Introduction In this experiment, we will study the behavior of simple electronic circuits whose response varies as a function of the driving frequency. One key feature of these circuits is that they exhibit linear response, i.e. the response only depends on the drive frequency and not the drive amplitude. During the course of this experiment, you will be introduced to two powerful experimental techniques which can be used to characterize the frequency response of these systems. They are: (1) Frequency Domain Spectroscopy (FDS) in which the circuit will be driven using a single frequency and the response will be recorded at each frequency, and () Time Domain Spectroscopy (TDS) in which we will apply a short pulse to the circuit and record its time response. We will find that for linear systems, both techniques yield the same information. We will find, however, that there are distinct advantages and disadvantages to both methods. Transients in RLC Circuits We will begin by studying the transient response in RLC circuits. We find that two qualitatively different transients are possible, a damped oscillation and an exponential decay. We then drive the RLC circuit with an external sinusoidal voltage and find that the response of the circuit depends on the driving frequency. We find that the response of the circuit is greatest at one frequency. We compare our observations to a simple model. A wide variety of physical systems are understood as examples of oscillating systems: the simple pendulum, the mass on a spring, the charged particle in a storage ring, and the series RLC circuit. In each of these physical systems we determine how a single variable, for example, the position of the mass on the spring or the charge on the capacitor, changes with time. There are many elements common to the description of all oscillating systems. For example, all of these systems are, in fact, described by equations of the same functional form. Consider the two systems shown in Figure 1. Figure 1: Two equivalent dynamical systems. For the mechanical system, x() t is the displacement of the mass and the equation of motion is d dx dx M r kx dt dt + dt + = F (1.1) 1
On the left hand side of this equation, M is the mass and the first term is then mass times acceleration; r is the coefficient of viscous friction represented schematically by the dashpot (the frictional force is proportional to velocity); and k is the spring constant (the spring force is kx ). On the right-hand side of the equation, F represents an external driving force. For the RLC circuit, equation q() t is the charge on the capacitor, and Kirchoff s voltage law gives the d dq dq 1 L + R + q = V () t (1.) dt dt dt C On the left-hand side of the equation, the first term is the voltage drop across the inductor. We have used the derivative of the charge on the capacitor for the current through the inductor. The second term is the voltage drop across the resistor, and the third term is the voltage drop across the capacitor. On the right-hand side V ( t ) is an externally applied voltage. We find in comparing Equation (1.1) and Equation (1.) that the mass behaves the same as an inductor, and the spring the same as an inverse capacitance. Displacement becomes charge, and the viscous friction is replaced by a resistor. Such correspondences can be found in other systems. The theory below first treats the case in which there is no externally applied voltage. The solution has two qualitatively different forms. Several important quantities are defined: natural (angular) frequency for oscillation, ω o, the quality factor, Q, and the logarithmic decrement,δ, which is also called the log decrement. The logarithm has the base e, but no one ever calls δ the ln decrement. The concept of critical damping is also introduced. The case in which there is an externally applied sinusoidal voltage is discussed only briefly. The concept of resonance is introduced. Solution to the RLC Equation of Motion Figure : RLC circuit with voltage on capacitor initially. The simplest RLC circuit where there is no externally applied voltage is shown schematically in Figure. At t = 0, the switch is closed to permit the initially charged capacitor to discharge through the inductor and resistor. (In the mechanical system the spring would be stretched initially, and the mass would be released att = 0.)
