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The strandbeest is a walking machine developed by Theo Jansen. Its cleverly designed legs consist of single-degree-of-freedom linkage mechanisms, actuated by the turning of a wind-powered crankshaft.

His working models are generally large - something of the order of the size of a bus. Look for videos on YouTube.  Commercially made small toy models are also available.  This one sells for under $10 and it's fun to assemble and works quite well. Beware that the kit consists of over 100 tiny pieces - so assembling it is not for the impatient type.

Here is a Maple worksheet that produces an animated strandbeest. Link lengths are taken from Theo Jansen's video (go to his site above and click on Explains) where he explains that he calculated the optimal link lengths by applying a genetic algorithm.

Here is a Maple animation of a single leg.  The yellow disk represents the crankshaft.

And here are two legs working in tandem:

Here is the complete beest, running on six legs. The crankshaft turns at a constant angular velocity.

The toy model noted above runs on twelve legs for greater stability.

Download the worksheet: strandbeest.mw

 

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This may be of interest to anyone curous about why the effective area of an isotropic antenna is λ^2/4π.


 

Friis Transmission Equation

NULL

Initialise

   

NULL``

The Hertzian Dipole antenna

 

The Hertzian Dipole is a conceptual antenna that carries a constant current along its length.

 

 

By laying a number of these small current elements end to end, it is possible to model a physical antenna (such as a half-wave dipole for example).  But since we are only interested in obtaining an expression for the effective area of an Isotropic Antenna (in order to derive The Friis Transmission Equation) the Hertzian Dipole will be sufficient for our needs.``

NULL

``

Maxwell's Equations

 

Since the purpose of a radio antenna is to either launch or to receive radio waves, we know that both the antenna, and the space surrounding the antenna, must satisfy Maxwell's Equations. We define Maxwell's Equations in terms of vector functions using spherical coordinates:

 

Maxwell–Faraday equation:

Maxwell_1 := Curl(E_(r, theta, `ϕ`, t)) = -mu*(diff(H_(r, theta, `ϕ`, t), t))

Physics:-Vectors:-Curl(E_(r, theta, varphi, t)) = -mu*(diff(H_(r, theta, varphi, t), t))

(3.1)

Ampère's circuital law (with Maxwell's addition):

Maxwell_2 := Curl(H_(r, theta, `ϕ`, t)) = J_(r, theta, `ϕ`, t)+epsilon*(diff(E_(r, theta, `ϕ`, t), t))

Physics:-Vectors:-Curl(H_(r, theta, varphi, t)) = J_(r, theta, varphi, t)+varepsilon*(diff(E_(r, theta, varphi, t), t))

(3.2)

Gauss' Law:

Maxwell_3 := Divergence(E_(r, theta, `ϕ`, t)) = rho/epsilon

Physics:-Vectors:-Divergence(E_(r, theta, varphi, t)) = rho/varepsilon

(3.3)

Gauss' Law for Magnetism:

Maxwell_4 := Divergence(H_(r, theta, `ϕ`, t)) = 0

Physics:-Vectors:-Divergence(H_(r, theta, varphi, t)) = 0

(3.4)

Where:

        E is the electric field strength [Volts/m]

        H is the magnetic field strength [Amperes/m]

        J is the current density (current per unit area) [Amperes/m2]

        ρ is the charge density (charge per unit volume) [Coulombs/m3]

        ε is Electric Permittivity

        μ is Magnetic Permeability

NULL

Helmholtz decomposition

 

The Helmholtz Decomposition Theorem states that providing a vector field, (F) satisfies appropriate smoothness and decay conditions, it can be decomposed as the sum of components derived from a scalar field, (Φ) called the "scalar potential", and a vector field (A) called the "vector potential".

 

F = -VΦ + V×A

 

And that the scalar (Φ) and vector (A) potentials can be calculated from the field (F) as follows (image from https://en.wikipedia.org/wiki/Helmholtz_decomposition):

 

Where:

        r is the vector from the origin to the observation point (P) at which we wish to know the scalar or vector potential.

        r' is the vector from the origin to the source of the scalar or vector potential (i.e. a point on the Hertzian Dipole antenna).

