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 ampli-tude. 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 fre-quency and the response will be recorded at each frequency, and (2) 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 qualita-tively 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 fre-quency. 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 dxM r kxdt 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 accel-eration; is the coefficient of viscous friction represented schematically by the dashpot (the fric-tional force is proportional to velocity); and is the spring constant (the spring force is

rk kx− ).

On the right-hand side of the equation, represents an external driving force. F For the RLC circuit, is the charge on the capacitor, and Kirchoff’s voltage law gives the equation

( )q t

( )1d dq dqL R qdt dt dt C⎛ ⎞ + + =⎜ ⎟⎝ ⎠

V t (1.2)

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 compar-ing Equation (1.1) and Equation (1.2) that the mass behaves the same as an inductor, and the spring the same as an inverse capacitance. Displacement becomes charge, and the viscous fric-tion 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 solu-tion has two qualitatively different forms. Several important quantities are defined: natural (an-gular) frequency for oscillation, oω , the quality factor, Q , and the logarithmic decrement,δ , which is also called the log decrement. The logarithm has the base , but no one ever calls e 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 in-troduced. Solution to the RLC Equation of Motion

The simplest RLC circuit where there is no externally applied voltage is shown schematically in Figure 2. At , 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 ini-tially, and the mass would be re-leased at

0t =

0t = .) Figure 2: RLC circuit with voltage on ca-pacitor initially.

2

Application of Kirchhoff’s law gives

0d dq dq qL Rdt dt dt C⎛ ⎞ + + =⎜ ⎟⎝ ⎠

, (1.3)

where dqdt

has been used for current. This equation is a homogeneous, linear, second-order dif-

ferential equation and has solutions of the form ( ) stq t Ae= , where is an arbitrary constant. Substituting this function into Equation

A(1.2) produces a quadratic equation for : s

2 1 0Rs sL LC

⎛ ⎞+ + =⎜ ⎟⎝ ⎠

, (1.4)

which must be satisfied for ( ) stq t Ae= to be a solution. The two roots of Equation (1.4) are

2

1 21,

2 2R Rs s a bL L LC

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − ± − = − ±⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(1.5)

where 2RaL

⎛= ⎜⎝ ⎠

⎞⎟ and

2 12RbL 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 (over-damped solution,

(critically damped solution), or

2 0b >2 0b = 2 0b < (under-damped, or oscillatory solution).

A. Over-damped solution 2 0b >

(If is positive, then the solution is periodic and 2b )q t falls to zero smoothly with no oscillations. The solution is of the form

(( ) )1 1at bt btq t e A e B e− −= +

1A 1

. (1.7) The constants and B are determined from the initial conditions. For the circuit in Figure 2, the charge on the capacitor at t is . Evaluating Equation (0= oq 0(1.7) at t = gives )1 1o B= +

0t =

( )

q A . At , the current is zero. Differentiating Equation (1.7) gives the current

( ) ( )1 1 1 1at bt bt at bt bti t ae A e B e be A e B e− − − −= − + + − . (1.8)

3

( ) ( )1 1 1 1 0a A B b A BEvaluating Equation (1.8) at t gives −0= + + − =

( )

. After some algebra we obtain

( ) ( )cosh sinh 1 1,2

a b tat oo

qa aq t q e bt bt e a b tb b

− −− ⎛ ⎞ ⎛ ⎞= + → + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )

(1.9)

( ) ( )2 2 2 2

sinh 1.2

a b tat oo

qa b a bi t q e bt e a b tb b

− −− ⎛ ⎞ ⎛ ⎞− −= − → − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2 0b > b>a b a− <

(1.10)

Since , from Equation (1.6) we find that a . Thus, at large times the charge and current decay with an exponential .

2 0b =B. Critically damped solution

(For , the solution is critically damped and 2 0b = )q t will fall to zero in the minimum time with-out oscillation. There are no longer two distinct solutions to Equation (1.4), and the form of the solution is now

( ) ( )2 2atq t A B t e −= +

0t = 2 oA q

(1.11)

Evaluating Equation (1.11) at gives = . Differentiating Equation (1.11) gives the cur-rent.