Application of Kirchhoff s law gives d dq dq q L + R + = 0, (1.3) dt dt dt C where dq has been used for current. This equation is a homogeneous, linear, second-order differential equation and has solutions of the form q( t) = Ae, where A is an arbitrary constant. dt st Substituting this function into Equation (1.) produces a quadratic equation for s : s R + s+ = 0, (1.4) L LC 1 st which must be satisfied for q() t = Ae to be a solution. The two roots of Equation (1.4) are R R 1 s1, s = ± = a± b L L LC (1.5) where R a = and L R 1 b = L LC (1.6) The nature of the solution depends on whether the first term under the radical is greater than, equal to, or less than the second term, or equivalently on whether b > 0 (over-damped solution, b = 0 (critically damped solution), or b < 0 (under-damped, or oscillatory solution). A. Over-damped solution b > 0 b is positive, then the solution is periodic and ( ) If The solution is of the form ( ) ( bt = bt 1 + 1 ) q t e Ae Be q t falls to zero smoothly with no oscillations.. (1.7) The constants A1 and B 1 are determined from the initial conditions. For the circuit in Figure, the charge on the capacitor at t = 0 is qo. Evaluating Equation (1.7) at t = 0 gives qo = ( A 1 + B1). At t = 0, the current is zero. Differentiating Equation (1.7) gives the current () ( 1 1 ) ( 1 1 ) i t ae A e B e be A e B e bt bt bt bt = + +. (1.8) 3
( ) ( ) Evaluating Equation (1.8) at t = 0 gives a A + B + b A B = obtain 1 1 1 1 0. After some algebra we a qo a ( a b) t q() t = qoe cosh bt + sinh bt 1 + e b b ( a b) t 1, a b qo a b ( a b) t i() t = qoe sinh bt e b b ( a b) t 1. (1.9) (1.10) b > Since 0, from Equation (1.6) we find that a > b. Thus, at large times the charge and current decay with an exponential a b< a. B. Critically damped solution b = 0 For b = 0, the solution is critically damped and ( ) q t will fall to zero in the minimum time without oscillation. There are no longer two distinct solutions to Equation (1.4), and the form of the solution is now ( ) = ( + ) q t A B t e (1.11) Evaluating Equation (1.11) at t = 0 gives A = qo. Differentiating Equation (1.11) gives the current. i t = B e a A + B t e (1.1) () ( ) Evaluation Equation (1.1) at t = 0 gives B a A = 0. After some algebra we obtain ( ) = q ( 1+ ) q t o at e, and (1.13) ( ) = i t a q t e o at. (1.14) The charge decays with an exponential a. The current starts at zero, goes through a maximum, and then also decays with an exponential a. C. Under-damped solution b < 0 The most interesting case is for b < 0. Then the two roots of Equation (1.4) are complex. It is convenient to make b real (change the signs of the factors under the radical in Equation (1.4)) and use j = 1. (We use j because i is the current.) Then the solution becomes 4
( ) at ( jbt jbt 3 3 ) q t = e Ae + B e (1.15) where in the above R 1 R a= and b= L LC L (1.16) Evaluating Equation (1.15) at t = 0 gives qo = ( A3 + B3). Differentiating Equation (1.15) gives the current. () ( ) ( ) i t = ae Ae + B e + jbe Ae B e jbt jbt jbt jbt 3 3 3 3 (1.17) ( ) ( ) Evaluating Equation (1.17) at t = 0 gives a A + B + jb A B = obtain 3 3 3 3 0. After some algebra we jbt ( φ) jbt ( φ) a + b e + e 1 q() t = qo e = qo e cos bt b cosφ ( φ ) (1.18) where tan φ = a/ b. Alternatively, a q() t = qoe cosbt + sin bt, (1.19) b () a + b i t = qoe sin bt. (1.0) b The angular frequency, ω 1, of the oscillation isb, and the frequency, 1 f, is f 1 ω1 1 1 R = =. (1.1) π π LC L The solution oscillates with frequency, f 1, and the amplitude of the oscillation decays with exponential a. Even in the presence of damping, the frequency can easily be determined by measuring the time between zeroes of charge. Note that with no damping, R = 0 ; this frequency of oscillation is the natural frequency of the LC oscillator. f o = ωo 1 1 π = π LC (1.) With damping, the frequency of oscillation is shifted to a smaller value. 5
