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CIRCUITS LABORATORY
EXPERIMENT 3
AC Circuit Analysis
3.1 Introduction
The steady-state behavior of circuits energized by sinusoidal sources is an important area of study for several reasons. First, the generation, transmission, distribution, and consumption of electric energy occur under essentially sinusoidal steady-state conditions. Second, an understanding of sinusoidal behavior makes possible the prediction of circuit behavior when nonsinusoidal sources are used through the use of techniques such as Fourier analysis and superposition. Finally, by specifying the performance of a circuit in terms of its steady-state sinusoidal behavior, the design of the circuit can often be simplified. Needless to say, the importance of sinusoidal steady-state behavior cannot be overemphasized, and many of the topics in future experiments are based on a thorough understanding of the techniques used to analyze circuits driven by sinusoidal sources. In this experiment, the behavior of several types of circuits will be examined to determine their behavior when excited by sinusoidal sources. First, the behavior of both RC and RLC circuits will be examined when driven by a sinusoidal source at a
given frequency. Subsequently, the frequency response of both a low-pass filter and a high-pass filter will be considered.
3.2 Objectives
At the end of this experiment, the student will be able to: (1) Determine the steady-state behavior of linear circuits driven by sinusoidal sources, (2) Use the oscilloscope to measure the phase difference between two sinusoidal signals, (3) Determine analytically the frequency response of a network, (4) Construct Bode plots relating the magnitude and phase response of the voltage ratio of a linear network as a function of frequency, and (5) Design primitive low- and high-pass filters using one resistor and one capacitor.
3.3 Theory
3.3.1 Sinusoidal Steady-State Analysis
As stated previously, the steady-state behavior of circuits that are energized by sin- usoidal sources is an important area of study. A sinusoidal voltage source produces a voltage that varies sinusoidally with time. Using the cosine function, we can write a sinusoidally varying voltage as follows
v ( t ) = Vm cos ( ω t + φ ) (3.1a)
and the corresponding sinusoidally varying current as
i(t) = I m cos ( ω t + θ). (3.1b)
Equation (3.3) is derived from the fact that the cosine function passes through a complete set of values each time its argument passes through 2π radians (360 o^ ). From Equation (3.3), we see that whenever t is an integral multiple of T , the argument ωt increases by an integral multiple of 2π radians. The maximum amplitude of the sinusoidal voltage shown in Figure 3.1 is given by the coefficient Vm and is called the peak voltage. It follows that the peak-to- peak voltage is equal to 2 Vm. The phase angle of the sinusoidal voltage is the angle φ. It determines the value of the sinusoidal function at t = 0. Changing the value of the phase angle φ shifts the sinusoidal function along the time axis, but has no effect on either the amplitude ( Vm ) or the angular frequency (ω). Note that φ is normally given in degrees. It follows that ω t must be converted from radians to degrees before the two quantities can be added. In summary, a major point to recognize is that a sinusoidal function can be completely specified by giving its maximum amplitude (either peak or peak-to-peak value), frequency, and phase angle. In general, there are three important criteria to remember regarding the steady-state response of a linear network (i.e., circuits consisting of resistors, inductors, and capacitors) excited by a sinusoidal source. (1) The steady-state response is a sinusoidal function. (2) The frequency of the response signal is identical to the frequency of the source signal. (3) The amplitude and phase angle of the response will most likely differ from the amplitude and phase angle of the source and are dependent on the values of the resistors, inductors, and capacitors in the circuit as well as the frequency of the source signal.
