How do I find the magnitude and phase of the frequency response function for a third order system using...
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A third order system is described by:
$$frac{{rm d}^3 y}{{rm d}t^3} + 2 frac{{rm d}^2 y}{{rm d}t^2} + 6 frac{{rm d} y}{{rm d}t} + 5y = u + 2 frac{{rm d} u}{{rm d}t}.$$ Determine the magnitude and phase of its frequency response function.
I know I have to change the equation into a transfer function but after that is where I get stuck because it doesn't simply very nicely. Any help would be appreciated, thanks in advance!
edit: This is the work I've done
Work I've done
ordinary-differential-equations fourier-analysis systems-of-equations
edited Jan 4 '17 at 16:34
Florian
1,4571721
asked Jan 4 '17 at 15:49
Fukuyama Fukuyama
113
$endgroup$
|
show 1 more comment
$begingroup$
A third order system is described by:
$$frac{{rm d}^3 y}{{rm d}t^3} + 2 frac{{rm d}^2 y}{{rm d}t^2} + 6 frac{{rm d} y}{{rm d}t} + 5y = u + 2 frac{{rm d} u}{{rm d}t}.$$ Determine the magnitude and phase of its frequency response function.
I know I have to change the equation into a transfer function but after that is where I get stuck because it doesn't simply very nicely. Any help would be appreciated, thanks in advance!
edit: This is the work I've done
Work I've done
ordinary-differential-equations fourier-analysis systems-of-equations
edited Jan 4 '17 at 16:34
Florian
1,4571721
asked Jan 4 '17 at 15:49
Fukuyama Fukuyama
113
$endgroup$
$begingroup$
You don't need to simplify. You get a Laplace domain function in the variable s that is rational (polynomial on top and bottom). Set $s = j omega$. You now have a complex number for a given choice of $omega$. This number has a magnitude and a phase, which you compute using complex arithmetic. If you are looking to plot this, look into the Bode plot, which is exactly a plot of magnitude and phase
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– Michael Stachowsky
Jan 4 '17 at 16:00
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@Moo I've edited my work in!
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– Fukuyama
Jan 4 '17 at 16:15
1
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@Fukuyama: Your $H(s)$ looks quite okay to me. Now substitute $s = jmath omega$ to get $G(omega)$. Then forming the magnitude is relatively straightforward complex arithmetic.
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– Florian
Jan 4 '17 at 16:21
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@Florian Thank you! And what would I do for the phase if I may ask?
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– Fukuyama
Jan 4 '17 at 16:23
1
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Your $G(omega)$ will be in the form $frac{A+jmath B}{C+jmath D}$. Expanding with $C-jmath D$ gives $frac{(A+jmath B)(C-jmath D)}{C^2+D^2} = frac{AC+BD +jmath (BC-AD)}{C^2+D^2}$. This separates real from imaginary part. The phase of a complex number $x+jmath y$ is basically $tan^{-1}frac{y}{x}$, i.e., in your case something like $tan^{-1}frac{BC-AD}{AC+BD}$. To get the four-quadrant phase, use atan2 if needed.
$endgroup$
– Florian
Jan 4 '17 at 16:31
|
show 1 more comment
$begingroup$
A third order system is described by:
$$frac{{rm d}^3 y}{{rm d}t^3} + 2 frac{{rm d}^2 y}{{rm d}t^2} + 6 frac{{rm d} y}{{rm d}t} + 5y = u + 2 frac{{rm d} u}{{rm d}t}.$$ Determine the magnitude and phase of its frequency response function.
I know I have to change the equation into a transfer function but after that is where I get stuck because it doesn't simply very nicely. Any help would be appreciated, thanks in advance!
edit: This is the work I've done
Work I've done
ordinary-differential-equations fourier-analysis systems-of-equations
edited Jan 4 '17 at 16:34
Florian
1,4571721
asked Jan 4 '17 at 15:49
Fukuyama Fukuyama
113
$endgroup$
A third order system is described by:
$$frac{{rm d}^3 y}{{rm d}t^3} + 2 frac{{rm d}^2 y}{{rm d}t^2} + 6 frac{{rm d} y}{{rm d}t} + 5y = u + 2 frac{{rm d} u}{{rm d}t}.$$ Determine the magnitude and phase of its frequency response function.
I know I have to change the equation into a transfer function but after that is where I get stuck because it doesn't simply very nicely. Any help would be appreciated, thanks in advance!
edit: This is the work I've done
Work I've done
ordinary-differential-equations fourier-analysis systems-of-equations
ordinary-differential-equations fourier-analysis systems-of-equations
edited Jan 4 '17 at 16:34
Florian
1,4571721
asked Jan 4 '17 at 15:49
Fukuyama Fukuyama
113
edited Jan 4 '17 at 16:34
Florian
1,4571721
asked Jan 4 '17 at 15:49
Fukuyama Fukuyama
113
edited Jan 4 '17 at 16:34
Florian
1,4571721
edited Jan 4 '17 at 16:34
Florian
1,4571721
edited Jan 4 '17 at 16:34
Florian
1,4571721
1,4571721
asked Jan 4 '17 at 15:49
Fukuyama Fukuyama
113
asked Jan 4 '17 at 15:49
Fukuyama Fukuyama
113
asked Jan 4 '17 at 15:49
Fukuyama Fukuyama
113
113
$begingroup$
You don't need to simplify. You get a Laplace domain function in the variable s that is rational (polynomial on top and bottom). Set $s = j omega$. You now have a complex number for a given choice of $omega$. This number has a magnitude and a phase, which you compute using complex arithmetic. If you are looking to plot this, look into the Bode plot, which is exactly a plot of magnitude and phase
$endgroup$
– Michael Stachowsky
Jan 4 '17 at 16:00
$begingroup$
@Moo I've edited my work in!
