin 0.6t\)

We know that:

\(\;{e^{ - bt}}sin\left( {at} \right) \leftarrow \to \frac{a}{{{{\left( {s + b} \right)}^2} + {a^2}}}\; \)

\(\frac{1}{6}{e^{ - 0.8t}}\sin 0.6t \leftarrow \to \frac{1}{6} \frac{0.6}{{{{\left( {s + 0.8} \right)}^2} + {0.6^2}}}\; \)

A.E. is given as:

s² + 1.6s + 1 = 0

Comparing this with equation (1), we get:

ω_n² = 1

ω_n = 1

2ξω_n = 1.6

ξ = 0.8

Hence, option 1 is correct.

Concept:

complex calculations in the time domain can be easily solved in the frequency domain and vice versa.

NOTE: Multiplication in one domain corresponds to convolution in the other domain.

These conclusions are very important.

\(x\left( t \right)*h\left( t \right) = y\left( t \right)\)

\(X\left( s \right) \times H\left( s \right) = Y\left( s \right)\)

Analysis:

Consider a complex exponential input

 \(x\left( t \right) = {e^{st}}\) 

to an LTI (linear time-invariant ) system with impulse response h(t).

The system output equation is given by

\(y\left( t \right) = h\left( t \right)*x\left( t \right)\)

\( = \mathop \smallint \limits_{ - \infty }^\infty h\left( \tau \right)x\left( {t - \tau } \right)d\tau \)                               

 \( = \mathop \smallint \limits_{ - \infty }^\infty h\left( \tau \right){e^{s\left( {t - \tau } \right)}}d\tau \)

 \( = {e^{st}}\mathop \smallint \limits_{ - \infty }^\infty h\left( \tau \right){e^{ - s\tau }}d\tau \)

\(y\left( t \right) = H\left( s \right){e^{st}}\) where

\(H\left( s \right) = \mathop \smallint \limits_{ - \infty }^\infty {e^{s\tau }}d\tau \)

or equivalently

\(H\left( s \right) = \mathop \smallint \limits_{ - \infty }^\infty h\left( t \right){e^{ - st}}dt\) ⋯ (i)

Calculation:

With the help of properties, we can solve all the transforms easily rather than solving the integral formula which nobody like.

Shifting in one domain corresponds to the multiplication with exponential in another domain

For example x(t) having the Laplace transform as X(s)

 \(x\left( t \right) \leftrightarrow X\left( s \right)\)

 \(x\left( {t - {t_0}} \right) \leftrightarrow {e^{ - s{t_0}}}X\left( s \right)\) ⋯ (i)

\({t^n}u\left( t \right) \leftrightarrow \frac{{n!}}{{{s^{n + 1}}}}\) ⋯ (ii)

\({t^n}{e^{ - at\;}}u\left( t \right) \leftrightarrow \frac{{n!}}{{{{\left( {s + a} \right)}^{n + 1}}}}\) ⋯ (iii)

The given h(t) = tu(t) and input x(t) = u(t-1)

Comparing h(t) with (ii) we get n = 1

From the above property (iii) Laplace transform of the h(t) will be

 \(H\left( s \right) = \frac{{1!}}{{{s^2}}} = \frac{1}{{{s^2}}}\)

Laplace transform of the input x(t), this is shifting in the time domain so in the frequency domain,

it will be multiplied by exponential according to property (i).

comparing with that t₀ is 1 so

 \(X\left( s \right) = \frac{{{e^{ - s}}}}{s}\)

Now multiplying X(s) and H(s) we get Y(s).

\(Y\left( s \right) = \frac{{{e^{ - s}}}}{s} \times \frac{1}{{{s^2}}} \)

\(Y\left( s \right) = \frac{{{e^{ - s}}}}{{{s^3}}}\)

Comparing Y(s) with property (iii) and (ii) t₀ = 1 and n = 2 so the output function in the time domain will be obtained as

 \(y\left( t \right) = \frac{{{{\left( {t - 1} \right)}^2}}}{2}u\left( {t - 1} \right)\)

Important properties:

These are valid for all transformations

• A shift in one domain corresponds to multiplication with exponential in another domain.
• Differentiation in one domain corresponds to multiplication with the variable in that domain
• Integration in one domain corresponds to the division in another domain with that variable
• Convolution in one domain corresponds to multiplication in another domain

x (t) = u (t - 1)

\(\therefore X\left( s \right) = \frac{{{e^{ - s}}}}{s}\)

Impulse response, h (t) = t u (t)

\(\therefore H\left( s \right) = \frac{1}{{{s^2}}}\)

Y (s) = H (s) X (s)

\(= \frac{{{e^{ - s}}}}{{{s^3}}}\)

Taking inverse Laplace,

\(y\left( t \right) = \frac{{{{\left( {t - 1} \right)}^2}}}{2}u\left( {t - 1} \right)\)

Concept :- 

A causal system is the one in which the output y(n) at time n depends only on the current input x(n) at time n, and its past input sample values such as x(n − 1), x(n − 2),…. Otherwise, if a system output depends on the future input values such as x(n + 1), x(n + 2),…, the system is non causal.

According to DTFT property Fourier transform of Sinc function in discrete time domain is rectangular pulse in frequency domain.

\(\rm x[n]=\frac{\sin (\hat {\omega}_bn)}{\pi n}\) \(\overset{DTFT}{\large\Leftrightarrow}\)

\(\rm X(e^{j\hat{\omega}})=\left\{\begin{matrix}1,&|\hat{\omega}|\le\hat{\omega}_b\\\ 0,&\hat{\omega}_b<|\hat{\omega}|\le \pi\end{matrix}\right.\)

Analysis:-

given function depends on the future input value such x[n+1] so that function is non causal function.

\(\rm {h[n] = \left\{ \begin{matrix} \frac{\omega_c}{\pi} & n = 0 \\\ \frac{\sin \omega_c n}{\pi n} & n \ne 0 \end{matrix} \right.}\)

this is Sinc function and for it , rectangular pulse generate in frequency domain which provide gain in only low frequency region so that, it is behave like low pass filter.

Assertion Explanation: A BIBO (bounded-input bounded-output) stable system is a system for which the outputs will remain bounded for all time, for any finite initial condition and input. So this statement is incorrect. For a system to be linear it should follow superposition theorem.

\(BIBO - stable \Leftrightarrow \mathop \smallint \nolimits_0^\infty \left| {h\left( t \right)} \right|dt\;finite\) 

Where h(t) = impulse resp.

= L^-1(H(s))

Reason Explanation:- A continuous-time linear time-invariant system is BIBO stable if and only if all the poles of the system have real parts less than 0 or negative or lie in the left half of plane. If system is BIBO stable then its response will have finite value everywhere and response due to initial condition will decay to zero as time(t) tends to infinity.

E.g. \( \propto et\;H\left( s \right) = \frac{1}{{st3}}\)

Since it has a pole s = -3 which is negative so it is stable system

\(\frac{1}{{stq}} \leftrightarrow {e^{ - atu\left( t \right)}}\) 

H(t) = e-^3t

As t → ∞ h(t) = 0

Concept:

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

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

Analysis:

\(h(n)=\frac{1}{2}(δ[n]+δ[n-2])\)

H(e^jω) = 0.5 + 0.5 e^-j2ω 

= e^-jω [0.5 e^jω + 0.5 e^-jω]

= e^-jω cosω 

The magnitude of the frequency response is: |cos ω|

Concept:

A transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, the transfer function is also known as the impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Analyses:-

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

\(h\left( t \right) = \frac{{ds\left( t \right)}}{{dt}}\)

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