V = Vm ej φ
V = Vm cos φ + jVm sinφ
It follows then the problem of finding the steady-state response is reduced to finding the maximum amplitude and phase angle of the response signal since the waveform and frequency of the response are the same as the source signal. In analyzing the steady-state response of a linear network, we use the phasor, which is a complex number that carries the amplitude and phase angle information of a sinusoidal function. Given the sinusoidal voltage defined previously in Equation (3.1a), we define the phasor representation as
(3.4a)
where V , is the maximum amplitude i.e., the peak voltage and φ is the phase angle
of the sinusoidal function. Equation (3.4) is the polar form of a phasor. A phasor can also be expressed in rectangular form. Thus, Equation (3.4a) can be written as
(3.4b)
In an analogous manner, the phasor representation of the sinusoidal current defined in Equation (3.1b) is defined as
I = I m e j^ θ^ = I m cos θ + jI m sin θ. (3.5)
We will find both the polar form and rectangular form useful in circuit applications of the phasor concept. Also, the frequent occurrence of the exponential function e j^ θ has led to a shorthand notation called the angle notation such that
Vm / φ ≡ Vm e j^ φ^ (3.6a)
and
I m / θ ≡ I m e j^ θ^ (3.6b)
Figure 3.2. Phasor Domain Equivalent Circuits
(b) The impedance (ZL) of an inductor is j ω L in rectangular form and ω L / 90o
in angle form. Again, from Equation (3.7), we see that the phasor voltage
at the terminals of an inductor equals j ω L times the phasor current. The
phasor-domain equivalent circuit for the inductor is shown in Figure 3.2(b).
If we assume I = I m / θ , then the voltage is given by
V = ( j ω L ) I = ( ω L / 90 o^ )( I m / θ ) = ω LI m / θ + 90 o^ (3.9)
from which we observe that the voltage and current will be out of phase by exactly 90 o^. In particular, the voltage will lead the current by 90o^ or, what is equivalent, the current will lag behind the voltage by 90o^.
(c) The impedance (ZC) of a capacitor is l/ j ω C (or - j/ ω C ) in rectangular form
and 1/ωC/ -90o^ in angle form. Equation (3.7) indicates that the phasor
voltage at the terminals of a capacitor equals l/ j ω C times the phasor current.
The phasor-domain equivalent circuit for the capacitor is shown in Figure 3.2(c).
ZR = R ZL = j ω L ZC = 1/j ω C = -j/ ω C
If we assume I = I m / θ , then the voltage is given by
V = ( - j/ ω C ) I = (1/ ω C / -90 o^ )( I m / θ ) = I m/ ω C / θ - 90 o^ (3.10)
where in this case, the voltage across the terminals of the capacitor will lag behind the capacitor current by 90o^. The alternative way to express the phase relationship is to say that the current leads the voltage by 90o^. It can be shown that Ohm's law and Kirchhoff's laws apply to the phasor do- main. Thus, all of the techniques used for analyzing dc resistive circuits can be used to find phasor currents and voltages. Thus, new analysis techniques are not needed to analyze circuits in the phasor domain. The basic tools of series-parallel simplifi- cations, source transformations, Thevenin-Norton equivalent circuits, superposition, node-voltage analysis, and mesh-current analysis can all be used in the analysis of circuits in the phasor domain in order to determine the steady-state response of a network to sinusoidal sources. The problem of learning phasor circuit analysis involves two parts. First, you must be able to construct the phasor-domain model of the circuit and, second, you must be able to algebraically manipulate complex numbers and/or quantities to arrive at a solution. We will demonstrate this by considering the example circuit shown in Figure 3.3, which consists of a resistor, inductor, and capacitor connected in series across the terminals of a sinusoidal voltage source. Assume the steady-state voltage source is a
sine wave that we can represent as v s ( t ) = 5 cos ωt at a frequency of f = 1000Hz.
The first step in the solution is to determine the phasor-domain equivalent circuit.