The logarithmic decrement, δ, is the natural log of the ratio of the charge (see equation 0.19) or current (see equation 0.0) between two successive maxima, which are separated in time by the damped period, T =1/ f. 1 1 q( t max max ) e =ln at ( max + T1 ) ( ) δ = ln = at1 q tmax + T1 e (1.3) Magnetic energy is stored in the inductor, and electric energy is stored in the capacitor. The energy is dissipated in the resistor. The Q of the circuit, or quality factor, is defined as total stored energy Q = π. (1.4) decrease in energy per period (To be very clear, Q here is not the charge.) For an RLC circuit Q is found to be ω L 1 π. (1.5) Q = = R δ (It takes some effort to derive this result.) For small R, the damping or logarithmic decrement is small, and the Q of the circuit is large. Note also for small damping (large Q ) that the damped frequency, f 1, and the natural frequency, f o, are approximately equal. (We introduce a third frequency, f, into the experiment when we drive the RLC circuit with a sinusoidal voltage. We also introduce another expression for Q.) Procedures for studying transient and steady-state response Figure 3 below is a circuit that closely approximates the idealized circuit of Figure. The significant difference is that the voltage source, the Wavetek, is coupled to the RLC circuit through a series output resistance of 50 Ω. You should use the hand held digital volt meter to R out R L 50 Ω C Vc Wavetek Figure 3: Connection of Wavetek to RLC circuit and oscilloscope. check the output impedance of the Wavetek function generator you use for your measurements. The Wavetek is used to generate a 8.0 V unipolar square wave of a period of 0.10 s (frequency of 10 Hz) and a duty factor of 50%, as shown in Figure 4. Duty factor is defined as the ratio of 6
length of the pulse (0.05 s here) to the period (0.10 s). Note, the figure shown below is an example, you should adjust the duty factor and the pulse period such that time between pulses is sufficiently long in order to resolve the transient response of the circuit. On the leading edge of the pulse, the capacitor charges. On the trailing edge of the pulse, the 8 V capacitor discharges. The capacitor discharges on the trailing edge of the pulse. 0.1 s Figure 4: Wavetek signal to excite RLC circuit. The transitions occur every 50 ms, and this time is long compared to the time constants of the transient behavior of the circuit. The voltage across the capacitor is observed with the TDS301B oscilloscope. It is very convenient to use this circuit to see changes in the response of the circuit as the resistance and capacitance values are changed. The inductor is actually two coils wound on the same bobbin. The two inner terminals of the double coil should be connected to obtain a larger inductance (by roughly a factor of 4) than one coil. The capacitor is a decade capacitor box, and the resistor is decade resistor box. Make a note of the number of the coil, and measure its series resistance with a DMM. In the exercises below you will change the values of the capacitance box and the resistance box. Your laboratory bench has an aluminum utility box with terminal posts that are set up for easy connection to the circuit elements. The box was built for an earlier version of the experiment which did not use the Wavetek. Even so, there should be little difficulty in making the circuit shown in Figure 3. The SYNC OUT of the Wavetek could be used to trigger the oscilloscope as an external trigger. But in our case. it may be convenient to use channel 1 as the trigger, so that it is possible to see if the oscilloscope is triggering on the leading or the trailing edge of the pulse. The Trigger >> Slope will make this selection. If channel 1 is used for the trigger and channel for the signal across the capacitor, you will have to adjust the vertical sensitivity of channel and Horizontal sweep rate (or Time/Div) to get an optimal view of the transient. Sometimes it will be useful to view only a portion of the transient and allow some portion of the transient to be off the screen either in the vertical or the horizontal. Recall that the accuracy of the time and voltage readings of the oscilloscope depend on the scale settings of the oscilloscope so recording the time per division and voltage per division are essential. Determine dependence of frequency on capacitance The goal of this part of the laboratory exercise is to observe the damped, oscillatory behavior of the circuit and measure how the frequency of the oscillation depends on capacitance. For t 7
