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AC Circuit Analysis in a Nutshell.
Christos Velissaris, Fall 2012
Written for the PHY3802L Intermediate Labs Physics Course.
Complex Numbers in a Nutshell.
The imaginary unit number j is defined according to the equation:
j^2 = − 1 ⇒ j = − 1
Any “number” z of the form z = x + j ⋅ y where j is the imaginary unit defined above and
x, y are Real numbers, is called a “complex number. The set of complex numbers is symbolized with the number C. The Real number x is called the Real part of the complex number z and symbolized as Re{z}. The Real number y is called the Imaginary part of complex number z and symbolized as Im{z}.
z = x + j ⋅ y x =Re { } z y =Im{ } z
This representation of the complex number z is called the “Cartesian representation”. Any complex number z can be represented graphically in a two dimensional Orthogonal Coordinate system where the horizontal axis is the Real axis and the perpendicular is the Imaginary one. The complex number is represented uniquely as the vector with coordinates (x,y). The “magnitude” of the complex number z is defined as:
z = x^2 + y^2
Except with the Cartesian, a complex number could be represented with its polar form as follows. Consider the following equations:
2 2 2 2
2 2 2 2
2 2
x y
y j x y
x z z
x y
y j x y
x z x j y z x y
If we look carefully at the complex number in the parenthesis the sum of the squares of its real and imaginary parts is equal to 1. Therefore its real part is the cosine and its imaginary part the sine of an angle. More specifically, geometrically, this angle θ is the one the real axis traverses until it meets the complex vector, if rotated counter clockwise.
( )
cos 2 2 sin 2 2
cos sin
x y
y x y
x
z x j y z j
θ θ
θ θ
Another way a complex number is represented in its polar form is the following:
z = z ⋅^ (^ cosθ + j sinθ)= z ⋅∠ θ
Consider the complex number exp{jx}=e jx^ where x is a real number. By Taylor expanding the exponential and using the equations j 2n^ = (-1) n^ and j 2n+1^ = j(-1) n^ we have:
( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( ) ∑ ∑ ∑ ∑ ∑
∞
=
∞ +
=
∞
=
∞ +
=
∞
= +
0
2 1
0
2
0
2 1
0
2
0 2 1!
! 2! 2 1! n
n n n
n n n
n
n
n
n
n jx n
x j n
x n
jx n
jx n
jx e
By looking at the Taylor expansion formulas for the sin(x) and cos(x) functions we can identify the Real part of the above equation as the cosine of x and the imaginary part as the sine of x. Therefore we arrive at the following very important relationships:
e jx^ = cos ( ) x + j sin( ) x e − jx =cos( ) x − j sin( ) x
( ) ( ) j
e e x e e x
jx jx jx jx
2
sin 2
cos
Therefore every complex number can be written as:
z = z ⋅ ( cosθ + j sin θ) = z ⋅ ej θ
The above discussion on complex number finds extensive application to ac circuit analysis. The usage of complex impedance greatly simplified the analysis, eliminates the need for the usage of “phasors” and reduces the ac circuit analysis into dc analysis with complex numbers.
Addition and subtraction of complex numbers:
1 2 (^12 )^ (^12 )
z z x x j y y
z x jy z x jy
Multiplication of complex numbers:
Let us use the expression: cos( x ) =Re{ e jx }
We have
( ) { }
C^ ( )^ {^ C jt }
jt C
jt j C
j t C C C
I t j CV e
I t CV t C V e C V e e C V je
ω
ω
π ω ω π
0
0 2 0
2 0 0
Re
Re Re Re 2
cos
^ =
^ +
Of course the ac current through the capacitor can have the general form:
IC ( t ) = IC 0 cos( ω t + φ) =Re { I (^) C 0 ej (^ ω t +φ)} =Re{ I (^) C 0 ej φ ej^ ω t }
Where φ is the phase difference between the ac voltage difference across the ends of the capacitor and its ac current. By comparing the last two expressions for the current I C (t) we get the very interesting expression:
j C
V
I (^) Cej j CVC C
0 0 = 0 =
Therefore by using the complex form of the capacitor’s reactance j C
X C
= and only
the amplitude of the applied ac voltage across the capacitor we get booth the amplitude of the ac current through the capacitor plus the phase difference between the ac voltage and the current.
