Tag: Antenna

An antenna is said to be resonant if its input impedance (or feed point impedance), Z=R+jX, is purely resistive, which means its reactance X is exactly zero.

The input impedance of a half-wave dipole is approximated by the equation:

where We and k are some constants, and delta is the excess length of an antenna over half the wavelength. Since both We and k are positive, we can make X to be zero by shortening the element slightly, usually a few percent depending on the diameter of the element.

The question here is what is the physical and practical meaning of the antenna being in resonant.

References:
[1] “Antenna and Radio propagation (in Japanese)” by Yasuto Mushiake,
[2] MMANA-GAL basic.

The above two figures are from
Mats Gustafsson et al., Forward scattering of loaded and unloaded antennas.

A receiving antenna interacts with an incoming electromagnetic wave. However, its interaction is not limited to absorption of the energy but includes some scattering from the antenna.

As is described in “Antenna and Radio propagation (in Japanese)” written by Yasuto Mushiake, pp 39-42, the scattering cross section of an (receiving) antenna loaded with Zl is given by

where Z=R+jX is the input impedance and Ae is the effective area of the antenna.

The scattering cross section As is a function of Zl, and obviously As(max)=4Ae when Zl=0. And if Zl is selected to be Zconj=R-jX, then As=Ae, which means the antenna emits the same amount of power as it supplies to the receiver.

Now, let’s try again with another DUT.

This time there is a phase difference, so we must determine complex impedance in the form Z = R + jX.

The voltage ratio is: V2/V1 = 1.713 [v] / 2.453 [v] = 0.698, and the phase difference is: 360.0 [deg] * (-22.80 [nS] / 142.32 [nS]) = -57.7 [deg], where 142.32 [nS] is the period for 7026kHz.

Assuming you know the answer that the DUT is 100 [ohm] // 1 [nF], the voltage ratio and the phase are obtained in the following manner. (This is a reverse way.)

(%i1) r6:50;
(%i2) r:100;                      <- here is your 100 [ohm]
(%i3) x:1/(%i*2*%pi*7026e3*1e-9); <- here is your 1 [nF]
(%i4) z:r*x/(r+x);
(%i5) v2:z/(z+r6);
(%i6) float(2*abs(v2));
(%o6) 0.74942127142784            <- voltage ratio
(%i7) float(carg(v2)*360/(2*%pi));
(%o7) -55.80120705365476          <- phase difference

In most cases you do not the answer in advance, so you must go like this:

(%i1) t:(-57.7/360.0)*2.0*3.16; <- measured phase difference
(%i2) v:0.698/2.0;              <- measured voltage ratio
(%i3) r:50.0;
(%i4) float(solve([(R*(r+R)+X^2)/((r+R)^2+X^2)=v*cos(t),r*X/((r+R)^2+X^2)=v*sin(t)],[R,X]));
(%o4) [[R=-50.0,X=0.0],[R=4.183382859001799,X=-19.67880950916058]]
(%i5) Z:4.1833-19.6788*%i;
(%i6) Y:1/Z;
(%i7) 1/realpart(Y);
(%o7) 96.75499446131043         <- Real part of Z (nominal 100 [ohm])
(%i8) 1/imagpart(Y);
(%o8) 20.56808181037462         <- Imaginary part of Z
(%i9) C:1/(2.0*3.1416*7026e3*1/imagpart(Y)); <- convert to capacitance
(%o9) 1.1013292647891441*10^-9  <- 1.1 [nF] (nomnal 1 [nF])

To verify the above result, we go again the reverse way:

(%i1) r6:50;
(%i2) r:96.755;                        <- R=96.755 [ohm]
(%i3) x:1/(%i*2*%pi*7026e3*1.1013e-9); <- C=1.1013 [nF]
(%i4) z:r*x/(r+x);
(%i5) v2:z/(z+r6);
(%i6) float(2*abs(v2));
(%o6) 0.69801459091613                <- voltage ratio
(%i7) float(carg(v2)*360/(2*%pi));
(%o7) -58.03754068937465              <- phase difference

Not bad?

With a signal generator (SG), you can make a measurement of various parameters of your device. One example is impedance measurement.

A fixed attenuator is inserted in front to improve the souce match. For 6dB attenuation, use R1=39 [ohm], R2=R3=150 [ohm]. “X” is the device under test (DUT), and R4=R5=R6=50 [ohm]. In the photograph, a 100 [ohm] resistor is employed as a DUT.

IC-7410 with tx frequency 7026kHz is used as a SG. RF power is minimized to 2 [W].　CH1 (in red) shows “V1”, and CH2 (in yellow) shows “V2”.

Since the DUT, a 100 [ohm] resistor, is pure resistive, there is no phase difference between the two waveforms.

By solving a simple equation, we conclude that the impedance of our DUT is 94.6 + j*0 [ohm].