HW3, EGME 431 (Mechanical Vibration) Date due and handed in April 8,2008

HW3, EGME 431 (Mechanical Vibration)
Date due and handed in April 8,2008

by Nasser M. Abbasi

June 16, 2014

1 Problem 3.2
2 Problem 3.8
3 Problem 3.11
4 Problem 3.16
5 Problem 3.21
6 Problem 3.29
7 Problem 3.38
8 Problem 3.44
9 Solving 3.44 using convolution
10 Problem 3.49
11 Problem 3.50

1 Problem 3.2

Problem

Calculate the solution to $¨x+ 2_x+ 3x= sint+ δ (t− π )$ with IC $x(0)= 0, _x(0)= 1$ and plot the solution.

Answer

Hence $√ -- ωn = 3$ and $2ξ ωn = 2$ , hence $1- ξ = √3$ $= 0.57735$ , hence this is underdamped system.

Since $x = xh+ xp$ , then

xh = e−ξωnt(Acosωdt+ B sin ωdt)

We have 2 particular solutions. The ﬁrst $xp1$ is due to $sint$ and the second $xp2$ is due to $δ(t− π)$ . When the forcing function is $sin t$ , we guess

xp = c1cost+ c2sint 1

and when the forcing function is $δ (t− π )$ the response is

-1--− ξωn(t−π) xp2 = ωdm e sinωd (t− π )Φ (t− π )

From $xp1$ we ﬁnd $_xp1$ and $¨xp1$ and plug these into $¨x+ 2_x+ 3x = sint$ to ﬁnd $c1$ and $c2$ , next we ﬁnd $A,B$ by using the IC, and then at the end we add the solution $xp 2$ . Notice that $xp 2$ do not enter into the calculation of $A,B$ since the impulse $δ (t− π)$ is not eﬀective at $t = 0$ .

Hence

Hence $(− 2c1+ 2c2) = 1$ and $(2c2+ 2c1)= 0$ . This results in

Hence

|----------------------| | | | xp1 = − 14cost+ 14sint | -----------------------

Therefore

xh+ xp1 = e−ξωnt(Acosωdt+ B sinωdt)− 1cost+ 1-sint 4 4

Now we use IC's to ﬁnd $A, B$ . At $t = 0$ we obtain

|--------| | | | A = 14 | ---------

And

At $t = 0$ we have

But $∘ ------ √ -∘ ----(--)2 √-∘ -- ωd = ωn 1− ξ2 = 3 1 − √13 = 3 23$ , Hence $√ -- ωd = 2$ then the above becomes

|---------| | √1- | | B = 2 | -----------

Hence the ﬁnal solution is

Substitute values for the parameters above we obtain

|----------------------------------------------------------------------------| | ( √ -- √ -) √ -- | | x(t) = e−t 1cos 2t+ 1√-sin 2t − 1cost+ 1sint+ √1e−(t−π)sin 2(t− π) Φ(t− π) | | 4 2 4 4 2 | ------------------------------------------------------------------------------

This is a plot of the solution superimposed on the forcing functions

2 Problem 3.8

The magnitude of the impulse resulting when the mass hits the ground is given by the change of momentum that occurs. Hence

^F = Ft = m (v − v) final 0

But assuming the mass is dropped from rest, hence $v0 = 0$ , and $vfinal = gt$ where $∘ -h- t = 2g$ where $h$ is the height that mass falls. Hence

Hence the equation of motion is

|----------------------------------| | mx¨(t)+ c_x(t)+ kx(t) = m√2gh-δ (t) | -----------------------------------|

Since underdamped, $x(t)= h(t)= -F^e−ξωntsinωdt mωd$ , hence the solution is

Taking $t = 0$ as time of impact.

3 Problem 3.11

Problem

Compute response of the system $3¨x(t)+ 6_x(t)+ 12x (t)= 3δ (t) − δ(t− 1)$ with IC $x(0)= 0.01m$ and $v(0) = 1m∕s$ . Plot the response.

Answer

Where $m = 1,ω2 = 4 n$ , hence $ω = 2 n$ and $2ξω = 2 n$ , hence $ξ = 1 2$ . This is an underdamped system.

