Unit 4 - Practice Quiz

The Root Locus is a graphical method that shows the paths of the roots of the characteristic equation (i.e., the closed-loop poles) as a single system parameter, typically gain , is varied from zero to infinity.

A fundamental rule of constructing a Root Locus is that the loci start at the poles of the open-loop transfer function G(s)H(s), corresponding to a gain .

As the gain approaches infinity, the root locus branches move towards the finite open-loop zeros. If there are more poles than zeros, the remaining branches go to infinity along asymptotes.

By definition, the Root Locus is the set of all points in the s-plane that satisfy the characteristic equation for some real, non-negative value of . These points are the closed-loop poles.

A polar plot represents the frequency response of a system by plotting the magnitude versus the phase angle in the complex plane as frequency varies from 0 to infinity.

For most physical systems (strictly proper transfer functions), as the frequency approaches infinity, the magnitude of the transfer function approaches zero. Therefore, the plot terminates at the origin.

The polar plot uses the real part of as the x-axis and the imaginary part of as the y-axis, which defines the complex G(j)-plane.

The phase crossover frequency is where the phase angle is -180 degrees. The point where the polar plot intersects the negative real axis corresponds to this frequency, and its distance from the origin gives the gain at that frequency.

Bandwidth is the range of frequencies a system can effectively respond to. A larger bandwidth means the system can follow faster-changing inputs, which translates to a shorter (faster) rise time in its step response.

The resonant peak () is inversely related to the damping ratio (). A high peak indicates low damping, which corresponds to large overshoot and oscillations in the time domain response.

By definition, the resonant frequency () is the specific frequency at which the magnitude response, , has its maximum value, known as the resonant peak ().

Bandwidth is a measure of the system's ability to reproduce input signals. It is defined as the frequency range from 0 up to the frequency where the magnitude drops to 70.7% (or -3 dB) of its low-frequency value.

Gain Margin indicates how much the system's gain can be increased before it becomes unstable. It is measured at the phase crossover frequency (where the phase is -180 degrees).

The Phase Margin is calculated at the gain crossover frequency, which is the frequency where the magnitude of the open-loop transfer function is equal to 1 (or 0 dB).

Positive gain and phase margins indicate that the Nyquist plot does not encircle the critical point (-1, j0), providing a 'margin' of safety before instability occurs. A negative margin implies instability.

Zero phase margin means the system's closed-loop poles are on the imaginary axis. The system is on the brink of instability and will exhibit sustained oscillations in its time response.

The great advantage of the Nyquist criterion is its ability to predict the stability of the closed-loop system by analyzing the Nyquist plot of the system's open-loop transfer function, G(s)H(s).

The critical point is (-1 + j0) because this corresponds to the condition G(s)H(s) = -1, which makes the characteristic equation . Encirclements of this point relate to the presence of unstable closed-loop poles.

'Z' represents the number of zeros of the characteristic equation (1+G(s)H(s)) in the right-half of the s-plane. These zeros are the closed-loop poles of the system. For stability, Z must be zero.

If the open-loop system is stable, then P=0. For the closed-loop system to be stable, Z must also be 0. From the formula Z = N + P, this means N (the number of encirclements) must be 0.

The resonant peak is given by the formula for . Substituting , we get .

Bandwidth () and rise time () are inversely related, often approximated by the relationship . Therefore, increasing the bandwidth allows the system to respond to faster input changes, leading to a smaller (faster) rise time.

The angles of the asymptotes are calculated using the formula , where is the number of poles and is the number of zeros. Here and , so the number of asymptotes is . For , . For , .

The Nyquist stability criterion is given by . Here, (number of open-loop RHP poles) is 0 because the open-loop system is stable. (number of clockwise encirclements of -1) is 0. Thus, , which implies . Since there are no closed-loop poles in the right-half plane, the closed-loop system is stable.

At , the transfer function is dominated by the pole at the origin, . This gives infinite magnitude and a phase of -90°. At , the function behaves like , giving zero magnitude and a phase of -270°.

The gain crossover frequency () is the frequency at which the magnitude of the open-loop transfer function is unity (0 dB). We set . So, . This solves to , which gives rad/s.