        V'·F(r')  is the Divergence of the vector field (F) at source position r'.

        V'×F(r')  is the Curl of the vector field (F) at source position r'.

 

 

Calculating the Scalar Potential for the magnetic Field, H

 

We know that the Divergence of the magnetic field (H) is zero:

Maxwell_4

Physics:-Vectors:-Divergence(H_(r, theta, varphi, t)) = 0

(4.1.1)

And so the magnetic field (H) must have a scalar vector potential of zero:

`Φ__H` := 0

0

(4.1.2)

 

Calculating the Vector Potential for the magnetic Field, H

 

We know that the Curl of the magnetic field (H) is equal to the sum of current density (J) and the rate of change of the electric filled (E):

Maxwell_2

Physics:-Vectors:-Curl(H_(r, theta, varphi, t)) = J_(r, theta, varphi, t)+varepsilon*(diff(E_(r, theta, varphi, t), t))

(4.2.1)

Since the Hertzian Dipole is a conductor, we need only concern ourselves with the current density (J) when calculating the vector potential (A). Integrating current density (J) over the volume of the antenna, is equivalent to integrating current along the length of the antenna (L).

 

We know that Maxwell's Equations can be solved for single frequency (monochromatic) fields, so we will excite our antenna with a single frequency current:

"`I__antenna`(t):=`I__0`*(e)^(j*omega*t);"

proc (t) options operator, arrow, function_assign; Physics:-`*`(I__0, exp(Physics:-`*`(I, omega, t))) end proc

(4.2.2)

We can simplify the integral for the vector potential (A) by recognising that:

 

1. 

Our observation point (P) will be a long way from the antenna and so (r) will be very large.

2. 

The length of the antenna (L) will be very small and so (r') will be very small.

 

Since |r|>>|r'|, we can substitute |r-r'| with r.

 

Because we have decided that the observation point at r will be a long way from the antenna, we must allow for the fact that the observed antenna current will be delayed.  The delay will be equal to the distance from the antenna to the observation point |r-r'| (which we have simplified to r), divided by the speed of light (c).  The time delay will therefore be approximately equal to r/c and so the observed antenna current becomes:

"`I__observed`(t):=`I__0`*(e)^(j*omega*(t-r/(c)));"

proc (t) options operator, arrow, function_assign; Physics:-`*`(I__0, exp(Physics:-`*`(I, omega, Physics:-Vectors:-`+`(t, -Physics:-`*`(r, Physics:-`^`(c, -1)))))) end proc

(4.2.3)

 

Since the length, L of the antenna will be very small, we can assume that the current is in phase at all points along its length.  Working in the Cartesian coordinate system, the final integral for the vector potential for the magnetic field is therefore:

A__H_ := (int(I__0*exp(I*omega*(t-r/c))*_k/r, z = -(1/2)*L .. (1/2)*L))/(4*Pi)

(1/4)*I__0*exp(I*omega*(t-r/c))*_k*L/(Pi*r)

(4.2.4)

 

We will now convert to the spherical coordinate system, which is more convenient when working with radio antenna radiation patterns:

The radial component of the observed current (and therefore vector potential), will be at a maximum when the observer is on the z-axis (that is when θ=0 or θ=π) and will be zero when the observer is in the x-y-plane:

A__Hr := (A__H_._k)*cos(theta)

(1/4)*I__0*exp(I*omega*t-I*omega*r/c)*L*cos(theta)/(Pi*r)

(4.2.5)

The angular component of the observed current (and therefore vector potential), in the θ direction will be zero when the observer is on the z-axis (that is when θ=0 or θ=π) and will be at a maximum when the observer is in the x-y-plane:

`A__Hθ` := -(A__H_._k)*sin(theta)

-(1/4)*I__0*exp(I*omega*t-I*omega*r/c)*L*sin(theta)/(Pi*r)

(4.2.6)

Since the observed current (and therefore vector potential) flows along the z-axis, there will be no variation in the ϕ direction.  That is to say, that varying ϕ will have no impact on the observed vector potential.

