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Help???? Steady-state output signal explanation

  • 14-11-2017 11:32pm
    #1
    Registered Users, Registered Users 2 Posts: 178 ✭✭


    Hi,

    I've been studying and studying for months now for this subject in electrical engineering and still dont know the method to find the steady-state output signal given the input signal. The subject in "Linear Control Systems".

    For example, one question is:

    For the system with the closed-loop transfer function :

    Y(s)/X(s) = (s+1)/[(s+2)(s^2+2s+2)]

    Determine the steady-state output signal, y(t) for an input signal x(t) = 5 Sin(2t) [Ans: y(t) = 0.88 Sin(2t–98.1degrees)]

    It seems simple enough, but the lecture slides are very confusing, and this type of question is preventing me from getting anywhere in the course because it's popping up all over the place.

    I'm aware that Y(s) is the Laplace transform of y(t) etc. , but have no idea where the degrees part comes from, of how to calculate it's value.

    Do you take the Laplace inverse of the numerator and denominator? Where does the angle come into play though? Am I missing an equation to use?

    Also, on a different note, can someone try to explain what jw (j omega) is exactly? I know, basic stuff but I have no idea.

    Any help would be amazing!!


Comments

  • Registered Users, Registered Users 2 Posts: 3,862 ✭✭✭mikhail


    You calculate the Laplace transform of x(t), yielding X(s).

    Then you multiply by the trasfer function: X(s) Y(s) / X(s) = Y(s).

    Take the inverse Laplace transform of Y(s) to get y(t), the steady state output.

    The 98.1 degrees is just a phase offset - a shift in time. It'll correspond to multiplication by an exp(-at) term in the Laplace transform.

    jw is the variable you use in the Fourier transform. Imagine you're plotting a Laplace transform, X(s). The s is a complex number, so you have to plot it on a 2D plane, where the x-axis is the real part of s, and the y-axis is the imaginary part. The Fourier transform is the bit of the Laplace transform that runs along the y-axis (i.e. where the real part of s is 0). So you can get the Fourier transform of a signal from its Laplace transform by swapping jw in for s.


  • Registered Users, Registered Users 2 Posts: 178 ✭✭thepikminman


    mikhail wrote: »
    You calculate the Laplace transform of x(t), yielding X(s).

    Then you multiply by the trasfer function: X(s) Y(s) / X(s) = Y(s).

    Take the inverse Laplace transform of Y(s) to get y(t), the steady state output.

    The 98.1 degrees is just a phase offset - a shift in time. It'll correspond to multiplication by an exp(-at) term in the Laplace transform.

    jw is the variable you use in the Fourier transform. Imagine you're plotting a Laplace transform, X(s). The s is a complex number, so you have to plot it on a 2D plane, where the x-axis is the real part of s, and the y-axis is the imaginary part. The Fourier transform is the bit of the Laplace transform that runs along the y-axis (i.e. where the real part of s is 0). So you can get the Fourier transform of a signal from its Laplace transform by swapping jw in for s.

    Wow...Mikhail to the rescue again, I was honestly considering PM'ing you about this but didn't want to be weird :P

    Thank you!! I don't really understand the jw bit, but the first bit is great I can solve the problem now thanks to it.

    You literally saved me two years of my life, this time and last; I would've had to repeat the year if I failed either of the exams.

    Thanks again!!!


  • Registered Users, Registered Users 2 Posts: 5 EEE


    Hi,

    I've been studying and studying for months now for this subject in electrical engineering and still dont know the method to find the steady-state output signal given the input signal. The subject in "Linear Control Systems".

    For example, one question is:

    For the system with the closed-loop transfer function :

    Y(s)/X(s) = (s+1)/[(s+2)(s^2+2s+2)]

    Determine the steady-state output signal, y(t) for an input signal x(t) = 5 Sin(2t) [Ans: y(t) = 0.88 Sin(2t–98.1degrees)]

    It seems simple enough, but the lecture slides are very confusing, and this type of question is preventing me from getting anywhere in the course because it's popping up all over the place.

    I'm aware that Y(s) is the Laplace transform of y(t) etc. , but have no idea where the degrees part comes from, of how to calculate it's value.

    Do you take the Laplace inverse of the numerator and denominator? Where does the angle come into play though? Am I missing an equation to use?

    Also, on a different note, can someone try to explain what jw (j omega) is exactly? I know, basic stuff but I have no idea.

    Any help would be amazing!!

    I registered to help you with this as I seen you posted two similar questions recently.

    Q: What is a transfer function?
    A: A transfer function function is like a black box, we pass a signal into it and we get a modified signal on the output.

