The Phasor Transform and Impedance

each student will experimentally verify the voltage-current relationship for
inductors and capacitors for sinusoidal steady-state excitation. see how the
phasor transform can simplify the voltage-current relationship, eliminating the need for
derivatives and integrals. , the voltage-current relationship in the phasor domain
looks just like Ohm’s law, where voltage equals current time a scaling constant. We call
the scaling constant impedance. It serves the same role as resistance, but in the phasor
domain. It is a constant like resistance, but turns out to be complex valued and varies
with the frequency of the signal involved. In this lab, we will also explore the basic
operation of an ideal transformer.
2. Background
2.1 What is the phasor transform?
Sinusoids are special signals. Note that the integral and derivative of a sinusoid is a
sinusoid. Thus, the voltage-current relationships for inductors and capacitors, which are
characterized by integrals and derivatives, tell us that a sinusoidal current produces a
sinusoidal voltage. The only difference between the sinusoidal voltage across and current
through these devices is possibly the amplitude and phase. The frequency of the current
will be the same as the frequency of the voltage. Thus, if we only consider sinusoidal
signals, all we need to keep track of is magnitude and phase of the voltages and currents.
This is where the phasor transform come in!
A complex number, unlike a real number, contains both a magnitude and a phase.
Thus, a complex number is a very convenient way to describe the various sinusoids in our
circuits. Here is how the phasor transform works. We begin by considering a sinusoid in
the so-called time domain (where our voltage and current equations are functions of time)
( ) cos( ) m x t  X t  . (1)
In the phasor domain, this signal is expressed as a single complex number X with a
magnitude, m X , equal to the sinusoid amplitude and a phase,  , equal to the sinusoid’s
phase. Any complex number can be written in two distinct ways, rectangular form and
polar form. While these forms looks very different, Euler proved that they are equivalent
with the famous theorem that bears his name. Consider a complex number Zwritten in
rectangular form and polar form respectively,
j X jY re  Z    . (2)
Here, X is the real part of Z, and Y is the purely imaginary part of Z. The variable r is
the magnitude of Z,  is the phase, j  1 , and e  2.71828182846…. The two forms
of a complex number can be found from one another using the following relationships:
2 2 r X Y, (3)
1 tan ( / ) YX    , (4)
cos Xr   (5)
and
Y  r sin . (6)
Note that the polar form is often written in shorthand notation like so
j re r   Z . (7)
This hides the rather bizarre j e part, and highlights the fact that the magnitude and phase
are the important pieces of information carried by a complex number. A scientific
calculator will do the conversions between the two forms (rectangular and polar) in (2).
The phasor domain representation for the sinusoid in (1) is given by
j
m Xe  X . (8)
We simply let the magnitude of the cosine equal the magnitude of the phasor (the
complex number) and the phase of the cosine becomes the phase term in the phasor.
Look at what we have done. We have replaced a function of time with a single
number (albeit, a complex number) that contains the only information we need to keep
track of, magnitude and phase. Performing a phasor transform means writing Equation (8)
from Equation (1). Performing an inverse phasor transform means writing Equation (1)
from Equation (8). Of course, in performing an inverse phasor transform, we must know
what the original frequency, , is. Here is an example. The voltage produce by a
standard AC outlet is described by
v(t) 169.7cos(120 t  60) , (9)
where we have arbitrarily set the phase to be 60 degrees. The phasor transform for this is
60 169.7 169.7 60 j e  V     . (10)
Be sure to mark the angular units (radians or degrees) the same in both your time domain
and phasor domain representations. Specify degrees with the  symbol, otherwise,
radians will be assumed.