( ) ( )2 2 2at ati t B e a A B t e − −= − +

0= 2 2 0B a A

(1.12) Evaluation Equation (1.12) at t gives = . After some algebra we obtain −

( ) ( )1 atoq at eq t , and (1.13) −= +

( ) 2 .at

oi t a q t e−= −

a

(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 .

C. Under-damped solution 2 0b <

2 0b <b

The most interesting case is for . Then the two roots of Equation (1.4) are complex. It is convenient to make real (change the signs of the factors under the radical in Equation (1.4)) and use 1.j = − j because i is the current.) Then the solution becomes (We use

4

(( ) )3 3at jbt jbtq t e A e B e− −= + (1.15)

where in the above

21 and 2 2R Ra bL LC L

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

0t =

(1.16)

(Evaluating Equation (1.15) at gives )3 3oq A B= +

( ) ( )

. Differentiating Equation (1.15) gives the current.

( )3 3 3 3at jbt jbt at jbt jbti t ae A e B e jbe A e B e− − − −= − + + −

0t =

(1.17)

( ) ( )3 3 3 3 0a A B jb A BEvaluating Equation (1.17) at gives − + + − =

( )

. After some algebra we obtain

( ) ( ) ( )

2 2 1 cos2 cos

j bt j btat at

o oa b e eq t q e q e bt

b

φ φ

φφ

− − −− −⎛ ⎞+ +

= = −⎜ ⎟⎝ ⎠

tan /a b

(1.18)

where φ =

( )

. Alternatively,

cos sinato

aq t q e bt btb

− ⎛ ⎞= +⎜ ⎟⎝ ⎠

( )

, (1.19)

2 2

sinato

a bi t q e btb

− ⎛ ⎞+= − ⎜ ⎟

⎝ ⎠

1

. (1.20)

ω , of the oscillation isb , and the frequency, The angular frequency, 1f , is

2

1 1 1 Rω ⎛ ⎞ ⎛ ⎞ ⎛= = − 1 2 2 2LC Lπ π⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

f ⎞ . (1.21)

The solution oscillates with frequency, 1f , and the amplitude of the oscillation decays with ex-ponential . Even in the presence of damping, the frequency can easily be determined by meas-uring the time between zeroes of charge. Note that with no damping, ; this frequency of oscillation is the natural frequency of the oscillator.

a0R =

LC

1 12 2

oof LC

ωπ π

= = (1.22)

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 0.20) between two successive maxima, which are separated in time by the damped period, T f 1 11/ .=

( )( )

( )

maxmax

1max 1max 1

ln =lnat

a t T

q t e aTq t T e

δ−

− +

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠

(1.23)

Magnetic energy is stored in the inductor, and electric energy is stored in the capacitor. The en-ergy is dissipated in the resistor. The Q of the circuit, or quality factor, is defined as

total stored energy

Q

2decrease in energy per period

Q π= . (1.24)

(To be very clear, Q here is not the charge.) For an RLC circuit is found to be

1LQRω π . (1.25) = =

δ (It takes some effort to derive this result.) For small R , the damping or logarithmic decrement is small, and the of the circuit is large. Note also for small damping (large Q ) that the damped frequency,

Q

1f , and the natural frequency, of , 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 2. 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

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

Figure 3: Connection of Wavetek™ to RLC circuit and oscilloscope.

Vc

Wavetek

L

C

RRout

50 Ω

6

length of the pulse (0.05 s here) to the period (0.10 s). Note, the figure shown below is an exam-ple, you should adjust the duty factor and the pulse period such that time between pulses is suffi-ciently long in order to resolve the transient response of the circuit. On the leading edge of the pulse, the capaci-tor charges. On the trailing edge of the pulse, the capacitor dis-charges. The ca-pacitor discharges on the trailing edge of the pulse. The transitions occur every 50 ms, and this time is long compared to the time con-stants of the tran-sient behavior of the circuit. The voltage across the capacitor is observed with the TDS3012B 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 ca-pacitance 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.

