Problem 13.3 - Steady-State Response of an RLC Circuit with Variable Damping
*
* The voltage across the capacitance in a series RLC circuit has a peak
* in the frequency response for circuits that have Q > 1.
* The normalized peak value is a function of the circuit damping ratio.
* In this problem, to relate peak response to Q and damping ratio, use
* PROBE to graph the voltage gain V(3)/V(1) as a function of frequency.
* After viewing the response curves produced by the program run, use
* the "X_axis" selection on the menu line to change the x-axis scale
* to 1k,100k for a better display of the important area of the graph.
* Make a hard-copy of the display. Using the same expanded scale, use
* PROBE to graph the phase response as a function of frequency for
* different values of circuit Q (or damping ratio). Also print a hard
* copy of the phase response graph.
*
* Use a variable resistance model to vary the circuit resistance from
* 250 to 2000 ohms in four steps, using the .STEP command with one step
* per octave. Use the frequency sweep range of Problem 13.2. Use PROBE
* to obtain a plot of all four frequency sweeps on one graph.
*
.OPT NOPAGE NOBIAS
V1 1 0 AC 1
R1 1 2 RVAR 1
L1 2 3 15.916M
C1 3 0 15.916N
.MODEL RVAR RES(R=250)
.AC ; Complete the .AC command. Use a
; minimum of 30 points per decade
; because of the shape of the curve.
; (See the note in Problem 13.2)
.STEP OCT RES RVAR(R) ; Complete the .STEP command. Note
; that in an OCT sweep, the variable
; doubles in each octave. In this
; case you want values of 250, 500, 1000,
; and 2000 for reasonably spaced values of
; damping ratio. One step/octave will do.
.PROBE
.END