Given a frequency of 1000 Hz, then ω = 2π f = 6283 radians/second. Therefore, the
ZT = ( 100 )^2 +( 156. 7 )^2
Figure 3.4: Phasor Domain Equivalent Circuit
since the resistor, inductor, and capacitor are connected in series. Substituting values for R , ZL , and ZC gives
ZT = 100 + j 188.5 - j 31.8 = 100 + j 156.7. (3.15)
Converting this to angle form, we have
/ tan -1^ (156.7/100) , (3.16)
which simplifies to
ZT = 185.9/ + 57.5o^. (3.17) Thus, the current is equal to I = VS / ZT = (5 / 0o^ )/(185.9 / 57.5 o^ ) = 0.0269 / -57.5° A. (3.18)
and the voltage across the capacitor is
VC = ZC I = (31.8 / -90o^ )(0.0269 / -57.5o^ ) = 0.855 / -147.5o^ V. (3.19)
We can now write the stead-state expressions for I ( t ) and vC ( t ) directly: i ( t ) = 26.9 cos (6283t - 57.5o^ ) mA (3.20) and v (^) C ( t ) = 0.855 cos (6283t - 147.5o^ ) V. (3.21)
3.3.2 Frequency Response of AC Circuits: Bode Diagrams As indicated above, network connections of linear circuit elements used to transmit signals generally result in an output signal with an amplitude and time dependence that differs from that of the input signal. If the input signal is sinusoidal, so too will be the output for a linear network. However, usually both the amplitude ratio and the phase difference between the input and the output signals will depend on the frequency. A particularly convenient graphical representation is obtained if logarithmic scales are used for both the frequency and the amplitude ratio while a linear scale is used for the phase difference. Also, extremely useful approximate graphical representations can be obtained for many networks that result in a series of straight lines on these graphs. Both representations are called Bode Diagrams. In this part of the experiment, the measured response of two networks will be compared to that theoretically predicted and the straight line approximations. In addition, decibel notation will be introduced. The frequency response of the elementary circuit shown in Figure 3.5, where v (^) i (t) corresponds to the input signal and v (^) o (t) to the output signal, will initially be determined. Sinusoidal signals will be assumed. In analyzing this circuit, the
A (ω )= VV^0 i = 1 + j^1 ω RC.
| A (ω )|= 1 +(^1 ω RC ) 2 ,
| ( )|^1
2 1 ⎟
A ω
result of unavoidable circuit and element capacitances, while the resistance, R , is the Thevenin equivalent resistance of the complex circuit. If we let A be defined as the ratio of the output voltage to the input voltage, then (3.25)
Now A(ω) can be expressed in terms of its magnitude and phase angle, where the magnitude is
(3.26) while the phase angle is
θ(ω) = -tan -1^ (ωRC). (3.27)
If we define the critical frequency ω 1 to be
ω 1 = 1/ RC (3.28)
then, upon substitution in Equation (3.26) and Equation (3.27), we find
(3.29)
and θ (ω) = - tan -1^ (ω/ω 1 ). (3.30) Now, the amplitude of the voltage ratio is normally specified in decibels (dB), and is defined as AdB = 20 log 10 | A (ω)| dB.
Bode plots relate the magnitude and phase angle of the voltage ratio of a network with frequency as the independent variable. Usually, frequency in Hertz or radians per second is used as the abscissa on a logarithmic scale and the amplitude of the voltage ratio in dB on one ordinate and the phase angle on a linear scale as another ordinate. An example for the low pass-filter is shown in Figure 3.6 where we have
plotted A dB versus log ω and θ versus log ω. Note that Bode plots are shown for
both the straight line approximation (solid) and the actual (dashed) values.
Figure 3.6: A Bode Plot for a Low-Pass Filter with Critical Frequency ω 1.
AdB
_______ (^) Approximate response
X ' − Y '= 20 ⎢⎣⎡log 10 ⎜⎜⎝⎛ωω^1 n ⎟⎟⎠⎞−log 10 ⎜⎜⎝⎛ 10 ω ω^1 n ⎟⎟⎠⎞⎥⎦⎤,
(1/√2)^2 , the critical frequency ω 1 (or f 1 when expressed in Hertz) is also known as the
half-power frequency. The phase angle for ω = ω 1 is calculated by substituting into
Equation (3.30) and is given by
θ = - tan -1^ (1) = - 45°. (3.37)
For ω >> ω 1 , we determine from Equation (3.29) that
| A ( ω 1 )| ≅ ω 1 / ω. (3.38)
Hence, in this region, | A ( ω)| varies inversely with frequency. The rate of change can
be determined by considering a decade of frequency. Let
X = | A ( ωn )| and X dB = 20 log 10 X , (3.39)
where X is the magnitude of the voltage ratio function at some frequency ωn , and
let
Y = | A (10 ωn )| and Y dB = 20 log 10 Y , (3.40)
where Y is the magnitude of the voltage ratio function at a frequency 10 times ωn.