these measurements, set the decade resistance box to zero, and set the decade capacitance box to C = 1.0 µf. Record the response of the RLC circuit using the Tektronix 301b oscilloscope. Data should be transferred directly from the scope to the computer (option data in the e-scope menu); do not use the screen shot. The quality of the data can be improved by using the averaging option of the scope. The oscillation frequency and decay rate should be obtained by importing the data to Origin fitting it to the solution of the underdamped oscillator. Figure 5 shows the data (green curve) as well as the fit (red curve) to the function t / τ Vc () t = Ae sin( ωt+ φ) + V0 (1.6) Repeat the measurement for capacitances between C = 1.0 µf and C = 0.1 µf in steps of 0.1 µf and C = 0.05 µf. Note that (1.6) assumes that t = 0 corresponds to the falling edge of the pulse. When performing the fit, it is a good idea to shift the time axis so that the oscillations start from t = 0. Time trace: input signal R=400oHms Signal (V) 0.4 0. 0.0 Data: Ch1R400_B Model: DampOsc Weighting: y No weighting Chi^/DoF = 1.9669E-6 R^ = 0.99976 P1 0.4906 ±0.00011 P 4.59648E-5 ±1.51871E-8 P3 80910.9913 ±7.85443 P4 1.14764 ±0.0006 P5-0.017 ±0.0000-0. -0.4 0.0 0.1 0. 0.3 time (ms) Figure 5: Transient in RLC circuit in the underdamped regime. Graph shows the data acquired using the Tektronix 301b oscilloscope and a fit to the data using Origin. Physics401 1 1 1 R = π T LC L (1.7) 8
Make a plot of 1/ T versus1/ C, and find the inductance of your coil from the slope of the line. (It should be about 50 mh.) Assuming that R in Equation (1.7) above is the series resistance of the coil and the parallel combination of the 50 Ω resistor and the input impedance of the Wavetek, show that the second term is, in fact, negligible. Since the second term is negligible, a plot of T versus C should also be a straight line. Make this plot (and the other plots described below) with Origin as you take the data. Print a copy to paste in your notebook. T = ( π ) LC (1.8) Also make a plot of T versus C. An impedance meter is available in the laboratory, the Z-meter, which can measure inductances and capacitances. Find the meter, read the brief instructions, and measure the inductance of your coil. Your inductance from the slope of the plot and the meter reading should agree. Determine the dependence of log decrement on resistance The goal of this part of the exercise is to find how the rate of damping of the transient depends on the resistance. With C = 1.0 μ F increase the resistance of the decade resistance box, R, from zero to 600 Ω in steps of 100Ω. Download the data to the computer and repeat the fit. this time, you should record the decay rate τ as a function of R. From the fit, make a plot of the log decrement δ vs. R. Since the oscillation period does not vary significantly with R, the plot should look like a straight line. Note, however, the x-intercept does not cross zero. The offset is due to additional resistance in the circuit, for example, the series resistance of the coil and the 50 Ω coupling resistor in parallel with the input impedance of the Wavetek. Determine the effective additional resistance in the circuit from the zero intercept of your plot. Determine the value of resistance for critical damping The goal of this part of the exercise is to observe critical damping of the circuit. From Equation (1.6), critical damping occurs when Rcritical 1 L b= = 0 Rcritical =. L LC C (1.9) Thus the critical resistance is proportional to 1 over the square root of the capacitance. For C = 1.0 µf, increase R until no oscillations are observed in the transient. The oscilloscope has a feature, Save/Recall >> Save Waveform that allows easy comparison of up to 4 oscilloscope traces in Ref1 to Ref4. After the traces are stored, use Recall Waveform>> Ref1 etc to superimpose on the existing display. It is interesting to see a series of traces for a range of values of R. The transient should go from damped oscillations to no oscillations. To remove the Reference waveforms from display, press REF button and select the waveform to erase. Press OFF button (under VERTICAL area) to remove it. Note that the waveforms can be recalled later from the volatile memory. A picture of the oscilloscope display can be transferred to the PC. Figure 6 below shows the response for three resistance values above, below and at the critical value. 9