$∘ ------ ∘ ---(-)2- ∘ -- ωd = ωn 1− ξ2 = 2 1− 12 = 2 34$ , Hence $√ -- ωd = 3$

xh = e−ξωnt(Acosωdt+ B sin ωdt)

The response due to the forcing function $δ (t)$ is given by

xp (t) = -1--e−ξωntsin(ωdt) 1 ωdm

The response due to the other forcing function $δ(t− 1)$ is given by

1 1 xp2(t)= − ------e−ξωn(t−1)sinωd (t− 1 )Φ(t− 1) 3 ωdm

Now we determine $A,B$ from IC's

Hence $A = 0.01$ Now to ﬁnd $B$

But $x_(0)+ x_ (0)= 1 h p1$ , hence from the above, and noting that $m = 1$

Hence

|-----------| | | | B = 1010√3- | -------------

Therefore

|------------------------------------| | ( ) | | xh = e−ξωnt 1100cosωdt + 1010√3sin ωdt | -------------------------------------|

Now we can combine the above solution to obtain the ﬁnal solution

Substitute numerical values for the above parameters, we obtain

|------------------------------------------------------------------------------| | −t( √-- √ --) (√ -) (√ -- ) | | x(t) = e100- cos 3t+ 1√3sin 3t + 1√3e−tsin 3t − 13 1√3e−(t−1)sin 3(t− 1) Φ (t− 1) | --------------------------------------------------------------------------------

This is a plot of the response

4 Problem 3.16

Let the response by $x(t)$ . Hence $x(t)= xh(t)+ xp(t)$ , where $xp (t)$ is the particular solution, which is the response due the the above forcing function. Using convolution

∫t xp(t)= f(τ)h (t− τ)d τ 0

Where $h (t)$ is the unit impulse response of a second order underdamped system which is

h(t)= --1-e−ξωntsinω t m ωd d

hence

Using $1 sin Asin B= 2[cos (A − B)− cos(A +B )]$ then

1 sin(τ)sin(ωd(t− τ))= 2-[cos(τ − ωd(t− τ))− cos(τ+ ωd(t− τ))]

Then the integral becomes

( ) Fe−ξωnt ∫t ∫t xp (t)= -0-----( eξωnτcos(τ− ωd (t− τ))dτ − eξωnτcos(τ+ ωd (t− τ))dτ) 2mωd 0 0

Consider the ﬁrst integral $I1$ where

∫t I1 = eξωnτcos(τ− ωd (t− τ)) dτ 0

Integrate by parts, where $∫ ∫ udv = uv− vdu$ , Let $ξωτ dv = eξωnτ → v= eξωnn-$ and let $u = cos(τ− ωd (t− τ))→ du = − (1 + ωd)sin (τ − ωd(t− τ))$ , hence

Integrate by parts again the last integral above, where $∫ ∫ udv= uv− vdu$ , Let $ξωnτ eξωnτ dv= e → v= ξωn$ and let $u = sin(τ − ωd(t− τ))→ du = (1+ ωd )cos(τ − ωd(t− τ))$ , hence

Substitute (2) into (1) we obtain

Hence

Now consider the second integral $I2$ where

∫t I2 = eξωnτcos(τ+ ωd (t− τ)) dτ 0

Integrate by parts, where $∫ udv = uv− ∫ vdu$ , Let $dv = eξωnτ → v= eξωnτ- ξωn$ and let $u = cos(τ+ ωd (t− τ))→ du = − (1 − ωd)sin (τ + ωd(t− τ))$ , hence

Integrate by parts again the last integral above, where $∫ udv= uv− ∫ vdu$ , Let $dv= eξωnτ → v= eξωnτ ξωn$ and let $u = sin(τ + ωd(t− τ))→ du = (1− ωd )cos(τ + ωd(t− τ))$ , hence

Substitute (4) into (3) we obtain

Hence

Using the above expressions for $I1,I2$ , we ﬁnd (and multiplying the solution by $(Φ(t)− Φ (t− π ))$ since the force is only active from $t = 0$ to $t = π$ , we obtain

Hence $xp(t)= (Φ (t)− Φ (t− π))$

$[ ( ) ] -F0-e−ξωnt ξωn[cos(t)eξωnt−cos(ωdt)]+(1+ωd)[sin(t)eξωnt+sin(ωdt)]-− ξωn[cos(t)eξωnt−cos(ωdt)]+(1−ωd)[sin(t 2mωd (ξωn)2+(1+ωd)2 (ξωn)2+(1−ωd)2$

But

and

(ξ ωn)2+ (1 − ωd)2 = 1− 2ωd + ω2n

Hence $xp(t)$ can now be written as

And

−ξωnt xh(t)= e (Acosωdt + Bsinωdt)