2.2 What is impedance?
For an inductor, capacitor, resistor, or combination, a sinusoidal current produces a
sinusoidal voltage of the same frequency (although, as we said, the magnitude and phase
may change). Thus, we have
( ) cos( ) m i i t  I t  (11)
and
( ) cos( ) m v v t V t  . (12)
In the phasor domain, we have
j i
m I e  I  (13)
and
j v
m Ve  V . (14)
Now to the amazing part… Note that the phasor voltage, V , can be written in
terms of phasor current, I , multiplied by a complex number (which scales the magnitude
as needed and adds to the phase as needed). This gives
 V IZ (15)
for some complex number j z
m Ze  Z , which we call the impedance. Again, the concept
of impedance applies to R’s, L’s and C’s and combinations thereof. Writing the complex
numbers in (15) in polar form yields
j v j i j z j ( i z )
m m m m m V e I e Z e I Z e        . (16)
Thus, the magnitude of the sinusoidal voltage across some impedance is the magnitude of
the current times the impedance magnitude. The phase of the voltage is the phase of the
current plus the phase of the impedance.
Solving for impedance, assuming we know the voltage and current, we get
( )
v
z v i
i
j
j m m j
m j
m m
V e V
Z e e
I e I

  

    
V
Z
I
. (17)
Thus, impedance can be determined by determining the relationship between voltage and
current magnitudes and phases, for a particular element (or combination of elements). In
particular, the impedance magnitude is given by
m
m
m
V
Z
I
 , (18)
and the impedance phase is given by
z v i    . (19)
You will find that impedance, for elements other than resistors, varies with the
frequency of the voltage or current source exciting the element. That is, the same
physical component will have different impedances when used with different source
frequencies.
Here is the real beauty. When the source or sources in your circuit are
sinusoidal, you can express all the voltages and currents in the circuit as complex
numbers (expressing them as phasors in the phasor domain). You can express all the
components (R, L & C’s) by their impedance (complex numbers). Now, because all the
voltage-current relationships in this phasor domain are given by (15), which is just like
Ohm’s law, the circuit can be analyzed like a DC resistive circuit (Chps. 3&4). One
solves for the phasor domain voltage or current that is of interest, and finally, performs
an inverse phasor transform to get the time domain voltage or current which is generally
the more intuitive form of the solution (the one you are accustomed to). No more
differential equations!!!
So what’s the catch? Well… one, we have to deal with complex numbers. And
two, the source in the circuit has to be sinusoidal. However, as you will see in ECE 303,
any waveform can be written in terms of sinusoids. Thus, the concept of impedance will
ultimately allow you to solve any RLC circuit with any input source without differential
equations.
It can be shown that the theoretical impedance for an ideal resistor is simply its
resistance, R. The theoretical impedance for an ideal inductor is
90 90 j jL Le L  Z      . (20)
The impedance for an ideal capacitor is given by
90 1 1 1
90 j j
e
jC C C C
  
Z       . (21)
Note that  is frequency in radians/second and 2 f    , where f is frequency in
cycles/second (Hz).
3. Procedure
3.1 Measuring an inductor’s impedance
Begin by constructing the circuit shown in Figure 1. Use a decade inductor for ‘L’. Also,
make the appropriate connections to the oscilloscope. Note that the voltage drop across
the resistor is the current in the inductor x 1000. Thus, by observing the voltage drop
across the inductor and the voltage drop across the resistor we may observe the voltagecurrent
relationship for an inductor. Note that you will want to invert the waveform you
observe on Channel 2 since the leads are connected in the opposite sense (press the
Channel 2 button near the Channel 2 input connector and select ‘Invert On’). Remember,
when you press ‘Autoscale’, the inversion is turned off. Pick 0.5H for ‘L’. Use the
waveform generator to generate a sinusoidal voltage signal (you may adjust the amplitude
to a suitable setting of your choice).
Figure 1: Inductor voltage and current measurement setup used to compute impedance.
1. Make a table containing the following quantities: frequency, voltage drop across
the inductor (Vp-p), voltage drop across the resistor (Vp-p), and the phase
difference between the voltage drop across the inductor and resistor. Take at least 10
measurements in the range of 100 Hz to 1 kHz. State the inductor value you used in
the table caption.
2. Use these data in Step 1 to compute the inductor impedance magnitude and phase at
each frequency using Equations (18) and (19).
3. With MATLAB, plot the experimental and theoretical impedance magnitude vs.
frequency on one figure. On a separate figure, plot both the experimental and
theoretical impedance phase. Use Equation (20) to compute the theoretical values.