0.1 s

8 V

t0.1 s

8 V

t

Figure 4: Wavetek™ signal to excite RLC circuit.

The SYNC OUT of the Wavetek™ could be used to trigger the oscilloscope as an external trig-ger. 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 2 for the sig-nal across the capacitor, you will have to adjust the vertical sensitivity of channel 2 and Horizon-tal 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 essen-tial.

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

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 3012b 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 averag-ing option of the scope. The oscillation frequency and decay rate should be obtained by import-ing 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

/0( ) sin( )t

cV t Ae t Vτ ω φ− + +

0t

(1.26) =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 (1.26) assumes that = 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

. 0t =

0.0 0.1 0.2 0.3

-0.4

-0.2

0.0

0.2

0.4

Data: Ch1R400_BModel: DampOscWeighting: y No weighting Chi^2/DoF = 1.9669E-6R^2 = 0.99976 P1 0.49206 ±0.00011P2 4.529648E-5 ±1.51871E-8P3 280910.29913 ±7.85443P4 1.14764 ±0.00026P5 -0.01272 ±0.00002

Sig

nal (

V)

time (ms)

Time trace: input signal

Physics401

R=400oHms

Figure 5: Transient in RLC circuit in the underdamped regime. Graph shows the data acquired using the Tektronix 3012b oscillo-scope and a fit to the data using Origin.

2 2

2

1 1 12 2

RT LC Lπ

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(1.27) ⎢ ⎥⎣ ⎦

8

Make a plot of 1/ versus1/ , and find the inductance of your coil from the slope of the line. (It should be about 50 mH.) Assuming that

2T CR in Equation (1.27) 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.

2

( )22 2T LCπ=

2 .

(1.28) 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 de-

pends on the resistance. With C = 1.0 Fμ increase the resistance of the decade resistance box, R, from zero to 600Ω 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 50Ω 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 Equa-

tion (1.6), critical damping occurs when

2 1 0 2 .2critical

criticalR Lb R

L LC C⎛ ⎞ ⎛ ⎞= − = → =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ (1.29)

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 super-impose 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 Refer-ence 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 de-cays with exponent ( )a b− , and Equation (1.12) shows that a critically damped RLC circuit decays with exponent . Thus the de-cay time for the over-damped 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 fea-ture is useful when trying to 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

a

=1.0, 0.5,C 0.1, 0.05, 0.01 Fμ , find the value of

Figure 6: Critically damped response of RLC circuit

0R at which critical damping occurs. Estimate the uncertainty in your 0R measurement. It should be possible to determine the resistance at which critical damping occurs to ~20% for each value of . Plot versus C criticalR 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 syn-chronous 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. Lock-in 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 defi-nite 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 Fig-ure 7. Specifically, we are interested in measuring the transfer function for the circuit.

( ) out

in

VHV

ω ≡ (1.30)

11

(a)

Vout

(b)

Vin

C

R e out ein

C

R

(c)

(d)

LR

R/2

RR

2C

C

eoute in

C

R/2

RR

2C

C

eoute in

C

C

LR

C Vc V0

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 com-plex impedance for a capacitor and inductor: cZ j C= − ω and LZ j L= ω .

As an example, let’s derive for the circuit shown in Figure 7(a), also known as a high-pass filter. In order to find the voltage across

( )H ωR , we first apply Kirchoff’s laws to find the current

passing through R .

( ) inVi tZ

= (1.31)

Here, Z is the impedance of the series combination of cZ j C= − ω and R .

jZ RC

= −ω

(1.32)

and

( )out inRCV i t R V

RC jω

= =ω −

(1.33)

thus

( ) out

in

V RCHV RC j

ωω = =

ω − (1.34)

2

( )( )

RCHRC 1ω

ω =ω +

(1.35)

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 1c RCω = , then 1out inV V < . Far above this frequency, 1out inV V → . Thus, the circuit filters out low frequencies and allows higher frequencies to pass. The magnitude of

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

( )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.