$endgroup$
– Fukuyama
Jan 4 '17 at 16:15
1
$begingroup$
@Fukuyama: Your $H(s)$ looks quite okay to me. Now substitute $s = jmath omega$ to get $G(omega)$. Then forming the magnitude is relatively straightforward complex arithmetic.
$endgroup$
– Florian
Jan 4 '17 at 16:21
$begingroup$
@Florian Thank you! And what would I do for the phase if I may ask?
$endgroup$
– Fukuyama
Jan 4 '17 at 16:23
1
$begingroup$
Your $G(omega)$ will be in the form $frac{A+jmath B}{C+jmath D}$. Expanding with $C-jmath D$ gives $frac{(A+jmath B)(C-jmath D)}{C^2+D^2} = frac{AC+BD +jmath (BC-AD)}{C^2+D^2}$. This separates real from imaginary part. The phase of a complex number $x+jmath y$ is basically $tan^{-1}frac{y}{x}$, i.e., in your case something like $tan^{-1}frac{BC-AD}{AC+BD}$. To get the four-quadrant phase, use atan2 if needed.
$endgroup$
– Florian
Jan 4 '17 at 16:31
|
show 1 more comment
$begingroup$
You don't need to simplify. You get a Laplace domain function in the variable s that is rational (polynomial on top and bottom). Set $s = j omega$. You now have a complex number for a given choice of $omega$. This number has a magnitude and a phase, which you compute using complex arithmetic. If you are looking to plot this, look into the Bode plot, which is exactly a plot of magnitude and phase
$endgroup$
– Michael Stachowsky
Jan 4 '17 at 16:00
$begingroup$
@Moo I've edited my work in!
$endgroup$
– Fukuyama
Jan 4 '17 at 16:15
1
$begingroup$
@Fukuyama: Your $H(s)$ looks quite okay to me. Now substitute $s = jmath omega$ to get $G(omega)$. Then forming the magnitude is relatively straightforward complex arithmetic.
$endgroup$
– Florian
Jan 4 '17 at 16:21
$begingroup$
@Florian Thank you! And what would I do for the phase if I may ask?
$endgroup$
– Fukuyama
Jan 4 '17 at 16:23
1
$begingroup$
Your $G(omega)$ will be in the form $frac{A+jmath B}{C+jmath D}$. Expanding with $C-jmath D$ gives $frac{(A+jmath B)(C-jmath D)}{C^2+D^2} = frac{AC+BD +jmath (BC-AD)}{C^2+D^2}$. This separates real from imaginary part. The phase of a complex number $x+jmath y$ is basically $tan^{-1}frac{y}{x}$, i.e., in your case something like $tan^{-1}frac{BC-AD}{AC+BD}$. To get the four-quadrant phase, use atan2 if needed.
$endgroup$
– Florian
Jan 4 '17 at 16:31
$begingroup$
You don't need to simplify. You get a Laplace domain function in the variable s that is rational (polynomial on top and bottom). Set $s = j omega$. You now have a complex number for a given choice of $omega$. This number has a magnitude and a phase, which you compute using complex arithmetic. If you are looking to plot this, look into the Bode plot, which is exactly a plot of magnitude and phase
$endgroup$
– Michael Stachowsky
Jan 4 '17 at 16:00
$begingroup$
You don't need to simplify. You get a Laplace domain function in the variable s that is rational (polynomial on top and bottom). Set $s = j omega$. You now have a complex number for a given choice of $omega$. This number has a magnitude and a phase, which you compute using complex arithmetic. If you are looking to plot this, look into the Bode plot, which is exactly a plot of magnitude and phase
$endgroup$
– Michael Stachowsky
Jan 4 '17 at 16:00
$begingroup$
@Moo I've edited my work in!
$endgroup$
– Fukuyama
Jan 4 '17 at 16:15
$begingroup$
@Moo I've edited my work in!
$endgroup$
– Fukuyama
Jan 4 '17 at 16:15
1
1
$begingroup$
@Fukuyama: Your $H(s)$ looks quite okay to me. Now substitute $s = jmath omega$ to get $G(omega)$. Then forming the magnitude is relatively straightforward complex arithmetic.
$endgroup$
– Florian
Jan 4 '17 at 16:21
$begingroup$
@Fukuyama: Your $H(s)$ looks quite okay to me. Now substitute $s = jmath omega$ to get $G(omega)$. Then forming the magnitude is relatively straightforward complex arithmetic.
$endgroup$
– Florian
Jan 4 '17 at 16:21
$begingroup$
@Florian Thank you! And what would I do for the phase if I may ask?
$endgroup$
– Fukuyama
Jan 4 '17 at 16:23
$begingroup$
@Florian Thank you! And what would I do for the phase if I may ask?
$endgroup$
– Fukuyama
Jan 4 '17 at 16:23
1
1
$begingroup$
Your $G(omega)$ will be in the form $frac{A+jmath B}{C+jmath D}$. Expanding with $C-jmath D$ gives $frac{(A+jmath B)(C-jmath D)}{C^2+D^2} = frac{AC+BD +jmath (BC-AD)}{C^2+D^2}$. This separates real from imaginary part. The phase of a complex number $x+jmath y$ is basically $tan^{-1}frac{y}{x}$, i.e., in your case something like $tan^{-1}frac{BC-AD}{AC+BD}$. To get the four-quadrant phase, use atan2 if needed.