Equation (1.9) shows that an over-damped RLC decays with exponent ( a b), and Equation (1.1) shows that a critically damped RLC circuit decays with exponent a. Thus the decay time for the overdamped circuit is longer! The digital oscilloscope is able to measure fall time directly when a signal passes from one level to another level. In Measure mean you should find the fall time option. This feature is useful when trying to Figure 6: Critically damped response of RLC circuit find the critical resistance. This measurement takes some effort. If the resistance is too small, there will be an undershoot. If the resistance is too large, the fall time will be too long. For C =1.0, 0.5, 0.1, 0.05, 0.01 μ F, find the value of R 0 at which critical damping occurs. Estimate the uncertainty in your R 0 measurement. It should be possible to determine the resistance at which critical damping occurs to ~0% for each value of C. Plot Rcritical versus 1/C. You should see a straight line. This exercise completes your investigation of the transient response of the RLC circuit. Set the utility box aside. Next you will study the response to a sinusoidal external voltage. Frequency Domain Measurement of Complex Impedance In this part of the lab, you will characterize the complex impedance of four different circuits as a function of frequency using a lock-in amplifier. Lock-in measurement, also known as synchronous detection, is a widely used technique for characterizing the frequency response of both linear and non-linear systems. Lock-in detection also enables detection of narrow band signals with great precision. When we characterize a physical system, we are often interested in knowing its response at different frequencies. One way to obtain the frequency response is to drive the system at a single frequency, then measure the response of the system at this frequency. We can then sweep the frequency and obtain the full frequency response of the system. Experimentally, we find, however, that the signal not only contains the response to the drive but also noise. We can think of noise as being made up of a bunch of different frequencies which add incoherently, that is with no phase relationship, to our signal. This is where lock-in amplifiers come in. Lockin detection is great at being allowing us to distinguish the signal, i.e. the response to the drive, from the rest of the noise. Lock-in detectors measure the component of the signal that has a definite phase relationship to a reference signal (the reference would be the signal used to drive the physical system.) Lock-in amplifiers have two outputs an in-phase or X output, which is the component of the input signal that is in-phase with the reference, and a quadrature or Y output, which is the component of the signal that is 90 out of phase with the reference. A great intro- 10
duction to lock-in detection is given in the manual for the Stanford Research Systems SR830 lock-in amplifier. The introductory section of the SR830 manual is included with this writeup. For this part of the lab, you will analyze the frequency response of the circuits shown in Figure 7. Specifically, we are interested in measuring the transfer function for the circuit. Vout H ( ω) (1.30) V (a) C (b) in R V in R V out e in C e out (c) (d) R R R L e in C R/ C C e out V 0 C V c Figure 7: Different RLC circuits for frequency analysis: (a) high-pass filter; (b) low-pass filter; (c) double-t notch filter; (d) series RLC resonance circuit. In your analysis, it will be helpful to recall the following expressions for the complex impedance for a capacitor and inductor: Zc = j ω C and ZL = jω L. 11
As an example, let s derive H ( ω) for the circuit shown in Figure 7(a), also known as a high-pass filter. In order to find the voltage across R, we first apply Kirchoff s laws to find the current passing through R. Vin it () = (1.31) Z Here, Z is the impedance of the series combination of Zc = j ω C and R. j Z = R (1.3) ω C and ωrc Vout = i() t R= Vin (1.33) ωrc j thus Vout ωrc H ( ω ) = = (1.34) Vin ω RC j ωrc H ( ω ) = (1.35) ( ω RC) + 1 We can see from (1.35) why this circuit is called a high-pass filter. If ω is smaller than a cutoff frequency given by ω c = 1 RC, then Vout V in < 1. Far above this frequency, Vout Vin 1. Thus, the circuit filters out low frequencies and allows higher frequencies to pass. The magnitude of H ( ω) tells us what happens to the amplitude of the output voltage and the phase tells us how the phase of the output voltage is shifted relative to the input voltage. In this exercise, we will use a lock-in amplifier to characterize both the real and imaginary parts of H ( ω ). To see how this is accomplished, it is convenient to represent the input voltage and output voltages as the real part of complex quantities. jωt Vin() t = Re[ V0e ] (1.36) jωt Vout () t = Re[ V0e H( ω)] Working through some algebra, we find Vin() t = V0 cosωt ωrc (1.37) Vout () t = V0 [( ωrc)cosω t+ sin ωt] ( ω RC) + 1 We see that while the input signal is only proportional to cos ω t, the output contains both cosω t and sin ωt terms. The in-phase or X output of the lock-in amplifier will thus be proportional to the amplitude of the c os ωt part and the quadrature or Y output will be proportional to the amplitude of the sin ωt part. V0 ( ωrc) VX = ( ω RC) + 1 (1.38) V0 ωrc VY = ( ω RC) + 1 The factor of 1 is there because the output of the lock-in amplifier gives root-mean-square (RMS) not peak amplitudes. H ( ω ) can be found directly from (1.38). 1