`A__Hϕ` := 0

0

(4.2.7)

And so the vector potential for the magnetic field (H) expressed using spherical coordinate system is:

A__H_ := A__Hr*_r+_theta*`A__Hθ`+`A__Hϕ`*`_ϕ`

(1/4)*I__0*exp(I*omega*t-I*omega*r/c)*L*cos(theta)*_r/(Pi*r)-(1/4)*I__0*exp(I*omega*t-I*omega*r/c)*L*sin(theta)*_theta/(Pi*r)

(4.2.8)

NULLNULL

Calculating the Magnetic Field components

 

The Helmholtz Decomposition Theorem states that providing a vector field (F) satisfies appropriate smoothness and decay conditions, it can be decomposed as the sum of components derived from a scalar field (Φ) called "scalar potential", and a vector field (A) called the vector potential.

 

F = -VΦ + V×A

NULL

And so the magnetic field, H will be:

NULL

H_ = -(Gradient(`Φ__H`))+Curl(A__H_)

H_ = ((1/4)*I)*I__0*L*exp(I*omega*t-I*omega*r/c)*sin(theta)*omega*_phi/(c*Pi*r)+(1/4)*I__0*L*exp(I*omega*t-I*omega*r/c)*sin(theta)*_phi/(Pi*r^2)

(4.3.1)

We see that the magnetic field comprises two components, one is inversely proportional to the distance from the antenna (r) and the other falls off with r2.  Since we are interested in the far-field radiation pattern for the antenna, we can ignore the r2 component and so the expression for the magnetic field reduces to:

H_ := I*omega*I__0*L*exp(I*omega*(t-r/c))*sin(theta)*_phi/(4*Pi*c*r)

((1/4)*I)*omega*I__0*L*exp(I*omega*(t-r/c))*sin(theta)*_phi/(Pi*c*r)

(4.3.2)

We can further simplify by substituting ω/c for 2π/λ:``

H_ := (I*2)*Pi*L*I__0*exp(I*omega*t-(I*2)*Pi*r/lambda)*sin(theta)*_phi/(4*Pi*lambda*r)

((1/2)*I)*L*I__0*exp(I*omega*t-(2*I)*Pi*r/lambda)*sin(theta)*_phi/(lambda*r)

(4.3.3)

````

````

 

Calculating the Poynting Vector

 

We know that the magnitude of the Poynting Vector (S) can be calculated as the cross product of the electric field vector (E) and the magnetic field vector (H) :

        S = -E x H which is analogous to a resistive circuit where power is the product of voltage and current: P=V*I.

We also know that the impedance of free space (Z) can be calculated as the ratio of the electric field (E) and magnetic field (H) vectors: Z = E /H = "sqrt((mu)/(`ε`))."

This is analogous to a resistive circuit where resistance is the ratio of voltage and current: R=V/I.

 

This provides two more methods for calculating the Poynting Vector (S):

        S = -E·E/Z which is analogous to a resistive circuit where power, P=V2/R, and:

        S = -H·H*Z which is analogous to a resistive circuit where power, P=I2R.

 

Since we have obtained an expression for the magnetic field vector (H), we can derive an expression for the Poynting Vector (S):

S_ = -(H_.H_)*Z*_r

S_ = (1/4)*L^2*I__0^2*(exp(I*omega*t-(2*I)*Pi*r/lambda))^2*sin(theta)^2*Z*_r/(lambda^2*r^2)

(5.1)

We can separate out the time variable part to yield:

S_ := S__0*(exp(I*omega*t-(I*2)*Pi*r/lambda))^2*_r

S__0*(exp(I*omega*t-(2*I)*Pi*r/lambda))^2*_r

(5.2)

Where:

S__0 := L^2*I__0^2*sin(theta)^2*Z/(4*lambda^2*r^2)

(1/4)*L^2*I__0^2*sin(theta)^2*Z/(lambda^2*r^2)