$endgroup$
– Florian
Jan 4 '17 at 16:31
$begingroup$
Your $G(omega)$ will be in the form $frac{A+jmath B}{C+jmath D}$. Expanding with $C-jmath D$ gives $frac{(A+jmath B)(C-jmath D)}{C^2+D^2} = frac{AC+BD +jmath (BC-AD)}{C^2+D^2}$. This separates real from imaginary part. The phase of a complex number $x+jmath y$ is basically $tan^{-1}frac{y}{x}$, i.e., in your case something like $tan^{-1}frac{BC-AD}{AC+BD}$. To get the four-quadrant phase, use atan2 if needed.
$endgroup$
– Florian
Jan 4 '17 at 16:31
|
show 1 more comment
1 Answer
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Laplace transform:
$$text{F}left(text{s}right)=mathcal{L}_tleft[text{f}left(tright)right]_{left(text{s}right)}:=int_0^inftytext{f}left(tright)e^{-text{s}t}spacetext{d}ttag1$$
So, we get:
$$text{n}_1cdottext{y}'''left(tright)+text{n}_2cdottext{y}''left(tright)+text{n}_3cdottext{y}'left(tright)+text{n}_4cdottext{y}left(tright)=text{n}_5cdottext{u}left(tright)+text{n}_6cdottext{u}'left(tright)tag2$$
Where $text{n}_text{a}$ is a constant for every $text{a}$
Now, when we take the Laplace transform of both sides, we need to know:
- $$mathcal{L}_tleft[text{n}_1cdottext{y}'''left(tright)right]_{left(text{s}right)}:=text{n}_1cdotint_0^inftytext{y}'''left(tright)e^{-text{s}t}spacetext{d}t=$$
$$text{n}_1cdotleft(text{s}^3cdottext{Y}left(text{s}right)-text{s}^2cdottext{y}left(0right)-text{s}cdottext{y}'left(0right)-text{y}''left(0right)right)tag3$$ - $$mathcal{L}_tleft[text{n}_2cdottext{y}''left(tright)right]_{left(text{s}right)}:=text{n}_2cdotint_0^inftytext{y}''left(tright)e^{-text{s}t}spacetext{d}t=text{n}_2cdotleft(text{s}^2cdottext{Y}left(text{s}right)-text{s}cdottext{y}left(0right)-text{y}'left(0right)right)tag4$$
- $$mathcal{L}_tleft[text{n}_3cdottext{y}'left(tright)right]_{left(text{s}right)}:=text{n}_3cdotint_0^inftytext{y}'left(tright)e^{-text{s}t}spacetext{d}t=text{n}_3cdotleft(text{s}cdottext{Y}left(text{s}right)-text{y}left(0right)right)tag5$$
- $$mathcal{L}_tleft[text{n}_4cdottext{y}left(tright)right]_{left(text{s}right)}:=text{n}_4cdotint_0^inftytext{y}left(tright)e^{-text{s}t}spacetext{d}t=text{n}_4cdottext{Y}left(text{s}right)tag6$$
- $$mathcal{L}_tleft[text{n}_5cdottext{u}left(tright)right]_{left(text{s}right)}:=text{n}_5cdotint_0^inftytext{u}left(tright)e^{-text{s}t}spacetext{d}t=text{n}_5cdottext{U}left(text{s}right)tag7$$
- $$mathcal{L}_tleft[text{n}_6cdottext{u}'left(tright)right]_{left(text{s}right)}:=text{n}_6cdotint_0^inftytext{u}'left(tright)e^{-text{s}t}spacetext{d}t=text{n}_6cdotleft(text{s}cdottext{U}left(text{s}right)-text{u}left(0right)right)tag8$$
Assuming that a initial conditions are equal to zero, we get:
$$text{n}_1cdottext{s}^3cdottext{Y}left(text{s}right)+text{n}_2cdottext{s}^2cdottext{Y}left(text{s}right)+text{n}_3cdottext{s}cdottext{Y}left(text{s}right)+text{n}_4cdottext{Y}left(text{s}right)=text{n}_5cdottext{U}left(text{s}right)+text{n}_6cdottext{s}cdottext{U}left(text{s}right)tag9$$
Simplifying it a little bit:
$$text{Y}left(text{s}right)cdotleft(text{n}_1cdottext{s}^3+text{n}_2cdottext{s}^2+text{n}_3cdottext{s}+text{n}_4right)=text{U}left(text{s}right)cdotleft(text{n}_5+text{n}_6cdottext{s}right)tag{10}$$
So, we also know that:
$$text{H}left(text{s}right)=frac{text{Y}left(text{s}right)}{text{U}left(text{s}right)}=frac{text{n}_5+text{n}_6cdottext{s}}{text{n}_1cdottext{s}^3+text{n}_2cdottext{s}^2+text{n}_3cdottext{s}+text{n}_4}tag{11}$$
For the respons you're looking for, we can use:
$$text{s}=text{j}omegatag{12}$$
Where $text{j}^2=-1$
So, we will also get:
$$text{H}left(text{j}omegaright)=frac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=frac{text{n}_5+text{n}_6cdotleft(text{j}omegaright)}{text{n}_1cdotleft(text{j}omegaright)^3+text{n}_2cdotleft(text{j}omegaright)^2+text{n}_3cdotleft(text{j}omegaright)+text{n}_4}tag{13}$$
Simplying it a little bit, gives us:
