Test: Bode Plot - 3 - Electrical Engineering (EE) MCQ

Given circuit,

The transfer function for series RLC circuit by taking output voltage acroos the capacitor is given by,

From the plot,
20 log|M_r| = 26 dB
|M_r| = 19.95
19.95 × 0.5R = 1
R = 0.1 Ω

According to figure
Gain = 20 dB (at phase cross over frequency)
Gian margin = -20 dB
System make to marginally stable G.M = 0 or Gain = 0 dB
i.e 20 log K = -20
log K = -1
K = 10^-1 = 0.1
So for K < 0.1 system will be stable

Phase margin = 180° + ∠GH(jω) at ω = ω_gc
ω_gc = Gain cross over frequency

ω_gc: Gain crossover frequency: It is defined as the frequency at which the gain of the system is unity.
Now, see the magnitude expression (1) of the system, it looks very complicated to calculate ωgc. Here comes, the role of the Bode plot.
Points to Remember:
Before drawing the Bode plot, always convert the transfer function into its time constant form.
Pole-zero form:

Time-constant form:

Bode plot:

Poles are at s = -0.1, -1 and -10
Initially, since it is a type zero system, the slope of the Bode plot is 0 dB/decade.
Magnitude, M = 20 log₁₀ k dB
M = 20 log₁₀ 10 = 20 dB

From Bode plot magnitude is 0 dB at ω = 1
0 dB = 20 log₁₀ M
M = 1 at ω = 1
Hence ω_gc = 1

At ω = ω_gc = 1, we get:

= - 45° - 84.289 – 5.710 = -135°  
Phase margin, PM = 180° + ∠GH(jω)  at ω = ω_gc
PM = 180° - 135°
PM = 45°

The phase of the system is given by 
ϕ = (-90º)^N- tan^-1(w/1) - tan^-1(w/10)
at w = 0
we can see in the graph we have -90º
∴ -90º = (-90º)^N- 0 - 0
∴ N = 1

Stability can be determined easily from a plot of the loop gain versus frequency.
The critical point is when the loop gain equals 0 dB (gain equals 1) because a circuit must have gain ≥ 1 to become unstable.
The phase margin, which is the difference between the measured phase angle and 180º, is calculated at the 0-dB point.
Bode plot is the plot of loop gain versus frequency.
Using Bode Plot we can easily find the gain margin and phase margin of the system.
We can also see the effect of resistor or capacitance on the Bode plot by analyzing the Bode plot.
We can get different parameters like Bandwidth, the cutoff frequency of op-amp by bode plot So the Bode plot is the best suitable for stability analysis.

Concept:
The gain margin is always calculated at the phase Crossover frequency (ω_pc)
The phase Crossover frequency (ω_pc) is calculated as:

Calculation:
Given:

The phase Crossover frequency (ω_pc) is calculated as:

On solving we'll get:

Concept:

• The slope of the bode plot decreases by 20 dB/decade at the pole
• The slope of the bode plot increases by 20 dB/decade at zero

Application:

Total poles: 6 = N_P
Total zeros: 3 = N_Z

d(t) = 0.1 sin (ωt)

as r = 0
y = - y L(s) + d
⇒ y = d/1+L(s)
From bode magnitude plot
At ω = 5 rad/sec, magnitude of L(s) = 40 dB
⇒ magnitude of L(s) = 100
From Bode phase plot
∠L(s) = 120° at ω = 5 rad/sec
L(s) at 5 rad/sec = 100 ∠-120°

⇒ |y| = 1 x 10^-3

Here starting slope is +20 db/dec
Hence s¹ is at numerator part of T.F.
Now from the straight line equation y = mx + c, we get
0 = 20 log (0.5) + c
∴ c = 6.0205
Now c = 20 log K
6.0205 = 20 log K
K = 2
We can find the value of ω₁ as
12 = 20 log (ω₁) + 6.0205
ω₁ = 1.99 ≈ 2
Now we have to find ω₂,  so that
0 = - 40 log (8) + c
c = 36.1236
∴ 12 = - 40 log (ω₂) + 36.1236
ω₂ = 4
Hence transfer function for given bode plot is

For a negative unity feedback system, if CLTF is given, the Open-loop transfer function is evaluated as:

The phase margin is defined as:
PM = 180^∘ − ϕ
where ϕ is the phase of the OLTF at ω_g
ω_gc (Gain crossover frequency) = frequency at which system gain is unity.
If the polar plot is not crossing the negative real axis, then GM = ∞
For a negative unity feedback system, the phase at ω = 0 is 0 degrees.
And phase at ω = ∞ is:
(p-z) × 90° 
p = number of open-loop poles and z = no. of open-loop zeros.

Analysis:

Replacing s with jω:

25× 10¹⁰ = ω⁴ + 990000ω² + 625× 10⁸On solving the above, we get:
gc = 403.32 rad/sec
Now we will calculate the phase of the system at ω_gc

PM = 180^∘ + ϕ 
= 180^o + (- 180^o + 34.37^o) = 34.37^o 

If we draw the polar plot of the above system, then we get to see that its polar plot will not cut the negative real axis, in that case:
GM = ∞