    Q: How is a transfer function defined?
    A: A transfer function is the output signal / input signal

    Q: What is 's'?
    A: 's' is a Laplace operator, a N multiplied by 's' is the same differentiating N.
    N * 1/s is the same as integration.

    Notation

    G(s) = Transfer Function
    X(t) = input signal
    Y(t) = output signal

    G(s) = G(jw)
    G(jw) = A*e^j(thetha)

    Step 1:

    G(s) = (s+1)/(s+2)(s^2+2s+2)

    Change all 's' values to jw (where j = sqrt(-1), w = omega = 2*pi*radians)

    G(jw) = (1+jw)/(2+jw)(-w^2+2jw+2)

    Note: (w^2 --> sqrt(-1)*w squared = -w^2)

    Step 2: Simplify and calculate magnitude/gain

    G(jw) = (1+j2) / (2+j2)(-2+j4) = sqrt(1^2+2^2) / (sqrt(2^2+2^2)*(sqrt(2^2+4^2)) = 0.177

    So 0.177 is the scaling of our output signal with respect to input signal.

    Next Calculate phase:

    phase = [tan^-(2) - tan^-1(1) - (180 degrees - Tan^-1(2)]
    = 63.4 deg - 45 deg - (180 deg - 63.4 deg) = -98.1 deg

    Step 3: Put your two answers together

    = 0.177e^-j98.1 deg

    So this tells us that the transfer function will scale our input by X * 0.177 and will introduce a phase shift of -98.1 degrees

    Step 4: Multiply input signal by Step 3

    x(t) = 5.sin(2t) => y(t) = 5(0.177)sin(2t-98.1 deg)

    == 0.88 sin(2t- 98.1 deg)

    Best of luck :)


  • Registered Users, Registered Users 2 Posts: 178 ✭✭thepikminman


    EEE wrote: »
    I registered to help you with this as I seen you posted two similar questions recently.

    Q: What is a transfer function?
    A: A transfer function function is like a black box, we pass a signal into it and we get a modified signal on the output.

    Q: How is a transfer function defined?
    A: A transfer function is the output signal / input signal

    Q: What is 's'?
    A: 's' is a Laplace operator, a N multiplied by 's' is the same differentiating N.
    N * 1/s is the same as integration.

    Notation

    G(s) = Transfer Function
    X(t) = input signal
    Y(t) = output signal

    G(s) = G(jw)
    G(jw) = A*e^j(thetha)

    Step 1:

    G(s) = (s+1)/(s+2)(s^2+2s+2)

    Change all 's' values to jw (where j = sqrt(-1), w = omega = 2*pi*radians)

    G(jw) = (1+jw)/(2+jw)(-w^2+2jw+2)

    Note: (w^2 --> sqrt(-1)*w squared = -w^2)

    Step 2: Simplify and calculate magnitude/gain

    G(jw) = (1+j2) / (2+j2)(-2+j4) = sqrt(1^2+2^2) / (sqrt(2^2+2^2)*(sqrt(2^2+4^2)) = 0.177

    So 0.177 is the scaling of our output signal with respect to input signal.

    Next Calculate phase:

    phase = [tan^-(2) - tan^-1(1) - (180 degrees - Tan^-1(2)]
    = 63.4 deg - 45 deg - (180 deg - 63.4 deg) = -98.1 deg

    Step 3: Put your two answers together

    = 0.177e^-j98.1 deg

    So this tells us that the transfer function will scale our input by X * 0.177 and will introduce a phase shift of -98.1 degrees

    Step 4: Multiply input signal by Step 3

    x(t) = 5.sin(2t) => y(t) = 5(0.177)sin(2t-98.1 deg)

    == 0.88 sin(2t- 98.1 deg)

    Best of luck :)

    WOW..... "'s' is a Laplace operator, a N multiplied by 's' is the same differentiating N. N * 1/s is the same as integration"...I ACTUALLY get it now, just because of this one sentence!! Why couldn't the lecturer/notes just say that....

    Thank you so much, esp. for registering just to help!!! I value your post so much, you have no idea how much it helped!

    There are just 4 things I'm still not fully sure about;

    1. I learned that the transfer function is output/input or c(t)/r(t), but now in other sections they're defining it as g(t)h(t)....is c(t)/r(t) the TF for open loop and g(t)h(t) the TF for closed loop?

    2. You mentioned "phase = [tan^-(2) - tan^-1(1) - (180 degrees - Tan^-1(2)]", can you tell me how you arrived at all the tan inverses...is there a formula? And why is it 180 when it's 2pi radians (and not 360)?