Hence the overall solution is

−ξωnt x (t)= e (A cosωdt+ B sinωdt)+ xp(t)

The above solution is a bit long due to integration by parts. I will not solve the same problem using Laplace transformation method. The diﬀerential equation is

x¨(t)+ 2ξωn _x(t)+ ω2nx(t) = f(t)

Take Laplace transform, we obtain (assuming $x(0)= x0$ and $_x(0)= v0$ )

Now we ﬁnd Laplace transform of $f(t)$

Integration by parts gives

[ −πs] F (s)= F0 1-+-e-- 1 + s2

(8)

Substitute (8) into (7) we obtain

Hence

Now we can use inverse Laplace transform on the above. It is easier to do partial fraction decomposition and use tables. I used CAS to do this and this is the result. I plot the solution $x(t)$ . I used the following values to be able to obtain a plot $ξ = 0.5,ωn = 2,F0 = 10,x0 = 1,v0 = 0$

5 Problem 3.21

The acceleration $¨x$ of the mass is measured w.r.t. to the inertial frame, but the spring length is measured relative to the ground which is moving with displacement $y(t)$ , hence the equation of motion of the mass $m$ is given by

mx¨(t)+ k(x(t)− y(y)) = 0

Therefore

m¨x(t)+ k x(t)= k y(t)

(1)

Where $y(t)$ is given as

( |||| 2.5t 0≤ t ≤ 0.2 ||| { y(t) = || 0.75 − 1.25t 0.2 < t ≤ 0.6 ||| ||( 0 0.6< t

The solution to (1) is given by $x(t) = x (t)+ x (t) h p$ where $x (t) p$ can be found using convolution, and $x (t) h$ is as usual given by

xh = Acosωn + Bsinωn

Let us ﬁrst ﬁnd $xp(t)$ . Note that the impulse response $h(t)$ to undamped system is given by

1 h (t)= mω--sin ωnt n

Hence for $0 ≤ t ≤ 0.2$ ,

Integration by parts, $∫ ∫ udv = uv− vdu$ where $u= τ$ , $dv= sinωn(t− τ)$ , hence $v= − cos(ω−nω(tn−τ))$ , therefore (2) becomes

For $0.2< t ≤ 0.6$

For the ﬁrst integral in (3), we obtain

For the second integral in (3) we obtain

For the third integral in (3) we obtain

Integration by parts gives

Hence

For $t > 0.6$

For the ﬁrst integral in (4), we obtain

2.5 2.5 I1 = 0.5cosωn(t− 0.2)+---sinωn(t− 0.2)− ---sinωnt ωn ωn

For the second integral in (4) we obtain

For the third integral in (4) we obtain

Integration by parts gives

Hence

Hence, the overall response is, assuming zero initial conditions, is given by

( ||| ( sinωnt) |||| 2.5 t− ωn 0≤ t ≤ 0.2 { x (t)= | 0.75 − 1.25t+ 3.ω75n sinωn(t− 0.2)− 2ω.5n sinωnt 0.2< t ≤ 0.6 |||| ||( 3.75sinωn(t− 0.2)− 2.5sinωnt− 1.25sin ωn(t− 0.6) t > 0.6 ωn ωn ωn

Noting that $∘ -- ∘ ---- ωn = km = 15500000 = 0.54772,$ the above becomes

( || ||| 2.5t− 4.5644 sinωnt 0≤ t ≤ 0.2 ||{ x(t)= 0.75− 1.25t+ 6.8466sin ω (t− 0.2) − 4.5644 sinω t 0.2< t ≤ 0.6 ||| n n |||| ( 6.8466sinωn (t− 0.2)− 4.5644sin ωnt− 2.2822 sinωn (t− 0.6) t > 0.6

This is a plot of the solution superimposed on top of the forcing function

6 Problem 3.29

Let $f(t)$ be the function shown above. Let $~ f(t)$ be its approximation using Fourier series. Hence

a0 ∞ ( 2π ) ( 2π ) f~(t)= --+ ∑ ancos ---nt + bnsin --nt 2 n=1 T T

Where $T$ is the period of $f(t)$ and

For $f (t)$ we see that $T = 2 π$ and $f (t) = tT$ for $0 ≤ t ≤ T$ , hence

And

Hence

These are few terms in the series

|----------------------------------------| | f~(t)= 1− 1sint− 1-sin 2t− 1-sin3t− ⋅⋅⋅ | | 2 π 2π 3π | -----------------------------------------

This is a plot of the above for increasing number of $n$

7 Problem 3.38

Problem

Solve the following system using Laplace transform $100¨x(t)+ 2000x(t) = 50δ(t)$ where the units are in Newtons and the initial conditions are both zero.