ωrc H( ω ) = ( VX + jvy) = ( ω RC+ j ) (1.39) V ( ω RC) + 1 0 0.7 0.6 0.5 hi-pass filter: 0nF, 0K FD (A) 80 hi-pass filter: 0nF, 0K (B) 0.4 60 (V x +V y )1/ 0.3 0. 0.1 f c 1 = πτ H ( ω ) = ωτ 1 + ( ωτ) τ = RC, ω = πf 10 10 3 10 4 10 f 5 c f (Hz) θ ο 40 0 0 10 10 3 10 4 10 5 f (Hz) Figure 8: (A) The green curve is the lock-in measurement of a high-pass filter showing the magnitude of the transfer function H ( ω ). The red curve is the fit to the calculated function displayed in the inset. (B) Curve showing the phase of H ( ω). You will need to derive H ( ω) for parts (b) (d), measure the magnitude and phase of the transfer function using the lock-in amplifier. You should then fit the measured response to the derived expressions in Origin to confirm your findings. The transfer function for part (c) is a bit tedious to derive, therefore, we shall give it here. 1 ( ωrc) H ( ω) = (1.40) 1 + jωrc ( ωrc) Note The SR830 comes with a sine wave function generator output which you will use to drive the circuit. When the lock-in reference is set to Internal, the reference is locked to the sine wave output of the lock-in. V out should be connected to the A-input of the SR830. In making these measurements, you will use the frequency scan program n01a. The sine wave generator of the SR830 has a 50 W output impedance, thus the magnitude of the impedance of your circuit should be greater than 500 W over the frequency range of the measurement in order to ensure sufficiently accurate results. All the components, such as resistors, capacitors, and inductors can be found in plastic boxes sorted by value on the TA table in the front of the room. For parts (a) and (b), pick three different R and C combinations with time constants between 1 khz to 0 khz and measure the response of the circuits between 100 Hz to 100 khz. For part (c), pick values that produce three different notch frequencies also in the 1 khz to 0 khz range. In part (d), pick a single resonance frequency in the same range and choose 10 different values of R that span the range from 0 W to 10 kw. For this part, analyze the frequency dependence of the resonant frequency on R. Make a graph of ω vs. R and compare to theory. Here, ω is the r LCR resonant frequency. Prior to your measurements, you should derive the transfer functions so that you can predict which values of R, L, and C to pick to produce the desired cutoff, notch, and resonance frequencies. r 13
SR830 BASICS WHAT IS A LOCK-IN AMPLIFIER? Lock-in amplifiers are used to detect and measure very small AC signals - all the way down to a few nanovolts! Accurate measurements may be made even when the small signal is obscured by noise sources many thousands of times larger. Lock-in amplifiers use a technique known as phase-sensitive detection to single out the component of the signal at a specific reference frequency AND phase. Noise signals at frequencies other than the reference frequency are rejected and do not affect the measurement. Why use a lock-in? Let's consider an example. Suppose the signal is a 10 nv sine wave at 10 khz. Clearly some amplification is required. A good low noise amplifier will have about 5 nv/ Hz of input noise. If the amplifier bandwidth is 100 khz and the gain is 1000, then we can expect our output to be 10 µv of signal (10 nv x 1000) and 1.6 mv of broadband noise (5 nv/ Hz x 100 khz x 1000). We won't have much luck measuring the output signal unless we single out the frequency of interest. If we follow the amplifier with a band pass filter with a Q=100 (a VERY good filter) centered at 10 khz, any signal in a 100 Hz bandwidth will be detected (10 khz/q). The noise in the filter pass band will be 50 µv (5 nv/ Hz x 100 Hz x 1000) and the signal will still be 10 µv. The output noise is much greater than the signal and an accurate measurement can not be made. Further gain will not help the signal to noise problem. Now try following the amplifier with a phasesensitive detector (PSD). The PSD can detect the signal at 10 khz with a bandwidth as narrow as 0.01 Hz! In this case, the noise in the detection bandwidth will be only 0.5 µv (5 nv/ Hz x.01 Hz x 1000) while the