    3. What actually IS phase/phase shift?

    4. In step 4, you said "x(t) = 5.sin(2t) => y(t) = 5(0.177)sin(2t-98.1 deg)". Is that always the form: multiply the number in front of sin in x(t) by the mag. of G(jw), and just add the phase value after the sin value from x(t), yielding the amplitude of y(t) and how much "out of phase" it is with x(t)?

    Last thing (lol I can't stop myself), just to make sure I'm understanding the other parts of your post, please correct me if I'm wrong:

    -The TF is what is "done" to the input signal to make the output signal.

    -Laplace is a way of using variables instead of differential values.

    -For the G(jw) value, 2pi radians is used just by default because that is the value at 0, 180, 360 etc i.e. the phase at important points(???).

    -For G(jw) = A*e^j(theta), the e^j(theta) just comes from some formula and the calculated phase value tells us how much the output signal's wave exceeds/lags the input signal's wave (lags in this problem), and A is the scaling value (= the mag. of G(jw)) which is multiplied the the amplitude of the input signal to give us the amplitude of the output signal at the chosen value of theta.

    Once again, thank you!!


  • Registered Users, Registered Users 2 Posts: 5 EEE


    WOW..... "'s' is a Laplace operator, a N multiplied by 's' is the same differentiating N. N * 1/s is the same as integration"...I ACTUALLY get it now, just because of this one sentence!! Why couldn't the lecturer/notes just say that....

    Thank you so much, esp. for registering just to help!!! I value your post so much, you have no idea how much it helped!

    There are just 4 things I'm still not fully sure about;

    1. I learned that the transfer function is output/input or c(t)/r(t), but now in other sections they're defining it as g(t)h(t)....is c(t)/r(t) the TF for open loop and g(t)h(t) the TF for closed loop?

    2. You mentioned "phase = [tan^-(2) - tan^-1(1) - (180 degrees - Tan^-1(2)]", can you tell me how you arrived at all the tan inverses...is there a formula? And why is it 180 when it's 2pi radians (and not 360)?

    3. What actually IS phase/phase shift?

    4. In step 4, you said "x(t) = 5.sin(2t) => y(t) = 5(0.177)sin(2t-98.1 deg)". Is that always the form: multiply the number in front of sin in x(t) by the mag. of G(jw), and just add the phase value after the sin value from x(t), yielding the amplitude of y(t) and how much "out of phase" it is with x(t)?

    Last thing (lol I can't stop myself), just to make sure I'm understanding the other parts of your post, please correct me if I'm wrong:

    -The TF is what is "done" to the input signal to make the output signal.

    -Laplace is a way of using variables instead of differential values.

    -For the G(jw) value, 2pi radians is used just by default because that is the value at 0, 180, 360 etc i.e. the phase at important points(???).

    -For G(jw) = A*e^j(theta), the e^j(theta) just comes from some formula and the calculated phase value tells us how much the output signal's wave exceeds/lags the input signal's wave (lags in this problem), and A is the scaling value (= the mag. of G(jw)) which is multiplied the the amplitude of the input signal to give us the amplitude of the output signal at the chosen value of theta.

    Once again, thank you!!

    No worries,

    # 1: It doesnt matter what notation is used, you could technically use any letter in the alphabet.

    There are differences in open-loop and closed-loop:
    Open-loop is input and output only, there is no feedback path (line going from output back to start to summing junction)

    (not realistic examples but to aid understanding)
    Open-loop system: A toaster, put the bread in and push the button, the toaster acts on a timer. If the bread starts to burn the toaster will keep applying heat until the timer is up. It has no way of knowing what the bread is like.

    Closed-loop system: a system which takes the output signal, compares it with input and and passes it back through the system. This is usually done in a summing junction at the input. If we pretend the input signal is a heat signal (C degrees) applied to the toaster. We turn the toaster on, after a period the bread starts to turn brown, this is recorded at the output and then fed back to the summing junction at the input. So now at our input we have the incoming signal controlling the temperature applied to the bread and we have the feedback signal which is the last actual recorded value of how toasted the bresad is. The summing junction usually subtracts the feedback from the input signal to produce an 'error' value that gets passed back through the transfer function (toaster). If we pretend in this case that the feedback signal is 75% then its is subtracted from the input signal 100% (I know we are subtracting power signal from temperature signal, just run with it :) ]. to produce an error of 25%. The error now tells us the difference between the desired input setpoint and output. The error is passed into the toaster and the toaster detects that we only need 25% more so it reduces its power output or reduces its time delay to pop...hope that explanation was not too confusing!

