Time Response of the First and Second Order System MCQ Quiz - Objective Question with Answer for Time Response of the First and Second Order System - 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:

Peak time:

It is the time required for the response to reach the peak time for the first time at t = tp, the first derivative of the response is zero.

\({t_p} = \frac{{nπ }}{{ω d}}\)

n = 1, 3, 5 …. For overshoot

2, 4, 6 …. For undershoot

Let n = 1

\({t_p} = \frac{π }{{ω d}}\)

ωd = damping frequency (rad/sec)

\({ω _d} = {ω _n}\;\sqrt {1 - {ξ ^2}}\)

ωn = natural frequency.

ξ = damping ratio

Calculation:

\({t_p} = \frac{{nπ }}{{ω d}}\)

Here the second peak is also the second overshoot

For second overshoot n = 3,

\({t_p} = \frac{{3π }}{{ω d}}\)

\(T(s)=\frac{25}{s^2+8s+25}\)

Compare it with standard transfer function we will get;
ωn² = 2⇒ ωn = 5

8 =2ξωn ⇒ ξ = 0.8

\(ω_d=ω_n\sqrt{1-ξ^2}\)  ⇒ ω_d = 3 rad/sec

The second peak in the response = \(\frac{{3π }}{{ω d}} = \frac{{3π }}{{3}} = π\) sec

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 

The correct answer is 404 sec

Concept:

The time at which peak is achieved is called peak time

\(Tp = \frac{n\pi}{\omega d}\)

For n = 1 we will get 1^st overshoot.

For n = 3 we will get 2^nd overshoot

∴ for Xth overshoot  n = 2x-1

Solution:

For n = 1 

\(Tp = \frac{\pi}{\omega d}\)

= 4 sec

for 51st overshoot 

\(Tp = \frac{(2x-1)\pi}{\omega d}\) put x = 51

=\(\frac{101 \pi}{\omega d}\)

=101× 4 

∴ Tp = 404

The correct answer is \(\frac{10s}{s^2 +2s + 200}\)

Solution:

From the graph, we can say it is the output of a 1st order system

let's say the transfer function of an LTI system

 \(G(s) = \frac{K}{(s + a)}\)

For the unit step input-output

 \(Y(s) = \frac{K}{s (s + a)}\)

It is given that a steady-state output value is

\(\lim_{S \rightarrow 0} \frac{S.K}{S(S + a)} = 5\)

∴ \(\frac{K}{a}= 5\)

Now the value of the output

 \(Y(s) = \frac{K}{s (s + a)} = \frac{2a}{s (s + a)}\)

Apply partial fraction

 \(∴\) \(\frac{5a}{s (s + a)} = \frac{A}{s} + \frac{B}{(s + a)}\)

after solving above partial fraction we get A = 5 and B = -5.

\(∴\)  \(Y(s) = \frac{5}{s} - \frac{5}{(s + a)}\)

Now take inverse Laplace you will get

y(t) = 5 ( 1 - e^-at) u(t)

from the graph, at t = 1.15-sec output is 4.5 so put on this value in above equation.

\(∴\) y(t) = 5 ( 1 - e^-a(1.15)) = 4.5

\(∴\) a = 2

So the overall transfer function of an LTI system is

= \(\frac{10}{(s + 2)}\)

Now we have to add feedback to it, which will make the block diagram as below.

The final transfer function of the above system is

T/F = \(\frac{\frac{10}{(s + 2)}}{1 + \frac{10}{(s + 2)}.\frac{20}{s}}\) 

= \(\frac{10 s}{s^2 +2s + 200}\)

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:-

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

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

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

\(G\left( s \right)H\left( s \right) = \frac{k}{{{{\left( {s + 1} \right)}^2}\left( {s + 2} \right)}}\)

From the given figure

steady state value of y(t) = 0.8

By using final value theorem,

steady state y(t) value of \(= \mathop {{\rm{lt}}}\limits_{s \to 0} s\;y\left( s \right)\)

\(\frac{T}{F} = \frac{{y\left( s \right)}}{{R\left( s \right)}} = \frac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\)

\(\frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{k}{{\frac{{{{\left( {s + 1} \right)}^2}\left( {s + 2} \right)}}{{1 + \frac{k}{{{{\left( {s + 1} \right)}^2}\left( {s + 2} \right)}}}}}}\)

\(\Rightarrow \frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{k}{{{{\left( {s + 1} \right)}^2}\left( {s + 2} \right) + k}}\)

As the input is unit step function,

R(s) = 1/s

\(\Rightarrow Y\left( s \right) = \frac{k}{{s\left( {{{\left( {s + 1} \right)}^2}\left( {s + 2} \right) + k} \right)}}\)

\(\mathop {{\rm{lt}}}\limits_{s \to 0} s\;Y\left( s \right) = 0.8\)

\(\Rightarrow \frac{k}{{\left( 1 \right)\left( 2 \right) + k}} = 0.8\)

⇒ k = 1.6 + 0.8k ⇒ k = 8

Concept:

Standard second-order closed-loop transfer function is given by:

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{ω _n^2}}{{{s^2} + 2\xi {ω _n}s + ω _n^2}}\) .... (1)

\(\frac{C(s)}{R(s)}=\frac{{G}}{{{s^2} + 2\xi {ω _n}s + ω _n^2}}\) ... (2)

ξ = damping ratio

ωn = undamped natural frequency

G = Gain

From equation (1) and (2),

G = ω_n²

∴ \(\omega _n=\sqrt{G}\) .... (3)

Now we have to find the pole of the system, and it will be located at,

\(s=-\zeta \omega_n\pm\omega_n \sqrt{1-\zeta^2}\)