(5.3)

And we can visualise this radiation pattern using Maple's plotting tools:

AntennaAxis := arrow(`<,>`(0, 0, -1), `<,>`(0, 0, 1), difference, color = "LightSteelBlue")``

AntennaPattern := plot3d(sin(theta)^2, phi = 0 .. 2*Pi, theta = 0 .. Pi, coords = spherical, scaling = constrained, size = [800, 800], labels = [x, y, z], title = ["The Electromagnetic radiation pattern of a Hertzian Dipole\n\nThe blue arrow represents the axis of the antenna", font = [Times, bold, 20]])
``

display(AntennaAxis, AntennaPattern, scaling = constrained, axes = frame)

 

So the Hertzian Dipole produces a electromagnetic radiation pattern with a pleasing doughnut shape :-)``

NULL``

Calculating Antenna Gain

 

We can calculate the total power radiated by the Hertzian Dipole by integrating the power flux density over all solid angles dΩ=sin(θ) dθ dφ.  Since we have expressed power flux density in terms of watts per square meter, we multiply the solid angle by r2 to convert the solid angle expressed in steradians into an area expressed in m2.

NULL

P__tx := int(int(S__0*r^2*sin(theta), theta = 0 .. Pi), phi = 0 .. 2*Pi)

(2/3)*L^2*I__0^2*Z*Pi/lambda^2

(6.1)

We can now use this power to calculate the power flux density that would be produced by an isotropic antenna by dividing the total transmitted power by the area of a sphere with radius r:

S__Isotropic := P__tx/(4*Pi*r^2)

(1/6)*L^2*I__0^2*Z/(lambda^2*r^2)

(6.2)

``

``

The Gain of the Hertzian Dipole is defined as the ratio between the maximum power flux density produced by the Hertzian Dipole and the maximum power flux density produced by the isotropic antenna:

G__HertzianDipole := S__0/S__Isotropic

(3/2)*sin(theta)^2

(6.3)

AntennaAxis := arrow(`<,>`(-1, 0), `<,>`(1, 0), difference, color = "LightSteelBlue")

Gain := polarplot(G__HertzianDipole, theta = -Pi .. Pi, axis[radial] = [color = "Blue"], angularorigin = top, angulardirection = clockwise, size = [800, 800], labels = [x, z], title = ["The Gain of a Hertzian Dipole over an isotropic antenna  \n\nThe blue arrow represents the axis of the antenna", font = [Times, bold, 20]])

display(AntennaAxis, Gain, scaling = constrained, axes = frame)

 

````

````

Calculating Radiation Resistance

 

The input impedance of the Hertzian Dipole will have both a real and a reactive part.  The reactive part will be associated with energy storage in the near field and will not contribute to the Poynting Vector in the far-field.  For an ideal antenna (with no resistive power loss) the real part will be responsible for the radiated power:

P__tx = I__0^2*R__rad

(2/3)*L^2*I__0^2*Z*Pi/lambda^2 = I__0^2*R__rad

(7.1)

R__rad := solve((2/3)*L^2*I__0^2*Z*Pi/lambda^2 = I__0^2*R__rad, R__rad)

