Solved Example of Nyquist Plot
1. Draw the Nyquist plot for the system whose open loop transfer function is given by
G(s)H(s) =
Also determine the value of ‘k’ for which the system is stable.
Solution
Step 1: Identify the poles and zeros
There are 3 poles and no zeros
Poles: 0, -2, -10
G(s)H(s) =
Step 2: Mapping of all the 4 section as seen in the given image
The value of ω in the region (1) is s = jω. Replacing the transfer function with s = jω
G(jω)H(jω) =
G(jω)H(jω) =
G(jω)H(jω) = —– equation 1
Now keeping the imaginary term 0 to find the phase crossover frequency.
ω(1 – 0.05ω2) = 0
1 – 0.05ω2 = 0
ω = 4.472 = phase crossover frequency
Putting the value of frequency in the real part of the equation 1.
G(jω)H(jω) =
G(jω)H(jω) =
G(jω)H(jω) = -0.00417K
The polar plot will start at -90 degrees and crosses the x axis at point -0.00417K as given below:
Step 3: Mapping the region 2 whose equation is given by: s = where, θ = +90o to -90o
If the transfer function is in the form G(s)H(s) = , then assume that (1 + sT) is equal to sT. It is given by:
G(s)H(s) =
G(s)H(s) =
G(s)H(s) =
Let s =
G(s)H(s) =
At , G(s)H(s) =
At , G(s)H(s) =
Thus, the region 2 varies from -270 degrees to 270 degrees. It is represented below:
Step 4: Now map the region 3 of the nyquist contour. The region 3 is simple inverse of region 1. Thus the graph will mirror image of region 1 graph which is shown below:
Step 5: Mapping the region 4 whose equation is given by: s = ,where, θ = -90o to +90o
If the transfer function is in the form G(s)H(s) = , then assume that (1 + sT) is equal to 1. It is given by:
G(s)H(s) =
G(s)H(s) =
Let s = s =
G(s)H(s) = ∞
At , G(s)H(s) = ∞
At , G(s)H(s) = ∞
So the graph will be circular arc of infinite radius as seen in the given image.
Step 6: Now we will perform the stability analysis. To find the value of ‘k’, we will check when our contour will pass through the point (-1+j0)
-0.00417K = -1
K = 1/0.00417
K = 240
Step 7: Final Nyquist Plot is shown below. It as been plotted by combining the mapping of all the four regions.
Now we will check foe which value of ‘k’, the system is stable.
Case 1: If k< 240
The point -1+j0 is not encircled. This means that there are no poles on the right half of the plane. This means the system is stable for k less than 240.
Case 2: k>240
The point -1+j0 is encircled two times in the clockwise direction. This means that Z>P and hence the system is unstable.
Stability condition: 0 < K < 240
Nyquist Plot
A Nyquist plot is a graphical representation used in control engineering. It is used to analyze the stability and frequency response of a system. The plot represents the complex transfer function of a system in a complex plane. The x-axis represents the real part of the complex numbers and the y-axis represents the imaginary part. Each point on the Nyquist plot reflects the complex value of the transfer function at that frequency.
- Nyquist Stability Criteria
- Important Terminologies of Nyquist Plot
- How to draw Nyquist Plot?
- Stability analysis using Nyquist Plot
- Solved Example of Nyquist Plot
- Advantages of Nyquist Plot
- Disadvantages of Nyquist Plot
- Application of Nyquist Plot
Contact Us