$$text{H}left(text{j}omegaright)=frac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=frac{text{n}_5+text{n}_6cdotomegacdottext{j}}{text{n}_4-text{n}_2cdotomega^2+text{j}cdotomegacdotleft(text{n}_3-text{n}_1cdotomega^2right)}tag{14}$$
For the magnitude we get:
$$left|text{H}left(text{j}omegaright)right|=frac{left|text{Y}left(text{j}omegaright)right|}{left|text{U}left(text{j}omegaright)right|}=frac{sqrt{text{n}_5^2+text{n}_6^2cdotomega^2}}{sqrt{left(text{n}_4-text{n}_2cdotomega^2right)^2+omega^2cdotleft(text{n}_3-text{n}_1cdotomega^2right)^2}}tag{15}$$
And for the phase, you can use:
$$argtext{H}left(text{j}omegaright)=argfrac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=argleft(text{Y}left(text{j}omegaright)right)-argleft(text{U}left(text{j}omegaright)right)tag{16}$$
Now, when $text{n}_5spacewedgespaceomegainmathbb{R}^+$:
$$argleft(text{Y}left(text{j}omegaright)right)=arctanleft(frac{text{n}_6cdotomega}{text{n}_5}right)tag{17}$$
answered Jan 4 '17 at 17:44
JanJan
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1 Answer
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$begingroup$
Laplace transform:
$$text{F}left(text{s}right)=mathcal{L}_tleft[text{f}left(tright)right]_{left(text{s}right)}:=int_0^inftytext{f}left(tright)e^{-text{s}t}spacetext{d}ttag1$$
So, we get:
$$text{n}_1cdottext{y}'''left(tright)+text{n}_2cdottext{y}''left(tright)+text{n}_3cdottext{y}'left(tright)+text{n}_4cdottext{y}left(tright)=text{n}_5cdottext{u}left(tright)+text{n}_6cdottext{u}'left(tright)tag2$$
Where $text{n}_text{a}$ is a constant for every $text{a}$
Now, when we take the Laplace transform of both sides, we need to know:
- $$mathcal{L}_tleft[text{n}_1cdottext{y}'''left(tright)right]_{left(text{s}right)}:=text{n}_1cdotint_0^inftytext{y}'''left(tright)e^{-text{s}t}spacetext{d}t=$$
$$text{n}_1cdotleft(text{s}^3cdottext{Y}left(text{s}right)-text{s}^2cdottext{y}left(0right)-text{s}cdottext{y}'left(0right)-text{y}''left(0right)right)tag3$$ - $$mathcal{L}_tleft[text{n}_2cdottext{y}''left(tright)right]_{left(text{s}right)}:=text{n}_2cdotint_0^inftytext{y}''left(tright)e^{-text{s}t}spacetext{d}t=text{n}_2cdotleft(text{s}^2cdottext{Y}left(text{s}right)-text{s}cdottext{y}left(0right)-text{y}'left(0right)right)tag4$$
- $$mathcal{L}_tleft[text{n}_3cdottext{y}'left(tright)right]_{left(text{s}right)}:=text{n}_3cdotint_0^inftytext{y}'left(tright)e^{-text{s}t}spacetext{d}t=text{n}_3cdotleft(text{s}cdottext{Y}left(text{s}right)-text{y}left(0right)right)tag5$$
- $$mathcal{L}_tleft[text{n}_4cdottext{y}left(tright)right]_{left(text{s}right)}:=text{n}_4cdotint_0^inftytext{y}left(tright)e^{-text{s}t}spacetext{d}t=text{n}_4cdottext{Y}left(text{s}right)tag6$$
- $$mathcal{L}_tleft[text{n}_5cdottext{u}left(tright)right]_{left(text{s}right)}:=text{n}_5cdotint_0^inftytext{u}left(tright)e^{-text{s}t}spacetext{d}t=text{n}_5cdottext{U}left(text{s}right)tag7$$
- $$mathcal{L}_tleft[text{n}_6cdottext{u}'left(tright)right]_{left(text{s}right)}:=text{n}_6cdotint_0^inftytext{u}'left(tright)e^{-text{s}t}spacetext{d}t=text{n}_6cdotleft(text{s}cdottext{U}left(text{s}right)-text{u}left(0right)right)tag8$$
Assuming that a initial conditions are equal to zero, we get:
$$text{n}_1cdottext{s}^3cdottext{Y}left(text{s}right)+text{n}_2cdottext{s}^2cdottext{Y}left(text{s}right)+text{n}_3cdottext{s}cdottext{Y}left(text{s}right)+text{n}_4cdottext{Y}left(text{s}right)=text{n}_5cdottext{U}left(text{s}right)+text{n}_6cdottext{s}cdottext{U}left(text{s}right)tag9$$
Simplifying it a little bit:
$$text{Y}left(text{s}right)cdotleft(text{n}_1cdottext{s}^3+text{n}_2cdottext{s}^2+text{n}_3cdottext{s}+text{n}_4right)=text{U}left(text{s}right)cdotleft(text{n}_5+text{n}_6cdottext{s}right)tag{10}$$
So, we also know that:
$$text{H}left(text{s}right)=frac{text{Y}left(text{s}right)}{text{U}left(text{s}right)}=frac{text{n}_5+text{n}_6cdottext{s}}{text{n}_1cdottext{s}^3+text{n}_2cdottext{s}^2+text{n}_3cdottext{s}+text{n}_4}tag{11}$$
For the respons you're looking for, we can use:
$$text{s}=text{j}omegatag{12}$$
Where $text{j}^2=-1$
So, we will also get:
$$text{H}left(text{j}omegaright)=frac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=frac{text{n}_5+text{n}_6cdotleft(text{j}omegaright)}{text{n}_1cdotleft(text{j}omegaright)^3+text{n}_2cdotleft(text{j}omegaright)^2+text{n}_3cdotleft(text{j}omegaright)+text{n}_4}tag{13}$$
Simplying it a little bit, gives us:
$$text{H}left(text{j}omegaright)=frac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=frac{text{n}_5+text{n}_6cdotomegacdottext{j}}{text{n}_4-text{n}_2cdotomega^2+text{j}cdotomegacdotleft(text{n}_3-text{n}_1cdotomega^2right)}tag{14}$$
For the magnitude we get:
$$left|text{H}left(text{j}omegaright)right|=frac{left|text{Y}left(text{j}omegaright)right|}{left|text{U}left(text{j}omegaright)right|}=frac{sqrt{text{n}_5^2+text{n}_6^2cdotomega^2}}{sqrt{left(text{n}_4-text{n}_2cdotomega^2right)^2+omega^2cdotleft(text{n}_3-text{n}_1cdotomega^2right)^2}}tag{15}$$
And for the phase, you can use:
$$argtext{H}left(text{j}omegaright)=argfrac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=argleft(text{Y}left(text{j}omegaright)right)-argleft(text{U}left(text{j}omegaright)right)tag{16}$$
Now, when $text{n}_5spacewedgespaceomegainmathbb{R}^+$:
$$argleft(text{Y}left(text{j}omegaright)right)=arctanleft(frac{text{n}_6cdotomega}{text{n}_5}right)tag{17}$$
answered Jan 4 '17 at 17:44
JanJan
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add a comment |
$begingroup$
Laplace transform:
$$text{F}left(text{s}right)=mathcal{L}_tleft[text{f}left(tright)right]_{left(text{s}right)}:=int_0^inftytext{f}left(tright)e^{-text{s}t}spacetext{d}ttag1$$
So, we get:
$$text{n}_1cdottext{y}'''left(tright)+text{n}_2cdottext{y}''left(tright)+text{n}_3cdottext{y}'left(tright)+text{n}_4cdottext{y}left(tright)=text{n}_5cdottext{u}left(tright)+text{n}_6cdottext{u}'left(tright)tag2$$
Where $text{n}_text{a}$ is a constant for every $text{a}$
Now, when we take the Laplace transform of both sides, we need to know:
- $$mathcal{L}_tleft[text{n}_1cdottext{y}'''left(tright)right]_{left(text{s}right)}:=text{n}_1cdotint_0^inftytext{y}'''left(tright)e^{-text{s}t}spacetext{d}t=$$
$$text{n}_1cdotleft(text{s}^3cdottext{Y}left(text{s}right)-text{s}^2cdottext{y}left(0right)-text{s}cdottext{y}'left(0right)-text{y}''left(0right)right)tag3$$ - $$mathcal{L}_tleft[text{n}_2cdottext{y}''left(tright)right]_{left(text{s}right)}:=text{n}_2cdotint_0^inftytext{y}''left(tright)e^{-text{s}t}spacetext{d}t=text{n}_2cdotleft(text{s}^2cdottext{Y}left(text{s}right)-text{s}cdottext{y}left(0right)-text{y}'left(0right)right)tag4$$
- $$mathcal{L}_tleft[text{n}_3cdottext{y}'left(tright)right]_{left(text{s}right)}:=text{n}_3cdotint_0^inftytext{y}'left(tright)e^{-text{s}t}spacetext{d}t=text{n}_3cdotleft(text{s}cdottext{Y}left(text{s}right)-text{y}left(0right)right)tag5$$
- $$mathcal{L}_tleft[text{n}_4cdottext{y}left(tright)right]_{left(text{s}right)}:=text{n}_4cdotint_0^inftytext{y}left(tright)e^{-text{s}t}spacetext{d}t=text{n}_4cdottext{Y}left(text{s}right)tag6$$
- $$mathcal{L}_tleft[text{n}_5cdottext{u}left(tright)right]_{left(text{s}right)}:=text{n}_5cdotint_0^inftytext{u}left(tright)e^{-text{s}t}spacetext{d}t=text{n}_5cdottext{U}left(text{s}right)tag7$$
- $$mathcal{L}_tleft[text{n}_6cdottext{u}'left(tright)right]_{left(text{s}right)}:=text{n}_6cdotint_0^inftytext{u}'left(tright)e^{-text{s}t}spacetext{d}t=text{n}_6cdotleft(text{s}cdottext{U}left(text{s}right)-text{u}left(0right)right)tag8$$
Assuming that a initial conditions are equal to zero, we get:
$$text{n}_1cdottext{s}^3cdottext{Y}left(text{s}right)+text{n}_2cdottext{s}^2cdottext{Y}left(text{s}right)+text{n}_3cdottext{s}cdottext{Y}left(text{s}right)+text{n}_4cdottext{Y}left(text{s}right)=text{n}_5cdottext{U}left(text{s}right)+text{n}_6cdottext{s}cdottext{U}left(text{s}right)tag9$$
Simplifying it a little bit:
$$text{Y}left(text{s}right)cdotleft(text{n}_1cdottext{s}^3+text{n}_2cdottext{s}^2+text{n}_3cdottext{s}+text{n}_4right)=text{U}left(text{s}right)cdotleft(text{n}_5+text{n}_6cdottext{s}right)tag{10}$$