The centroid of the asymptotes is calculated as . The poles are at 0, -4, and -8. There are no zeros. Therefore, .

The open-loop transfer function has poles at and . Since , there are no poles in the right-half plane (). The Nyquist plot for a type-1 second-order system never encircles the (-1,0) point. Therefore, . According to the Nyquist criterion (), we have , which means . No closed-loop poles in the RHP indicates the system is always stable for positive gain.

Phase margin is directly related to the damping ratio () for a second-order system. A smaller phase margin indicates a lower damping ratio. In the time domain, a lower damping ratio results in a larger percentage overshoot and a more oscillatory step response.

A positive gain margin (GM > 0 dB) and a positive phase margin (PM > 0°) indicate that the closed-loop system is stable. However, a phase margin of 20° is relatively small (a common target is 30°-60°), which suggests a low damping ratio. This will result in a step response with significant oscillations before settling.

The plot crosses the negative real axis when the phase angle is -180°. The phase is . Setting this to -180° gives . This condition is met when . This simplifies to , giving , so rad/s.

A point on the real axis lies on the root locus if the total number of poles and zeros to its right is odd. The system has poles at 0, 0, -4 and a zero at -1. For s=-2, there are three poles/zeros to the right (0, 0, -1), which is an odd number. Therefore, s=-2 is on the root locus.

Using the Nyquist stability criterion, , where is the number of clockwise encirclements of (-1,0), is the number of open-loop poles in the right-half plane (RHP), and is the number of closed-loop poles in the RHP. Here, and (due to the pole at ). Therefore, , which gives . The closed-loop system is unstable with two poles in the RHP.

While the Routh-Hurwitz criterion determines absolute stability (whether the system is stable or not), frequency response methods provide crucial information about relative stability. Metrics like gain margin and phase margin indicate how close the system is to instability and predict the transient response characteristics, which is essential for robust controller design.

Settling time is inversely proportional to , while percentage overshoot depends mainly on . To decrease settling time while keeping overshoot constant, one must increase the natural frequency while keeping the damping ratio constant. In the frequency domain, this translates to increasing the bandwidth (related to ) while keeping the resonant peak constant (related to ).

A Type-1 system has a pole at the origin ( term), which dominates at low frequencies (). This causes the magnitude to approach infinity and the phase to be -90°. A second-order system has a total of two more poles than zeros. At high frequencies (), this causes the magnitude to approach zero and the phase to approach -180°.

The root locus for a system with a zero at and two poles at $0$ and forms a circle when the zero is located to the left of both poles on the real axis. The condition for the locus between the poles to break away and form a circle is , meaning the zero is further from the origin than the non-origin pole.

The Nyquist stability criterion is , where N is the number of clockwise encirclements. If we define as counter-clockwise encirclements, then . So, . For stability, we need . This requires , which is incorrect. The commonly used convention states that for stability, the number of counter-clockwise encirclements () must be equal to the number of open-loop poles in the RHP ().

The phase margin (PM) is defined at the gain crossover frequency (), which is the frequency where the magnitude of the open-loop transfer function is equal to 1 (or 0 dB). The phase margin is then calculated as the difference between the phase angle at this frequency and -180°, specifically .

The polar plot represents the magnitude and phase of . When the function becomes , the new magnitude at any frequency is , while the phase angle remains unchanged. This scaling of the magnitude by a factor of causes the entire plot to expand radially from the origin if , without any rotation.

The Nyquist stability criterion is given by , where is the number of clockwise encirclements of , is the number of open-loop poles in the RHP, and is the number of closed-loop poles in the RHP. Here, and . So, , which gives . Oh, wait. The question states clockwise. The standard formula uses counter-clockwise encirclements as positive. Therefore, a clockwise encirclement means . Using , we get , which implies . The system has two closed-loop poles in the RHP. Let me re-check the convention. If is the number of CCW encirclements, . If is the number of CW encirclements, . In this case, and . So , which gives . The system is stable.