    Better example - ABS braking in a car, without it the car skids until it stops (open-loop). With ABS when activated the onboard computer is feeding a signal to the computer to say the car is still moving and the break is pressed, if wheels are locked and car is still in motion release breaks, apply breaks, check is car still in motion. This is feedback so a closed loop system. p.s. i don't know anything about ABS, that's just another example so you can think in the right direction.

    # 2: If you think of an x-y plot and the we have the vector starting at x=0,y=0 and finishing at x=2,y=2. If we want to know the magnitude of the vector its sqrt(x^2+y^2) which is Pythagorean theory. To calculate the phase (angle) its tan^-1(y/x). This is were that comes from.

    "tan^-(2) - tan^-1(1) - (180 degrees - Tan^-1(2)"
    tan^-(2) == tan^-1(2/1) = tan^-1(y1/x1)
    tan^-1(1) = tan^-1(1/1) = tan^-1(y2/x2)..... etc

    The -180 degrees -> look at the quadrant the angle (tan^-1(4/-2)) falls within with reference to the positive x-axis. The -180degree is used to correct the value.


    #3: Phase shift as per what mikhail is the difference in angle between input and output signal. If you think of a sinusoidal we can say it lags the reference point by 90 degrees etc. If we input a signal which 1 in magnitude at 90 degrees the output could be 0.5 magnitude but -10 degrees from the input signal. If we wanted to move a robotic arm say 120 degrees we would need to know that when we input 120 degrees that the robotic arm is going to be 10 degrees off our input (desired) point so we should compensate for this by inputting 130 degrees.

    #4: This is just notation, your interpretation is correct. you can multiply 5 *0.177 but don't touch the ' sin(2t-98.1 deg)' part as this in its final form already.

    Transfer function - yes its changing the input to a certain output, we need to understand whats happening:

    Think y(t) = 2*x(t)

    x(t) = input
    y(t) = output
    g(t) = transfer function (2*x(t) in this case)

    if i told you x(t) = 5 and y(t) = 10, what is the transfer function? It would be that the input signal is being multiplied by 2 (output/input = 10/5 = 2).

    Laplace - yes, without Laplace we would need to solve differential equations. With Laplace it transforms differential equations into algebra.

    2*pi -> in electrical engineering, frequency is the number of cycles per-second. If the signal repeats 2 times a second then the frequency is 2 Hz. If the signal repeats 100 times a second its 100 Hz. Think of a sinusoidal waveform, its got a positive and negative, its a full circle -> 2*pi before the signal does another full sweep of the circle...

    For G(jw) = A*e^j(theta) - yes you are correct. The form A^e^-j(thetha) is just the mathematically correct way of writing our answer. The only thing we really care about is the Magnitude(A) and the phase shift(thetha). We then write the answer in the form A*e^j(theta).


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  • Registered Users, Registered Users 2 Posts: 178 ✭✭thepikminman


    EEE wrote: »
    No worries,

    # 1: It doesnt matter what notation is used, you could technically use any letter in the alphabet.

    There are differences in open-loop and closed-loop:
    Open-loop is input and output only, there is no feedback path (line going from output back to start to summing junction)

    (not realistic examples but to aid understanding)
    Open-loop system: A toaster, put the bread in and push the button, the toaster acts on a timer. If the bread starts to burn the toaster will keep applying heat until the timer is up. It has no way of knowing what the bread is like.

    Closed-loop system: a system which takes the output signal, compares it with input and and passes it back through the system. This is usually done in a summing junction at the input. If we pretend the input signal is a heat signal (C degrees) applied to the toaster. We turn the toaster on, after a period the bread starts to turn brown, this is recorded at the output and then fed back to the summing junction at the input. So now at our input we have the incoming signal controlling the temperature applied to the bread and we have the feedback signal which is the last actual recorded value of how toasted the bresad is. The summing junction usually subtracts the feedback from the input signal to produce an 'error' value that gets passed back through the transfer function (toaster). If we pretend in this case that the feedback signal is 75% then its is subtracted from the input signal 100% (I know we are subtracting power signal from temperature signal, just run with it :) ]. to produce an error of 25%. The error now tells us the difference between the desired input setpoint and output. The error is passed into the toaster and the toaster detects that we only need 25% more so it reduces its power output or reduces its time delay to pop...hope that explanation was not too confusing!

    Better example - ABS braking in a car, without it the car skids until it stops (open-loop). With ABS when activated the onboard computer is feeding a signal to the computer to say the car is still moving and the break is pressed, if wheels are locked and car is still in motion release breaks, apply breaks, check is car still in motion. This is feedback so a closed loop system. p.s. i don't know anything about ABS, that's just another example so you can think in the right direction.