0

0

( ) Re[ ]

( ) Re[ ( )]

j tin

j tout

V t V e

V t V e H

− ω

− ω

=

= ω (1.36)

Working through some algebra, we find

0

0 2

( ) cos

( ) [( ) cos sin ]( ) 1

in

out

V t V tRCV t V RC t t

RC

= ω

ω= ω ω +

ω +ω

(1.37)

We see that while the input signal is only proportional to cos tω , the output contains both cos tω and sin terms. The in-phase or X output of the lock-in amplifier will thus be proportional to the amplitude of the c part and the quadrature or Y output will be proportional to the ampli-tude of the sin part.

tωos tω

tω

20

2

02

( )( )2

( )2

X

Y

V RCVRC

V RCVRC

ω=

1

1

ω +ω

=ω +

(1.38)

The factor of 1 2 is there because the output of the lock-in amplifier gives root-mean-square (RMS) not peak amplitudes. can be found directly from ( )H ω (1.38).

12

20

2( ) ( ) ( )( ) 1X Y

RCH V jV RCV RC

ωω = + = ω +

ω +j (1.39)

102 103 104 1050.1

0.2

0.3

0.4

0.5

0.6

0.7

12cf πτ

=

hi-pass filter: 20nF, 20K

(V2 x +

V2 y)1/

2

f (Hz)

FD

fc

2( )

1 ( ), 2

H

RC f

ωτω =

+ ωτ

τ = ω = π

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

(A) (B)

102 103 104 1050

20

40

60

80

hi-pass filter: 20nF, 20K

θο

f (Hz)

You will need to derive for parts (b) – (d), measure the magnitude and phase of the trans-fer 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.

( )H ω

2

2

1 ( )( )1 (

RCH)j RC RC

−=

+ −ωω

ω ω (1.40)

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 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.

out

For parts (a) and (b), pick three different R and C combinations with time constants between 1 kHz to 20 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 20 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 2

rω vs. 2R and compare to theory. Here, rω is the 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.

13

SR830 BASICS

Lock-in amplifiers are used to detect and measurevery small AC signals - all the way down to a fewnanovolts! Accurate measurements may be madeeven when the small signal is obscured by noisesources many thousands of times larger.

Lock-in amplifiers use a technique known asphase-sensitive detection to single out the compo-nent of the signal at a specific reference frequencyAND phase. Noise signals at frequencies otherthan the reference frequency are rejected and donot affect the measurement.

Why use a lock-in? Let's consider an example. Suppose the signal is a10 nV sine wave at 10 kHz. Clearly some amplifi-cation is required. A good low noise amplifier willhave about 5 nV/√Hz of input noise. If the amplifierbandwidth is 100 kHz and the gain is 1000, thenwe 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 havemuch luck measuring the output signal unless wesingle out the frequency of interest.

If we follow the amplifier with a band pass filterwith a Q=100 (a VERY good filter) centered at10 kHz, any signal in a 100 Hz bandwidth will bedetected (10 kHz/Q). The noise in the filter passband will be 50 µV (5 nV/√Hz x √100 Hz x 1000)and the signal will still be 10 µV. The output noiseis much greater than the signal and an accuratemeasurement can not be made. Further gain willnot help the signal to noise problem.

Now try following the amplifier with a phase-sensitive detector (PSD). The PSD can detect thesignal at 10 kHz with a bandwidth as narrow as0.01 Hz! In this case, the noise in the detectionbandwidth will be only 0.5 µV (5 nV/√Hz x √.01 Hzx 1000) while the signal is still 10 µV. The signal tonoise ratio is now 20 and an accurate measure-ment of the signal is possible.

What is phase-sensitive detection?Lock-in measurements require a frequency refer-ence. Typically an experiment is excited at a fixedfrequency (from an oscillator or function generator)and the lock-in detects the response from the

experiment at the reference frequency. In the dia-gram below, the reference signal is a square waveat frequency ωr. This might be the sync outputfrom a function generator. If the sine output fromthe function generator is used to excite the experi-ment, the response might be the signal waveformshown below. The signal is Vsigsin(ωrt + θsig)where Vsig is the signal amplitude.

