Frequency Domain Specifications MCQ Quiz - Objective Question with Answer for Frequency Domain Specifications - Download Free PDF

Concept:

DC gain = 5

$M_r=\frac{10}{\sqrt{3}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}$

100 [4ε² (1 – ε²)] = 3

400 ε²– 400 ε⁴ = 3

400x² – 400 x + 3 = 0  

x² – x + 0.0075 = 0

ξ² = 0.99244 , ξ² = 0.00755

ξ = 0.99 , ξ = 0.08689

If resonant peak > 1 then

 $\zeta \le \frac{1}{\sqrt{2}}=0.707$

∴ ξ = 0.08689

$5\sqrt{2}={\omega }_n\sqrt{1-2{\xi }^2}$

$5\sqrt{2}={\omega }_n\sqrt{1-2\times 0.00755}$

ωn = 7 rad/s

$s^2+2\xi {\omega }_{n}s+{\omega }_n^2$

= s² + 1.21 s + 49

Now ${\frac{K}{s^2+1.21s+49}|}_{s=0}=5$ 

K = 49 × 5 = 245

So the required transfer function = $\frac{245}{\left(s^2+1.21s+49\right)}$

The bode magnitude plot is shown

At resonant frequency, (ωp), magnitude is maximum and hence slope is 0.

System 1: $G\left(s\right)=\frac{1}{2s+1}$

Time constant (τ₁) = 2 sec.

System 2: $G\left(s\right)=\frac{1}{5s+1}$

Time constant (τ₂) = 5 sec

Bandwidth is inversely proportional to time constant.

In signal processing and control theory the bandwidth is the frequency at which the closed-loop system gain drops 3 dB below peak.

From the given options, the system gain must be less than 10 dB.

First-order system:

The transfer function of the standard first-order system is given by

$TF=\frac{k}{\left(1+\tau s\right)}$

Where,

τ = time constant of the system

The time constant can be defined as the negative reciprocal of the pole of the system.

Bandwidth(BW):

It is the range of frequencies over which the system acts satisfactorily.

For the first-order system, bandwidth is given by

BW = 1/τ 

Calculation:

Given that

System 1: $TF=\frac{10}{\left(s+4\right)}$

System 2: $TF=\frac{10}{\left(s+10\right)}$

In standard form, we can write the above TF's as 

System 1: $TF=\frac{10}{4\left(\frac{1}{4}s+1\right)}$

System 2: $TF=\frac{10}{10\left(\frac{1}{10}s+1\right)}$

On comparing with a standard first-order transfer function, we get

The time constant of system 1 is equal to τ₁ = 1/4 

The time constant of system 2 is equal to τ₂ = 1/10

The bandwidth of the system 1 is = BW₁ = 1/τ₁ = 1/(1/4) = 4 rad/sec

The bandwidth of the system 2 is = BW₂ = 1/τ₂ = 1/(1/10) = 10 rad/sec

We know that the characteristic equation for a unity feedback system is written as:

1 + G(s) = 0

$\Rightarrow 1+\frac{\mathrm{K}}{\mathrm{s}\left(\mathrm{s}+2\right)}=0$ 

⇒ s² + 2s + k = 0

Comparing this with the standard 2^nd order characteristic equation, we get:

ω_n² = k

ω_n = √k

Also 2ξω_n = 2

$\xi =\frac{1}{\sqrt{\mathrm{k}}}$

${\mathrm{M}}_{\mathrm{r}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}=2$

$\xi \sqrt{1-{\xi }^2}=\frac{1}{4}$

${\xi }^2=.933,.067$

For resonance peak ξ < 0.707.

So, we can neglect 0.933

${\xi }^2=.067=\frac{1}{\mathrm{k}}$ 

Data:

Ten signals with each having 3000 Hz frequency.

Multiplexed onto a single channel. 

Explanation

If there are n signals multiplexed onto a single channel. Then the number of guard bands i.e. gap between each signal required = n -1

Here, guard bands required = 10 - 1 = 9 (each 300 Hz wide).

Minimum bandwidth required for the multiplexed channel = 3000 × 10 + 300 × 9 

= 30000 + 2700

= 32700 Hz

The transfer function of the standard 2nd order system is defined as

$T.F=\frac{{\omega }_n^2}{s^2+2\xi {\omega }_{n}s+{\omega }_n^2}$

The resonant frequency ω0 is given by

${\omega }_0={\omega }_n\sqrt{1-2{\xi }^2}$

Where ωn = undamped natural frequency

ξ = Damping ratio

Resonant peak M0 is given by:

$M_0=\frac{1}{2\xi \sqrt{1-{\xi }^2}},0\le \xi \le \frac{1}{\sqrt{2}}$

Resonant frequency (ωr) is the frequency at which the magnitude characteristic of frequency response curve has its peak value.

The above equation is meaningful only for 1 – 2ζ2 ≥ 0, i.e.

