With a Cable

withCable

If your DUT is an antenna and if there is a cable between the antenna and the impedance bridge, you must adjust the phase of the V(reflected) according to the electrical length of the cable. Suppose the electrical length is alpha times the wavelength, then the vector V(reflected) must be advanced for twice alpha times twice pi [rad] to obtain the feed point impedance of the antenna.

It is a counter clockwise movement of the vector, and is said to be “Towards the Lord” in Smith Chart terminology.

Smith Chart

Now let us graphically obtain Z(DUT) for three cases. Note that the reflection coefficient, gamma, is a complex number defined as V(reflected)/V(forward). I forgot to annotate V(R2) as V(forward). Please see, for example, this figure.

vectorBridge12

This is the case in which abs(gamma)=1/2 and arg(gamma)=90 deg. We see graphically, or by measuring the length of each vector, that Z(DUT)=40 + j30 [ohm]. Note that since abs(gamma)=0.5, VSWR=(1+abs(gamma))/(1-abs(gamma))=3.

vectorBridge13

Next, this is the case abs(gamma)=1/3 and arg(gamma)=45 deg. Again, we see graphically that Z(DUT)=69.5 + j36.8 [ohm], and VSWR=(1+1/3)/(1-1/3)=2.

vectorBridge14

Finally, this is the case abs(gamma)=3/5 and arg(gamma)=135 deg. We see that Z(DUT)=14.5 + j19.2 [ohm], and VSWR=(1+3/5)/(1-3/5)=4.

Notice that the length of V(R3) differs in three figures, although the length corresponds to 50 [ohm] in each case.

Now, here is the fun part.

Let’s see the Smith Chart with three Lord impedances, which are the Z(DUT)s in the three cases. (Note the green circle is a SWR=3 circle.)

smithbridge

Do you see how these three red points in the Smith Chart relate to the three vector diagrams above? Yes, the red pionts are the end points of the vector V(reflected).

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So if you observe V1 and V2 with your oscilloscope and draw a vector V(reflected), or equivalently a vector V(DUT) on a Smith Chart, you will directly get Z(DUT) just by reading a chart. Cool?

Painlessly

Now you understand that if you observe two waveforms V1 and V2 with your impedance bridge, you will obtain Z(DUT) after some simple calculations, which you can do it manually or with a short program as is shown in my previous post, Eureka.

But, how can you do it painlessly, assuming that you do not like calculations very much, by just drawing a vector diagram? Here is the answer, namely a graphical solution:

vectorBridge10

First, you draw two vectors V(R2) and V(DUT), which you observed as V1 and V2, respectively. In this example, V2/V1=1.118 and at V2 is advanced in relative to V1 with 26.56 deg. This is all the information you need to get Z(DUT), but you can add some more vectors to fully understand the situation with your bridge.

vectorBridge11

First, you will immediately notice that by symmetry of the diagram, the length of V(DUT) is equal to that of V(R3), and since R3=50 [ohm] and the same current goes through both the DUT and R3, abs(Z(DUT)) is also 50 [ohm], which means R^2+X^2=50 [ohm], where R and X are defined as Z(DUT)=R+jX.

Now, the next step is to determine R and X.

vectorBridge12

For that, you will decompose the vector V(DUT) into two orthogonal vectors, the blue and the purple, by extending the vector V(R3) until it hits the outer circle. Then the length of the blue vector tells you how large X is, and the purple, R.

Noting that R3 is 50 [ohm], you will know that R (purple) is 30 [ohm] and X (blue) is 40 [ohm]

by just measuring the length of the three vectors, blue, purple, and V(R3).

Eureka

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Are you sure that Z(DUT)=30+j40 [ohm]? Can you tell by just observing the waveforms?

Let’s check with a python program.

# given waveforms
t=arange(0,1000)
v1=1.0*sin(t*(2.0*np.pi/444.0)+0.0)
v2=sqrt(5.0)/2.0*sin(t*(2.0*np.pi/444.0)+math.atan(1.0/2.0))
# Note that sqrt(5.0)/2.0=1.118 and atan(1.0/2.0)=26.56deg.