So, we also know that:
$$text{H}left(text{s}right)=frac{text{Y}left(text{s}right)}{text{U}left(text{s}right)}=frac{text{n}_5+text{n}_6cdottext{s}}{text{n}_1cdottext{s}^3+text{n}_2cdottext{s}^2+text{n}_3cdottext{s}+text{n}_4}tag{11}$$
For the respons you're looking for, we can use:
$$text{s}=text{j}omegatag{12}$$
Where $text{j}^2=-1$
So, we will also get:
$$text{H}left(text{j}omegaright)=frac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=frac{text{n}_5+text{n}_6cdotleft(text{j}omegaright)}{text{n}_1cdotleft(text{j}omegaright)^3+text{n}_2cdotleft(text{j}omegaright)^2+text{n}_3cdotleft(text{j}omegaright)+text{n}_4}tag{13}$$
Simplying it a little bit, gives us:
$$text{H}left(text{j}omegaright)=frac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=frac{text{n}_5+text{n}_6cdotomegacdottext{j}}{text{n}_4-text{n}_2cdotomega^2+text{j}cdotomegacdotleft(text{n}_3-text{n}_1cdotomega^2right)}tag{14}$$
For the magnitude we get:
$$left|text{H}left(text{j}omegaright)right|=frac{left|text{Y}left(text{j}omegaright)right|}{left|text{U}left(text{j}omegaright)right|}=frac{sqrt{text{n}_5^2+text{n}_6^2cdotomega^2}}{sqrt{left(text{n}_4-text{n}_2cdotomega^2right)^2+omega^2cdotleft(text{n}_3-text{n}_1cdotomega^2right)^2}}tag{15}$$
And for the phase, you can use:
$$argtext{H}left(text{j}omegaright)=argfrac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=argleft(text{Y}left(text{j}omegaright)right)-argleft(text{U}left(text{j}omegaright)right)tag{16}$$
Now, when $text{n}_5spacewedgespaceomegainmathbb{R}^+$:
$$argleft(text{Y}left(text{j}omegaright)right)=arctanleft(frac{text{n}_6cdotomega}{text{n}_5}right)tag{17}$$
answered Jan 4 '17 at 17:44
JanJan
21.8k31240
$endgroup$
add a comment |
$begingroup$
Laplace transform:
$$text{F}left(text{s}right)=mathcal{L}_tleft[text{f}left(tright)right]_{left(text{s}right)}:=int_0^inftytext{f}left(tright)e^{-text{s}t}spacetext{d}ttag1$$
So, we get:
$$text{n}_1cdottext{y}'''left(tright)+text{n}_2cdottext{y}''left(tright)+text{n}_3cdottext{y}'left(tright)+text{n}_4cdottext{y}left(tright)=text{n}_5cdottext{u}left(tright)+text{n}_6cdottext{u}'left(tright)tag2$$
Where $text{n}_text{a}$ is a constant for every $text{a}$
Now, when we take the Laplace transform of both sides, we need to know:
- $$mathcal{L}_tleft[text{n}_1cdottext{y}'''left(tright)right]_{left(text{s}right)}:=text{n}_1cdotint_0^inftytext{y}'''left(tright)e^{-text{s}t}spacetext{d}t=$$
$$text{n}_1cdotleft(text{s}^3cdottext{Y}left(text{s}right)-text{s}^2cdottext{y}left(0right)-text{s}cdottext{y}'left(0right)-text{y}''left(0right)right)tag3$$ - $$mathcal{L}_tleft[text{n}_2cdottext{y}''left(tright)right]_{left(text{s}right)}:=text{n}_2cdotint_0^inftytext{y}''left(tright)e^{-text{s}t}spacetext{d}t=text{n}_2cdotleft(text{s}^2cdottext{Y}left(text{s}right)-text{s}cdottext{y}left(0right)-text{y}'left(0right)right)tag4$$
- $$mathcal{L}_tleft[text{n}_3cdottext{y}'left(tright)right]_{left(text{s}right)}:=text{n}_3cdotint_0^inftytext{y}'left(tright)e^{-text{s}t}spacetext{d}t=text{n}_3cdotleft(text{s}cdottext{Y}left(text{s}right)-text{y}left(0right)right)tag5$$
- $$mathcal{L}_tleft[text{n}_4cdottext{y}left(tright)right]_{left(text{s}right)}:=text{n}_4cdotint_0^inftytext{y}left(tright)e^{-text{s}t}spacetext{d}t=text{n}_4cdottext{Y}left(text{s}right)tag6$$
- $$mathcal{L}_tleft[text{n}_5cdottext{u}left(tright)right]_{left(text{s}right)}:=text{n}_5cdotint_0^inftytext{u}left(tright)e^{-text{s}t}spacetext{d}t=text{n}_5cdottext{U}left(text{s}right)tag7$$
- $$mathcal{L}_tleft[text{n}_6cdottext{u}'left(tright)right]_{left(text{s}right)}:=text{n}_6cdotint_0^inftytext{u}'left(tright)e^{-text{s}t}spacetext{d}t=text{n}_6cdotleft(text{s}cdottext{U}left(text{s}right)-text{u}left(0right)right)tag8$$
Assuming that a initial conditions are equal to zero, we get:
$$text{n}_1cdottext{s}^3cdottext{Y}left(text{s}right)+text{n}_2cdottext{s}^2cdottext{Y}left(text{s}right)+text{n}_3cdottext{s}cdottext{Y}left(text{s}right)+text{n}_4cdottext{Y}left(text{s}right)=text{n}_5cdottext{U}left(text{s}right)+text{n}_6cdottext{s}cdottext{U}left(text{s}right)tag9$$