The amplitude of the current is simply the amplitude of the voltage divided by the magnitude of the complex reactance (Ohm’s law) and the negative of the phase of the complex reactance gives us the phase difference between the capacitor voltage and current. Clearly:
(^02) 2
0 0 (^0 ) (^11)
π π
φ
ω ω ω
C^ j j
j C C C e
C
V
e C
V
j C
V
I e = = = −
Therefore:
C
V
I C C
0 0 =^ and^2
If one needs the complete time dependence of the current simply multiply by ej^ ω t^ and take the Real part of the complex number.
( ) { } (^)
= ^ +
cos 1 1
Re 0 Re^020
ω
π φ ω t
C
V
e e
C
V
I t I e e jt C j jt C j C C
as expected.
Complex Reactance of an Inductor
Consider an Inductor L connected with a power supply feeding the inductor with an ac voltage V L (t) = V L0 cos(ωt). Let us calculate the ac current provided by the power supply.
( ) ( ) ( ) ( t ) d t L
V
V t I t dt
dI L dt
dI V t L
t L L L
L L L =^ ⇒ = ⇒ = ∫ ′ ′ 0
0
0 cosω cos^ ω
We have assumed that at t=0 when the ac voltage is applied across the inductor the ac current through the inductor is also 0:
( ) ( ) (^)
= = ^ −
(^0) sin 0 cos
t L
V
t L
V
I (^) Lt L L
From the expression for the current we see that:
1. The current lags the voltage by π 2 radians.
- The relationship between the voltage and current amplitudes is given by the
formula: VL 0 = XLIL 0 where X L = ω L the capacitors Reactance.
Let us use the expression: cos( x ) =Re{ e jx }
We have:
( )
( )
⇒
^ =
= ^ − −
(^) −
L j^ t L
L L j t L jt j L jt L
e j L
V I t
e L j
V e e L
V e L
V t L
V I t
ω
ω
π ω ω π
ω
ω ω ω
π ω ω
0
0 0 2 0 2 0
Re
1 Re Re Re 2
cos
Complex Impedance analysis of an R-C Circuit.
Consider now a Resistance R connected in series with a capacitor C and powered by an ac power supply providing a voltage V(t) = V 0 cos(ωt). Let us calculate the current I(t) provided by the power supply as well as the voltage drops across the capacitor V C (t) and the resistor V R (t).
According to Kirchoff second rule:
V (^) 0 cos ( ω t ) = VC ( t ) + VL ( t )
The difficulty with this expression is that we cannot add the voltage expressions for the resistor and the capacitor because they are not in phase (neither with each other nor with the power supply). The only thing we know is that the V R (t) is in phase with the current and V C (t) lags the current by π/2. Let us make an assumption about the ac current:
I ( t ) = I 0 cos(ω t + φ)
where the current amplitude I 0 and the phase difference φ between the current and the voltage provided by the power supply are to be determined. Therefore we can write:
( ) (ω φ)
V t = V ^ t + − V t C R cos 2 0 cos^0 cos 0
We also know how to relate the amplitude of the voltage drop across the capacitor and the resistor with the amplitude of the current through these elements:
V I R
C
V C 0 I 0 R 0 0
Therefore :
( ) ( )
{ } { }
{ }
φ φ
ω φ ω
ω ω φ ω φ
ω
ω
ω
ω
ω φ π ω φ ω
ω
j j
jt j jt
jt jt j jt j
I e
j C
R
V
e j C
V I R
e e j C
Ve I R
e e I R e e C j
V e I
t I R t C
V t I
0
0 0 0
0 0
0 0 0
0 0 0
Re Re
Re
Re
Re
cos 2
cos
cos
^ ⇒
+ +^ ⇒
Therefore by using the complex total impedance seen by the ac power supply we can get not only the amplitude of the current but also the phase difference of the ac current from the ac voltage provided by our generator.
