4.2 Impedance and Admittance
157
If the electrical network shown in Figure 4.2-8 was purely resistive, we would have no phase
difference between voltage and current, if it was purely capacitive we would have the current
leading the voltage by 90 , but, since we have a combination of these two elements, we may still
expect a current lead but a phase difference somewhat less than 90 ! This is shown in Figure 4.2-9.
Figure 4.2-9 Current and voltage waveform for a 1000 MHz stimulus
As we mentioned in section 1.5.5, we can work out the impedance of an electrical network purely
from the voltage and current waveforms. Consider the markers in rectangular frames in Figure 4.2-9,
which show us the voltage and current peaks. From these measurements we may write
Ω
Now consider the markers in circular frames in Figure 4.2-9, which show us the time interval by
which voltage lags current. Remembering that (section 1.5.4.2) and that we are in a
sinusoidal steady state at a frequency of 1000 MHz, we may write
|
| | | | |
( )
Since the current leads the voltage in this case i.e.
, ( ) must be negative and may
be written in polar form as
This, in Cartesian form, translates to
[
( ) ( )
]
Ω
Let us now calculate the impedance algebraically!
0 0.5 1 1.5 2
Time (ns)
series_R_C
-1
-0.5
0
0.5
1
-20
-10
0
10
20
p2
1.168 ns
17.39 mA
1.248 ns
0.9999 V
0.5 ns
0 V
0.4177 ns
0 mA
Vtime(M_PROBE.VP1,1)[*] (L, V)
series R_C
Itime(ACVS.V1,1)[*] (R, mA)
series R_C
p1: Freq = 1000 MHz
p2: Freq = 1000 MHz
Conquer Radio Frequency
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