Ministry of Science and Education of Russian Federation

Ministry of Science and Education of Russian Federation

National Research Tomsk Polytechnic University

Institute of Power Engineering

Department of Electric Drive and Equipment

Bachelor’s Degree Programme:

13.03.02 Electric Power Engineering and Electrical Engineering

Report

Laboratory work №1

Investigation of control systems’ typical dynamic links characteristics

Student of group___________ _____________

Supervisor_______________ _____________

Tomsk – 2015

The aim of the work: To investigate characteristics of the control systems’ typical dynamic links.

The basic differential equation for all typical dynamic links is as follows:

. (1)

By substituting the coefficients a ₀, a ₁, a ₂, b ₀, b ₁, we get the differential equation of any typical dynamic link (see table 2 [1]).

The simulation model is represented in fig. 0.

Fig. 0. Simulation model of the link under study

1. Proportional link

1.1. Differential equation:

or ,

where – input value, – output one, t – time.

1.2. Operator equation is derived from the differential equation (see paragraph 1.1) by substitution , , :

.

1.3. Transfer function is derived from the operator equation (see paragraph 1.2) as the relation between output and input:

.

1.4. Frequency response equation is deduced from the transfer function (see paragraph 1.3) by substituting

.

1.4.1. Magnitude frequency response is derived from the frequency response equation (see paragraph 1.4) as follows:

.

1.4.2. Logarithmic magnitude frequency response can be obtained using magnitude frequency response (see paragraph 1.4.1) as follows:

.

1.4.3. Phase frequency response is deduced from the frequency response equation (see paragraph 1.4) as follows:

,

where – the phase of frequency response numerator, – the phase of frequency response denominator.

Proportional link characteristics are represented in fig. 1.1–1.3.

Fig. 1.1. Step response

The output value changes synchronously with the input value. The output value is ten times greater than input one (fig. 1.1). Transfer coefficient K = b ₁/ a ₂ = 10.

Fig 1.2. Zero–pole map

As it can be seen from fig. 1.2 there are no both zeros and poles for the link under consideration since our link is of zero order.

Fig. 1.3. Logarithmic magnitude and phase characteristics (Bode plot)

Transfer coefficient K = 10 corresponds to L (ω) = 20 dB. There are no phase shift between input signal and output one as φ(ω) = 0.

2. Aperiodic link

2.1. Differential equation:

or .

2.2. Operator equation (see paragraph 2.1):

or

or .

2.3. Transfer function (see paragraph 2.2):

.

2.4. Frequency response equation is deduced from the transfer function (see paragraph 2.3) by substituting

.

2.4.1. Magnitude frequency response is derived from the frequency response equation (see paragraph 2.4) as follows:

.

2.4.2. Logarithmic magnitude frequency response can be obtained using magnitude frequency response (see paragraph 2.4.1) as follows:

.

2.4.3. Phase frequency response is deduced from the frequency response equation (see paragraph 2.4) as follows:

.

Proportional link characteristics are represented in fig. 2.1–2.4.

Fig. 2.1. Step response

The output value changes by exponent. The steady state output value is ten times greater than input one (fig. 1). Transfer coefficient K = b ₁/ a ₂ = 10.

Fig 2.2. Zero–pole map

Our link is of the first order. Therefore, there is one pole. The pole has the coordinates (–10; j 0). How to find the pole. Let’s solve the characteristic equation:

.

Fig. 2.3. Frequency response plot

Fig. 2.4. Logarithmic magnitude and phase characteristics (Bode plot)

Steady state transfer coefficient K = 10 corresponds to L ₀ = 20 dB. Suppression rate is –20 dB/dec. Maximum phase shift φ_max = –π/2. Cutoff frequency is the frequency when x (j ω) = y (j ω).

3. Individual task

Variant 14

Estimate the parameter T ₂ influence on the transient response for the oscillatory link.

Initial value: T ₂ = 0.04 s.

Another value: T ₂’ = 0.02 T ₂ = 0.0008 s.

For creating the model we will use matlab software. It is convenient for the task performed and more versatile. The system under consideration is shown in fig. 3.1.

Fig. 3.1. Initial system

After varying the parameter T ₂ the system under study is represented in fig. 3.2.

Fig. 3.2. New system with another parameter of T ₂ = 0.0008 s

The characteristics for the couple of links are represented in fig. 3.3–3.6.

Fig. 3.3. Transients for oscillatory link with different T2:

1 – T ₂ = 0.04 s; 2 – T ₂’ = 0.0008 s

As it can be seen from fig. 3.3 the initial link when decreasing parameter T ₂ is converted into the second order aperiodic link. The response time is decreased. There is no overshoot for the new system.

Fig. 3.4. Zero-pole map for the links under study

Link with the initial parameter T ₂ = 0.04 s has complex conjugate poles whereas the link with T ₂’ = 0.0008 s has two poles with only real components.

Fig. 3.5. Bode diagram for the two links under consideration

When we decrease the parameter T ₂ the cutoff frequency is decreased. Moreover there is no resonance peak for the link with T ₂’ = 0.0008 s.

Question:

14. What is the difference between the characteristics of aperiodic, oscillatory and integral links.

Consider the step response plots. Step response of aperiodic link has no oscillations whereas the response of oscillatory link has oscillations. The peculiarity of integral link lies in the absence of steady state value. When we consider the zero–pole map, all the links under consideration have no zeros. Poles of aperiodic link have only real components whereas the poles of oscillatory link are complex conjugate. Considering the bode plots for the links under investigation it can be concluded that the magnitude characteristic of aperiodic and integral links has no resonance peak, which is the feature of oscillatory link. Both aperiodic and oscillatory links’ bode magnitude plots have both corner and cutoff frequencies. Bode plot of the integral link have only cutoff frequency. The slope (suppression rate) of the bode magnitude characteristic of integral link is –20 dB/dec within all the frequency range. When the frequency of input signal is increased the transfer gain for the aperiodic and oscillatory links start to decrease at the corner frequency (slope is equal to –20 dB/dec). Phase lag under these circumstances is increased. For the integral link the phase shift is constant at all frequencies and equals to –π/2.

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