(2/3)*L^2*Z*Pi/lambda^2

(7.2)

````

``

Calculating the power received by a Hertzian Dipole

 

If an electromagnetic field (E) is incident on the Hertzian Dipole antenna, it will generate an Electro-Motive Force (EMF) at the antenna terminals.  The EMF will be at a maximum when the transmitter is on the x-y-plane (that is when θ=π/2) and will be zero when the transmitter is on the z-axis.

 

For and incident E-field:

E := E__0*exp(I*omega*t)

E__0*exp(I*omega*t)

(8.1)

The z-axis component will be:

E__z := E*sin(theta)

E__0*exp(I*omega*t)*sin(theta)

(8.2)

The z-axis component of the E-field will create an EMF at the antenna terminals that will draw charge out of the receiver to each tip of the antenna. We can calculate the work done per unit charge by integrating the z-axis component of (E) over the length of the antenna (L):

V__emf := int(E__z, z = -(1/2)*L .. (1/2)*L)

E__0*exp(I*omega*t)*sin(theta)*L

(8.3)

In order to extract the maximum possible power from the antenna, we will form a conjugate match between the impedance of the antenna and the load.  This means that the load resistance must be the same as the radiation resistance of the antenna.  The voltage developed across the load resistance will therefore be half of the open circuit EMF:

V__pd := (1/2)*V__emf

(1/2)*E__0*exp(I*omega*t)*sin(theta)*L

(8.4)

And so the power delivered to the load will be:

P__rx := V__pd^2/R__rad

(3/8)*E__0^2*(exp(I*omega*t))^2*sin(theta)^2*lambda^2/(Pi*Z)

(8.5)

NULL

``

Calculating the Effective area of an Isotropic Antenna

 

We can also calculate the power received by the Hertzian Dipole by multiplying the power flux density arriving at the antenna with the effective area of an isotropic antenna and the gain of the Hertzian Dipole relative to an isotropic antenna:

P__rx = G__HertzianDipole*A__Isotropic*S__rx

(3/8)*E__0^2*(exp(I*omega*t))^2*sin(theta)^2*lambda^2/(Pi*Z) = (3/2)*sin(theta)^2*A__Isotropic*S__rx

(9.1)

We can express the incident power flux density in terms of electric field strength and wave impedance:

subs(S__rx = E^2/Z, (3/8)*E__0^2*(exp(I*omega*t))^2*sin(theta)^2*lambda^2/(Pi*Z) = (3/2)*sin(theta)^2*A__Isotropic*S__rx)

(3/8)*E__0^2*(exp(I*omega*t))^2*sin(theta)^2*lambda^2/(Pi*Z) = (3/2)*sin(theta)^2*A__Isotropic*E__0^2*(exp(I*omega*t))^2/Z

(9.2)

Rearranging, we obtain an expression for the effective area of the isotropic antenna:

A__Isotropic := solve((3/8)*E__0^2*(exp(I*omega*t))^2*sin(theta)^2*lambda^2/(Pi*Z) = (3/2)*sin(theta)^2*A__Isotropic*E__0^2*(exp(I*omega*t))^2/Z, A__Isotropic)

(1/4)*lambda^2/Pi

(9.3)

NULL

``

The Friis Transmission Equation

 

P__tx := 'P__tx'

We can calculate the power flux density that would be produced by an isotropic antenna at a distance r from the antenna by dividing the total transmitted power Ptx by the area of a sphere with radius r:

P__tx/(4*Pi*r^2)

(1/4)*P__tx/(Pi*r^2)

(10.1)

And so the power flux density that would be produced by an antenna with gain Gtx is:

S__tx := (1/4)*G__tx*P__tx/(Pi*r^2)

(1/4)*G__tx*P__tx/(Pi*r^2)

(10.2)

We can calculate the power received by an isotropic antenna by multiplying the power flux density incident onto the antenna with the effective area of an isotropic antenna:

S__tx*A__Isotropic

(1/16)*G__tx*P__tx*lambda^2/(Pi^2*r^2)

(10.3)

And so the power that would be received by an antenna with gain Grx is:

P__rx := (1/16)*G__rx*G__tx*P__tx*lambda^2/(Pi^2*r^2)

(1/16)*G__rx*G__tx*P__tx*lambda^2/(Pi^2*r^2)

(10.4)

The free space path loss is defined as the ratio between the received power and the transmitted power:

P__rx/P__tx

(1/16)*G__rx*G__tx*lambda^2/(Pi^2*r^2)

(10.5)

And so:

PathLoss := G__tx*G__rx*(lambda/(4*Pi*r))^2

(1/16)*G__rx*G__tx*lambda^2/(Pi^2*r^2)

(10.6)

````

````

NULL


 

Download Friis_Transmission_Equation.mw



How to do a simple vector plot?

Maple asked by cky1946 5 September 17

How can I solve two equations?

Maple 2020 asked by 9009134 90 September 18