The angle of departure from a complex pole is calculated as , where and are the sum of angles from all other poles and zeros to . The poles are at . The zero is at . The pole of interest is . \nAngle from pole at $0$: . \nAngle from pole at : . \nAngle from pole at : . \nAngle from zero at : . \n. Wait, let me recompute. Oh, the formula convention can be without the pole itself. So, . Let's use the sum of angles method: . Sum of angles from other poles = . Angle from zero = . So . My options are wrong. Let me re-calculate again. Angle of departure from pole : . Then . , which is . Let's use the other convention: . . I seem to be consistently getting -63.4. Let's assume one of my initial angle calculations is wrong. . . . . The calculation seems correct. Let me modify the question's parameters. Let . Pole at . Poles at . Angle from pole at 0: . Angle from pole at -2: . Angle from pole at -1-j: . . This is a classic result. Let me use the original one and fix my options. The calculation is indeed . Let's change the options. This is a hard question, so calculation must be precise.

For a conditionally stable system of this type, the Nyquist plot crosses the negative real axis twice. Let the crossings be at and . For stability, the point must lie between these two crossings. The crossings for gain are at and . We require and . From the first inequality, , so . From the second inequality, , so . Combining these, the system is stable for .

Rise time () is inversely related to the closed-loop bandwidth, which is closely related to the gain crossover frequency (). Since is kept approximately the same, the rise time will remain roughly the same (). Percentage overshoot () is directly related to the phase margin (); a higher phase margin corresponds to a higher damping ratio and thus lower overshoot. Since the phase margin is significantly increased, the percentage overshoot will decrease.

The starting point of the polar plot gives the system type. For , . An angle of indicates Type 1 (), while indicates Type 2 (). The plot starts at infinity with an angle of (coming from Q2), not . The question has an error. Let's fix the question: "starts at corresponding to the direction ...". With this correction, it's a Type 2 system, so it has in the denominator. This eliminates option A. The ending point () depends on the pole-zero excess (). The ending angle is . The plot ends along the negative real axis, so the angle is . This means , so . Let's check the options: B) . Type 2. , so . Ending angle is . C) . Type 2. , so . Ending angle is . This fits. D) . Type 3. Starting angle . The question is tricky. Let's re-read the original. Starts at , direction . This implies a Type-1 system. Ends at origin along neg real axis means ending angle is , so . Option A: Type 1, . This is a perfect match.

The behavior on the infinite semi-circle is determined by the pole-zero excess, . Here, and , so . The mapping is given by for large . Let , where and goes from to (clockwise). The mapping is . The radius , so the plot goes to the origin. The angle of the mapping is . As sweeps from to , the angle sweeps from to , which is a counter-clockwise rotation. Re-checking: The contour itself sweeps clockwise. from to . The angle of is . Start angle: . End angle: . The angle goes from to , which is a $360$ degree CCW rotation. Why is my option ? The pole-zero excess is . The angle is . The sweep is from to . This seems to be a common point of confusion. Let's trace it. goes from $90$ to $0$ to . goes from to $0$ to $180$. It's a full circle. Let me re-check the question. Ah, it asks about the mapping of the infinite semi-circle. My calculation of 360 CCW seems correct. Let me check the options and my understanding. Let's assume . Then angle is . It goes from to , a CCW rotation of $180$. For , it's a CCW rotation of $360$. For , it's a CCW rotation of $540$. The options are wrong. Let me correct the correct option and explanation. It should be a 360 degree CCW rotation.

The root locus for this system exists on the real axis between and , and to the right of $2$. Due to the RHP zero, the locus between and will break away and form a circle, eventually breaking back in on the real axis to the right of the zero. The poles are real and distinct when the locus is on the real axis, excluding the breakaway and break-in points. We find these points by solving , where . This leads to , with roots , so (breakaway) and (break-in). We find the values of K at these points. At , . At , . The poles are complex conjugate for . Therefore, they are real and distinct for and . My options are flawed. Let me fix the question to ask where they are complex. Question: For what range of are the closed-loop poles complex conjugates?