Simplifying it a little bit:
$$text{Y}left(text{s}right)cdotleft(text{n}_1cdottext{s}^3+text{n}_2cdottext{s}^2+text{n}_3cdottext{s}+text{n}_4right)=text{U}left(text{s}right)cdotleft(text{n}_5+text{n}_6cdottext{s}right)tag{10}$$
So, we also know that:
$$text{H}left(text{s}right)=frac{text{Y}left(text{s}right)}{text{U}left(text{s}right)}=frac{text{n}_5+text{n}_6cdottext{s}}{text{n}_1cdottext{s}^3+text{n}_2cdottext{s}^2+text{n}_3cdottext{s}+text{n}_4}tag{11}$$
For the respons you're looking for, we can use:
$$text{s}=text{j}omegatag{12}$$
Where $text{j}^2=-1$
So, we will also get:
$$text{H}left(text{j}omegaright)=frac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=frac{text{n}_5+text{n}_6cdotleft(text{j}omegaright)}{text{n}_1cdotleft(text{j}omegaright)^3+text{n}_2cdotleft(text{j}omegaright)^2+text{n}_3cdotleft(text{j}omegaright)+text{n}_4}tag{13}$$
Simplying it a little bit, gives us:
$$text{H}left(text{j}omegaright)=frac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=frac{text{n}_5+text{n}_6cdotomegacdottext{j}}{text{n}_4-text{n}_2cdotomega^2+text{j}cdotomegacdotleft(text{n}_3-text{n}_1cdotomega^2right)}tag{14}$$
For the magnitude we get:
$$left|text{H}left(text{j}omegaright)right|=frac{left|text{Y}left(text{j}omegaright)right|}{left|text{U}left(text{j}omegaright)right|}=frac{sqrt{text{n}_5^2+text{n}_6^2cdotomega^2}}{sqrt{left(text{n}_4-text{n}_2cdotomega^2right)^2+omega^2cdotleft(text{n}_3-text{n}_1cdotomega^2right)^2}}tag{15}$$
And for the phase, you can use:
$$argtext{H}left(text{j}omegaright)=argfrac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=argleft(text{Y}left(text{j}omegaright)right)-argleft(text{U}left(text{j}omegaright)right)tag{16}$$
Now, when $text{n}_5spacewedgespaceomegainmathbb{R}^+$:
$$argleft(text{Y}left(text{j}omegaright)right)=arctanleft(frac{text{n}_6cdotomega}{text{n}_5}right)tag{17}$$
answered Jan 4 '17 at 17:44
JanJan
21.8k31240
$endgroup$
Laplace transform:
$$text{F}left(text{s}right)=mathcal{L}_tleft[text{f}left(tright)right]_{left(text{s}right)}:=int_0^inftytext{f}left(tright)e^{-text{s}t}spacetext{d}ttag1$$
So, we get:
$$text{n}_1cdottext{y}'''left(tright)+text{n}_2cdottext{y}''left(tright)+text{n}_3cdottext{y}'left(tright)+text{n}_4cdottext{y}left(tright)=text{n}_5cdottext{u}left(tright)+text{n}_6cdottext{u}'left(tright)tag2$$
Where $text{n}_text{a}$ is a constant for every $text{a}$
Now, when we take the Laplace transform of both sides, we need to know:
- $$mathcal{L}_tleft[text{n}_1cdottext{y}'''left(tright)right]_{left(text{s}right)}:=text{n}_1cdotint_0^inftytext{y}'''left(tright)e^{-text{s}t}spacetext{d}t=$$
$$text{n}_1cdotleft(text{s}^3cdottext{Y}left(text{s}right)-text{s}^2cdottext{y}left(0right)-text{s}cdottext{y}'left(0right)-text{y}''left(0right)right)tag3$$ - $$mathcal{L}_tleft[text{n}_2cdottext{y}''left(tright)right]_{left(text{s}right)}:=text{n}_2cdotint_0^inftytext{y}''left(tright)e^{-text{s}t}spacetext{d}t=text{n}_2cdotleft(text{s}^2cdottext{Y}left(text{s}right)-text{s}cdottext{y}left(0right)-text{y}'left(0right)right)tag4$$
- $$mathcal{L}_tleft[text{n}_3cdottext{y}'left(tright)right]_{left(text{s}right)}:=text{n}_3cdotint_0^inftytext{y}'left(tright)e^{-text{s}t}spacetext{d}t=text{n}_3cdotleft(text{s}cdottext{Y}left(text{s}right)-text{y}left(0right)right)tag5$$
- $$mathcal{L}_tleft[text{n}_4cdottext{y}left(tright)right]_{left(text{s}right)}:=text{n}_4cdotint_0^inftytext{y}left(tright)e^{-text{s}t}spacetext{d}t=text{n}_4cdottext{Y}left(text{s}right)tag6$$
- $$mathcal{L}_tleft[text{n}_5cdottext{u}left(tright)right]_{left(text{s}right)}:=text{n}_5cdotint_0^inftytext{u}left(tright)e^{-text{s}t}spacetext{d}t=text{n}_5cdottext{U}left(text{s}right)tag7$$
- $$mathcal{L}_tleft[text{n}_6cdottext{u}'left(tright)right]_{left(text{s}right)}:=text{n}_6cdotint_0^inftytext{u}'left(tright)e^{-text{s}t}spacetext{d}t=text{n}_6cdotleft(text{s}cdottext{U}left(text{s}right)-text{u}left(0right)right)tag8$$