Answer

Divide the equation by $50$ we obtain

2¨x(t)+ 40x(t)= δ(t)

Let $m = 2,k = 40,$ hence the equation becomes

m ¨x(t)+ kx(t)= δ(t)

Applying Laplace transform

m (s2X (s)− sx − v )+ kX (s) = 1 0 0

But due to zero initial conditions, the above simpliﬁes to

From tables, the inverse Laplace transform of $-α2-2 s+α$ is $sin αt$ , but

( ∘ --) 1 k ---1---= ---m--= 1-∘1--( ---m-) ms2 +k s2+ mk m k s2 + km m

Hence, letting $∘ -k α = m$ we see that inverse laplace transform of $--1- ms2+k$ is the same as the inverse laplace transform of $( ) 1m 1α s2α+α2$ which is $1m 1α sin αt$

But $α = ω n$ , hence

|------------------| | | | x(t)= m1ωn sinωnt | -------------------

8 Problem 3.44

Problem

Calculate the response spectrum of an undamped system to the forcing function

$( |||{ πt F0sin t1 0 ≤ t ≤ t1 F (t)= || |( 0 t > t1$ assuming zero initial conditions.

Answer

Solution sketch: Find the response $x(t)$ of the system to the above input. Then ﬁnd $t$ where this response is maximum, call this $xmax$ , then plot $( ) xmax kF 0$ vs. $tω2nπ-$

The system EQM is

′′ 2 F-(t) x (t)+ ω nx(t) = m

For $0< t ≤ t1$ ,

Guess $xp(t) = c1cos ωt+ c2sin ωt$ , hence $x′p(t)= − ωc1sin ωt+ ωc2 cosωt$ and $x′′p(t)= − ω2c1cosωt − ω2c2sinωt$ , hence substitute these into the EQM and compare, we obtain

(− ω2c1cos ωt− ω2c2sinωt) +ω2 (c1cosωt+ c2sinωt)= F0sin πt n m t1

The input is half sin where $ωt = πt1t$ , hence $ω = πt1-$ $,$ hence the above becomes

( ) ( ) F − ω2c1+ ω2nc1 cosωt + − ω2c2 + ω2nc2 sinωt = -0sin ωt m

Hence $c1 = 0$ and $c2(− ω2 + ω2) = F0 n m$ or $c2 =--F0m-- ω2n−ω2$ $,$ Then the solution becomes

Fm0 x1(t)= A cos ωnt+ Bsinωnt+ ω2-−-ω2 sinωt n

And since $x(0)= 0$ then $A = 0$ and take derivative we obtain

′ Fm0 x1(t) = ωnBcosωnt +ω ω2-−-ω2 cosωt n

And since $′ x(0)= 0$ then the above results in

Hence the solution becomes

Hence

F0 ( ω ) x1(t)= [----k(--)2] sin ωt− ---sin ωnt 0 < t ≤ t1 1 − ωω- ωn n

(1)

Now we need to ﬁnd where the maximum is. Take derivative, and set it to zero, we obtain

x′(t) = F0[----1----](ω cosωt − ω cos ω t) = 0 1 k ( ω-)2 n 1 − ωn

For $ω ⁄= ωn$ , we need to solve

cosωt− cosω t = 0 n

Using $cosA − cosB= − 2sin(A+B)sin(A−B) 2 2$ , then the above becomes

Hence, either $(ω+ωn)tp= nπ 2$ or $(ω−ωn)tp= nπ 2$ for $n = ±1,±2, ⋅⋅⋅$ or the time $t p$ which makes the maximum $x(t)$ is one of the following

(| ||{ -2nπ tp = ω+ωn n= ±1, ±2,⋅⋅⋅ ||| -2nπ ( ω−ωn

We now need to ﬁnd which one of the above 2 solution gives a larger maximum. Using the ﬁrst solution $-2nπ- tp = ω+ωn$ , then (1) becomes

And at $tp = 2ωn−πω, n$ then (1) becomes

Need now to ﬁnd which of the above is larger. Let us take the diﬀerence and see if the result is positive or negative (is there an easier way?)

