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## Notes on Bode Plot for GATE || simple ways

Bode Plot is a very important topic of GATE ECE. This article will share the tips on how to prepare Bode Plot for GATE 2017.

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## What is Bode Plot?

The Bode Plot provides a graphical method for calculating the stability of a control system based on a sinusoidal frequency response.Bode Plot serves as a graphical representation of the magnitude and phase of  G(j * ω).

## Tips to Prepare Bode Plot for GATE 2017

1. Magnitude Plot

This is a plot of the 20 log10|G (jω)| versus logω.

2. Phase Plot

It is a plot of the phase shift (in degrees) versus logω (frequency).

3. Corner Frequency

It is termed as the frequency at which the slope of the asymptotic log magnitude plot changes.

4. Initial Slope of the Bode Plot

The value of the initial slope is determined from the type of the system and is different for different types of systems.

5. The Minimum Phase Transfer function is given by

G1(s) = 1 + sT1/1 + sT2

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6. Non-minimum phase system is given by

G2(s) = 1 – sT1/ 1 + sT2

important article :   Experts tips to draw Signal Flow Graph for GATE without mistakes

7. If the number of poles of the transfer function at the origin is zero, then the initial slope of the LM plot will be zero and the magnitude will be 20 log Kp up to the lowest corner frequency.

8. Equation of the All pass Transfer Function

G3(s) = 1 – sT/1 + sT

9. If a single pole exists at the origin of the open loop transfer function, then the initial slope of LM plot will be equal to -20 dB/decade up to the lowest corner frequency.

On extension, this line will intersect the frequency axis at ω = kv

10. If two poles exist at the origin of the open loop transfer function, then the initial slope of the LM plot will be equal to -40 dB/decade up to the lowest corner frequency.

On extension, this line will intersect the frequency axis at

ω = √Ka

11.Multiplication of the transfer function by a gain factor will be analogous to shifting the LM plot vertically up by a proportionate gain in dB.

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## Solved Examples

Q1.  A system has a transfer function Y(s)/X(s) = (s/s + p) with an output y(t) = cos (2t - ∏/3) for the input signal x(t) = p cos (2t - ∏/2). Then calculate the system parameter ‘p’?

Options

1. √3
2. 2/√3
3. 1
4. √3/2

Soln.

The transfer function is (s/s + p) where x(s) is the input and y(s) is the output, x(t) is the input time domain and y(t) is the output time domain.

y(t)/x(t) = cos (2t - ∏/3) / p cos (2t - ∏/2 )

Then, the Magnitude will be

Modulus of  [y(t)/x(t)] = 1/p

Phase Angle of y(t)/x(t) = - ∏/3 + ∏/2 = - 60º + 90º = 30º

Even if we write in the s domain, the magnitude and the phase angle will remain the same.

The parameter ‘p’ can be calculated in two ways:

• By equating the magnitudes of the time domain and the s domain
• By equating the phase angles of the time domain and the s domain.

Next, we will write the magnitude and phase angle of ‘s’ domain transfer function.

s/s + p = can also be written as jω/jω + p

The Magnitude = Modulus of (ω/√ω2 + p2)

Phase angle = 90º - tan-1(ω/p)

tan-1 (ω/p) will appear since this a pole.

Ist Method of Computing Parameter ‘p’

• Equating Magnitudes

ω/ √ω2 + p2 = 1/p

ω value will be 2 by referring to the question:

since y(t) = cos (ωt - ∏/3)

Comparing this with y(t) = cos (2t - ∏/3)

ω = 2

2/√22+ p2 = 1/p

2/ √4 + p2 = 1/p

2p = √4 + p2

Sqauring both sides, we get

4p2 = 4 + p2

3p2 = 4

p = √4/3

p = 2/√3

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2nd Method of Computing Parameter ‘p’

• Equating Phase Angles

90º - tan-1(2/p) = 30º

60º = tan-1 (2/p)

Taking tan both sides, we get

tan 60º = 2/p

√3 = 2/p

p = 2/√3

Therefore, the correct answer is (b)

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