$zeta\le \frac{1}{\sqrt{2}}or\zeta \le 0.707$

Note: If ζ > 0.707, then ωr = 0

Concept:

DC gain = 5

$M_r=\frac{10}{\sqrt{3}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}$

100 [4ε² (1 – ε²)] = 3

400 ε²– 400 ε⁴ = 3

400x² – 400 x + 3 = 0  

x² – x + 0.0075 = 0

ξ² = 0.99244 , ξ² = 0.00755

ξ = 0.99 , ξ = 0.08689

If resonant peak > 1 then

 $\zeta \le \frac{1}{\sqrt{2}}=0.707$

∴ ξ = 0.08689

$5\sqrt{2}={\omega }_n\sqrt{1-2{\xi }^2}$

$5\sqrt{2}={\omega }_n\sqrt{1-2\times 0.00755}$

ωn = 7 rad/s

$s^2+2\xi {\omega }_{n}s+{\omega }_n^2$

= s² + 1.21 s + 49

Now ${\frac{K}{s^2+1.21s+49}|}_{s=0}=5$ 

K = 49 × 5 = 245

So the required transfer function = $\frac{245}{\left(s^2+1.21s+49\right)}$

In signal processing and control theory the bandwidth is the frequency at which the closed-loop system gain drops 3 dB below peak.

From the given options, the system gain must be less than 10 dB.

DC gain = 5

$M_r=\frac{10}{\sqrt{3}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}$ 

100 [4ε² (1 – ε²)] = 3

400 ε² – 400 ε⁴ = 3

400 x² – 400 x + 3 = 0  

x² – x + 0.0075 = 0

ξ² = 0.99244 , ξ² = 0.00755

ξ = 0.99 , ξ = 0.08689

If resonant peak > 1 then

 $\zeta \le \frac{1}{\sqrt{2}}=0.707$

∴ ξ = 0.08689

$5\sqrt{2}={\omega }_n\sqrt{1-2{\xi }^2}$ 

$5\sqrt{2}={\omega }_n\sqrt{1-2\times 0.00755}$ 

ω_n = 7 rad/s

$s^2+2\xi {\omega }_{n}s+{\omega }_n^2$ 

= s² + 1.21 s + 49

Now ${\frac{K}{s^2+1.21s+49}|}_{s=0}=5$ 

K = 49 × 5 = 245

Transform coding, in conjunction with psycho-acoustic modeling, is a common technique used to compress digital audio. But the use of such a technique can lead to a coding artifact known as a "pre-echo".

• Pre-echoes typically appear in transient situations, situations where silence is broken by a sharp attack. For example, firecrackers, chimes, and castanets all produce sharp attacks that can generate pre-echoes.
• This is a measure of perceptually relevant information contained in any audio record. Expressed in bits per sample, PE represents a theoretical limit on the compressibility of a particular signal.
• Premasking has been exploited in conjunction with adaptive block-size transform coding to compensate for pre-echo distortions.

The PE estimation process is accomplished as follows:

• The signal is first windowed and transformed to the frequency domain.
• A masking threshold is then obtained using perceptual rules.
• Finally, a determination is made on the number of bits required to quantize the spectrum without injecting perceptible noise.
• The PE measurements are obtained by constructing a PE histogram over many frames and then choosing a worst-case value as the actual measurement.

Concept:

The closed-loop transfer function for a standard 2^nd order system is defined as

$CLTF=\frac{{\omega }_n^2}{s^2+2\xi {\omega }_{n}s+{\omega }_n^2}$

The Resonant Peak is one of the frequency domain specifications for a 2^nd order system, which is defined at the peak (maximum) value of the magnitude of the CLTF and is calculated at resonant frequency ω_r, given by

${\omega }_r={\omega }_n\sqrt{1-2{\xi }^2}$

Where, ω_n = Natural frequency and

ξ = Damping ratio

Calculation:

The given feedback is a unity feedback system,

$CLTF=\frac{G\left(s\right)}{1+G\left(s\right)}=\frac{\frac{16}{s(s+4)}}{1+\frac{16}{s(s+4)}}$

$=\frac{16}{s(s+4)+16}=\frac{16}{s^2+4s+16}$

The characteristic equation is given by;

s² + 4s + 16 = 0

Comparing this with the standard 2^nd order characteristic equation $s^2+2\xi {\omega }_{n}s+{\omega }_n^2=0$, we see that,

${\omega }_n^2=16$.

So, ω_n = 4

And, 2ξω_n = 4

So, $\xi =0.5=\frac{1}{2}$

So, the resonant frequency is given by;

${\omega }_r={\omega }_n\sqrt{1-2{\xi }^2}$

${\omega }_r=4\sqrt{1-2{\left(\frac{1}{2}\right)}^2}$

We know that the characteristic equation for a unity feedback system is written as:

1 + G(s) = 0

$\Rightarrow 1+\frac{\mathrm{K}}{\mathrm{s}\left(\mathrm{s}+2\right)}=0$ 

⇒ s² + 2s + k = 0

Comparing this with the standard 2^nd order characteristic equation, we get:

ω_n² = k

ω_n = √k

Also 2ξω_n = 2

$\xi =\frac{1}{\sqrt{\mathrm{k}}}$

${\mathrm{M}}_{\mathrm{r}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}=2$

$\xi \sqrt{1-{\xi }^2}=\frac{1}{4}$

${\xi }^2=.933,.067$

For resonance peak ξ < 0.707.

So, we can neglect 0.933

${\xi }^2=.067=\frac{1}{\mathrm{k}}$ 

Data:

Ten signals with each having 3000 Hz frequency.

Multiplexed onto a single channel. 

Explanation

If there are n signals multiplexed onto a single channel. Then the number of guard bands i.e. gap between each signal required = n -1

Here, guard bands required = 10 - 1 = 9 (each 300 Hz wide).

Minimum bandwidth required for the multiplexed channel = 3000 × 10 + 300 × 9 

= 30000 + 2700

= 32700 Hz