2 2
2
0 0 0
C
R
V
Z
V
I
ω
For the phase φ notice that the complex impedance can be written as:
2 2
2 2 2
2
2 2
2 1
C
R
j C
C
R
R
C
Z Zej R
θ
therefore the angle θ is defined by:
RC
C
R
C
C
R
R
tan 1
sin 1
cos
2 2
2 2 2
2
Therefore we have:
2 2 2
0 0 R L
V
I
and the phase difference φ id given by the relationships:
R
L
R L
L
R L
R ω
cos = sin tan 2 2 2 2 2 2
We see that since the angle φ is negative the current lags the voltage.
Exercise 1.
For an R-L-C series circuit calculate the current provided by the ac generator as well as the phase difference between these two. Plot the current magnitude and the phase difference as a function of angular frequency ω.
Exercise 2.
For the low pass filter shown calculate the ratio of the amplitude of the output voltage over the amplitude of the input voltage. Assume the input voltage to be of the form:
Vin ( t ) = V 0 cos( ω t )
Plot that ratio (called otherwise magnitude of the complex Transfer Function T(ω)) as a function of frequency.
Find expressions for the phase difference φ between the input and output signal. Plot the phase difference as a function of frequency.
Important note: Engineers and electronics technicians in practice plot the logarithm of the transfer function magnitude times 20 vs the logarithm(base 10) of the frequency. They use it as the Gain G (Attenuation if negative) of the circuit and they assign the arbitrary unit dB to it.
GdB = 20 log 10 ( T ( f ))
The frequency RC
f
0 =^ is called the “corner
frequency” of the circuit. In this plot the gain vs frequency is 1 and almost constant below the corner frequency. Passed the corner frequency, the gain drops linearly and looses 20 dB per decade (-20dB every time the frequency is multiplied by 10). At the corner frequency the signal has lost 3 dB (the gain is –3dB) that is why the corner frequency is also called the 3dB frequency. The 3dB frequency (corner frequency) is also the “Bandwidth” (BW) of the low pass filter. Practically as seen in the figure the corner frequency the dB gain of the low pass filter is 0 and after the corner frequency it decreases linearly loosing 20dB per decade that is every time it hits 10f 0 , 100 f 0 , 1000 f 0 etc.
The phase difference of the output from the input signal follows the approximate figure shown (and used for practical reasons to describe the behavior of the low pass filter).It is 0 up to 0.1f 0 , and then it drops linearly to – 90^0 at 10f 0. It practically stays –90^0 for higher values of frequency. At f 0 it is –45^0_._
T r and the corner frequency f 0 (which is also the Bandwidth BW) of the low pass filter?
From this discussion describe how would you practically measure the bandwidth of an electronic device (e.g. amplifier) using a square wave generator and observing the rise time of the output signal?
Answer: You should find that Trise = 2. 2 RCand if you use the formula for the corner frequency
f 2 π RC
0 =^1 you will end up with the very important expression for the Bandwidth:^ Trise
BW =^0.^35
Hint: You should set up and solve the differential equation describing the time dependence of the circuit current:
( ) ( ) ( ) ( ) ( ) ( ) 0 =^ + ⇒ 0 = + ⇒ + =^0 t ≥^0 R
V
RC
q t t dt
dq C
q t t dt
dq V R C
q t V ItR
with q(t) is the Capacitor Charge, I(t) the current through the circuit and ( ) ( ) t dt
dq I t =
Solve the above differential equation and derive an expression for the Capacitor charge and electric
current as a function of time with the initial conditions: I ( 0 ) = 0 q ( 0 ) = 0
Exercise 5.
Consider the low pass filter shown below. It is loaded with a resistance R L. Find the output voltage on the load if you decide not to ignore the R L resistance of the load. Do the analysis using complex impedance and plot the ratio
( ω )
T
in
load (^) =
as a function of ω. Find an expression for that function. Find an expression for the phase difference between the load and input voltages and plot it as a function of ω. Assume an input signal: Vin ( t ) = V 0 cos( ω t )
What is the new corner frequency of the circuit? What is the condition for R L in order not to alter significantly the corner frequency?
Exercise 6.
For the low pass circuit shown calculate the ratio of the amplitude of the output voltage over the amplitude of the input voltage. Assume the input voltage to be of the form:
Vin ( t ) = V 0 cos( ω t )
What is happening in the case where R 1 =R 2 =R and C 1 =C 2 =C?