    # 2: If you think of an x-y plot and the we have the vector starting at x=0,y=0 and finishing at x=2,y=2. If we want to know the magnitude of the vector its sqrt(x^2+y^2) which is Pythagorean theory. To calculate the phase (angle) its tan^-1(y/x). This is were that comes from.

    "tan^-(2) - tan^-1(1) - (180 degrees - Tan^-1(2)"
    tan^-(2) == tan^-1(2/1) = tan^-1(y1/x1)
    tan^-1(1) = tan^-1(1/1) = tan^-1(y2/x2)..... etc

    The -180 degrees -> look at the quadrant the angle (tan^-1(4/-2)) falls within with reference to the positive x-axis. The -180degree is used to correct the value.


    #3: Phase shift as per what mikhail is the difference in angle between input and output signal. If you think of a sinusoidal we can say it lags the reference point by 90 degrees etc. If we input a signal which 1 in magnitude at 90 degrees the output could be 0.5 magnitude but -10 degrees from the input signal. If we wanted to move a robotic arm say 120 degrees we would need to know that when we input 120 degrees that the robotic arm is going to be 10 degrees off our input (desired) point so we should compensate for this by inputting 130 degrees.

    #4: This is just notation, your interpretation is correct. you can multiply 5 *0.177 but don't touch the ' sin(2t-98.1 deg)' part as this in its final form already.

    Transfer function - yes its changing the input to a certain output, we need to understand whats happening:

    Think y(t) = 2*x(t)

    x(t) = input
    y(t) = output
    g(t) = transfer function (2*x(t) in this case)

    if i told you x(t) = 5 and y(t) = 10, what is the transfer function? It would be that the input signal is being multiplied by 2 (output/input = 10/5 = 2).

    Laplace - yes, without Laplace we would need to solve differential equations. With Laplace it transforms differential equations into algebra.

    2*pi -> in electrical engineering, frequency is the number of cycles per-second. If the signal repeats 2 times a second then the frequency is 2 Hz. If the signal repeats 100 times a second its 100 Hz. Think of a sinusoidal waveform, its got a positive and negative, its a full circle -> 2*pi before the signal does another full sweep of the circle...

    For G(jw) = A*e^j(theta) - yes you are correct. The form A^e^-j(thetha) is just the mathematically correct way of writing our answer. The only thing we really care about is the Magnitude(A) and the phase shift(thetha). We then write the answer in the form A*e^j(theta).

    Thank you, you're a saint!! The examples were very helpful, it's so much easier to know what the hell is going on with real examples.

    ONE last question:

    For the phase shift, when you got "phase = [tan^-(2) - tan^-1(1) - (180 degrees - Tan^-1(2)]
    = 63.4 deg - 45 deg - (180 deg - 63.4 deg) = -98.1 deg"
    , where are the values of 2 and 1 and 2 in the tan's coming from? Like what do I sub in, i can't see a pattern??

    Is there a formula like "tan^-1(value) - tan^-1(other value)" or, you know, something like that??


  • Registered Users, Registered Users 2 Posts: 5 EEE


    Thank you, you're a saint!! The examples were very helpful, it's so much easier to know what the hell is going on with real examples.

    ONE last question:

    For the phase shift, when you got "phase = [tan^-(2) - tan^-1(1) - (180 degrees - Tan^-1(2)]
    = 63.4 deg - 45 deg - (180 deg - 63.4 deg) = -98.1 deg"
    , where are the values of 2 and 1 and 2 in the tan's coming from? Like what do I sub in, i can't see a pattern??

    Is there a formula like "tan^-1(value) - tan^-1(other value)" or, you know, something like that??

    If we think back to the vector from (0,0) to (2,2) its in the form (x,y). So starting at 0, move two points out to x=2. Now move up two points to y=2. We now have a straight line at 45 degrees going from (0,0) to (2,2).

    The values are always in the form x,y.

    So (1+2j) is the same as saying x=1, y=2. The j is a complex operator and is defined as sqrt(-1).

    To find the phase of that vector (0,0) to (2,2) we can calculate tan^-1(y/x) = tan^-1(2/2) = 45 degrees.

    The formula tan^-1(y/x) is from trigonometry and is used in all forms of engineering.

    if: G(jw) = (1+j2) / (2+j2)(-2+j4)

    (1+j2) -> tan^-1(2/1) = 63.4 deg
    (2+j2) -> tan^-1(2/2) = 45 deg etc


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