Then, using Equation (3.38), the difference in voltage ratio (in dB) over a decade of frequency is given by
(3.41)
which simplifies to X ' - Y ' = 20 log 10 (10) = 20 dB. (3.42)
Thus, the gain at 10 ωn is 20 dB less than it was at ωn. It follows then, that the rate
of fall of the amplitude ratio, where ω >> ω 1 , is -20 dB/decacde.
Finally, from the expression for the phase angle in Equation (3.30), we observe
that as^ ω^ Æ^ ∞,^ θ^ Æ^ -90°, but at^ ω^ = 10^ ω 1 ,
θ = - tan -1^ (10) = -84.289 ≈ - 84°. (3.43)
As indicated previously, Figure 3.6 is a Bode plot for a low-pass filter with
critical frequency ω 1. Notice that the approximate Bode plot makes use of the
straight-line segments to approximate the actual response (shown as a dotted line around the critical frequency). Also, note that the maximum amplitude rolloff is - dB/decade and the maximum phase shift is - 90°. Equally important to the low-pass filter just considered is a network in which the positions of the capacitor and the resistor are interchanged as shown in Figure 3.7. In this case, since the capacitance is in series with the input voltage source and the
output terminals of the network, the voltage v 0 becomes very small as the frequency is reduced. For zero frequency (i.e., "dc"), the output voltage is zero since a capacitor blocks dc voltages. At very high frequencies, the capacitor's impedance approaches
| ( )|^1
1 2 +⎜⎝⎛ ⎟⎠⎞
A ω
θ(ω )= tan−^1 ⎜⎝⎛ωω^1 ⎟⎠⎞.
Again, in terms of the critical frequency ω 1 = 1/ RC , we have
and
(3.49)
A Bode plot for the high-pass filter is shown in Figure 3.8. Notice in this case that
Figure 3.8: A Bode plot for a High-Pass Filter with Critical Frequency ω.
_______ (^) Approximate response
the magnitude and phase angle of the voltage ratio is again 3 dB down and +45°,
respectively, at the critical frequency. The rate of increase for ω << ω 1 , is
20 dB/decade and as ω Æ 0, θ Æ + 90 °, while at ω = 0.1 ω 1 , θ = + 84°. For
ω >> ω 1 , the magnitude of the voltage ratio is 1 (0 dB) and the phase angle is 0°.
Obtaining values for A ( ω) and θ( ω) from Equation (3.48) and Equation (3.49), as
shown previously for the low-pass filter, is left to the student.
3.4 Advanced Preparation The following advanced preparation is required before coming to the laboratory: (1) Thoroughly read and understand the theory and procedures. (2) Solve for the steady-state voltage across C 2 in the circuit shown in Figure 3. at both frequencies given to you by your instructor. (3) Perform a PSpice Transient simulation to find the steady-state voltage across C 2 in the circuit shown in Figure 3.9. Note that the simulation time must be long enough for the transient response to reach steady-state. Make a copy of a segment of the steady-state voltage. Do the simulation for both frequencies. (4) Solve for the steady-state voltage across R 4 in the circuit shown in Figure 3.10 at both frequencies given to you by your instructor. (5) Perform a PSpice Transient simulation to find the steady-state voltage across R 4 in the circuit shown in Figure 3.10. Note that the simulation time must be long enough for the transient response to reach steady-state. Make a copy of a segment of the steady-state voltage. Do the simulation for both frequencies. (6) Calculate the critical frequency in Hertz (using the values given to you by your instructor) for the low-pass and high-pass filter test circuits shown in Figure 3.11 and Figure 3.12, respectively.