The SR830 generates its own sine wave, shownas the lock-in reference below. The lock-in refer-ence is VLsin(ωLt + θref).

The SR830 amplifies the signal and then multipliesit by the lock-in reference using a phase-sensitivedetector or multiplier. The output of the PSD issimply the product of two sine waves.

Vpsd = VsigVLsin(ωrt + θsig)sin(ωLt + θref)

= 1/2 VsigVLcos([ωr - ωL]t + θsig - θref) -1/2 VsigVLcos([ωr + ωL]t + θsig + θref)

The PSD output is two AC signals, one at the dif-ference frequency (ωr - ωL) and the other at thesum frequency (ωr + ωL).

If the PSD output is passed through a low passfilter, the AC signals are removed. What will beleft? In the general case, nothing. However, if ωrequals ωL, the difference frequency componentwill be a DC signal. In this case, the filtered PSDoutput will be

Vpsd = 1/2 VsigVLcos(θsig - θref)

WHAT IS A LOCK-IN AMPLIFIER?

3-1

θ ref

Reference

Signal

Lock-in Reference

sigθ

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SR830 Basics

This is a very nice signal - it is a DC signal propor-tional to the signal amplitude.

Narrow band detectionNow suppose the input is made up of signal plusnoise. The PSD and low pass filter only detect sig-nals whose frequencies are very close to the lock-in reference frequency. Noise signals at frequen-cies far from the reference are attenuated at thePSD output by the low pass filter (neither ωnoise-ωref nor ωnoise+ωref are close to DC). Noise at fre-quencies very close to the reference frequency willresult in very low frequency AC outputs from thePSD (|ωnoise-ωref| is small). Their attenuationdepends upon the low pass filter bandwidth androll-off. A narrower bandwidth will remove noisesources very close to the reference frequency, awider bandwidth allows these signals to pass. Thelow pass filter bandwidth determines the band-width of detection. Only the signal at the referencefrequency will result in a true DC output and beunaffected by the low pass filter. This is the signalwe want to measure.

Where does the lock-in reference come from?We need to make the lock-in reference the sameas the signal frequency, i.e. ωr = ωL. Not only dothe frequencies have to be the same, the phasebetween the signals can not change with time, oth-erwise cos(θsig - θref) will change and Vpsd will notbe a DC signal. In other words, the lock-in refer-ence needs to be phase-locked to the signalreference.

Lock-in amplifiers use a phase-locked-loop (PLL)to generate the reference signal. An external refer-ence signal (in this case, the reference squarewave) is provided to the lock-in. The PLL in thelock-in locks the internal reference oscillator to thisexternal reference, resulting in a reference sinewave at ωr with a fixed phase shift of θref. Sincethe PLL actively tracks the external reference,changes in the external reference frequency donot affect the measurement.

All lock-in measurements require a reference signal. In this case, the reference is provided by the exci-tation source (the function generator). This iscalled an external reference source. In many situa-tions, the SR830's internal oscillator may be usedinstead. The internal oscillator is just like a func-tion generator (with variable sine output and a TTL

sync) which is always phase-locked to the refer-ence oscillator.

Magnitude and phaseRemember that the PSD output is proportionalto Vsigcosθ where θ = (θsig - θref). θ is the phasedifference between the signal and the lock-in refer-ence oscillator. By adjusting θref we can make θequal to zero, in which case we can measure Vsig(cosθ=1). Conversely, if θ is 90°, there will be nooutput at all. A lock-in with a single PSD is called asingle-phase lock-in and its output is Vsigcosθ.