Assuming that a initial conditions are equal to zero, we get:
$$text{n}_1cdottext{s}^3cdottext{Y}left(text{s}right)+text{n}_2cdottext{s}^2cdottext{Y}left(text{s}right)+text{n}_3cdottext{s}cdottext{Y}left(text{s}right)+text{n}_4cdottext{Y}left(text{s}right)=text{n}_5cdottext{U}left(text{s}right)+text{n}_6cdottext{s}cdottext{U}left(text{s}right)tag9$$
Simplifying it a little bit:
$$text{Y}left(text{s}right)cdotleft(text{n}_1cdottext{s}^3+text{n}_2cdottext{s}^2+text{n}_3cdottext{s}+text{n}_4right)=text{U}left(text{s}right)cdotleft(text{n}_5+text{n}_6cdottext{s}right)tag{10}$$
So, we also know that:
$$text{H}left(text{s}right)=frac{text{Y}left(text{s}right)}{text{U}left(text{s}right)}=frac{text{n}_5+text{n}_6cdottext{s}}{text{n}_1cdottext{s}^3+text{n}_2cdottext{s}^2+text{n}_3cdottext{s}+text{n}_4}tag{11}$$
For the respons you're looking for, we can use:
$$text{s}=text{j}omegatag{12}$$
Where $text{j}^2=-1$
So, we will also get:
$$text{H}left(text{j}omegaright)=frac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=frac{text{n}_5+text{n}_6cdotleft(text{j}omegaright)}{text{n}_1cdotleft(text{j}omegaright)^3+text{n}_2cdotleft(text{j}omegaright)^2+text{n}_3cdotleft(text{j}omegaright)+text{n}_4}tag{13}$$
Simplying it a little bit, gives us:
$$text{H}left(text{j}omegaright)=frac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=frac{text{n}_5+text{n}_6cdotomegacdottext{j}}{text{n}_4-text{n}_2cdotomega^2+text{j}cdotomegacdotleft(text{n}_3-text{n}_1cdotomega^2right)}tag{14}$$
For the magnitude we get:
$$left|text{H}left(text{j}omegaright)right|=frac{left|text{Y}left(text{j}omegaright)right|}{left|text{U}left(text{j}omegaright)right|}=frac{sqrt{text{n}_5^2+text{n}_6^2cdotomega^2}}{sqrt{left(text{n}_4-text{n}_2cdotomega^2right)^2+omega^2cdotleft(text{n}_3-text{n}_1cdotomega^2right)^2}}tag{15}$$
And for the phase, you can use:
$$argtext{H}left(text{j}omegaright)=argfrac{text{Y}left(text{j}omegaright)}{text{U}left(text{j}omegaright)}=argleft(text{Y}left(text{j}omegaright)right)-argleft(text{U}left(text{j}omegaright)right)tag{16}$$
Now, when $text{n}_5spacewedgespaceomegainmathbb{R}^+$:
$$argleft(text{Y}left(text{j}omegaright)right)=arctanleft(frac{text{n}_6cdotomega}{text{n}_5}right)tag{17}$$
answered Jan 4 '17 at 17:44
JanJan
21.8k31240
edited Jan 13 '17 at 16:48
answered Jan 4 '17 at 17:44
JanJan
21.8k31240
answered Jan 4 '17 at 17:44
JanJan
21.8k31240
answered Jan 4 '17 at 17:44
JanJan
21.8k31240
21.8k31240
add a comment |
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$begingroup$
You don't need to simplify. You get a Laplace domain function in the variable s that is rational (polynomial on top and bottom). Set $s = j omega$. You now have a complex number for a given choice of $omega$. This number has a magnitude and a phase, which you compute using complex arithmetic. If you are looking to plot this, look into the Bode plot, which is exactly a plot of magnitude and phase
$endgroup$
– Michael Stachowsky
Jan 4 '17 at 16:00
$begingroup$
@Moo I've edited my work in!
$endgroup$
– Fukuyama
Jan 4 '17 at 16:15
1
$begingroup$
@Fukuyama: Your $H(s)$ looks quite okay to me. Now substitute $s = jmath omega$ to get $G(omega)$. Then forming the magnitude is relatively straightforward complex arithmetic.
$endgroup$
– Florian
Jan 4 '17 at 16:21
$begingroup$
@Florian Thank you! And what would I do for the phase if I may ask?
$endgroup$
– Fukuyama
Jan 4 '17 at 16:23
1
$begingroup$
Your $G(omega)$ will be in the form $frac{A+jmath B}{C+jmath D}$. Expanding with $C-jmath D$ gives $frac{(A+jmath B)(C-jmath D)}{C^2+D^2} = frac{AC+BD +jmath (BC-AD)}{C^2+D^2}$. This separates real from imaginary part. The phase of a complex number $x+jmath y$ is basically $tan^{-1}frac{y}{x}$, i.e., in your case something like $tan^{-1}frac{BC-AD}{AC+BD}$. To get the four-quadrant phase, use atan2 if needed.
$endgroup$
– Florian
Jan 4 '17 at 16:31