Not sure how to continue. Now let us look at $t > t1.$ The solution here is

x2(t)= A cosωnt+ B sinωnt

But with IC given by $x1(t1)$ and $x′1(t1),$ hence from (1)

F ( ) -----0k----- -ω- x1(t1)= [ ( ω)2] sinωt1− ωn sin ωnt1 1− ωn

and

′ F0 -----1----- x1(t1)= k [ ( )2] (ω cosωt1− ω cosωnt1) 1− ωωn

Hence

( ) x2(t1)= A cosωnt1+ B sin ωnt1 = [--F0(∕k-)-] sinωt1 − ω-sinωnt1 1− ω- 2 ωn ωn

(3)

And

x′2(t1) = − ωnA sin ωnt+ Bωn cos ωnt = [-F0(∕k)-](ω cosωt1− ω cosωnt1) 1 − ω- 2 ωn

(4)

We need to solve (3) and (4) for $A$ and $B.$ Combining (3) and (4) we obtain

⌊ ⌋⌊ ⌋ ⌊ ( ) ⌋ | [-F0∕k-2] sin ωt1− ω-sin ωnt1 | || cosωnt1 sinωnt1 |||| A|| || 1−(ωωn) ωn || |⌈ |⌉|⌈ |⌉ = | | − ωn sinωnt ωn cosωnt B ⌈[--F0∕k2](ω cosωt1− ω cosωnt1)⌉ 1−( ωωn)

This is in the form $Ax = b$ , solve for $x,$ we obtain

⌊ ⌋ ⌊ ⌋⌊ ⌋( ) ( ω ) ||A|| 1 || ωncosωnt1 − sinωnt1|||| sinωt1− ωn sinωnt1 ||| F0∕k | |⌈ |⌉ = ω-|⌈ |⌉|⌈ |⌉|( [----(--)2]|) B n ωn sinωnt cosωnt (ω cosωt1− ω cosωnt1) 1 − ωωn

Hence

And

Ask about the above, why can't I get the answer shown in notes?

9 Solving 3.44 using convolution

To ﬁnd the response $x(t)$ use convolution. Since this is an undamped system, then the impulse response is

1 h (t)= ----sin ωnt mωn

Hence, for $0 ≤ t ≤ t1$

Using $1 sin Asin B= 2(cos(A− B)− cos(A + B))$ , then

A ◜◞π◟τ◝ ◜--B◞◟--◝ 1 ( πτ ) 1 ( πτ ) sin ---sin ωn(t− τ)= -cos ---− ωn(t− τ) − -cos ---+ ωn(t− τ) t1 2 t1 2 t1

Hence the convolution integral becomes

And for $t > t1$

As was done earlier, perform integration by parts, we obtain

But $sin(π − α) = sinα$ and $sin(π + α)= − sinα$ , hence the above becomes

{ } -F0-- sin-ωn(t−-t1) sin(ωnt)- sinωn-(t−-t1) sin-(ωnt)- x(t) = 2mωn πt+ ωn + πt+ ωn + πt-− ωn + πt− ωn 1 1 1 1

Therefore, the ﬁnal solution is

( { } |||{ -F0-- -sinπtt1- sinωnt- sinπt1t- sinωnt 2mωn πt1+ωn + πt1+ωn − πt1−ωn + tπ1−ωn 0 ≤ t ≤ t1 x (t)= || { } |( F2m0ωn- sinωnπ(+t−ωt1n)+ sinπ(ω+nωt)n + sinωπn−(tω−nt1)+ sinπ(ω−nωt)n t > t1 t1 t1 t1 t1

(1)

We can simplify the above more as follows

( || { ( ) ( )} |{ 2Fm0ω- sinπtt -π1+ω-− π−1ω-- + sinωnt π+1ω--+ π1−ω-- 0 ≤ t ≤ t1 x(t)= n 1 t1 n t1 n t1 n t1 n ||| -F0-{ ( -1--- --1--) (--1-- -1--)} ( 2mωn sinωn(t− t1) πt1+ωn + πt1−ωn + sin(ωnt) πt1+ ωn + πt1−ωn t > t1

(1)

Hence

( | { (( π−ω)−(π+ω )) ((π −ω)+(π+ω ))} ||{ 2mF0ωn- sin πtt -(t1π--n)(t1π--n)- + sinωnt -t(1π-n-)(-t1π--n)- 0≤ t ≤ t1 x(t) = { 1 (t1+(ωn t1)−ω(n )) t1+(ω(n t1−)ωn( ))} ||| -F0-- -πt1−ωn-+-πt1+ωn- -tπ1−ωn+-πt1+ωn- ( 2mωn sin ωn(t− t1) (tπ+ωn)(πt−ωn) + sin(ωnt) (πt+ωn)(πt−ωn) t > t1 1 1 1 1