This phase dependency can be eliminated byadding a second PSD. If the second PSD multi-plies the signal with the reference oscillator shiftedby 90°, i.e. VLsin(ωLt + θref + 90°), its low pass fil-tered output will be

Vpsd2 = 1/2 VsigVLsin(θsig - θref)

Vpsd2 ~ Vsigsinθ

Now we have two outputs, one proportional tocosθ and the other proportional to sinθ. If we callthe first output X and the second Y,

X = Vsigcosθ Y = Vsigsinθ

these two quantities represent the signal as avector relative to the lock-in reference oscillator. Xis called the 'in-phase' component and Y the'quadrature' component. This is because whenθ=0, X measures the signal while Y is zero.

By computing the magnitude (R) of the signalvector, the phase dependency is removed.

R = (X2 + Y2)1/2 = Vsig

R measures the signal amplitude and does notdepend upon the phase between the signal andlock-in reference.

A dual-phase lock-in, such as the SR830, has twoPSD's, with reference oscillators 90° apart, andcan measure X, Y and R directly. In addition, thephase θ between the signal and lock-in reference,can be measured according to

θ = tan-1 (Y/X)

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SR830 Basics

So what exactly does the SR830 measure?Fourier's theorem basically states that any inputsignal can be represented as the sum of many,many sine waves of differing amplitudes, frequen-cies and phases. This is generally considered asrepresenting the signal in the "frequency domain".Normal oscilloscopes display the signal in the"time domain". Except in the case of clean sinewaves, the time domain representation does notconvey very much information about the variousfrequencies which make up the signal.

What does the SR830 measure? The SR830 multiplies the signal by a pure sinewave at the reference frequency. All componentsof the input signal are multiplied by the referencesimultaneously. Mathematically speaking, sinewaves of differing frequencies are orthogonal, i.e.the average of the product of two sine waves iszero unless the frequencies are EXACTLY thesame. In the SR830, the product of this multiplica-tion yields a DC output signal proportional to thecomponent of the signal whose frequency is exact-ly locked to the reference frequency. The low passfilter which follows the multiplier provides the aver-aging which removes the products of the referencewith components at all other frequencies.

The SR830, because it multiplies the signal with apure sine wave, measures the single Fourier (sine)component of the signal at the reference frequen-cy. Let's take a look at an example. Suppose theinput signal is a simple square wave at frequencyf. The square wave is actually composed of manysine waves at multiples of f with carefully relatedamplitudes and phases. A 2V pk-pk square wavecan be expressed as

S(t) = 1.273sin(ωt) + 0.4244sin(3ωt) + 0.2546sin(5ωt) + ...

where ω = 2πf. The SR830, locked to f will singleout the first component. The measured signal willbe 1.273sin(ωt), not the 2V pk-pk that you'd meas-ure on a scope.

In the general case, the input consists of signalplus noise. Noise is represented as varying signalsat all frequencies. The ideal lock-in only respondsto noise at the reference frequency. Noise at other

WHAT DOES A LOCK-IN MEASURE?

frequencies is removed by the low pass filter fol-lowing the multiplier. This "bandwidth narrowing" isthe primary advantage that a lock-in amplifier pro-vides. Only inputs at frequencies at the referencefrequency result in an output.

RMS or Peak?Lock-in amplifiers as a general rule display theinput signal in Volts RMS. When the SR830 dis-plays a magnitude of 1V (rms), the component ofthe input signal at the reference frequency is asine wave with an amplitude of 1 Vrms or2.8 V pk-pk.

Thus, in the previous example with a 2 V pk-pksquare wave input, the SR830 would detect thefirst sine component, 1.273sin(ωt). The measuredand displayed magnitude would be 0.90 V (rms)(1/√2 x 1.273).

Degrees or Radians?In this discussion, frequencies have been referredto as f (Hz) and ω (2πf radians/sec). This isbecause people measure frequencies in cyclesper second and math works best in radians. Forpurposes of measurement, frequencies as meas-ured in a lock-in amplifier are in Hz. The equationsused to explain the actual calculations are some-times written using ω to simplify the expressions.

Phase is always reported in degrees. Once again,this is more by custom than by choice. Equationswritten as sin(ωt + θ) are written as if θ is inradians mostly for simplicity. Lock-in amplifiersalways manipulate and measure phase indegrees.