signal is still 10 µv. The signal to noise ratio is now 0 and an accurate measurement of the signal is possible. What is phase-sensitive detection? Lock-in measurements require a frequency reference. Typically an experiment is excited at a fixed frequency (from an oscillator or function generator) and the lock-in detects the response from the experiment at the reference frequency. In the diagram below, the reference signal is a square wave at frequency ω r. This might be the sync output from a function generator. If the sine output from the function generator is used to excite the experiment, the response might be the signal waveform shown below. The signal is V sig sin(ω r t + θ sig ) where V sig is the signal amplitude. The SR830 generates its own sine wave, shown as the lock-in reference below. The lock-in reference is V L sin(ω L t + θ ref ). Reference Signal θ sig θ ref Lock-in Reference The SR830 amplifies the signal and then multiplies it by the lock-in reference using a phase-sensitive detector or multiplier. The output of the PSD is simply the product of two sine waves. V psd = V sig V L sin(ω r t + θ sig )sin(ω L t + θ ref ) = 1/ V sig V L cos([ω r - ω L ]t + θ sig - θ ref ) - 1/ V sig V L cos([ω r + ω L ]t + θ sig + θ ref ) The PSD output is two AC signals, one at the difference frequency (ω r - ω L ) and the other at the sum frequency (ω r + ω L ). If the PSD output is passed through a low pass filter, the AC signals are removed. What will be left? In the general case, nothing. However, if ω r equals ω L, the difference frequency component will be a DC signal. In this case, the filtered PSD output will be V psd = 1/ V sig V L cos(θ sig - θ ref ) 3-1
SR830 Basics This is a very nice signal - it is a DC signal proportional to the signal amplitude. Narrow band detection Now suppose the input is made up of signal plus noise. The PSD and low pass filter only detect signals whose frequencies are very close to the lockin reference frequency. Noise signals at frequencies far from the reference are attenuated at the PSD output by the low pass filter (neither ω noise - ω ref nor ω noise +ω ref are close to DC). Noise at frequencies very close to the reference frequency will result in very low frequency AC outputs from the PSD ( ω noise -ω ref is small). Their attenuation depends upon the low pass filter bandwidth and roll-off. A narrower bandwidth will remove noise sources very close to the reference frequency, a wider bandwidth allows these signals to pass. The low pass filter bandwidth determines the bandwidth of detection. Only the signal at the reference frequency will result in a true DC output and be unaffected by the low pass filter. This is the signal we want to measure. Where does the lock-in reference come from? We need to make the lock-in reference the same as the signal frequency, i.e. ω r = ω L. Not only do the frequencies have to be the same, the phase between the signals can not change with time, otherwise cos(θ sig - θ ref ) will change and V psd will not be a DC signal. In other words, the lock-in reference needs to be phase-locked to the signal reference. Lock-in amplifiers use a phase-locked-loop (PLL) to generate the reference signal. An external reference signal (in this case, the reference square wave) is provided to the lock-in. The PLL in the lock-in locks the internal reference oscillator to this external reference, resulting in a reference sine wave at ω r with a fixed phase shift of θ ref. Since the PLL actively tracks the external reference, changes in the external reference frequency do not affect the measurement. All lock-in measurements require a reference signal. In this case, the reference is provided by the excitation source (the function generator). This is called an external reference source. In many situations, the SR830's internal oscillator may be used instead. The internal oscillator is just like a function generator (with variable sine output and a TTL sync) which is always phase-locked to the reference oscillator. Magnitude and phase Remember that the PSD output is proportional to V sig cosθ where θ = (θ sig - θ ref ). θ is the phase difference between the signal and the lock-in reference oscillator. By adjusting θ ref we can make θ equal to zero, in which case we can measure V sig (cosθ=1). Conversely, if θ is 90, there will be no output at all. A lock-in with a single PSD is called a single-phase lock-in and its output is V sig cosθ. This phase dependency can be eliminated by adding a second PSD. If