How do the experimental and theoretical plots compare?
Note that impedance magnitude goes to zero as frequency goes to zero. This is
consistent with the notion that an inductor is a short circuit to DC ( 0  ).
3.2 Measuring a capacitor’s impedance
Next, replace the decade inductor with a decade capacitor. Pick 0.5uF for ‘C’. Use the
waveform generator to generate a sine wave (you may adjust the amplitude to a suitable
setting of your choice).
4. Make a table containing the following quantities: frequency, voltage drop across
the capacitor (Vp-p), voltage drop across the resistor (Vp-p), and the phase
difference between the voltage drop across the capacitor and resistor. Take at least
10 measurements in the range of 100 Hz to 1 kHz. State the capacitor value you used
in the table caption.
5. Use the data from Step 4 to compute the capacitor impedance magnitude and phase
for each frequency using Equations (18) and (19).
6. With MATLAB, plot the experimental and theoretical impedance magnitude on one
figure. On a separate figure, plot both the experimental and theoretical impedance
phase. Use Equation (21) to compute the theoretical values. How do the
experimental and theoretical plots compare?
Note that impedance magnitude goes to infinity as frequency goes to zero. This
is consistent with the notion that a capacitor is an open circuit to DC ( 0  ).
3.3 Using phasors to solve a circuit problem
Figure 2: Series RLC circuit.
Construct the circuit in Figure 2 using R=1.5 k, L=1 H, and C=0.01 F. Note that here
we are declaring the output to be the voltage drop across the capacitor. Let the input
voltage, Vin(t), be a 4 Vp-p sinusoidal signal at a frequency of 1000 Hz. That is,
Vin (t)  2cos(2000 t  0) . (22)
In the phasor domain, we have
0 2 2 0 2 j
in e  V     . (23)
7. Observe the input (Ch. 1) and output (Ch. 2) on the oscilloscope. Adjust the
oscilloscope to allow for a nice comparison and perform a screen capture (Capture
should include Vpp measurements for both channels and phase difference
measurement).
8. Theoretically calculate the output voltage, Vout(t), using phasor transform
techniques. Hint: think voltage divider in the phasor domain. Is this result
consistent with the observed output waveform? (“Lab_10_RCL_2.m” on the
isidore helps to calculate magnitude and phase of Z. Then use magnitude and
phase to calculate theoretical output voltage).
9. Repeat the two steps above using a frequency of 1600 Hz and then 2000Hz. Note
that this 2nd order circuit has a resonant frequency near 1600 Hz. Do you notice
something special happening near this frequency? (Observe Vpp’s changes.)
3.4 Ideal transformer basics
An ideal transformer uses two separate inductors with a common metal core to link the
magnetic fields. Thus, current in one of the inductors induces a voltage in itself (as any
inductor does) and in the adjacent inductor (mutual inductance) and vice versa. To make
a long story short, an ideal transformer has these input/output properties in the phasor
domain
2
1
out in
N
N
VV , (24)
where 1 N is the number of wire turns on the inductor on the input side and 2 N is the
number of turns on the output side. The current is given by
1
2
out in
N
N
II . (25)
In this way, the voltage may be stepped-up, but the current is then stepped down and vice
versa. The power is the same on the input and the output, so there is no loss of power in
the conversion (for an ideal transformer).
10. Connect the transformer to the waveform generator and generate a sinusoidal voltage
(1kHz, 4 Vp-p) as shown in Figure 3. The input side has two leads while the output
has three. Based on the amplitudes of the input and output voltages, determine the
turn ratio (N2/N1) for the transformer where the output is taken between one output
lead and the center tap lead.
11. Now, determine the turn ratio between the input and the output, where the output is
taken between the two outputs leads (excluding the center tap lead).
Figure 3: Transformer circuit.
NOTE: Please do NOT reverse the transformer as it would act as a voltage step-up transformer and would be capable of generating potentially dangerous voltages.
(+1 Bonus) What is the Nyquist theorem?