The decaying exponential has a time constant equal to,

\(1=\frac{1}{\zeta \omega_n}\)

∴ \(\zeta =\frac{1}{\omega_n}\) .... (4)

From equation (3) and (4),

\(\large{\zeta=\frac{1}{\sqrt{G}}}\)

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 

\(G\left( s \right) = \frac{{ - s + 1}}{{s + 1}}\)

r(t) = u(t)

\(\Rightarrow r\left( s \right) = \frac{1}{s}\)

Output, C(s) = R(s) G(s)

\(= \frac{1}{s}\left( {\frac{{1 - s}}{{1 + s}}} \right)\)

\(\Rightarrow c\left( s \right) = \frac{1}{s} - \frac{2}{{1 + s}}\)

By applying inverse Laplace transform,

⇒ c(t) = u(t) – 2e^-t u(t)

At t = 1.5 sec,

The standard equation for a step input is,

c(t) = 1 – e^-t/τ

Where τ is time constant.

At c(t) = 0.9 and t = 30s

⇒ 0.9 = 1 – e^-30/τ

Effect of addition of zero:

• It attracts root locus branches away from the jω-axis, due to which the system becomes more stable.
• Relative stability improves.
• The system becomes less oscillatory.
• As the rise time is inversely proportional to speed, the rise time will decrease.
• Also, because the bandwidth is inversely proportional to the rise time, so the bandwidth increases.

Effect of addition of poles:

• It attracts root locus branches towards the jω-axis due to which the system becomes less stable.
• The relative stability reduces.
• The system becomes more oscillatory.
• As the rise time is inversely proportional to speed, so the rise time will increase.
• Also, because the bandwidth is inversely proportional to the rise time, so the bandwidth decreases.

Concept:

Y(s) = X(s) . G(s)

X(s): Applied input.

G(s): transfer function

Y(s): Output response

A unit step input is defined as u(t). Its Laplace transform is given by:

\(x\left( t \right) \leftrightarrow X\left( s \right) = \frac{1}{s}\)

Calculation:

The transfer function is given as:

\(G\left( s \right) = \frac{1}{{1 + \tau s}}\)

\(X\left( s \right) = \frac{1}{s}\)

The unit step response is the response when the input is a unit step function, i.e.
\(Y\left( s \right) = \frac{1}{s} \cdot \frac{1}{{\left( {1 + \tau s} \right)}}\)

Using Partial fraction, we get:

\(\frac{1}{s} \cdot \frac{1}{{1 + \tau s}} = \frac{A}{s} + \frac{B}{{1 + \tau s}}\)

On solving it, we’ll get:

\(Y\left( s \right) = \frac{1}{s} - \frac{\tau }{{1 + \tau s}}\)

\(Y\left( s \right) = \frac{1}{s} - \frac{1}{{s + \frac{1}{\tau }}}\)

Taking the Inverse Laplace Transform, we get:

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

We have, \({\rm{G}}\left( {\rm{s}} \right) = \frac{{{\rm{s}} - 2}}{{\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}}\)

Now, \(\rm Y(s) = G(s). X(s)\)

We have, \(\rm X(s) = 1/s\)

\(\Rightarrow {\rm{Y}}\left( {\rm{s}} \right) = \frac{{\left( {{\rm{s}} - 2} \right)}}{{{\rm{s}}\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}}\)

Using properties of Laplace transform we have

\(\frac{{{\rm{dy}}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}\mathop \leftrightarrow \limits^{{\rm{LT}}} {\rm{s}}.{\rm{Y}}\left( {\rm{s}} \right)\)

Thus, \(\frac{{{\rm{dy}}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}\mathop \leftrightarrow \limits^{{\rm{LT}}} \frac{{{\rm{s}} - 2}}{{\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}}\)

Taking inverse Laplace of \(\frac{{{\rm{s}} - 2}}{{\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}}\) assuming right sided ROC, as we need to calculate \(\frac{{{\rm{dy}}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}\) at \(\rm t = 0^+\), we have

\(\frac{{\left( {{\rm{s}} - 2} \right)}}{{\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}} = \frac{{ - \frac{3}{2}}}{{\left( {{\rm{s}} + 1} \right)}} + \frac{{\frac{5}{2}}}{{\left( {{\rm{s}} + 3} \right)}}\)

Thus, \(\frac{{{\rm{dy}}\left( {\rm{t}} \right)}}{{{\rm{dt}}}} = - \frac{3}{2}{{\rm{e}}^{ - {\rm{t}}}}{\rm{u}}\left( {\rm{t}} \right) + \frac{5}{2}{{\rm{e}}^{ - 3{\rm{t}}}}{\rm{u}}\left( {\rm{t}} \right)\)

\(\Rightarrow {\left. {\frac{{{\rm{dy}}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}} \right|_{{\rm{t}} = {0^ + }}} = - \frac{3}{2} + \frac{5}{2} = \frac{2}{2} = 1\)

\(T\left( s \right) = \frac{1}{{1 + s\tau }}\)

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{1}{{1 + s\tau }}\)

\(R\left( s \right) = \frac{1}{s}\)

\(\Rightarrow C\left( s \right) = \frac{1}{{s\left( {1 + s\tau } \right)}} = \frac{1}{s} - \frac{\tau }{{1 + s\tau }}\)

\( \Rightarrow C\left( s \right) = \frac{1}{s} - \frac{1}{{s + \frac{1}{\tau }}}\)

By applying inverse Laplace transform,

C(t) = 1 – e^-t/τ

at t = 10s,

C(t) = 0.5

⇒ e^-10/τ = 0.5