the second PSD multiplies the signal with the reference oscillator shifted by 90, i.e. V L sin(ω L t + θ ref + 90 ), its low pass filtered output will be V psd = 1/ V sig V L sin(θ sig - θ ref ) V psd ~ V sig sinθ Now we have two outputs, one proportional to cosθ and the other proportional to sinθ. If we call the first output X and the second Y, X = V sig cosθ Y = V sig sinθ these two quantities represent the signal as a vector relative to the lock-in reference oscillator. X is called the 'in-phase' component and Y the 'quadrature' component. This is because when θ=0, X measures the signal while Y is zero. By computing the magnitude (R) of the signal vector, the phase dependency is removed. R = (X + Y ) 1/ = V sig R measures the signal amplitude and does not depend upon the phase between the signal and lock-in reference. A dual-phase lock-in, such as the SR830, has two PSD's, with reference oscillators 90 apart, and can measure X, Y and R directly. In addition, the phase θ between the signal and lock-in reference, can be measured according to θ = tan -1 (Y/X) 3-
SR830 Basics WHAT DOES A LOCK-IN MEASURE? So what exactly does the SR830 measure? Fourier's theorem basically states that any input signal can be represented as the sum of many, many sine waves of differing amplitudes, frequencies and phases. This is generally considered as representing the signal in the "frequency domain". Normal oscilloscopes display the signal in the "time domain". Except in the case of clean sine waves, the time domain representation does not convey very much information about the various frequencies which make up the signal. What does the SR830 measure? The SR830 multiplies the signal by a pure sine wave at the reference frequency. All components of the input signal are multiplied by the reference simultaneously. Mathematically speaking, sine waves of differing frequencies are orthogonal, i.e. the average of the product of two sine waves is zero unless the frequencies are EXACTLY the same. In the SR830, the product of this multiplication yields a DC output signal proportional to the component of the signal whose frequency is exactly locked to the reference frequency. The low pass filter which follows the multiplier provides the averaging which removes the products of the reference with components at all other frequencies. The SR830, because it multiplies the signal with a pure sine wave, measures the single Fourier (sine) component of the signal at the reference frequency. Let's take a look at an example. Suppose the input signal is a simple square wave at frequency f. The square wave is actually composed of many sine waves at multiples of f with carefully related amplitudes and phases. A V pk-pk square wave can be expressed as frequencies is removed by the low pass filter following the multiplier. This "bandwidth narrowing" is the primary advantage that a lock-in amplifier provides. Only inputs at frequencies at the reference frequency result in an output. RMS or Peak? Lock-in amplifiers as a general rule display the input signal in Volts RMS. When the SR830 displays a magnitude of 1V (rms), the component of the input signal at the reference frequency is a sine wave with an amplitude of 1 Vrms or.8 V pk-pk. Thus, in the previous example with a V pk-pk square wave input, the SR830 would detect the first sine component, 1.73sin(ωt). The measured and displayed magnitude would be 0.90 V (rms) (1/ x 1.73). Degrees or Radians? In this discussion, frequencies have been referred to as f (Hz) and ω (πf radians/sec). This is because people measure frequencies in cycles per second and math works best in radians. For purposes of measurement, frequencies as measured in a lock-in amplifier are in Hz. The equations used to explain the actual calculations are sometimes written using ω to simplify the expressions. Phase is always reported in degrees. Once again, this is more by custom than by choice. Equations written as sin(ωt + θ) are written as if θ is in radians mostly for simplicity. Lock-in amplifiers always manipulate and measure phase in degrees. S(t) = 1.73sin(ωt) + 0.444sin(3ωt) + 0.546sin(5ωt) +... where ω = πf. The SR830, locked to f will single out the first component. The measured signal will be 1.73sin(ωt), not the V pk-pk that you'd measure on a scope. In the general case, the input consists of signal plus noise. Noise is represented as varying signals at all frequencies. The ideal lock-in only responds to noise at the reference frequency. Noise at other 3-3