(1)

Hence

( { ( ) ( ) } || F πt −2ω 2tπ ||{ 2m0ωn sin t1- (π)2n−ω2 + sinωnt (-π)21−ω2 0 ≤ t ≤ t1 x(t)= { t(1 n ) t1( n ) } ||| -F0- --2πt1--- ---2πt1-- |( 2mωn sinωn (t− t1) (πt)2−ω2n + sin(ωnt) ( πt)2−ω2n t > t1 1 1

(1)

( | ( ) { } |||{ (π1)2--- 2Fm0ωn − 2ωn sinπtt+ 2πt-sin ωnt 0≤ t ≤ t1 x(t)= ( t1 −ω2n ) 1 1 || ||( (π)ω2--2 Fm0ωn {sin ωn(t− t1)− sin (ωnt)} t > t1 t1 −ωn

(1)

Hence

( ( ) ||| ---1--- F0-{ πt π- } |{ (πt)2−ω2n mωn − ωnsint1 + t1 sinωnt 0 ≤ t ≤ t1 x(t)= ( 1 ) |||| (-)ω--- F0-{sinω (t− t)− sin ω t} t > t ( πt1 2−ω2n mωn n 1 n 1

(1)

To ﬁnd where $xmax$ is, we need to ﬁnd $xmax$ . Take the derivative, we obtain

$( ( ) || 1 F { π πt π } ||{ (π)2−ω2 m0ωn − ωnt1 cost1-+ t1ωncos ωnt 0 ≤ t ≤ t1 _x(t) = ( t1 n) ||| ---ω--- -F0- |( (πt )2−ω2n mωn {ωn cos ωn(t− t1)− ωncosωnt} t > t1 1$

Now let $x_(t)= 0$ for $t > t1$ to ﬁnd $tpeak$ .

But

cosωn(tp− t1)= cos ωntpcosωnt1+ sinωntpsinωnt1

Substitute the above into (2) we obtain

(cosωntpcosωnt1+ sinωntpsinωnt1)− cosωntp = 0

Divide by $cosωntp$

Hence, the hypotenuse is $∘ -----------2-----2---- ∘ ------------- (1− cosωnt1)+ sin ωnt1 = 2(1− cosωnt1)$ and so $∘ 1------------ sin ωntp = − 2(1− cosωnt1)$ and $cosωntp = √-− sinωnt1- 2(1−cosωnt1)$ and using these into (1) we ﬁnd $xmax$ when $t > t1$ as

( ) xmax = |( (-)ω-----|) -F0-{sinωn(tp− t1)− sinωntp} π-2− ω2 m ωn t1 n

But $sin ωn(tp − t1)= sinωntpcosωnt1− cosωntpsinωnt1$ , hence

Hence

|-----------------------------| | x k-= (--r) 1√1-−-cosω-t- | | maxF0 r2−1 2 n 1 | ------------------------------

Where $ω- r = ωn$

A plot of $xmax kF0$ vs. $r$ gives the response spectrum

10 Problem 3.49

Problem Calculate the compliance transfer function for a system described by $ax′′′′+ bx′′′+ cx′′+ dx′+ ex = f(t)$ where $f (t)$ is the input and $x (t)$ is the displacement.

Answer

Take Laplace transform (assuming zero IC) we obtain

as4X(s)+ bs3X(s)+ cs2X (s)+ dsX(s)+ eX (s) = F (s)

Hence

[ ] X (s) as4+ bs3+ cs2+ ds+ e = F (s)

Hence

|---------------------------| | X(s) | | H (s)= F(s) = as4+bs31+cs2+ds+e | -----------------------------

11 Problem 3.50

Problem

Calculate the frequency response function for the system of problem 3.49 for $a = 1,b= 4,c = 11,d = 16,e = 8$

Answer

Let $s = jω$

Hence

and

( ) −1 --16ω-−-4ω3--- Phase(H (jω))= − tan ω4− 11ω2 + 11

This is a plot of the magnitude and phase

EDU>> num=1;
EDU>> den=[1 4 11 16 11]
EDU>> sys=tf(num,den)

Transfer function:
1
--------------------------------
s^4 + 4 s^3 + 11 s^2 + 16 s + 11

EDU>> w = logspace(-1,1);
EDU>> freqs(num,den,w)