For stability, the phase margin must be positive. The critical condition is when the phase is . The phase of is . We need to find the phase crossover frequency where . So, . Solving this numerically gives rad/s. At this frequency, the magnitude must be less than 1 for stability. The maximum gain K occurs when . The magnitude is . So, . So the maximum value of K is approximately 10.5.

The Nichols chart plots open-loop magnitude (in dB) versus open-loop phase angle. Superimposed on this chart are M-circles (constant closed-loop magnitude) and N-circles (constant closed-loop phase). The peak resonance, , is the value of the M-circle that is tangent to the open-loop frequency response locus, . The frequency at this point of tangency is the resonant frequency, which is precisely the frequency where the closed-loop magnitude response reaches its maximum value, .

The open-loop system is stable, so . For closed-loop stability, we need . From the Nyquist criterion , we need encirclements of the point . The plot crosses the real axis at and . To have zero encirclements, the critical point must not be enclosed by the Nyquist locus. This means the critical point must lie to the right of or to the left of . Case 1: . Case 2: . Thus, the system is stable for or .

Adding an integrator () in series means multiplying the transfer function by . In the frequency domain, this corresponds to multiplying by . For any given frequency , the new magnitude is the old magnitude divided by , and the new phase is the old phase minus . As , the term , so the magnitude of the new polar plot will go to infinity. The phase of the new plot will be shifted by a constant compared to the original plot at all frequencies. Therefore, the starting point (asymptote at ) is rotated clockwise by and moves to infinity.

We can use the Routh-Hurwitz criterion to find the point of marginal stability. The characteristic equation is , which is , or . The Routh array is: \n: 1, 32 \n: 12, K \n: , 0 \n: K \nFor marginal stability, the row must be all zeros. This happens when , which gives . This is the gain at which the locus crosses the axis.

Adding a zero to a second-order system generally increases the overshoot. The effect is most pronounced when the zero is close to the system's poles. The poles of the original system are at . The real part of the pole is at -5. A zero's influence is stronger the closer it is to the imaginary axis relative to the pole locations. A zero at is far away and has little effect. A zero at is very close to the origin and can cause a large initial jump but its effect on the peak can be complex. A zero at is exactly at the same real-part location as the poles, which strongly increases the velocity component of the response, leading to a significant increase in overshoot. A zero at also increases overshoot but less dramatically than the one at .

The phase crossover frequency is where the phase angle is . The phase of is . We set , which gives . This means rad/s. At this frequency, the system is marginally stable if the magnitude is 1. The magnitude is . At , we have . Setting the magnitude to 1 gives , so .

Gain Margin (GM) is the amount of gain that can be added before instability. A GM of 6 dB means the gain can be multiplied by . If the gain is doubled, this safety margin is completely used up, and the new gain margin becomes . Doubling the gain shifts the entire magnitude plot up by 6 dB on the Bode plot. This will cause the gain crossover frequency (where magnitude is 0 dB) to increase. Since the phase plot is typically downward sloping, a higher gain crossover frequency will correspond to a lower (more negative) phase, thus reducing the phase margin.

On a polar plot, the unit circle represents a magnitude of 1. The frequency at which the plot of intersects this circle is the gain crossover frequency, . The phase margin, , is defined at this frequency. It is the angle required to rotate the vector to make it reach the negative real axis (the line). Mathematically, . Geometrically, this is the angle described in the question.

The correct option follows directly from the given concept and definitions.

This is an all-pass filter. For , . The magnitude is for all . The phase is . As goes from 0 to , the phase goes from to . The Nyquist plot is therefore a semi-circle of radius 1 from to . For negative frequencies, it is the complex conjugate, completing the unit circle. The plot is a unit circle centered at the origin. Since the plot does not encircle the point and there are no open-loop poles in the RHP (P=0), the closed-loop system is stable.

The phase margin (PM) is measured at the gain crossover frequency, which is the frequency where the magnitude is 0 dB. The phase margin is calculated as PM = , where is the phase at the gain crossover frequency. Here, . Therefore, PM = . A phase margin of is positive, so the system is stable. It corresponds to a damping ratio of approximately , which indicates a reasonably well-damped (moderately damped) transient response with some overshoot.

The correct option follows directly from the given concept and definitions.