# start computing
z0=50.0+0.0j
cv1=1.0+0.0j
cv2=vratio*(cos(phase2advance)+sin(phase2advance)*1.0j)
cvr=2.0*cv1-cv2
cz=z0*(cv2/cvr)
gamma=(cz-z0)/(cz+z0)
swr=(1+abs(gamma))/(1-abs(gamma))

print "vratio=", vratio
print "phase2advance=", phase2advance*360.0/(2.0*np.pi)
print "Z(DUT)=", cz
print "gamma=",gamma
print "swr=", swr

And the results are:

vratio= 1.11803398875
phase2advance= 26.5650511771
Z(DUT)= (30+40j)
gamma= (-1.02119201983e-13+0.5j)
swr= 3.0

So you are correct!

bridge45deg

Actually, the DUT was 30 [ohm] + 909.5 [nH], which is at 7026kHz 30 [ohm] + j40 [ohm].

A Simple Example

bridge45deg1

Suppose that you have some unknown DUT, a green box, and observed the waveforms V(v1) and V(v2) with your oscilloscope. Looking closely, you see that V(v2)/V(v1)=1.118 and that V(v2) is advanced in relative to V(v1) with 26.56 deg.

You will also see that by comparing the green trace, V(v2)-V(v1), and the red trace, V(v1), the amplitude of the green is just half of the red, and the green is exactly 90 deg in advance. (Note that the red and the yellow cross each other, which means the green is zero, at the peak of the red.)

Now that you have enough information to know Z(DUT), you will close your eyes and think very hard…

After a few seconds, you will say, “Eureka! Z(DUT)=30+j40 [ohm].”

Reflection Coefficient (2)

vectorBridge6

Let’s now think of a case, in which the reflection coefficient, gamma, is 0.5, or abs(gamma)=0.5 to be more exact. By definition, it follows that abs(V(reflected))/abs(V(forward))=0.5, but the phase of the reflected wave against the forward wave could be anything. Suppose the DUT is your antenna with some length of a coax cable between the impedance bridge and the antenna. Depending of the electrical length of the cable, the phase could be 0 degrees:

vectorBridge8

or 180 degrees:

vectorBridge7

or anything between zero and 360 degrees as is typically shown in the first figure.

In the 0 degrees case, V(DUT):V(R3)=3:1, which means Z(DUT)=50*3 [ohm] since Z(R3)=50 [ohm], and in the 180 degrees case, V(DUT):V(R3)=1:3, and Z(DUT)=50/3 [ohm].

Or equivalently, if Z(DUT)=50*3 [ohm], then gamma=(50*3-50)/(50*3+50)=0.5, and if Z(DUT)=50/3 [ohm], then gamma=(50/3-50)/(50/3+50)=-0.5, both from the well-known equation:

gamma=(cz-z0)/(cz+z0)

Reflection Coefficient

The reflection coefficient, gamma, defined as Vreflected/Vforward, is obtained as follows:

z0=50.0+0.0j # define system impedance
cv1=1.0+0.0j # V1 as a complex number
cv2=vratio*(cos(phase2advance)+sin(phase2advance)*1.0j) # V2 as a complex number
cvr=2.0*cv1-cv2 # V(R3)
cz=z0*(cv2/cvr) # the same current for both DUT and R3
gamma1=(cz-z0)/(cz+z0) # a well-known formula
swr=(1+abs(gamma))/(1-abs(gamma)) # another well-known formula
# another method
cvfwd=cv1
cvrfl=cv2-cv1
gamma2=cvrfl/cvfwd # check yourself that gamma1 and gamma2 are the same
#
# checking now
#
gamma1=(cz-z0)/(cz+z0) # a well-known formula
      =(cz/z0-1)/(cz/z0+1)
      =(cv2/cvr-1)/(cv2/cvr+1)
      =(cv2-cvr)/(cv2+cvr)
      =(cv2-(2.0*cv1-cv2))/(cv2+(2.0*cv1-cv2))
      =(2.0*cv2-2.0*cv1)/(2.0*cv1)
      =(cv2-cv1)/cv1
      =cvrfl/cvfwd
      =gamma2

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